Problems Chapter 9

Quantum Mechanics
                      K. Konishi, G. Paffuti

Problem

Consider the Hamiltonian H = (Subscript[H, 0])+(Subscript[H, 1]), where

(Subscript[H, 0])= ((p^2)/(2m))+ (1/2)m (ω^2)(x^2);   (Subscript[H, 1])= g (ω/2)(x p+ p x);     g∈ ℝ, |g|<1.

Solution

1

We can get rid of the perturbation by a unitary transformation

p → P + γ x;  x→ x.

This transformation leaves invariant the commutation relations

[x,P]=[x,p] = i ℏ.

The unitary operator implementing the transformation is

U = Exp[-i (γ/(2ℏ))(x^2)];    p→ P = U p (SuperDagger[U]) ; x→ U x (SuperDagger[U])= x .

The transformation law () can be easily checked in Schrödinger picture

U p (SuperDagger[U])ψ[x]= (e^(-i (γ/(2ℏ))(x^2))) (ℏ/i)(∂_x)((e^(+i (γ/(2ℏ))(x^2)))ψ[x]) =  (γ x ψ[x] +(ℏ/i)(∂_x)ψ[x]).

The Hamiltonian changes with

H[x,p]→ H' = U H[x,p](SuperDagger[U])= H[x,p+γ x] = ((p^2)/(2m)) + (g (ω/2)+ (γ/(2m)))(x p+ p x) + ((1/2)m (ω^2)+ ((γ^2)/(2m))+ g ω γ)(x^2).

With γ = - g m ω the mixed term cancels and we have

H'= ((p^2)/(2m)) + (1/2)m (ω^2)(1-(g^2))(x^2).

This is anharmonic oscillator with frequency

ω'= ω (Sqrt[1-(g^2)]).

For |g|<1 the spectrum is discrete and

(Subscript[E, n])= ω (Sqrt[1-(g^2)]) ℏ (n + (1/2));   n = 0,1,2,3...

For | g | > 1 the system in unstable (has no ground state) and the spectrum is continuous.

For |g| = 1 the system is equivalent to a free particle.

We leave to the reader to check that the situation is exactly the same for the classical version of the system. Let us note that the system is not a dissipative system, as a naïve concept of "force" - ∂V/∂x = - m (ω^2)x - g ω p could suggest.

2

The perturbation expansion is  easily computed using annihilation and creation operators

x = (Sqrt[(ℏ/(2m ω))])(a+(SuperDagger[a])) ;   p = -i (Sqrt[((m ω ℏ)/2)])(a-(SuperDagger[a]));

The term x p + p x reads

x p + p x = (1/i)ℏ ((a^2)- ((SuperDagger[a])^2)).

Let us consider for instance  the ground state. The first order correction vanish, at second order

(Subscript[δE, 0])=(∑_n)⟨0|(Subscript[H, I])|n⟩(1/((Subscript[E, 0])-(Subscript[E, n])))<n|(Subscript[H, I])|0⟩=- ((ℏ^2)/(2ℏ ω))(((((g ω)/2)))^2)⟨0|(a^2)|2⟩⟨2|((SuperDagger[a])^2)|0⟩= - (ℏ/( ω))(((((g ω)/2)))^2)= - ℏ ω ((g^2)/4);

in agreement with previous result:

(Subscript[E, 0]) = ((ℏ ω)/2)(Sqrt[1-(g^2)]) ≃(( ℏ ω)/2)(1 - ((g^2)/2)+... ).

Problem

A harmonic oscillator with mass m and frequency ω is subjected to an external perturbation. Compute the corrections to first and second order in perturbation theory to energy eigenvalues for the following perturbations:

V =:    (Subscript[g, 1])q;   (1/2)(Subscript[g, 2])(q^2);  (1/2)(Subscript[g, 3])(q^3);  (1/2)(Subscript[g, 4])(q^4).

Solution

Hamiltonian and operators

The unperturbed Hamiltonian is

(Subscript[H, 0])= ((p^2)/(2m))+ (1/2)m (ω^2)(q^2).

Matrix elements for perturbations can be computed or performing integrals with Hermite polynomials or using the decomposition in annihilation and creation operators:

q = (Sqrt[(ℏ/(2m ω))])(a + (SuperDagger[a]));  p = -i (Sqrt[((m ω ℏ)/2)])(a - (SuperDagger[a]));  (Subscript[H, 0])= ℏ ω ((SuperDagger[a])a + (1/2)).

Using ket[n] for the state |n⟩

(SuperDagger[a]) ({{|n⟩}})  = (Sqrt[n+1])({{|n+1⟩}}) ;    a|n⟩ = (Sqrt[n])|n-1⟩;

and the decomposition () it is easy to show that

q|n⟩ = ((((ℏ/(2m ω))))^(1/2))( (Sqrt[n+1])({{|n+1⟩}}) + (Sqrt[n])({{|n-1⟩}}) )

(q^2)|n⟩ = ((ℏ/(2m ω)))( (Sqrt[(n+1)(n+2)])({{|n+2⟩}}) + (2n+1)({{|n⟩}})+ (Sqrt[n(n-1)])({{|n-2⟩}}))

(q^3)|n⟩ = ((((ℏ/(2m ω))))^(3/2))( (Sqrt[(n+1)(n+2)(n+3)])({{|n+3⟩}}) + 3(((n+1))^(3/2))({{|n+1⟩}})+ 3(n^(3/2))({{|n-1⟩}})+(Sqrt[n(n-1)(n-2)])({{|n-3⟩}}))

(q^4)|n⟩ = ((((ℏ/(2m ω))))^2)( (Sqrt[(n+1)(n+2)(n+3)(n+4)])({{|n+4⟩}}) + (4n+6)(Sqrt[(n+1)(n+2)])({{|n+2⟩}})+ (6(n^2)+6n+3)({{|n⟩}})+(4n-2)(Sqrt[n(n-1)])({{|n-2⟩}})+ (Sqrt[n(n-1)(n-2)(n-3)])({{|n-4⟩}}))

Interactions  (q^2)and (q^4)produce a first order effect on levels. This effect can be read from diagonal matrix elements of these operators

(Subscript[ΔE, n]) = (1/2)(Subscript[g, 2]) ⟨n|(q^2)|n⟩ = (Subscript[g, 2])((ℏ/(2m ω)))(n+(1/2));

(Subscript[ΔE, n]) = (1/2)(Subscript[g, 4]) ⟨n|(q^4)|n⟩ = (Subscript[g, 4])(3/2)((((ℏ/(2m ω))))^2)(1+2n+2(n^2)).

Operators q and (q^3)produce a shift at second order. The first order contribution is zero by parity conservation and this is confirmed from absence of diagonal elements intheir matrix elements. At second order the shift in energy levels is

((ΔE)^((2)))= (∑_s^(')) ⟨(Subscript[ψ, 0])|V|s⟩ (1/((Subscript[E, 0])-(Subscript[E, s])))⟨s| V |(Subscript[ψ, 0])⟩.

Only a finite number of intermidiate states contribute to the sum.

Operator q

(Subscript[ΔE, n]^((2)))= (Subscript[g, 1]^2)(((|⟨n|q|n-1⟩(|^2))/((Subscript[E, n])- (Subscript[E, n-1])))+((|⟨n|q|n+1⟩(|^2))/((Subscript[E, n])- (Subscript[E, n+1])))) =- (1/2)((Subscript[g, 1]^2)/(m (ω^2))).

This result is obvious, a linear term in q canbe reabsorbed by a translation of the quadratic potential:

(1/2)m (ω^2)(q^2) +(Subscript[g, 1])q = (1/2)m (ω^2)(q + ((Subscript[g, 1])/(m (ω^2)))) - (1/2)((Subscript[g, 1]^2)/(m (ω^2))).

Operator (q^3)

The only contributions different from zero are

(Subscript[ΔE, n]^((2)))= (1/4)(Subscript[g, 3]^2)(((|⟨n|(q^3)|n-1⟩(|^2))/((Subscript[E, n])- (Subscript[E, n-1])))+((|⟨n|(q^3)|n+1⟩(|^2))/((Subscript[E, n])- (Subscript[E, n+1])))+((|⟨n|(q^3)|n-3⟩(|^2))/((Subscript[E, n])- (Subscript[E, n-3])))+((|⟨n|(q^3)|n+3⟩(|^2))/((Subscript[E, n])- (Subscript[E, n+3])))) =- ((Subscript[g, 3]^2)/(4ℏ ω))((((ℏ/(2m ω))))^3)(30(n^2)+30n+11).

Operator (q^4)

The only contributions different from zero come from transitions to intermidiate states n, n±2, n±4. A computation similar to the one above gives, for second order contribution:

(Subscript[ΔE, n]^((2)))- ((Subscript[g, 4]^2)/(4ℏ ω))((((ℏ/(2m ω))))^4)(68(n^3)+102(n^2)+118n+42).

Problem

Consider the proton as a small uniformly charged sphere of radius R. Compute how the hydrogen levels are affected by this hypothesis.

Solution

Hamiltonian

Inside an omogeneous sphere of radius R and charge e  the electrostatic potential is

(Subscript[V, sph])[r] = e ((3/(2R))-(1/2)((r^2)/(R^3))),

while outside is identical to usual Coulomb potential (Subscript[V, C]). The difference between the to potential can be considered as a small perturbation to e/r potential

(Subscript[H, I])[r] = - e ((Subscript[V, sph])- (e/r)) = - (e^2)((3/(2R))-(1/2)((r^2)/(R^3))-(1/r));    r≤ R.

The perturbation is radial, so energy corrections are computed by

(Subscript[ΔE, nL])= (∫_0^R)(Subscript[H, I])[r](((Subscript[R, nL])[r])^2)(r^2)dr.

Using radial wave functions ((Subscript[a, B]) is the Bohr radius)

(Subscript[R, 1s])[r] = ((2 (e^(-(r/(Subscript[a, B])))))/(Subscript[a, B]^(3/2))) ;  (Subscript[R, 2s])[r] = (((e^(-(r/(2 (Subscript[a, B]))))) (2-(r/(Subscript[a, B]))))/(2 (Sqrt[2]) (Subscript[a, B]^(3/2)))) ;

it follows , up to order (((R/(Subscript[a, B])))^3),

(Subscript[ΔE, 1s]) = (2/5)((((R/(Subscript[a, B]))))^2) ((e^2)/(Subscript[a, B])) ≡ (2/5)((((R/(Subscript[a, B]))))^2) a.u.;         (Subscript[ΔE, 2s]) = (1/20)((((R/(Subscript[a, B]))))^2) ((e^2)/(Subscript[a, B])) ≡ (1/20)((((R/(Subscript[a, B]))))^2) a.u.  .

Energy shift for states with higher L are depressed, since radial wave functions decrease as (r^L). For p states  we expect ΔE∼(((R/(Subscript[a, B])))^4)a.u., as is easily confirmed for example computing the correction to 2p state:

(Subscript[R, 2p])[r] =(((e^(-(r/(2 (Subscript[a, B]))))) r)/(2 (Sqrt[6]) (Subscript[a, B]^(5/2)))) ;   (Subscript[ΔE, 2p])= (((e^2) (R^4))/(1120 (Subscript[a, B]^5)))+((O[R])^5)≃ (1/(1120 ))((((R/(Subscript[a, B]))))^4) a.u.

A different approach

Let us call  e ρ the charge density of the nucleus, or in general Z e ρ for a nucleus with Z protons. If V is the electrostatic potential we can write the generic energy shift as

ΔE = - e∫(𝕕^3)r ((|ψ[r]|)^2)(V - (Ze/r))≃ - e((|ψ[0]|)^2)∫(𝕕^3)r (V - (Ze/r)).

Last approximate equality take care of the fact that V differ from Z/r only at very small distances, and clarify that in practice only s-states are affected by the perturbation. Using the identity

1 = (1/6)Δ (r^2),

and integrating by parts eq () can be rewritten as

ΔE = - e((|ψ[0]|)^2)∫(𝕕^3)r (1/6)(r^2)Δ(V - (Ze/r))=  - e((|ψ[0]|)^2)(1/6)∫(𝕕^3)r (r^2) (4π e ρ - Z e 4 π (δ^3)[r])= (2/3)π Z((|e|)^2)((|ψ[0]|)^2)⟨(r^2)⟩.

The measure of ΔE would be a direct measure of mean square radius of the (charge of the) nucleus.

Problem

Suppose that Coulomb potential has a behaviour 1/(r^(1+ε)), with ε ≪ 1. Compute the effect on hydrogen levels and derive a bound on ε comparing the 2p-2s induced energy gap with the observed Lamb shift correction. The theory agrees with experiment at leat up to  δE/h = (10^4) kHz.

Solution

Hamiltonian and perturbation

In this problem we will use atomic units |e|= ℏ=m = 1. We can consider the difference between the supposed Coulomb law and usual -1/r law as a perturbation

(Subscript[H, I]) = - ( (1/(r^(1+ε))) - (1/r)).

Energy corrections are given by the mean value of (Subscript[H, I])on eigenfunctions of hydrogen atom. Radial wave functions for 1s, 2s, 2p states are, in atomic units:

(Subscript[R, 1s])[r] = 2 (e^(-r)) ;    (Subscript[R, 2s])[r] = (((e^(-r/2)) (2-r))/(2 (Sqrt[2]))) ; (Subscript[R, 2p])[r] =(((e^(-r/2)) r)/(2 (Sqrt[6]))) .

A simple integration gives (energies in atomic units)

(Subscript[ΔE, 1s])= (2^ε) Γ(2-ε)-1  ;  (Subscript[ΔE, 2s])=(1/8) (((ε^2)-ε+2) Γ(2-ε)-2);    (Subscript[ΔE, 2p])=(1/24) (Γ(4-ε)-6).

Γ is the Euler Gamma function. For small ε a Taylor expansion gives

(Subscript[ΔE, 2s]) - (Subscript[ΔE, 2p]) ≃ (ε/12)a.u. = (ε/6)Ry.

The agreement between theory and experiment for the computed Lamb shift between 2(Subscript[s, 1/2]) and 2(Subscript[p, 1/2]) levels imply the bound

|ε|< 6·(10^4)kHz h /Ry ≃ 1.8·(10^(-11)).

It can be useful to note the following conversion factors (h is Planck constant, i.e. 2 π ℏ)

(Ry/h) = 3289841.9680 GHz;   (Ry/(h c))≃((1.0973731568549`*^7)/Meter)= ((1.09737×(10^5))/cm).

Problem

Compute the effect of an electric field on the levels of a charged harmonic oscillator. Compute the induced electric dipole and verify the Feynman-Hellmann theorem.

Solution

Hamiltonian and perturbation

This is the simplest model for studying the effect of an electric field on a quantum system. Let us denote by ℰ the external electric field and by e the charge of the oscillator. The Hamiltonian is

H = ((p^2)/(2m))+ (1/2)m (ω^2)(q^2) - e ℰ q = ((p^2)/(2m))+ (1/2)m (ω^2)((( q - ((e ℰ)/(m (ω^2)))))^2) - (1/2)(((e^2) (ℰ^2))/(m (ω^2)))

The change of variable

P = p ;  Q = q - ((e ℰ)/(m (ω^2)))

is a unitary (and classically canonical) transformation, then the physics do not change and we can consider as Hamiltonian

H = ((P^2)/(2m))+ (1/2)m (ω^2)(Q^2)- (1/2)(((e^2) (ℰ^2))/(m (ω^2)));

which differ from unperturbed hamiltonian only by an additive constant. All levels are shifted by the same ℰ dipendent constant

ΔE = - (1/2)(((e^2) (ℰ^2))/(m (ω^2))).

The reader can check the result in perturbation theory as an exercise (see also prob.[]). Let us make some comments on this seemingly trivial exercise.

The unperturbed system is invariant under parity, q → - q and the system, as must be well known by the reader, has no dipole moment in stationary states.

The perturbed system is not parity invariant and the system has a dipole in stationary states (classically an induced dipole). The Hamiltonian () is invariant under the transformation Q → - Q, which is not the parity of the system. By this symmetry the mean value of Q on stationary states of () is zero. On exact stationary states, i.e. eigenstates of (), we have the induced dipole (use the definition of Q):

⟨ψ|d|ψ⟩ ≡ e⟨ψ|q|ψ⟩ = e⟨ψ|Q|ψ⟩ + e⟨ψ|((e ℰ)/(m (ω^2)))|ψ⟩= (((e^2) ℰ)/(m (ω^2))).

The coefficient of propotionality between induced dipole and external field has dimension of a volume and is called polarizability of the system. In our case

d = α ℰ;       α = (((e^2) )/(m (ω^2))).

Interaction lower energy of the system, see (), and is written in the same form as classical dipole induced interaction:

ΔE = - (1/2)(((e^2) (ℰ^2))/(m (ω^2)))= - (( 1)/2)α (ℰ^2)

We can verify in this simple system Feynman - Hellmann theorem, which reads in this case

((∂ΔE)/(∂ℰ))= ⟨ψ|((∂H)/(∂ℰ))|ψ⟩.

The l.h.s. of () is -α ℰ. The r.h.s. is:

⟨ψ|((∂H)/(∂ℰ))|ψ⟩ = - e ⟨ψ|q|ψ⟩ = -  (((e^2) ℰ)/(m (ω^2))) = - α ℰ,

verifying the theorem.

Problem

Solve the two-level system

H = (Subscript[E, 0])+(({{-(Subscript[E, 1]), -δ}, {-(SuperStar[Δ]), (Subscript[E, 1])}}))≡ (Subscript[E, 0]) - σ·B

and compare with the results of perturbation theory as B→ 0 . Discuss how the phases of eigenstates depend on B.

Solution

Exact solution

Our problem has only two states, conventionaly denoted by |+⟩ and |-⟩. The constant (Subscript[E, 0]) given in the text of the exercise is just an additive constant to the energy and will be disregarded from now on.  The Hamiltonian can be considered in a form

H = (Subscript[H, 0])+ V;  (Subscript[H, 0])= (({{-(Subscript[E, 1]), 0}, {0, (Subscript[E, 1])}})) ;  V = (({{0, -δ}, {-(SuperStar[δ]), 0}}));

and studied exactly or with perturbation theory. The eigenvalues are trivially found solving the quadratic equation det(λ - H) = 0, but as an exercise we want to follow another more instructive path. System () can be considered as the Hamiltonian of a spin 1/2 interacting with a magnetic field. Magnetic interaction for spin 1/2  is described by -μ σ·B where μ is the magnetic moment, B the magnetic field and σ the Pauli matrices. With the substitution B → μ B our system is indeed a magnetic Hamiltonian.  If we write

δ = Δ (e^(-i φ));   H = -σ·B;      (Subscript[σ, x])=( ({{0, 1}, {1, 0}})); (Subscript[σ, y]) = ( ({{0, -i}, {i, 0}}));  (Subscript[σ, z]) = ( ({{1, 0}, {0, -1}})).

we have

H = - ( ({{(Subscript[E, 1]), (Δ Cos[φ] - i Δ Sin[φ])}, {(Δ Cos[φ] + i Δ Sin[φ]), -(Subscript[E, 1])}})),

and we can identify the components of our magnetic field with

(Subscript[B, x]) = Δ Cos[φ];  (Subscript[B, y]) = Δ Sin[φ]; (Subscript[B, z])= (Subscript[E, 1]);   B = |B|= (Sqrt[(Δ^2)+(Subscript[E, 1]^2)]).

In our magnetic analogy the perturbation treatment would consist in adding a transverse magnetic field of strength Δ to a magnetic field (Subscript[E, 1])along z (the direction of the magnetic field is what identify z direction). In this language it is clear that we can choose the x direction along the transverse field and put φ = 0, i.e. kill y-component of B field. From now on we adopt this convention and write Δ∈ ℝ instead of δ. Let us note that we can also choose φ = π, than have in effect both positive and negatives vaules for Δ. A priori the two choices are distinct and this will have some consequence below.

The reader now has 3 choices, he can forget this discussion and assume from the beginning a real Hamiltonian, or he can read the note at the end of this subsection to understand for a generic system (without a real magnetic field) what it means the choice δ real. The third (better) choice is to try to understand by himself the question.

In the real form H is given by

H = - ( ({{(Subscript[E, 1]), Δ}, {Δ , -(Subscript[E, 1])}}))= - (Subscript[σ, x])(Subscript[B, x]) - (Subscript[σ, z])(Subscript[B, z]) = - B ( ({{Cos[θ], Sin[θ]}, {Sin[θ] , -Cos[θ]}})).

(Subscript[B, x])= BSin[θ];  (Subscript[B, z])= B Cos[θ];    Sin[θ] = (Δ/B); B = (Sqrt[(Subscript[E, 1]^2)+ (Δ^2)])

and it is apparent that we can diagonalize the Hamiltonian by bringing z axis along B. The rotation of reference frame is an anticlockwise rotation around y axis, i.e.

R = Exp[i (θ/2)(Subscript[σ, y])] = Cos[(θ/2)] + i (Subscript[σ, y]) Sin[(θ/2)] = ( ({{Cos[(θ/2)], +Sin[(θ/2)]}, {-Sin[(θ/2)] , Cos[(θ/2)]}}))

In the rotated basis the Hamiltonian will be diagonal, and in effect the reader can check that

R H (SuperDagger[R]) = - B (Subscript[σ, z]).

The new basis vectors will be

|g⟩ = (SuperDagger[R]) ({{|+⟩}}) = ( ({{Cos[(θ/2)], -Sin[(θ/2)]}, {+Sin[(θ/2)] , Cos[(θ/2)]}}))(({{1}, {0}})) = (({{Cos[(θ/2)]}, {Sin[(θ/2)]}})); ({{|e⟩}})=(SuperDagger[R]) ({{|-⟩}}) =  (({{-Sin[(θ/2)]}, {Cos[(θ/2)]}})).

N.B. We remember that basis vectors transform as the inverse rotation, see notebook [**] for more details.

The reader can verify directly that

H |g⟩= - B |g⟩;      H |e⟩=  B |e⟩;

| g⟩ and | e⟩ are respectively the ground state and the excited state of the system with energy eigenvalues ∓ B = ∓ (Sqrt[(Subscript[E, 1]^2)+(Δ^2)])

The choice of initial phases

The reader can skip this subsection in a first reading.

The thesis is that we can always choose the phases of our states in such a way that δ is real. This is trivially true in the sense that if we choose as basis vectors

|a⟩ = (e^(-i φ/2))({{|+⟩;  }})|b⟩ = (e^(+i φ/2))({{|-⟩;  }})

the Hamiltonian haa the same diagonal matrix elements

⟨a|H|a⟩ =⟨+|H|+⟩ = - (Subscript[E, 1]); ⟨b|H|b⟩ =⟨-|H|-⟩ =  (Subscript[E, 1]);

and real off diagonal elements

⟨a|H|b⟩ = (e^(i φ)) ⟨+|H|-⟩= - Δ;    ⟨b|H|a⟩ = (e^(-i φ)) ⟨-|H|+⟩= - Δ.

The choice of the phase of basis functions can always been made by a unitary transformation so the physics is not changed. Let us note that | a⟩ and | b⟩ are obtained from | +⟩ and | -⟩ states by the unitary transformation Exp[-i φ/2 (Subscript[σ, z])]:

|a⟩ = Exp[-i φ (Subscript[σ, z])/2]({{|+⟩;  }})|b⟩ =Exp[-i φ (Subscript[σ, z])/2]({{|-⟩;  }})

and this is exactly how a rotation of an angle φ acts on spinors in the magnetic analogy.

From a physical point of view we can state the same conclusions with a different perspective. In two dimensions we have only 4 independent hermitian matrices, the identity and the four Pauli matrices, each matrix canbe written as a linear superposition of these matrices. The "maximal abelian set" of observables, neglecting identity, is formed by only one element, one of the Pauli matrices, by convention let us choose (Subscript[σ, z]). Our states are labeled by eigenvalues of this observable, nothing else, and in effect the two states |a⟩ and |+⟩ for example belong to the same eigenvalue of (Subscript[σ, z]). To pass from rays to vectors in  Hilbert space we have to make a choice for phases, and this means to choice |a⟩ or |+⟩. To avoid misunderstandings this does not means that we can change the phases in such a way that every linear combination  |ψ> = (Subscript[c, 1])|+⟩ + (Subscript[c, 2])|-⟩ has real coefficients: once we have choosen a phase for basis vectors this choice is valid for every linear combination.

There is one more subtle point. When we write σ matrices in the usual form we have in effect done a choice of phases, for example (Subscript[σ, y ])has purely imaginary elements, how this combine with the above freedom in phases? In effect when we have stated in the main text that the usual representation of spin 1/2 is The representation of SU(2) of dimension 2 it was understood up to unitary tansformations. Which means again that we can always choose the phases in such a way the σ matrices have the usual form. Now if we do a unitary transformation on a basis which satisfy usual convention on the phases we obtain again a representation with the correct phases. In effect the only point is the state |-⟩ which is defined, without additional phases, as ((Subscript[σ, x])- (Subscript[iσ, y]))/2 |+⟩. If we perform a unitary transformation both matrices and states change. Under the transformation R = Exp[-i φ (Subscript[σ, z])/2]for example:

(Subscript[σ, i])→(Subscript[Σ, i]) = R (Subscript[σ, i]) (SuperDagger[R])

and under a unitary transformation all linear relation, and commutation relations, remain invariant, in particular the new |-⟩ state, which is |b⟩, is obtained by applying the new sigma matrices, Σ, to |a⟩ and everything works in exactly the same way.

In effect after the transformation we had to write Σ instead of σ for matrices and spin operators, but as these operators on the new basis  have the same matrix elements as old operators inthe old basis we use the same symbols, as is always more or less  implicitily done.

Non degenerate perturbation theory

Simple perturbation theory on non degenerate levels can be applied for Δ ≪ (Subscript[E, 1]).  In absence of perturbation -B Sin[θ] (Subscript[σ, x]) ground and excited state are |+⟩ and  |-⟩. As ⟨+|(Subscript[σ, x])|+⟩=0=⟨-|(Subscript[σ, x])|-⟩  corrections to eigenvalues start from second order and we have

(Subscript[ΔE, g]) = - (Subscript[E, 1]) + (Δ^2) ⟨+|(Subscript[σ, x])|-⟩(1/(-2 (Subscript[E, 1])))⟨-|(Subscript[σ, x])|+⟩ = - (Subscript[E, 1])(1 + (1/2)((Δ^2)/(Subscript[E, 1]^2)));
(Subscript[ΔE, e]) =  (Subscript[E, 1]) + (Δ^2) ⟨+|(Subscript[σ, x])|-⟩(1/(2 (Subscript[E, 1])))⟨-|(Subscript[σ, x])|+⟩ =  (Subscript[E, 1])(1 + (1/2)((Δ^2)/(Subscript[E, 1]^2))).

These results are in agreement with Taylor expansion of exact results (), as for small Δ

B =(Sqrt[(Subscript[E, 1]^2)+(Δ^2)]) ≃ (Subscript[E, 1])(1 + (1/2)((Δ^2)/(Subscript[E, 1]^2))).

The eigenstates have a first order correction. Using the general formula

δ |n⟩ = (∑_s^(')) (1/((Subscript[E, n])- (Subscript[E, s])))({{|s⟩}})⟨s|V|n⟩,

we have for first order eigenstates with V = - Δ (Subscript[σ, x]):

|g⟩≃ |+⟩ + (Δ/(2 (Subscript[E, 1]))) ({{|-⟩}}) ≃ ({{|+⟩}}) + (( θ)/2)({{|-⟩}})  ;  e = ({{|-⟩}})  - (Δ/(2 (Subscript[E, 1]))) ({{|+⟩}})≃ ({{|-⟩}})  -  (( θ)/2) ({{|+⟩}});

which are first order expansion of ().

Degenerate perturbation theory

As (Subscript[E, 1])→0 , and in general for |(Subscript[E, 1])|≪ Δ previous formulae do not apply and in this regime we have to switch to degenerate perturbation theory. To see exactly wy perturbation theory on degenerate levels is different let us consider the exact solution (). As Tan[θ] = Δ/(Subscript[E, 1])we see that the vaule of θ in the degeneration limit (Subscript[E, 1])→0depends on the sign of Δ

(lim_((Subscript[E, 1])→0)) θ = |({{    π/2;  if   Δ/(Subscript[E, 1])>0}, {-π/2;  if   Δ/(Subscript[E, 1])<0}})

Accordingly for ground state, as an example

(lim_((Subscript[E, 1])→(SuperPlus[0]))) ({{|g⟩}}) =  |({{(|+⟩ + |-⟩)/(Sqrt[2]) ≡ |S⟩ ;  if   Δ>0}, {(|+⟩ - |-⟩)/(Sqrt[2]) ≡ |A⟩ ;  if   Δ<0}})

i.e. when we switch off the perturbation Δ we do not know in advance  which state of the two dimensional space spanned by | +⟩ and | -⟩ correspond to ground and excited state.

States |S⟩ ans |A⟩ diagonalize the degenerate hamiltonian but we see that |g⟩ is not analytic in Δ,  precisely is discontinuous for Δ = 0.  A similar thing happens for eigenvalues. In the degenerate limit eigenvalue are ± |Δ| which are continuous but not derivable in Δ = 0.

This picture has some relevant consequences. If we imagine to vary smoothly the external field, the eigenvectors follow the variation except if we pass through the singular point (Subscript[E, 1])= 0, Δ = 0, where there is an abrupt change. This is wy adiabatic theorem does not work if in the evolution the system goes through a degeneracy point.

This breaking of adiabatic theorem is even more clear if we switch on the y component of the magnetic field. In the general case

(Subscript[B, x])= B Sin[θ]Cos[φ]; (Subscript[B, y])= B Sin[θ]Sin[φ]; (Subscript[B, z])= B Cos[θ];   B = (Sqrt[(Subscript[E, 1]^2)+(Δ^2)]);

As already noted the y component can be reabsorbed by an anticlockwise rotation of φ around z axis of the reference frame. After rotation the new unperturbed eigenstates will be

(|+⟩)'= (Subscript[R, z]^†)[φ] ({{|+⟩}}) = Exp[-i (φ/2)(Subscript[σ, z])]({{|+⟩}})= Exp[-i (φ/2)]({{|+⟩}}) ;  ({{(|-⟩)'}}) = Exp[+i φ/2 ] ({{|-⟩}}).

For these states everything works as before the only difference being that now Δ, the modulus of transverse field, is necessarly positive. In the degerate limit

|g⟩ = (1/(Sqrt[2]))(({{(|+⟩)'}}) + ({{(|-⟩)'}}) ) = (1/(Sqrt[2]))(({{ Exp[-i (φ/2)]}, { Exp[+i (φ/2)]}})).

The ground state has now a continuous range of variation, not only a double determination as before. Which of the infinite set corrspond to the true ground state depends on the actual value of φ.

The general solution, see (),  is

|g⟩ = Cos[(θ/2)]({{(|+⟩)'}}) + Sin[(θ/2)]({{(|-⟩)'}}) = (({{Cos[(θ/2)] Exp[-i (φ/2)]}, {Sin[(θ/2)] Exp[+i (φ/2)]}})).

We note that | g⟩ change its sign as φ varies from 0 to 2π, like spinor (we remember that in principle our system has nothing to do with a real spinor). This effect is a particularly simple form of Berry phase.

Consider a system in which we vary adiabatically the transverse field around a circle, i.e. φ varies from 0 to 2π. In principle we can locally adjust the phase of our eigenstates in such a way that the ground state follows the variation of the field, why we obtain an additive minus sign after a cicle? Let us consider the infinitesimal variation of the state under a change of a set of parameters (Subscript[λ, i])

({{⟨g|}})(∂/(∂(Subscript[λ, i])))({{|g⟩}})

if we can reset to zero expression () with a change of phase, then we can  choose our phases in such a way that they follow variations of external field. In the present case our parameters are θ and φ and we have from ()

({{⟨g|}})(∂/(∂θ))({{|g⟩}})= 0;   ({{⟨g|}})(∂/(∂φ))({{|g⟩}}) = -(( i)/2)Cos[θ].

Consider now a totation of the transverse field at fixed θ. To reabsorb the phase variation we would need a change of phase

|g⟩ → Exp[i α]|g⟩;      α[φ] = (1/2)Cos[θ]φ,

but the function α is not monodrome in φ, i.e. is not a function, so it is not allowed, and we cannot reabsorb the phase variation in a closed loop by a global defined α. The amount of phase change under a cyclic variation of external condition is substantially the Berry phase of the system.

Problem

A two-level system as that of problem [] can be considered as a simplified model of an ammonia molecule, (Subscript[NH, 3]). The Nitrogen position with respect the the plane of 3 hydrogen atoms corresponds, classically, to two equivalent configurations. Define a parity operator for this model and discuss its property. Study the effect of an electric field on the system.

Solution

Description of the model

An ammonia molecule can be described by three coplanar hydrogen atoms, (x,y) plane, and a nitrogen outside the plane. The system admits a specular configuration obtained by a reflection across (x,y) plane,  i.e. z→-z, then a second position for nitrogen atom, classically equivalent to the first one, exists. Classically the two position are separated by a potential barrier, and are local minima of the potential energy. Tunnel effect allow a mixing between these two configurations. The situation is described pictorially inthe folowing figure:

Neglecting tunnel effects an effective Hamiltonian for the system can be taken as

(Subscript[H, 0])=( ({{(Subscript[E, 0]), 0}, {0, (Subscript[E, 0])}}));     P = ( ({{0, 1}, {1, 0}})).

The basis states correspond to classically localized states, |+⟩ and |-⟩.  P is the operator which exchange these states, i.e. parity. Tunnel effect produce an amplitude different from zero for the transition between the two states and can be added to (Subscript[H, 0]) in the form of off diagonal terms

H = ( ({{(Subscript[E, 0]), -Δ}, {-Δ,  (Subscript[E, 0])}})).

This is exactly  the Hamiltonian considered in previuos exercize, with (Subscript[E, 1])= 0,and the two eigenstates are

({{|g⟩= (1/(Sqrt[2]))}})(({{|+⟩ +}})({{|-⟩ )}})≡ ({{|S⟩;   }})({{|e⟩= (1/(Sqrt[2]))}})(({{|+⟩ -}})({{|-⟩ )}})≡ ({{|A⟩;   }})

Let us notice that for a one dimensional problem we now that ground state is symmetric, then in effect |g⟩ is the ground state. This state is even under parity, P|g⟩ = |g⟩ while excited state is odd.

A question for the reader: the sign of Δ depends onthe choice of the relative phase between |+⟩ and |-⟩ states. How this is consistent withthe assertion that ground state for a one dimensional potential must be symmetric?

Electric field

The charge distribution in configurations |+⟩ and |-⟩ is asymmetric, due to greater electronegativity of Nitrogen. Classically this  that these states possess an electric dipole. We can describe this effect with

d = ⟨+|e z|+⟩ ;  - d =  ⟨-|e z|-⟩ ; ⟨-|e z|+⟩ = 0;

We note that the first equality is a definition, while the other two equalities are a consequence of our model.

Let us assume that our states are (relatively ) real, e.g. are approximate real solution of one dimensional Schrödinger equation. The parity operator P must acts on z as P z (P^(-1))≡ P z P = - z (we used (P^2)= 1), then

⟨-|z|-⟩ = ⟨+|P z P|+⟩ = - ⟨+|z|+⟩ = -d.

For the third equality,using reality of matrix elements:

⟨-|z|+⟩ = ⟨+|P z P|-⟩ = - ⟨+|z|-⟩ = - (SuperStar[(⟨-|z|+⟩)]) = - ⟨-|z|+⟩  ⇒   ⟨-|z|+⟩ = 0.

Let us note that states | +⟩ and | -⟩ are not eigenstates of Hamiltonian () so the general theorem about absence of electric dipole moment on stationary states do not apply. If we neglect tuneling the states are eigenstates of () but the system is degenerate and again the theorem does not apply.

If we switch on an electric field ℰ, this field will couple with dipole mpoment and the Hamiltonian becomes

H =(Subscript[E, 0])+ ( ({{-d ℰ, -Δ}, {-Δ,  d ℰ}})).

The Hamiltonian () is identical to that of problem [] and we can write at once anergy levels and eigenstates:

E = (Subscript[E, 0]) ∓ (Sqrt[(Δ^2)+ (d^2)(ℰ^2)]);  ({{|(Subscript[g, ℰ])⟩}})= (({{Cos[(θ/2)] }, {Sin[(θ/2)] }}));  ({{|(Subscript[e, ℰ])⟩}})= (({{-Sin[(θ/2)] }, {  Cos[(θ/2)] }}));           Tan[θ] = (Δ/(d ℰ)).

Equation () explain in a simple way how in quantum mechanics we can have approximatively a linear Stark effect, which mimics the classical interaction - d ℰ. In the limit of large fields with respect the transition matrix element, or energy splitting, dℰ ≫ Δ, we have θ → 0 and

E = (Subscript[E, 0]) ∓ d ℰ;   ({{(Subscript[|g⟩, ℰ])}})≃ (({{1 }, {0 }}));  ({{(Subscript[|e⟩, ℰ])}})= (({{0 }, {  1 }}));

We have classical linear dependency on electric field wich mimics a permanent dipole of the system. Eigenstates are "frozen" toward classical localized states. In these states, see eq.()

⟨(Subscript[g, ℰ])|e z|(Subscript[g, ℰ])⟩≃ d ;    ⟨(Subscript[e, ℰ])|e z|(Subscript[e, ℰ])⟩≃ - d .

In effect permanent static dipole, polarizability etc, are defined in the limit of vanishing extenal field. Let us assume d ℰ → (SuperPlus[0]), then θ → π/2 and we have

E = (Subscript[E, 0]) ∓ (Δ  + (1/2) (((d^2)(ℰ^2))/Δ));

({{|(Subscript[g, ℰ])⟩}})=(1/(Sqrt[2]))((1+ ((d ℰ)/(2Δ)) )({{|+⟩ + }})(1- ((d ℰ)/(2Δ)) )({{|-⟩  }}));
 ({{|(Subscript[e, ℰ])⟩}})=(1/(Sqrt[2]))((1- ((d ℰ)/(2Δ)) )({{|-⟩ - }})(1+ ((d ℰ)/(2Δ)) )({{|+⟩  }}));

From eq.() we see that Stark effect is indeed quadratic with polarizations

E - (Subscript[E, 0])=- (1/2)α (ℰ^2);        (Subscript[α, g])= (( (d^2))/Δ);    (Subscript[α, e])= -(( (d^2))/Δ).

We note that for excited states polarizability has a negative sign, so the atoms behave at the opposite of usual dielectrics (they are repelled in an external gradient field).

These result are confirmed by taking the mean value of the dipole operator on exact eigenstates  inthe limit of small field:

⟨(Subscript[g, ℰ])|e z|(Subscript[g, ℰ])⟩ ≃ (( (d^2))/Δ)ℰ ;            ⟨(Subscript[e, ℰ])|e z|(Subscript[e, ℰ])⟩ ≃  -(( (d^2))/Δ)ℰ .

Problem

Study the effect of an electric field (Lo Surdo-Stark effect) on the n=2 states of hydrogen atom, neglecting the effects of spin.

Solution

General properties

In the non relativistic approximation the n=2 level of hydrogen atom has degeneration 4, (we neglect spin degrees of freedom): there is one 2s state and three 2p states. Let us choose as quantization axis (z axis) the direction of the electric field ℰ. The Hamiltonian is

H = ((p^2)/(2m)) - ((e^2)/r) - e z ℰ

e is the electron charge. H is invariant for rotations around z axis, then (Subscript[L, z ])can be diagonalized simultaneously with H. In perturbation theory this is reflected by the fact that the term e z ℰ has no matrix elements between states with different eigenvalues of (Subscript[L, z]).

The Hamiltonian is also invariant under reflection ℛ across a plane containing z axis, ex. y → - y. Classically the porjections of angular momentum along z change sign under this operation, the same is true in quantumtheory. In angular coordinates this symmetry is equivalent to φ → - φ and with our choice of phase for spherical harmonics

(Subscript[Y, L]^m)(ϑ,-φ)==(((-1))^m) (Subscript[Y, L]^(-m))(ϑ,φ);  ⇒    ℛ | n, L, m⟩ =  (((-1))^m)| n, L,- m⟩.

As H commutes with ℛ the states with opposite m (eigenvalue of (Subscript[L, z]))will remain degenerate even in presence of the perturbation. The same conclusion can be drawn using time reversal symmetry and the relation

(Subscript[Y, L]^(-m))(ϑ,φ)==(((-1))^m) (((Subscript[Y, L]^m)(ϑ,φ))^*).

Perturbation theory

Perturbation theory on n = 2 level is a typical example of degenerate perturbation theory and we have to diagonalize matrix elements of the perturbation on the subspace spanned by states 2s, 2p. We have seen some selection rules in previous section, to this we add that z is odd with respect parity, so only matrix elements between different parity can be different from zero.

Denoting the states with |2s⟩, |2p,m⟩, where m is the eigenvalue of (Subscript[L, z]), at first order the selection rule on (Subscript[L, z]) imply the following form block diagonal form  for perturbation:

-e ℰ(({{0, ⟨2s|z|2p,0⟩, 0, 0}, {⟨2p,0|z|2s⟩, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}}))= -e ℰ(({{0, a, 0, 0}, {a, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}})).

We see that energy of states |2p, ± 1⟩ are left unchanged by the perturbation, its only effect is a mixing between |2s⟩ and |2p,0⟩ states. Wave functions for these states are

(Subscript[ψ, 2s]) = (Subscript[R, 20])(Subscript[Y, 00])= (1/((a^(3/2))(Sqrt[8π])))Exp[-(r/(2 a))](1- (r/(2a)));                (Subscript[ψ, 2p,0]) =(Subscript[R, 21])(Subscript[Y, 20]) =(1/(a^(3/2))) (1/(2(Sqrt[6]))) (r/a)Exp[-(r/(2a))](Sqrt[(3/(4π))]) Cos[θ];

a is the Bohr radius. An easy integral gives

⟨2s| - e ℰ z |2p,0⟩ = 3 e ℰ a.

The rlevant part of the perturbation to be diagonalized is just the 2×2 matrix

3e ℰ a (({{0, 1}, {1, 0}}))

Eigenvalues and eigenvectors are

(Subscript[E, A]) = - 3 e ℰ a;   (Subscript[ψ, A])=  (1/(Sqrt[2]))( (Subscript[ψ, 2s]) - (Subscript[ψ, 2p,0]));     (Subscript[E, B]) = + 3 e ℰ a;   (Subscript[ψ, B])=  (1/(Sqrt[2]))( (Subscript[ψ, 2s]) + (Subscript[ψ, 2p,0])).

We have a linear Stark effect due the degeneracy of the n = 2 level. States (Subscript[ψ, A])and (Subscript[ψ, B]) are not parity eigenstates and admit a dipole moment. Fromprevious matrix elements it follow

⟨(Subscript[ψ, A])| z |(Subscript[ψ, A])⟩ = + 3 a ;      ⟨(Subscript[ψ, B])| z |(Subscript[ψ, B])⟩ = + 3 a .

Problem

A particle moves in a one-dimensional potential U. Let us suppose that we know all eigenfunctions for this system and that a perturbation V[x] acts on the system. The computation of the first order correction to eigenvalues is, in usual cases, reduced to an integral. The computation of first order correction (ψ^((1))) to wave functions involves generically a sum over infinite terms. Show that the problem of finding (ψ^((1)))can be recast in the form of solving an inhomogenous partial differential equation.  Set (ψ^((1)))=f[x](ψ^((0)))[x] for first order eigenfunctions and write an equation for f.  In particular consider the case of a central potential and a perturbation in the form of a constant electric field. This method of solution is known as Dalgarno-Lewis' method.

Solution

The method

The complete stationary Schrödinger equation for this problem has the form:

-((ℏ^2)/(2m))Δ ψ + U ψ + V ψ = E ψ.

Let (Subscript[ψ, 0])a given eigenstate solution of the unperturbed equation, with eigenvalues (Subscript[E, 0]). Let us suppose that the level is non degenerate or that  V is diagonal with respect quantum numbers used to classify the eigenstates of the degenerate subspace belonging to eigenvalue (Subscript[E, 0.])In these cases the first order correction to energy and eigenstate are given by

(Subscript[ϵ, 1])= ∫ |(Subscript[ψ, 0])(|^2) V dx ;    (Subscript[ψ, 1])= (∑_(s≠0))(Subscript[f, s])[x] (1/((Subscript[E, 0])-(Subscript[E, s])))⟨s|V|0⟩;

(Subscript[f, s])is the generic name for eigenfunctions. In general is very difficult to compute exactly the infinite sum in (), in this problem we study a method which reduce this problem to a solution of a differential equation, usually easier to manage. Having computed (Subscript[ψ, 1]) perturbation theory allows an easy computation of second and third order corrections to energy levels.

let us write ψ = (Subscript[ψ, 0])+ (Subscript[ψ, 1])and E = (Subscript[E, 0])+(Subscript[ϵ, 1]) with (Subscript[ψ, 1])and (Subscript[ϵ, 1]) of first order in the perturbation. Neglecting second order quantities in V and using the fact that (Subscript[ψ, 0])satisfies unperturbed equation, Schrödinger equation for ψ can be written as

-((ℏ^2)/(2m))Δ (Subscript[ψ, 1]) + U (Subscript[ψ, 1]) + V (Subscript[ψ, 0]) = (Subscript[ϵ, 1])(Subscript[ψ, 0]) + (Subscript[E, 0])(Subscript[ψ, 1]);   or   ((Subscript[H, 0])- (Subscript[E, 0]))(Subscript[ψ, 1])= ((Subscript[ϵ, 1])-V)(Subscript[ψ, 0])

This is an inhomogenous equation for (Subscript[ψ, 1]). Let ϕ one particular solution of this equation, as the operator (Subscript[H, 0])- (Subscript[E, 0]) has a nontrivial kernel of dimension 1 (we assume non degeneracy) then the general solution is of the form

(Subscript[ψ, 1]) = ϕ + c (Subscript[ψ, 0]).

In perturbation theory the correction to a state is fixed by requiring  ⟨(Subscript[ψ, 1])|(Subscript[ψ, 0])⟩= 0, this fixes the constant and the solution is

(Subscript[ψ, 1])= ϕ - ⟨(Subscript[ψ, 0])|ϕ⟩ (Subscript[ψ, 0]).

The equation () can be simplified with the position (Subscript[ψ, 1])= f (Subscript[ψ, 0]). Expanding the derivatives and using equation for (Subscript[ψ, 0])we have:

-((ℏ^2)/(2m))(Subscript[ψ, 0])Δ f  - ((ℏ^2)/m)∇f·∇(Subscript[ψ, 0]) + (V-(Subscript[ϵ, 1]))(Subscript[ψ, 0])=0.

This equation does not contain explicitely the unperturbed potential U.

Electric field and central potential

Consider now the particularly important case of a central potential U[r] and a perturbationdue to an electric field V = - e ℰ z . We want to compute corrections to the wave function of fundamental state. (Subscript[ψ, 0])is spherically symmetric and the perturbation can mix only to states with L=1,(Subscript[L, z])=1, odd states, i.e. functions of the form z Φ[r]. The first order correction (Subscript[ϵ, 1])is zero for parity. We then put in the general equation ()

f = z F[r].

Using (the prime denote a derivative with respect to r)

(∂_i) (Subscript[ψ, 0])= ((Subscript[x, i])/r)(Subscript[ψ, 0]^(')) ;  (∂_i) (z F) = (Subscript[δ, i3])F + z ((Subscript[x, i])/r)F';    ⇒    (∂_i) (Subscript[ψ, 0])(∂_i)(z F)= (Subscript[ψ, 0]^('))((z/r)F + z F'); Δ(zF) = 4(z/r)F'+z F'';

the equation for F becomes

-((ℏ^2)/(2m))(4(z/r)F'+z F'') - ((ℏ^2)/m) ((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))((z/r)F + z F') - e ℰ z = 0;

i.e.

((1/2)F''+ (2/r)F') + ((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))((F/r)+ F')+ e ℰ (m/(ℏ^2))= 0.

This equation is valid for any central potential, but clearly (Subscript[ψ, 0]) depends on the problem at hand.

Electric field and central potential

Consider now the particularly important case of a central potential U[r] and a perturbationdue to an electric field V = - e ℰ z . We want to compute corrections to the wave function of fundamental state. (Subscript[ψ, 0])is spherically symmetric and the perturbation can mix only to states with L=1,(Subscript[L, z])=1, odd states, i.e. functions of the form z Φ[r]. The first order correction (Subscript[ϵ, 1])is zero for parity. We then put in the general equation ()

f = Cos[θ] F[r];    (Subscript[ψ, 1])= Cos[θ] F[r](Subscript[ψ, 0])[r].

In polar coordinates the Laplace operator is (ℒ is the angular momentum operator)

Δ = (2/r)(∂_r)+ (∂_r^2) -  ((ℒ^2)/(r^2));   and    ℒ Cos[θ] = 2 Cos[θ]

Cos[θ] is proportional to (Subscript[Y, 10])then (ℒ^2) Cos[θ] = 2 Cos[θ] and equation ()  takes the form (with (Subscript[ϵ, 1])=0)

Cos[θ]F(((ℏ^2)/(2m))(Subscript[Δψ, 0]) + ((Subscript[E, 0]) - U)(Subscript[ψ, 0])) + ((ℏ^2)/(2m))Cos[θ]( (2/r)F'(Subscript[ψ, 0])+ 2 F'(Subscript[ψ, 0]^('))+(Subscript[ψ, 0])F''- (2/(r^2))F (Subscript[ψ, 0]))+ e ℰ r Cos[θ](Subscript[ψ, 0])=0

Using Schrödinger equation for (Subscript[ψ, 0])

((1/2)F''+ ((F')/r)-(F/(r^2)))+ ((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))F'+ ((e ℰ m)/(ℏ^2))r = 0.

This equation is valid for any central potential, but clearly (Subscript[ψ, 0]) depends on the problem at hand. Solving this equation we can compute (Subscript[ψ, 1])and second order correction to energy is simply an integral

(Subscript[ϵ, 2]) = ⟨(Subscript[ψ, 0])|V|(Subscript[ψ, 1])⟩ = - e ℰ ⟨(Subscript[ψ, 0])|r Cos[θ]|(Subscript[ψ, 1])⟩ = -  ((e ℰ)/3)∫dr (r^2) (Subscript[ψ, 0]^2) F[r]r.

In the last formula we have used the equality ⟨((Cos[θ])^2)⟩= 1/3.

Higher orders

Stark effect contain only even powers of angular momentum, and even power of electric field in energy shift, for parity reasons. Equation () then generalizes as

ψ = (Subscript[ψ, 0])+ Cos[θ] (Subscript[F, 1])[r](Subscript[ψ, 0])[r] + ((Subscript[P, 2])[Cos[θ]](Subscript[F, 22])[r]+ (Subscript[F, 20])[r])(Subscript[ψ, 0])+ …

(Subscript[P, L])is Legendre polynomial, (ℒ^2)(Subscript[P, L])= - L(L+1)(Subscript[P, L.])By substitution in () and using Schrödinger equation for (Subscript[ψ, 0]) we have now, up to second order and remembering that (Subscript[ϵ, 1])=0:

((ℏ^2)/(2m))Cos[θ]( (2/r)(Subscript[F, 1]^('))(Subscript[ψ, 0])+ 2 (Subscript[F, 1]^('))(Subscript[ψ, 0]^('))+(Subscript[ψ, 0])(Subscript[F, 1]^(''))- (2/(r^2))(Subscript[F, 1]) (Subscript[ψ, 0]))+((ℏ^2)/(2m))(Subscript[P, 2])( (2/r)(Subscript[F, 22]^('))(Subscript[ψ, 0])+ 2 (Subscript[F, 22]^('))(Subscript[ψ, 0]^('))+(Subscript[ψ, 0])(Subscript[F, 22]^(''))- (6/(r^2))(Subscript[F, 22]) (Subscript[ψ, 0]))

+((ℏ^2)/(2m))( (2/r)(Subscript[F, 20]^('))(Subscript[ψ, 0])+ 2 (Subscript[F, 20]^('))(Subscript[ψ, 0]^('))+(Subscript[ψ, 0])(Subscript[F, 20]^('')))+ e ℰ r Cos[θ](Subscript[ψ, 0]) + e ℰ r Cos[θ]( Cos[θ] (Subscript[F, 1])[r](Subscript[ψ, 0])) + (Subscript[ϵ, 2])(Subscript[ψ, 0])=0

We can use

((Cos[θ])^2) = ((2 (Subscript[P, 2])[Cos[θ]] + 1)/3)

to separate different angular momenta, obtaining, equating similar orders in ℰ:

( (1/2)(Subscript[F, 1]^(''))+(1/r)(Subscript[F, 1]^('))- (1/(r^2))(Subscript[F, 1])+((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))(Subscript[F, 1]^(')))+ ((e ℰ m)/(ℏ^2))r = 0;
( (1/2)(Subscript[F, 22]^(''))+(1/r)(Subscript[F, 22]^('))- (3/(r^2))(Subscript[F, 22])+((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))(Subscript[F, 22]^(')))+ (2/3)((e ℰ m)/(ℏ^2))r (Subscript[F, 1])= 0;
( (1/2)(Subscript[F, 20]^(''))+(1/r)(Subscript[F, 20]^('))+((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))(Subscript[F, 20]^(')))+ (1/3)((e ℰ m)/(ℏ^2))r (Subscript[F, 1]) + (Subscript[ϵ, 2])(m/(ℏ^2))=0

Problem

Use results of problem [] to compute the approximate polarizability for a particle moving in a one dimensional potential well of width a, with the condition a χ ≪ 1, where χ = (Sqrt[2m|E|/(ℏ^2)]) , |E| is the binding energy.

Solution

Unidimensional case

Outside the potential well the unperturbed wave function behaves as

(Subscript[ψ, 0]) = A Exp[- χ |x|]

The condition a χ ≪ 1 imply a slow variation inside the well and (Subscript[ψ, 0])∼ A for |x| ≤ a. In this conditions the contribution to electrostatic energy inside the well canbe neglected and the wave functioncanbe taken everywhere in the form ().  In this approximation A = (Sqrt[χ]).

We look at first oder corrections to (Subscript[ψ, 0]) due to a perturbation - e ℰ x,  ℰ is an external electric field. If we write (Subscript[ψ, 1])= f (Subscript[ψ, 0])it has been shown in problem [] that f[x] satisfy the differential equation

(1/2)f''+ ((Subscript[ψ, 0]^('))/(Subscript[ψ, 0])) f'+  ((m e ℰ)/(ℏ^2)) x = 0  ⇒ (1/2)f''-χ ε[x] f'+((m e ℰ)/(ℏ^2)) x = 0.

With ε[x] = Sign[x]. A particular solution of this equation is the odd function:

f[x] = (1/2)((m e ℰ)/(ℏ^2))( (x/(χ^2))+ (((x^2)ε[x])/χ))

The odd function f (Subscript[ψ, 0])is automatically orthogonal to the even function (Subscript[ψ, 0]), then the first order correction to ground state is

(Subscript[ψ, 1])[x] = (1/2)((m e ℰ)/(ℏ^2))( (x/(χ^2))+ (((x^2)ε[x])/χ)) (Sqrt[χ]) Exp[-χ |x|].

We can compute second order Stark effect with

ΔE = ⟨(Subscript[ψ, 0])|V|(Subscript[ψ, 1])⟩ = - e ℰ ⟨(Subscript[ψ, 0])|x|(Subscript[ψ, 1])⟩.

Performing the integral  we have for energy shift and polarizability:

ΔE = - (1/2)(5/4)((m (e^2) (ℰ^2))/(ℏ^2)) (1/(χ^4)) ;  ΔE = -(1/2)α (ℰ^2)   ⇒  α = (5/4)((m (e^2))/(ℏ^2)) (1/(χ^4)).

Three dimensional well

In the same approxiamation the wave function of the ground state can be taken as

(Subscript[ψ, 0]) = (Sqrt[2 χ]) (1/r)Exp[- χ r](1/(Sqrt[4π]))

We put (Subscript[ψ, 1])= Cos[θ] F (Subscript[ψ, 0]), see problem [], and F must satisfy

((1/2)F''+ ((F')/r)-(F/(r^2)))+ ((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))F'+ ((e ℰ m)/(ℏ^2))r = 0 ⇒ ((m ℰ e)/(ℏ^2))r-((F[r])/(r^2))-χ F'[r]+((F''[r])/2)=0.

If we look for a polynomial solution we find

F[r] = ((m  ℰ e)/(χ (ℏ^2)))((r^2)/2)

and

(Subscript[ψ, 1])= Cos[θ]  ((m ℰ e)/(ℏ^2))((r^2)/(2χ)) (Subscript[ψ, 0]).

The second order energy shift is

ΔE = -  ((m (ℰ^2) (e^2))/(ℏ^2))(1/(2χ))⟨(Subscript[ψ, 0])|(r Cos[θ])(r^2)Cos[θ] (Subscript[ψ, 0])⟩ = - (1/2) ((m (ℰ^2) (e^2))/(ℏ^2))(1/(4(χ^4)));       α = ((m (ℰ^2) (e^2))/(4 (ℏ^2)(χ^4))).

Problem

Use results of problem [] to compute the  polarizability for the ground-state hydrogen atom. Give an estimate of the contribution of continuum states.

Solution

Polarizability

The polarizability α is defined through the expression for energy shift in a constant electric field ℰ:

ΔE = -(1/2)α (ℰ^2).

For parity conservation in unperturbed Hamiltonian, the first order correction to energy level is zero. The second order correction is computed by the general formula

ΔE = ⟨(Subscript[ψ, 0])|- e ℰ z | (Subscript[ψ, 1])⟩;

where (Subscript[ψ, 0])and (Subscript[ψ, 1])are respectively the unperturbed wave function and its first order correction.

In the following we use atomic units, with e = ℏ=m=1. To come back to usual unities it is sufficient to note that the adimensional expansion parameter is (a is the Bohr radius).

ℱ = e ℰ a/((e^2)/a).

The wave function of ground state hydrogen atom is

(Subscript[ψ, 0])=  2 Exp[-r]/(Sqrt[4π]).

To compute (Subscript[ψ, 1])we use the results of problem []. With

(Subscript[ψ, 1]) = Cos[θ] F[r] (Subscript[ψ, 0])

F must satisfy, in atomic units,

0 = ((1/2)F''+ ((F')/r)-(F/(r^2)))+ ((Subscript[ψ, 0]^('))/(Subscript[ψ, 0]))F'+ ℰ r = ((1/2)F''+ ((F')/r)-(F/(r^2)))- F'+ ℰ r

A particular regular solution of this equation is

F[r] = ℰ (r + (1/2)(r^2)).

Inserting this solution in () we have (Subscript[ψ, 1.])This function is orthogonal to (Subscript[ψ, 0])behaving as a angular momentum 1, then no more computations are necessary.

(Subscript[ψ, 1])= Cos[θ] ℰ (r + (1/2)(r^2)) (Subscript[ψ, 0]).

Inserting in () we obtain easily

ΔE =- ((9 )/4)(ℰ^2),

and for the polarizability

α = (9/2);          α = (9/2)(a^3)  (usual unities)

Contribution of continuum states

From

ΔE = (ℰ^2) (∑_(n≠1s))(((|⟨1s|z|n⟩|)^2)/((Subscript[E, 1s])-(Subscript[E, n]))),

it follows

α = 2(∑_(n≠1s))(((|⟨1s|z|n⟩|)^2)/((Subscript[E, n])-(Subscript[E, 1s])))

We see that all terms give a positive contribution to α. Angular momentum and parity conservation imply that to the sum only discrete p-states and continuum states can contribute.

It is easy to give an upper bound to α considering the lowest possible denominator: (Subscript[E, 2p])- (Subscript[E, 1s])= 3/8 and using completeness:

α <(2/(3/8))(∑_(n≠1s))((|⟨1s|z|n⟩|)^2) = (16/3) ⟨1s|(z^2)|1s⟩ = (16/3).

A lower bound is obtained inserting only the |2p;m=0⟩ state into the sum

α > (16/3)((|⟨1s|z|2p;0⟩|)^2).

The angular part of 2 p state wave function is (Subscript[Y, 10])[θ,φ]then previuous angular integral  is

∫(Subscript[Y, 00])(Subscript[Y, 10])Cos[θ] dΩ = ∫((dΩ)/(4π))(Sqrt[3])((Cos[θ])^2)= (1/(Sqrt[3]))

α > (16/9)((|⟨1s‖r‖2p;0⟩|)^2)

where ⟨1 s ‖r‖ 2 p; 0⟩ stands for reduced matrix element, i.e. matrix element between radial waves functions. From

(Subscript[R, 1s])= 2 Exp[-r];  (Subscript[R, 2p])= (1/(2(Sqrt[6]))) r Exp[-r/2]

we have

((|⟨1s‖r‖2p;0⟩|)^2) = ((2^14)/(3^8))(2/3);              α >(( (2^19))/(3^11))≃ 2.95.

Previous computations extend immidiately to all np discrete states, which are the whole discrete contribution to α:

(Subscript[α, disc]) = (2/3)(∑_(n=2)^∞)(2/(1-(1/(n^2))))((|⟨1s‖r‖np;0⟩|)^2).

The radial integrals can be computed,

((|⟨1s‖r‖np;0⟩|)^2) =  (2^8) (n^7) ((((n-1))^(2n-5))/(((n+1))^(2n+5))),

and the series evaluated numerically

(Subscript[α, disc]) = (2/3)(∑_(n=2)^∞)(2/(1-(1/(n^2)))) (2^8) (n^7) ((((n-1))^(2n-5))/(((n+1))^(2n+5))) ≃ 3.663

this means that continuum contribution is

(Subscript[α, c])= 9/2 - (Subscript[α, disc])≃ 0.837.

We leave to the reader the following exercise : let us approximate continuous states by plane waves, show that the extimated continuum contribution will be 7/3, i.e. a bad approximation in excess of the true result.

Problem

Spin interactions split n=2 degenerate levels of hydrogen atom in two levels (2(Subscript[s, 1/2]),2(Subscript[p, 1/2]))and 2(Subscript[p, 3/2]). Study the effect of an electric field (Lo Surdo-Stark effect) taking into account this separation. The level  (2(Subscript[s, 1/2]),2(Subscript[p, 1/2])) is split due to radiative corrections (Lamb shift), the separation being about (10^(-6))eV.Which are the effects of this separation on the Lo Surdo-Stark effect?

Solution

Orders of magnitude

In problem [] we have studied an idealized form of Lo Surdo - Stark effect on hydrogen atom, neglecting spin effects, fine structure etc. In this problem our aim is to be more close to reality and see quantitatively the interplay of different interactions.

We remember some numbers, a.u. stands for atomic unit, Ry for Rydberg, eV for electron Volt and Δ (Subscript[E, FS]) for fine structure energy splitting. a is the Bohr radius:

1 a.u. =2 Ry ≃ 27.21 eV;      (Subscript[ΔE, FS]) ≃ (α^2) a.u. ≃ (10^(-4)) a.u.≃ 2.7 (10^(-3))eV;     a = 5.29 (10^(-9 ))cm.

If we measure the electric field in Volt/cm the typical energy involved is

ℱ = e ℰ a ∼ 5.29 (10^(-9)) eV ℰ

We see that up to field of the order of  (10^5)-(10^6)Volts/cm the fine structure is much bigger than electric field effects, so the previous problem was quite unrealistic for small fields.

Another energy scale is given by the separation induced by radiative corrections on 2(Subscript[s, 1/2])and 2(Subscript[p, 1/2 ])levels. Infrequency this separation is about 1057 MHz in energy

(Subscript[δE, Lamb]) ≃ 1.057 (10^9) h = 4.37 (10^(-6)) eV

This means that up to field of  (10^3)Volts/cm this splitting is  higher than electric field effects, i.e. the levels appear not degenerate. If we put real numbers for fine structure of n=2 level we find in effect only 1 order of magnitude between Lamb shift and fine structure, (Subscript[δE, Lamb]) ∼ 0.1 (Subscript[ΔE, FS]) (n=2).

In real life things are complicated by the width of the states, which we neglect in this problem, and by hyerfine structure, i.e. splitting due to nuclear magnetic moment.

Fine structure

In this section we take into account the fine structure but neglect Lamb shift, so the solution is reasonable only for ℰ ≫ (10^3)Volts/cm.

Level n=2 of hydrogen atom consists of 8 states, taking into account the spin of the electron. We know, see text, that spin orbit interaction commute only with total angular momentum j = L + S. A level with a given L (S = 1/2 is understood for a single electron) split according to the values of j, with |L-S| ≤ j ≤ |L + S|. Compatible quantum numbers are E, j,(Subscript[j, z]), and in a perturbative analysis a state can be identifyed by |n,L; j,(Subscript[j, z])⟩, n being the principal quantum number and L the angular momentum of the unperturbed state. As we work exclusively with n=2 states we omit this number inthe notation of states, which can be denoted by

({{|s;(1/2),±(1/2)⟩}}) ,  ({{|p;(1/2),±(1/2)⟩}}) ,  ({{|p;(3/2),(±(1/2),±(3/2))⟩}}).

The accidental degeneracy s - p survives spin orbit and relativistic corrections and 2(Subscript[s, 1/2]) 2(Subscript[p, 1/2]) levels remain degenerate. Radiative corections produce an energy shift (Subscript[Δ, L]) between these two states. If we call (Subscript[Δ, F]) the fine structure splitting between 2(Subscript[p, 3/2])and 2(Subscript[p, 1/2]) the energies of the levels above can be written, in order

(Subscript[E, 2(Subscript[p, 1/2])]) + ( (Subscript[Δ, L]) , 0, (Subscript[Δ, F]) ).

In what follow we neglect the common energy factor and consider only splittings.

Selection rules

As discussed on problem [] the Hamiltonian in presence of an external electric field, directed along z by convention, is invariant under the reflection ℛ: y → - y, or the azimuthal angle φ → - φ, and this imply a degenracy for levels with opposite (Subscript[j, z]). In effect we discussed the simmety only in absence of spin, but the operation ℛ can be implemented also on spinors without major changes. Even in presence of ℰ the levels have a double degeneracy and we can limit ourselves to the study of positive (Subscript[j, z]), i.e to four states.

z commutes with (Subscript[j, z]), so states with different (Subscript[j, z]) cannot mix via first order perturbation,i.e. state |p; 3/2,3/2⟩ is separated from other states.

z is behaves like a component of a vector, i.e. is odd under parity. This, and the previous point, imply that the only possible matrix element for  |p; 3/2,3/2⟩ (the diagonal one) vanishes. We are left with only three mixing states,  |s; 1/2,3/2⟩,  |p; 1/2,1/2⟩,  |p; 3/2,1/2⟩, and for parity only s-p matrix elements can be different from zero.

Matrix elements

We know from problem  [] that the only orbital matrix element different from zero was

⟨2s| - e ℰ z |2p,m=0⟩ = 3 e ℰ a.

to compute the full matrix of perturbation we have only to write decomposition of states into orbital and spinpart, i.e. use Clebsch-Gordan coefficients. Denoting by (Subscript[ψ, Lm])the orbital part of the state and by |+⟩ and |-⟩ the spinors, we have

({{|p;(1/2),(1/2)⟩}})= (Sqrt[(2/3)])(Subscript[ψ, 2p,1]) |-⟩- (Sqrt[(1/3)])(Subscript[ψ, 2p,0]) |+⟩;  ({{|p;(1/2),(1/2)⟩}})= (Sqrt[(1/3)])(Subscript[ψ, 2p,1]) |-⟩+ (Sqrt[(2/3)])(Subscript[ψ, 2p,0]) |+⟩; ({{|s;(1/2),(1/2)⟩ = (Subscript[ψ, 2s,0]) ({{|+⟩;  }})}})

Using () matrix elements for the perturbation, both Stark and Hyperfine splitting, are given by (the order of rows is the one in which states are written in previous equation):

V = (({{(Subscript[Δ, F]), 0, -(Sqrt[2])x}, {0, 0, x}, {-(Sqrt[2])x, x, (Subscript[Δ, L])}}));  where   x ≡ (Sqrt[3]) e ℰ a.

Solutions of the secular equation  det(V-λ) = 0, gives energy levels.

Let us now consider different regimes

x ≫ (Subscript[Δ, F]) ≫ (Subscript[Δ, L]) ( i.e. ℰ ≫ (10^9) V/cm )

In this regime approximatively we have to solve

det(({{-λ, 0, -(Sqrt[2])x}, {0, -λ, x}, {-(Sqrt[2])x, x, -λ}}))= 0  ⇒ - (λ^3) + 3 (x^2)λ = 0;  ⇒ λ = (0, ± (Sqrt[3])x) = (0, ± 3 e ℰ a)

We recover the result of problem  []. Remembering the decoupled and unaffected |p;3/2,3/2⟩ state and the doubling (Subscript[j, z])→ - (Subscript[j, z])  we have 4 states at energy 0, 2 states at energy 3 eℰa and 2 states at energy - (3 ℰ e a). The doubling in the degeneracy with respect to problem [] is due to spin. We have as expected a linear Stark effect.

x ≪ (Subscript[Δ, L]) ≪ (Subscript[Δ, F]) ( i.e. ℰ ≪ (10^3) V/cm )

It is a useful exercise to compute approximate eigenvalues of () using perturbation theory to this matrix, x - terms elements are the perturbation, δV. There are no diagonal elements, so first order effect is zero, and correction start from (x^2).At second order consider for example the first state. We have as is well known

(Subscript[ΔE, 1])= ⟨1|δV |n⟩ (1/((Subscript[E, 1])- (Subscript[E, n])))⟨n|δV|1⟩ = ⟨1|δV |3⟩ (1/((Subscript[E, 1])- (Subscript[E, 3])))⟨3|δV|1⟩ = ((2(x^2))/((Subscript[Δ, F])-(Subscript[Δ, L])));  (Subscript[E, 1])= (Subscript[Δ, F])+ (Subscript[ΔE, 1])

For the other two levels we have respectively

(Subscript[ΔE, 2]) = - (( (x^2))/(Subscript[Δ, L]));  (Subscript[ΔE, 3]) = (x^2)((1/(Subscript[Δ, L])) - (2/((Subscript[Δ, F])-(Subscript[Δ, L]))));  (Subscript[E, 3])= (Subscript[Δ, L])+ (Subscript[ΔE, 3]).

(Subscript[Δ, L]) ≪ x ≪ (Subscript[Δ, F])

This case is a bit academic as for n=2 levels (Subscript[Δ, L])∼ 0.1 (Subscript[Δ, F]) but can be a useful exercize for the reader. Neglecting (Subscript[Δ, L ])the matrix () becomes

V = (({{(Subscript[Δ, F]), 0, -(Sqrt[2])x}, {0, 0, x}, {-(Sqrt[2])x, x, 0}}));

The last two states,  |p;1/2,1/2⟩ and |s;1/2,1/2⟩ are degenerate and we have to apply for the submatrix 2×2 degenerate perturbation theory. Eigenstates nd eigenvalues are obvious

|A⟩ = (1/(Sqrt[2]))(|2⟩+ |3⟩);  eig. +x;     |B⟩ = (1/(Sqrt[2]))(|2⟩ - |3⟩);  eig. - x.

In the new basis, |1⟩, | A⟩, | B⟩ the matrix V reads

V = (({{  (Subscript[Δ, F]), -x,   x}, {-x,   x,   0}, {  x,   0, -x}})).

Now we can apply non degenerate perturbation theory to first state, i.e. formula () obtaining

(Subscript[E, 1])= (Subscript[Δ, F]) + 2 ((x^2)/(Subscript[Δ, F])).

Then, as expected, the two degenerate states  |p;1/2,1/2⟩ , |s;1/2,1/2⟩ have a linear Stark effect, while the third state, here  |p;3/2,1/2⟩ has a quadratic Stark effect.

Inthe figure below we present energy levels as a function of x/(Subscript[Δ, F]), with (Subscript[Δ, L])= 0.1 (Subscript[Δ, F]). We note that linear Stark effect is quite precocius and start at about x/(Subscript[Δ, F])∼0.6.

A useful exercise that we leave for the reader is to consider Stark effect as the leading term and fine structure as a perturbation, i.e. compute higher orders in ().

Problem

Parabolic coordinates  (ξ, η, φ) are  defined by

x = (Sqrt[ξ η]) Cos[φ]; y = (Sqrt[ξ η]) Sin[φ]; z = (1/2)(ξ-η);

In notebook problems [NB-6.4-ParabolicCoordinates.nb] it has been shown that the Schrödinger equation for a hydrogen atom is separable in these coordinates. Study the Lo Surdo - Stark effect in parabolic coordinates. Compute first and second order effect for an arbitrary state.

Solution

The Schrödinger equation

In this notebook we will use atomic units, ℏ = m = |e| = 1.

In notebook [*] the results obtained here are derived within Mathematica and extended to an arbitrary order in perturbation theory.

In notebook [*] the Schrödinger equation for a coulombic potential has been solved in parabolic coordinates. We report here the basic connection formulae between cartesian and parabolic coordinates and the for of the laplacian:

x = (Sqrt[ξ η]) Cos[φ]; y = (Sqrt[ξ η]) Sin[φ]; z = (1/2)(ξ-η);
ξ = r+z; η = r-z; φ=ArcTan[(y/x)];  r = (1/2)(ξ+η); 
Δ = (4/(ξ+η))(∂/(∂ξ))(ξ(∂/(∂ξ))) + (4/(ξ+η))(∂/(∂η))(η(∂/(∂η)))+ (1/(ξ η))((∂^2)/(∂(φ^2))).

The range for these coordinates is

0≤ (ξ, η ) ≤ ∞ ;   0≤ φ ≤ 2π

and volume element is

(𝕕^3)x = ((ξ+η)/4) 𝕕ξ 𝕕η 𝕕φ.

Interaction with an electric field ℰ, directed along z by convention,  is decribed by an interacion - e z ℰ = + |e| z ℰ. We put F for the product |e| ℰ a in atomic units (a is the Bohr radius) and write the interaction Hamiltonian as (see eq. () )

V = (F/2)(ξ - η).

Using () the Schrödinger equation takes the form:

-(( 1)/2)((4/(ξ+η))(∂/(∂ξ))(ξ(∂/(∂ξ))) + (4/(ξ+η))(∂/(∂η))(η(∂/(∂η)))+ (1/(ξ η))((∂^2)/(∂(φ^2))))ψ - ((2 Z)/(ξ+η))ψ +(F/2)(ξ - η)ψ =  E ψ.

or

((∂/(∂ξ))(ξ(∂/(∂ξ))) + (∂/(∂η))(η(∂/(∂η)))+ ((ξ+η)/(4ξ η))((∂^2)/(∂(φ^2))))ψ + (Z - (F/4)((ξ^2)-(η^2))+ ((ξ+η)/2)E)ψ = 0.

The sistem is invariant under rotation around z axis, then one of the "good" quantum numbers will be m, the eigenvalue of (Subscript[L, z]). We look for a solution with separate variables, in  the form

ψ = (Subscript[f, 1])[ξ] (Subscript[f, 2])[η] Exp[i m φ].

Substitution in (), after division by ψ, gives

(1/(Subscript[f, 1]))((d/(d ξ))(ξ((Subscript[df, 1])/(d ξ)))-((m^2)/(4ξ))(Subscript[f, 1])+ (E/2)ξ (Subscript[f, 1]) - (F/4)(ξ^2)(Subscript[f, 1])) + (1/(Subscript[f, 2]))((d/(d η))(η((Subscript[df, 2])/(d η)))-((m^2)/(4η))(Subscript[f, 2])+ (E/2)η (Subscript[f, 2])+ (F/4)(η^2)(Subscript[f, 2]))+Z = 0.

The first term depends only on ξ, the second only on η, since their sum is constant there must be two constant (Subscript[Z, 1]) and (Subscript[Z, 2]) such that

(d/(d ξ))(ξ((Subscript[df, 1])/(d ξ)))+((E/2)ξ-((m^2)/(4ξ)) - (F/4)(ξ^2)+ (Subscript[Z, 1]))(Subscript[f, 1])= 0;
(d/(d η))(η((Subscript[df, 2])/(d η)))+((E/2)η-((m^2)/(4η)) + (F/4)(η^2)+ (Subscript[Z, 2]))(Subscript[f, 2])= 0;
(Subscript[Z, 1])+ (Subscript[Z, 2])= Z.

We can simplify a bit these equations by the change of variables

ϵ= (Sqrt[-2 E]);  x = ϵ ξ;  y = ϵ η;  (Subscript[β, i]) =((Subscript[Z, i])/ϵ);  ℱ = (F/(ϵ^3)).

(d/(d x))(x((Subscript[df, 1])/(d x)))+(-(1/4)x-((m^2)/(4x)) - (ℱ/4)(x^2)+ (Subscript[β, 1]))(Subscript[f, 1])= 0;
(d/(d y))(y((Subscript[df, 2])/(d y)))+(-(1/4)y-((m^2)/(4y)) + (ℱ/4)(y^2)+ (Subscript[β, 2]))(Subscript[f, 2])= 0;
(Subscript[β, 1])+ (Subscript[β, 2])= (Z/ϵ).

As is apparent from () the spectrum depends only on |m|, and, as we know from general arguments, states ±m will be degenerate.

The procedure to find eigenvalues, as in the case ℰ = 0, is the following

Perturbative solution at first order

At zero order the solution of equations  () has been found in notebook [*]. Normalizability of the solutions impose

(Subscript[β, 1])= (Subscript[n, 1])+ ((|m|+1)/2);  (Subscript[β, 2])= (Subscript[n, 2])+ ((|m|+1)/2);       ⇒          (Subscript[β, 1])+(Subscript[β, 2]) = (Subscript[n, 1])+(Subscript[n, 2])+ (|m|+1) ≡ n;

then

(Subscript[β, 1])+(Subscript[β, 2]) = (Z/ϵ)       ⇒           ϵ = (Z/n)        ⇒        E = - (1/(2 (n^2)))

At fixed n, 0 ≤ |m| ≤n-1 and with fixed n and m, one of the two numbers, e.g. (Subscript[n, 1]), can vary as 0 ≤ (Subscript[n, 1])≤ n - |m| - 1, the other being fixed by ().

Normalized eigenfunctions are given by

(Subscript[f, (Subscript[n, i]),m])[x] =(Sqrt[((n!)/((n+|m|)!))]) (x^(|m|/2)) Exp[-(x/2)](Subscript[L, n]^(|m|))(x)≡(Subscript[c, (Subscript[n, i]),m])  (x^(|m|/2)) Exp[-(x/2)](Subscript[L, n]^(|m|))(x)
(∫_0^∞)(Subscript[f, (Subscript[n, 1]),m])(Subscript[f, (Subscript[n, 2]),m])dx = (Subscript[δ, (Subscript[n, 1])(Subscript[n, 2])]).

We will comment below on the normalization coefficients.

The perturbative calculation is done independently on the two equations in (), each of them is a kind of Schrödinger equation in which (Subscript[β, i]) has the role of energy eigenvalue. It is sufficient to do coputations for (Subscript[β, 1]), the other one is obtained by ℱ→ - ℱ.

To compute perturbative corrections  we need matrix elements of (x^2). We start from recursive relation

x (Subscript[L, n]^m)(x)=-(n+m) (Subscript[L, n-1]^m)(x)-(n+1) (Subscript[L, n+1]^m)(x)+(2 n+m+1) (Subscript[L, n]^m)(x).

Denoting the diagonal matrix element by ⟨ (Subscript[f, n])| x^2|(Subscript[f, n])⟩,using () to right and left functions, and inserting normalization coefficients, we have immediately (by orthogonality and omitting the fixed index m):

⟨(Subscript[f, n])|(x^2)|(Subscript[f, n])⟩ = ((Subscript[c, n]^2)/(Subscript[c, n+1]^2))(((n+1))^2)+  ((Subscript[c, n]^2)/(Subscript[c, n-1]^2))(((n+m))^2) + (2n+m+1) = 6(n^2)+6n(m+1)+ (m+1)(m+2).

Up to first order then

(Subscript[β, 1])= (Subscript[n, 1])+ ((|m|+1)/2) + (ℱ/4)( 6(Subscript[n, 1]^2)+6(Subscript[n, 1])(m+1)+ (m+1)(m+2) );  (Subscript[β, 2])= (Subscript[n, 2])+ ((|m|+1)/2) + (ℱ/4)( 6(Subscript[n, 2]^2)+6(Subscript[n, 2])(m+1)+ (m+1)(m+2) ).

and, see the definition of n in ():

(Subscript[β, 1])+(Subscript[β, 2])= n + ℱ ( (3/2)((Subscript[n, 1]^2) - (Subscript[n, 2]^2)) + (3/2)((Subscript[n, 1])-(Subscript[n, 2]))(m+1)) = n + (3/2) ℱ ((Subscript[n, 1])- (Subscript[n, 2]))n.

Imposing the constraint () , using ℱ = F/(ϵ^3) and the zeroth order result ϵ = Z/n

(Z/ϵ) = n + (3/2) (F/(ϵ^3)) ((Subscript[n, 1])- (Subscript[n, 2]))n ≃ n + (3/2) (F/(Z^3)) ((Subscript[n, 1])- (Subscript[n, 2]))(n^4)   ⇒  (ϵ/Z) ≃ (1/n)(1 - (3/2) (F/(Z^3)) ((Subscript[n, 1])- (Subscript[n, 2]))(n^3)) ⇒  ϵ ≃ (Z/n) - (3/2)F ((((n/Z)))^2)((Subscript[n, 1])-(Subscript[n, 2])),

and finally for energy, E = - (ϵ^2)/2

E = -(1/2)((Z^2)/(n^2)) + (3/2)F (n/Z)((Subscript[n, 1])-(Subscript[n, 2])).

This gives the first order Stark effect on all levels. It is easy to check low level to compare with more elementary results (prob. [] ), we consider hydrogen, Z=1:

n=1; m=0; ((Subscript[n, 1]),(Subscript[n, 2])) = 0 ⇒ δE = 0;
n=2; m=0; ((Subscript[n, 1]),(Subscript[n, 2])) = (1,0),(0,1) ⇒ δE = ∓3 F ;
n=2; m=±1; ((Subscript[n, 1]),(Subscript[n, 2])) = (0,0),(0,1) ⇒ δE = 0 ;

We have the known Stark effect  on level n = 2, to compare with problem [] use F = e ℰ a.

The normalization of wave functions

The eigenfunctions () are a complete and orthonormal set with respect the measure dx. The eigenfunctions used in notebook [*] where similar but not identical. In ξ variable they were

(Subscript[ψ, (Subscript[n, 1])(Subscript[n, 2])m])[ξ,η,φ]:= (Subscript[f, (Subscript[n, 1])m])[ξ] (Subscript[f, (Subscript[n, 2])m])[η]((e^(i m φ))/(Sqrt[2π]));
(Subscript[f, (Subscript[n, 1])m])[ξ]= (2^(1/4))ε (Sqrt[(( n!)/((n+|m|)!))]) (((ε ξ))^(|m|/2))(e^(- ε ξ/2))(Subscript[L, n]^(|m|))(ε ξ)

Functions () are orthogonal with respect the three dimensional measure induced by parabolic coordinates, i.e.

∫(Subscript[ψ, (Subscript[n, 1])(Subscript[n, 2])m])*(Subscript[ψ, (Subscript[k, 1])(Subscript[k, 2])m'])((ξ+η)/4)dξ 𝕕η 𝕕φ = (Subscript[δ, (Subscript[n, 1])(Subscript[k, 1])])(Subscript[δ, (Subscript[n, 2])(Subscript[k, 2])])(Subscript[δ, m m'])

Functions  () are not a complete set with respect to this measure, we clearly miss continuum spectrum. In this problem eigenfunctions ()  are used to solve the auxiliary problem (). For negative energies we look at real eigenvalues (Subscript[β, i])and in this framework (at ℱ = 0 ) functions () are a complete set. If we make the change of variables

f[x] = (x^(-1/2))g[x]

the first equation () becomes

((d^2)/(dx^2))g[x] + ((1/(4 (x^2)))-(1/4)x-((m^2)/(4x)) - (ℱ/4)(x^2)+ (Subscript[β, 1]))g[x],

i.e. a "usual" one dimensional Schrödinger equation with potential

2 V[x] = (1/4) x + ((m^2)/(4x))-(1/(4 (x^2))) + (ℱ/4)(x^2)

and have only a bound spectrum for ℱ ≥ 0. For ℱ = 0 both equations () admit only a discrete spectrum, the continuous spectrum of the coulomb potential comes from E > 0 and ϵ imaginary. The constraint in this case force to look for imaginary eigenvalues.

When ℱ ≠ 0 one of the two equations has a potential non bounded below at infinity, for example the second one if ℱ > 0, the energy levels will be instable for tunneling, this phenomenon will be analysed with WKB methods, but do not concerne us as far as we consider perturbation theory.

Perturbative solution at second (and third) order

It is not difficult extend the solution to second order, First we  write implicitely the correction (ψ^((1))) to wave functions in  the two eigenvalue problems ()

(Subscript[ψ, n]^((1)))[x]= (∑_s^('))(Subscript[f, s])[x] ((⟨(Subscript[f, s])|V|(Subscript[f, n])⟩)/((Subscript[β, n])-(Subscript[β, s])));      V = (ℱ/4) (x^2);

then we use the general formulae

(Subscript[δE, n]^((2)))= ⟨(Subscript[ψ, 0])|V|(Subscript[ψ, n]^((1)))⟩,  (Subscript[δE, n]^((3)))= ⟨(Subscript[ψ, n]^((1)))|V|(Subscript[ψ, n]^((1)))⟩- (Subscript[δE, n]^((1)))⟨(Subscript[ψ, n]^((1)))|(Subscript[ψ, n]^((1)))⟩.

Iterating recursion relation () we obtain

(x^2)(Subscript[L, n]^m)=(n+2)(n+1)(Subscript[L, n-2]^m)-2(n+1)(2n+m+2)(Subscript[L, n-1]^m)+ (6n(n+m+1)+(m+2)(m+1))(Subscript[L, n]^m)-2(n+m)(m+2n)(Subscript[L, n+1]^m)+(m+n)(n+m-1)(Subscript[L, n+2]^m).

Let us call (Subscript[d, k])(k = -2,..+2 the coefficients of this relation. Using normalization constants (Subscript[c, k]) this relation an be transformed in a recursive relation for basis functions:

(x^2) (Subscript[f, k])[x] = (∑_(k=-2)^2)((Subscript[c, n])/(Subscript[c, n+k]))(Subscript[d, k]) (Subscript[f, n+k])[x].

Using orthogonality of basis functions the first order correction () is

(Subscript[ψ, n]^((1)))[x]= (ℱ/4)(∑_(k=-2,k≠0)^2) (1/((Subscript[β, n]^((0)))-(Subscript[β, n+k]^((0)))))((Subscript[c, n])/(Subscript[c, n+k]))(Subscript[d, k]) (Subscript[f, n+k])[x]

Now we have only to insert () in () and using () and orthogonality to perform integrals. The result with ℱ → - ℱ will give correction to (Subscript[β, 2]). Imposing the constraint () will give the corrections to enery levels. We report here the final result of this strightforward but tedious computation

E =- ((Z^2)/(2(n^2)))+ (3/2)F (n/Z)((Subscript[n, 1])-(Subscript[n, 2])) - (1/16)(F^2)((((n/Z)))^4)[17(n^2) - 3((((Subscript[n, 1])-(Subscript[n, 2])))^2)-9(m^2)+19]+
        +(3/32)(F^3)((((n/Z)))^7)((Subscript[n, 1])-(Subscript[n, 2]))[23(n^2)-((((Subscript[n, 1])-(Subscript[n, 2])))^2)+11(m^2)+39].

A method to perform the computation to all orders is given in notebook [9.3]. In particular for the ground state of hydrogenic atoms (n=1, (Subscript[n, 1])=(Subscript[n, 2])=0, m=0)one obtains

E = - ((Z^2)/(2(n^2))) - (9/4)(F^2)(1/(Z^4)).

From previuous result the polarizability α follows

α F =- (( ∂E)/(∂F)) = (9/2)(F/(Z^4))

which reproduces the known result of hydrogen atom, see problem[11].

Problem

In a helium atom let us consider the electron - electron interaction as a small perturbation. Compute to lowest order the ground state energy of helium atom. Compute 3-dimensional Fourier transform for functions

(1/r);  ((Exp[-μ r])/r);  Exp[-μ r];

and show how these results can be used to solve the problem.

Solution

Introduction

This problem is intended as an introducion to the next one, prob.[].

In this problem we will use atomic units. Units for length and energy are

(Subscript[a, B]) = ((ℏ^2)/(m (e^2)))  (Bohr radius);     (Subscript[E, B])=((m (e^4))/(ℏ^2))= ((e^2)/(Subscript[a, B]))≃ 27.2114 eV

In these units the non relativistic Hamiltonian for Helium atom is

H =- (1/2)(Subscript[Δ, 1]) - (1/2)(Subscript[Δ, 2]) - (Z/(Subscript[r, 1])) - (Z/(Subscript[r, 2]))+ (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))≡ (Subscript[H, 0])+ (1/(|(Subscript[r, 1])-(Subscript[r, 2])|)).

Suffixes 1,2 refer to the electrons. (Subscript[r, 1]), (Subscript[r, 2]) are their position with respect to nucleus, (Subscript[r, 1])=|(Subscript[r, 1])| etc. Z is the nuclear charge, Z = 2 for helium otherwise we will describe ions of Berillium, Z=3, Boron, Z=4 etc.

In this problem we are considering electron-electron interaction as a perturbation. We do not expect good quantitative results as its order of magnitude, (e^2)/(Subscript[a, B])in usual units, is similar toother terms in the Hamiltonian.

(Subscript[H, 0])is a sum of independent hamiltonians, then its ground state is written at once

(Subscript[Ψ, 0])((Subscript[r, 1]),(Subscript[r, 2]))= (Subscript[ψ, 1s])((Subscript[r, 1]))(Subscript[ψ, 1s])((Subscript[r, 2]));   (Subscript[E, 0])= (Subscript[E, 1s])+ (Subscript[E, 1s]) =- ((Z^2)/2)((1/1)+(1/1))= - (Z^2).

Perturbation

The perturbative contribution to the energy of the ground state is

ΔE = ⟨1s;1s|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|1s;1s⟩= ∫(d^3)(Subscript[r, 1])(d^3)(Subscript[r, 2])((((|(Subscript[ψ, 1s])((Subscript[r, 1]))|)^2)((|(Subscript[ψ, 1s])((Subscript[r, 2]))|)^2))/(|(Subscript[r, 1])-(Subscript[r, 2])|)).

The most simple and general way to compute integrals like () is to use multipole expansion for Coulomb potential:

(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))= (∑_(ℓ=0)^∞)(1/(Subscript[r, >]))(((((Subscript[r, <])/(Subscript[r, >]))))^ℓ)(Subscript[P, ℓ])[Cos(γ)];

with  (Subscript[P, ℓ]) Legendre polynomials and

(Subscript[r, >])=Max[(Subscript[r, 1]),(Subscript[r, 2])];  (Subscript[r, <])=Min[(Subscript[r, 1]),(Subscript[r, 2])];   (Subscript[r, 1])·(Subscript[r, 2]) = (Subscript[r, 1])(Subscript[r, 2])Cos[γ].

In the actual case our distribution are spherically symmetric, only ℓ=0 term in the sum survives angular integration and we have, splitting the integral:

ΔE = ∫(d^3)(Subscript[r, 1])(d^3)(Subscript[r, 2])ρ((Subscript[r, 1]))(1/(Subscript[r, >]))ρ((Subscript[r, 2]))= 
(((4π))^2)((∫_0^∞)d(Subscript[r, 1]) (Subscript[r, 1]^2)ρ((Subscript[r, 1]))(1/(Subscript[r, 1]))(∫_0^(Subscript[r, 1]))ρ((Subscript[r, 2]))(Subscript[r, 2]^2)d(Subscript[r, 2])+(∫_0^∞)d(Subscript[r, 2]) (Subscript[r, 2]^2)ρ((Subscript[r, 2]))(1/(Subscript[r, 2]))(∫_0^(Subscript[r, 2]))ρ((Subscript[r, 1]))(Subscript[r, 1]^2)d(Subscript[r, 1])) = 2 (((4π))^2)(∫_0^∞)d(Subscript[r, 1]) (Subscript[r, 1]^2)ρ((Subscript[r, 1]))(1/(Subscript[r, 1]))(∫_0^(Subscript[r, 1]))ρ((Subscript[r, 2]))(Subscript[r, 2]^2)d(Subscript[r, 2]).

Using the known wave functions

(Subscript[ψ, 1s])= (Subscript[Y, 00]) (Subscript[R, 1s])[r] = (1/(Sqrt[4π]))2 (Z^(3/2))Exp[-Z r].

it is easy to compute

ΔE  = 2 (5/16)Z = (5/8)Z.

For the energy the prediction is (for Z = 2)

E = - (Z^2)+(5/8)Z -> - 2.75 a.u.;     (Subscript[E, exp]) ≃ - 2.90 a.u.

Usually what is reported is ionization energy, i.e. the energy necessary to extract one electron. After ionization we have a one electron atom, with energy - (Z^2)/2then ionization energy is defined as (we neglect relativistic corrections)

I = - (( (Z^2))/2)- (Subscript[E, 0])≃ (( (Z^2))/2)-(5/8)Z = 0.75 a.u. = 1.5 Ry= 20.41 eV.

to be compared with

(Subscript[I, exp])≃0.9035 a.u. = 1.807 Ry = 24.5855 eV.

Different methods to compute an integral

The multipole expansion () is a standard way to compute integrals of the form (), nevertheless it can be useful to have alternative strings in one's bow.

Electrostatic analogy

The integral  () is the douvle of  an electrostatic self - energy for a charge density ρ:

ΔE= 2 U;    U =  (1/2)∫(d^3)(Subscript[r, 1])(d^3)(Subscript[r, 2])((ρ((Subscript[r, 1]))ρ((Subscript[r, 2])))/(|(Subscript[r, 1])-(Subscript[r, 2])|))

Let us divide the space in shells of width dr, with charge ρ(r) 4(π^2)dr.  The self energy can be obtained by summing potential energy of these shells, the potential being created by charge inside sphere of radius r. The total charge up to radius r, Q(r) creates a potential Q(r)/r at limiting radius r, hen the self energy is

U =  (∫_0^∞)4π (r^2)dr ρ(r) (1/r)Q(r)

Once multiplied by the additional factor of two this expression coincides with ().

Fourier transforms

In this book, as often in Physics literature, the fourier transforms are defined in an asymmetric way:

f[x] = ∫ (((d^n)k)/(((2π))^n))(Subscript[ℱ, f])[k] (e^(i k x));        (Subscript[ℱ, f])[k] = ∫ (d^n)x (Subscript[ℱ, f])[k] (e^(-i k x)).

This method is based on two well known facts about Fourier transforms :

∫ (d^n)x (SuperStar[f])[x] g[x] = ∫ (((d^n)k)/(((2π))^n))(Subscript[ℱ, f]^( * ))[k](Subscript[ℱ, g])[k];  (Subscript[ℱ, f*g])[k]= (Subscript[ℱ, f])[k](Subscript[ℱ, g])[k].

For real functions (Subscript[ℱ, f]^( * ))[k]= (Subscript[ℱ, f])[-k]. f*g is the convolution of two functions, i.e.

f*g[x]=∫ (d^n)y f[x-y] g[y].

Direct application of () gives for the integral ():

ΔE =  ∫ (((d^n)k)/(((2π))^n))(Subscript[ℱ, ρ]^( * ))[k](Subscript[ℱ, 1/r])[k](Subscript[ℱ, ρ])[k].

By direct computation, or by taking the Fourier transforms of the equations

- Δ (1/r) = 4π (δ^((3)))(r);  (- Δ + (μ^2))((e^(-μ r))/r) = 4π (δ^((3)))(r);

we have

(Subscript[ℱ, 1/r])[k] =((4π)/(k^2));        ℱ[((e^(-μ r))/r)][k] = ((4π)/((k^2)+(μ^2))).

From the second of these relations follows

ℱ[(e^(-μ r))][k]=- (( ∂)/(∂μ)) ℱ[((e^(-μ r))/r)][k]= ((8π μ)/((((k^2)+(μ^2)))^2)).

In our computation :

ρ=((|(Subscript[ψ, 1s])|)^2)=  ((Z^3)/π)Exp[-2 Z r];   (Subscript[ℱ, ρ])[k]=16 ((Z^4)/((((k^2)+4 (Z^2)))^2));

and

ΔE = ((16^2)/(((2π))^3))4π ∫ (d^3)k ((Z^8)/((((k^2)+4 (Z^2)))^4))(1/(k^2))= ((2 ·(16^2))/π)(∫_0^∞)((Z^8)/((((k^2)+4 (Z^2)))^4))dk = (4/π)Z(∫_0^∞)(1/((((x^2)+1))^4))dx = (5/8) Z

The last integral can be computed by method of residues. This use of Fourier transforms may seem complicated but the reader must note that it automatically transform a doubleintegral like () in a simpe integral,like the one just computed.

Problem

In a helium atom let us consider the electron - electron interaction as a small perturbation. Compute to lowest order the energy of the first excited S state of helium atom. Discuss how the Pauli principle selects possible states.

Solution

Introduction

In this notebook we will use atomic units. Units for length and energy are

(Subscript[a, B]) = ((ℏ^2)/(m (e^2)))  (Bohr radius);     (Subscript[E, B])=((m (e^4))/(ℏ^2))= ((e^2)/(Subscript[a, B]))≃ 27.2114 eV

In these units the non relativistic Hamiltonian for Helium atom is

H =- (1/2)(Subscript[Δ, 1]) - (1/2)(Subscript[Δ, 2]) - (Z/(Subscript[r, 1])) - (Z/(Subscript[r, 2]))+ (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))≡ (Subscript[H, 0])+ (1/(|(Subscript[r, 1])-(Subscript[r, 2])|)).

Suffixes 1,2 refer to the electrons. (Subscript[r, 1]), (Subscript[r, 2]) are their position with respect to nucleus, (Subscript[r, 1])=|(Subscript[r, 1])| etc. Z is the nuclear charge, Z = 2 for helium otherwise we will describe ions of Berillium, Z=3, Boron, Z=4 etc.

In this problem we are considering electron-electron interaction as a perturbation. We do not expect good quantitative results as its order of magnitude, (e^2)/(Subscript[a, B])in usual units, is similar toother terms in the Hamiltonian, but the perturbative computations will give us some hints on the general properties of the spectrum.

Quantum numbers

We are neglecting spin interacions, albeit as we will see spin enter in the determination of energies. Hamiltonian () is invariant under global rotations, i.e. L and (Subscript[L, z])are conserved. L is total angular momentum:

L = (Subscript[ℓ, 1])+ (Subscript[ℓ, 2]).

Energy in general will depend on L, and, always neglecting spin and other effects (as Pauli principle) we expect at least a degeneracy 2L+1 for each level. To know how levels are organized we need more dynamical information, this is wy we use perturbation theory.

Hamiltonian (Subscript[H, 0])has a much bigger symmetry, it is symmetric under independent rotations of first and second electron. Each rotation invariance imply a degeneracy (2ℓ+1) then for (Subscript[H, 0])we have at least a degeneracy (2(Subscript[ℓ, 1])+1)(2(Subscript[ℓ, 2])+1).In effect we have more degeneracy as (Subscript[H, 0])is symmetric under the exchange of the two electrons, then given a state with quantum numbers (a,b) (and a≠b) then the state (b,a) will have the same energy. Finally (Subscript[H, 0])is of coulomb type and the accidental degeneracy typical of hydrogen-like systems is at work, i.e. energies depend only on principal quantum number and not on ℓ, this will be a minor complication.

(Subscript[H, 0]) is a sum of independent hamiltonians then eigenstates and energies are written at once:

(Subscript[Ψ, ab])((Subscript[r, 1]),(Subscript[r, 2]))= (Subscript[ψ, a])((Subscript[r, 1]))(Subscript[ψ, b])((Subscript[r, 2]));   (Subscript[E, a,b])= (Subscript[E, a])+ (Subscript[E, b]) =- ((Z^2)/2)((1/(Subscript[n, a]^2))+(1/(Subscript[n, b]^2))).

(Subscript[ψ, a]),(Subscript[ψ, b])are one electron states in coulomb potential and a,b the relative quantum numbers, i.e. n, ℓ, m.

Pauli principle

Taking into account the overall spin degeneracy, 4 spin states, and neglecting the accidental coulombic degeneracy, on the basis of  () we have the following list of levels. We write individual quantum numbers of the electrons to select the level, and write possible total angular momentum and spin using the known rules for summing angular momenta:

({{State, deg., (L,S), (L,S) observed}, {(1s,1s), 4, (0,0),(0,1), (0,0) deg. 1}, {(1s,2s) or (2s,1s), 4×2=8, 2(0,0),2(0,1), ({{(0,0) deg.1}, {(0,1) deg.3}}) 4 states}, {(1s,2p)or (2p,1s), 2×4×3=24, 2×(1,0), 2(1,1), ({{(1,0)deg.3}, {(1,1) deg.9}}) 12 states}})

We see that the observed states are half of the predicted ones, this is a consequence of Pauli principle, or in general of the antisymmetry of total wave function. Each function of two variables can always be decomposed in a symmetric and one antisymmetric part: Spin Statistics theorem imply that only antisymetric states are physical states, and this explain the halving of states. We want to stress some points:

Functions (Subscript[Ψ, a,b])and (Subscript[Ψ, b,a]) are different functions, as an example

(Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 2])] ≠ (Subscript[ψ, 1s])[(Subscript[r, 2])](Subscript[ψ, 2s])[(Subscript[r, 1])]

they are even orthogonal!. This is wy the factor 2 due to symmetry exchange 1↔2 appears in the degeneracy.

Last column levels are those permitted by statistics, symmetry of spin function will force the symmetry of the corresponding orbital function, so to a symmetric S=1 spin state will correspond an antisymmetric orbital function, while to anantisymmetric S=0 spin state will correspond a symmetric orbital wave function.

Statistics explain the total number of states, do not explain the splitting implicitely exposed in the last column. In effect it is safe to say that if two level can split they will split,  due to a pertubation. it is important that the reader appreciate that in this problem (and for a general atom) the perturbation is the electrostatic repulsion between electron, which has nothing to do with spin. Then while statistics (and hence spin) is responsible for the halving of the states, their actual splitting is of electrostatic origin, then quite large on an atomic scale.

We have classified the final levels with (L,S) which are good quantum numbers of the whole Hamiltonian, not simply of (Subscript[H, 0]), then this is a good classification, and could be changed only if we add some more terms to H.

Energy of excited s - states

Helium states can be classified according to their total spin, S=0 states constitue the parahelium spectrum, S=1 the orthoHelium spectrum. S=0 and S=1 are respectively antisymmetric and symmetric in spin. The opposite symmetry must be own by the orbital par of the wave function. We note that symmetry can be read at lowest order in perturbation theory, being the perturbation symmetric, higher order corrections do not change symmetry properties.

1s1s ≡ 1(s^2)state

At lowest order in perturbation theory its wave function is necessarily symmetric, being the product of the same single orbital wave function:

(Subscript[Ψ, 1s1s])((Subscript[r, 1]),(Subscript[r, 2]))= (Subscript[ψ, 1s])((Subscript[r, 1]))(Subscript[ψ, 1s])((Subscript[r, 2]))

This means that Pauli principle imply S=0 for the spinpart, i.e. the ground state is a parahelum state.

1s2s state

This is the first excited state. Spin statistics theorem imply

({{S=0,         (Subscript[Ψ, 1s2s])((Subscript[r, 1]),(Subscript[r, 2])) antisymmetric}, {S=1, (Subscript[Ψ, 1s2s])((Subscript[r, 1]),(Subscript[r, 2])) symmetric}})

At lowest order in perturbation theory symmetric and antisymmetric functions can be easily constructed from the two degenerate states (Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 2])] and  (Subscript[ψ, 2s])[(Subscript[r, 1])](Subscript[ψ, 1s])[(Subscript[r, 2])].

(Subscript[Ψ, 1s2s]) = (1/(Sqrt[2]))( (Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 2])] ∓ (Subscript[ψ, 2s])[(Subscript[r, 1])](Subscript[ψ, 1s])[(Subscript[r, 2])])

These two wave function, that will be denoted by (Ψ^A)and (Ψ^S) are orthogonal and do not mix under perturbation, i.e. coulomb interaction is diagonal in this basis. We have:

⟨A|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|A⟩= ∫ ((| (Subscript[ψ, 1s])[(Subscript[r, 1])]|)^2) (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))((| (Subscript[ψ, 2s])[(Subscript[r, 2])]|)^2) -  ∫ | (Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 1])]| (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))((| (Subscript[ψ, 1s])[(Subscript[r, 2])](Subscript[ψ, 2s])[(Subscript[r, 2])]|)^2)

⟨S|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|S⟩= ∫ ((| (Subscript[ψ, 1s])[(Subscript[r, 1])]|)^2) (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))((| (Subscript[ψ, 2s])[(Subscript[r, 2])]|)^2) +  ∫ | (Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 1])]| (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))((| (Subscript[ψ, 1s])[(Subscript[r, 2])](Subscript[ψ, 2s])[(Subscript[r, 2])]|)^2)

For obvious reasons the first integral is named direct integral, and will be denoted by K, and the second exchange integral, and will be denoted by J. K is very similar to the integral met in ground state computation, J determines the splitting between parahelium and orthohelium. We write:

⟨A|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|A⟩ = K - J;  ⟨S|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|S⟩ = K + J.

K, and J canbe easily computed on the basis built on hydrogenic functions

(Subscript[ψ, 1s]) = (1/(Sqrt[4π]))(Z^(3/2)) 2 Exp[-Z r];     (Subscript[ψ, 2s]) = (1/(Sqrt[4π]))(Z^(3/2)) (((2 - Zr))/(2(Sqrt[2])))Exp[-Z r/2];

using multipole expansion, already seen in problem []:

(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))= (∑_(ℓ=0)^∞)(1/(Subscript[r, >]))(((((Subscript[r, <])/(Subscript[r, >]))))^ℓ)(Subscript[P, ℓ])[Cos(γ)];

An easy computation gives

K = ((17 Z)/81);   J = ((16 Z)/729)

Then, at first order in perturbation theory (we report also approximate experimental data):

E[1s2s; S=0] = -(( (Z^2))/2)(1+(1/4)) + (K + J) = -(5/8)(Z^2) + (169/729)Z ≃ - 2.036 a.u.; exp: -2.14577 a.u.
E[1s2s; S=1] = -(( (Z^2))/2)(1+(1/4)) + (K - J) = -(5/8)(Z^2) + (137/729)Z ≃- 2.124 a.u.;  exp: -2.17503 a.u.
E[1s2s; S=0]- E[1s2s; S=1]= ((32 Z)/729)≃ 0.088 a.u.;    exp: 0.02926 a.u.

We see that while ordering of levels is correct, energy differences are badly reproduced.

Hystorical remarks

It would be very difficult to overemphasize the imprtance of the problem of Helium atom in Quantum Mechanics.

The disagreement between theory and experiment for highly excited Helium states marked the end of old quantum theory. In the words of Born [1], who carried the calculations with Heisenberg:

In two seminal papers on helium atom [2] Heisenberg put forward the idea of the connection between spin and symmetry of the wave function, idea almost simultaneously advanced by Dirac. The observation is that if one work on unperturbed functions, and without knowing about spin statistics connection, one should treat electrostatic interaction as a perturbation on a a degenerate level. In the basis (Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 2])],(Subscript[ψ, 1s])[(Subscript[r, 2])](Subscript[ψ, 2s])[(Subscript[r, 1])] the matrix willbe written as

(Subscript[H, I])=(({{K, J}, {J, K}}))

and we know that eigenvalues are K ± J, while eigenfunctions are symmetric and antisymmetric combinations. This structure agree with experimental known helium spectrum, whose spin can be detected by interaction with a magnetic field. In the second paper Heisenberg elaborate a quite sophisticated form of perturbation theory trying to explain in the new Quantum Mechanics what Old Quantum Theory was enable to explain.

The definitive word on the argument was given by Hylleraas [3], were a variational technique were developed and a spectacular agreement with experimental data was found. We will study later variational comptations.

The reader who want to go deeper in some of these aspects is referred to the book of Bethe and Salpeter [4].

References

[1] M. Born : Vorlesungen u¨ber Atommechanik, Springer Verlag, Berlin, 1925; english translation: Mechanics of the Atom, Bell, london, 1927

[2] W. Heisenberg: Z. Physik, 38 (1926) 411 and 39 (1926) 499.

[3] E.A. Hylleraas: Z. Physik, 54, (1929) 347.

[4] H.A. Bethe, E.E. Salpeter: Quantum Mechanics of One-And Two-Electron Atoms, Plenum Publishing Corporation; (1977)

Problem

Consider the helium atom in a generic (1s,nL) state. Interactions between electrons are taken into account by perturbation theory. As the unperturbed Hamiltonian for excited states, take an asymmetric form with potential Z/r for one electron and (Z-1)/r for the other.

Solution

Introduction

For highly excited states we expect that one electron is tightly bound to the necleus, while the other is subject substantially to a screened potential, (Z-1)/r. Levels must be approximated by hydrogenoid terms:

E ≃ -((Z^2)/2) - ((((Z-1))^2)/(2(n^2)))

and the spectrum (frequency are given by the difference of energies) must be similar to hydrogen. Energy levels () will be affected by corrections and it is usual to parametrize these corrections in the form of Rydberg corrections, or Rydberg defects

E ≃ -((Z^2)/2) - ((((Z-1))^2)/(2(((n + δ))^2)))

To make sense this formula the parameter δ must be independent on n, while a dependence on L can be present. This parametrization works quite well for alkaline atoms, and for excited helium states. In helium we expect to defects, one for parahelium and another for orthoelium, and levels can be parametrized as

E ≃ -((Z^2)/2) - ((((Z-1))^2)/(2(((n + (Subscript[δ, c])-(Subscript[δ, A])))^2)));   E ≃ -((Z^2)/2) - ((((Z-1))^2)/(2(((n + (Subscript[δ, c])+(Subscript[δ, A])))^2))).

The first perturbative analysis of this problem within Quantum Mechanics has been given by Heisenberg in the papers quoted in problem []. To this aim Heisenberg elaborated an unortodox form of perturbation theory, its simplest form will be reviewed below.

The unsymmetrical perturbation theory

The idea is that in a given problem a priori one is free to choose what consider unperturbed hamiltonian and what consider perturbation, the only computational requiriment (just to simplify our work) is that unperturbed Hamiltonian must be exactly soluble.

Suppose that in a system λ is a small parameter and we can split the Hamiltonian in two different ways

H = (Subscript[H, a]^((0)))+λ (Subscript[V, a]) = (Subscript[H, b]^((0)))+λ (Subscript[V, b]).

λ in general will be contained also in zero order H, but the two Hamiltonian are equal for λ=0.  Suppose that we want to study a degenerate level, (Subscript[E, a]^((0)))=(Subscript[E, b]^((0)))=(Subscript[E, 0]).We will have two different wave functions (Subscript[φ, a])and (Subscript[φ, b]) which we can suppose orthogonal at least for λ=0, when the two unperturbed hamiltonian are equal. This choice will be surely possible in a degenerate level. As our aim will be to handle the two decomposition onthe same footing we will search for a first order correction to energy which reduces to a symmetric combination of the two eigenfunctions for λ→0, i.e. our eigenfunction will be of the form

ψ = (1/(Sqrt[2]))((Subscript[φ, a]) + (Subscript[φ, b])) +λ (Subscript[ψ, 1.])

(Subscript[φ, a]) and (Subscript[φ, b]) satisfy the equations

( (Subscript[H, a]^((0)))-(Subscript[E, 0]))(Subscript[φ, a]) = 0;        ( (Subscript[H, b]^((0)))-(Subscript[E, 0]))(Subscript[φ, b])=0.

ψ will be the solution of the Schrödinger equation with a corrected eigenvalue:

(H - E)ψ = 0;        E=(Subscript[E, 0])+λ (Subscript[E, 1])

Substituting eq.() and using (), we have, at first order in λ:

(Sqrt[2])(H-E)(Subscript[ψ, 1]) + ((Subscript[V, a]) (Subscript[φ, a]) +(Subscript[V, b]) (Subscript[φ, b]) ) - (Subscript[E, 1])((Subscript[φ, a]) + (Subscript[φ, b])) = 0.

As in usual perturbation theory we can now multiply at the left by zero order ψ, i.e. ((Subscript[φ, a]) + (Subscript[φ, b])) and integrate. The first term, already at first order, vanishes by the equations  (). At the same order (Subscript[φ, a]) and (Subscript[φ, b]) are orthogonal and we have

(Subscript[E, 1]) = (1/2)∫((Subscript[φ, a]) + (Subscript[φ, b]))((Subscript[V, a]) (Subscript[φ, a]) +(Subscript[V, b]) (Subscript[φ, b])) .

This is the correction we are looking forward. An identical deduction could be done for antisymmetric combinations, the plus signs in () would be replaced by minus signs.

Excited helium states

In the two electron problem of helium-like systems a quite general decomposition could be (we will use atomic units)

H = -(1/2)(Subscript[Δ, 1]) - (1/2)(Subscript[Δ, 2]) + V[(Subscript[r, 1])] + V[(Subscript[r, 2])] + W;   W =(- (Z/(Subscript[r, 1]))-V[(Subscript[r, 1])])+ (- (Z/(Subscript[r, 2]))-V[(Subscript[r, 2])]) + (1/(|(Subscript[r, 1])-(Subscript[r, 2])|)).

With the aim of reproducing the hydrogen like levels () it is natural to make an asymmetric choice

V[(Subscript[r, 1])] = - (Z/(Subscript[r, 1]));   V[(Subscript[r, 2])]=-(((Z-1))/(Subscript[r, 2]));      ⇒    W = (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 2])).

This will be our a - choice, in the sense of eq.(). A state (1s,nL) and the perturbation will be described by:

(Subscript[ψ, a])= (Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])];   (Subscript[E, a]^((0)))= -(( (Z^2))/2) - ((((Z-1))^2)/(2(n^2)));   (Subscript[V, a])= (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 2])).

As b - system we take the same system but with role of 1 and 2 exchanged

(Subscript[ψ, b])= (Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])];   (Subscript[E, a]^((0)))= -(( (Z^2))/2) - ((((Z-1))^2)/(2(n^2)));   (Subscript[V, b])= (1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 1])).

The two states have the same unperturbed energy. The two limiting states of opposite simmetry will be

(Subscript[ψ, ±]) = (1/(Sqrt[2]))((Subscript[ψ, a])±(Subscript[ψ, b]))= (1/(Sqrt[2]))( (Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])]±(Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])]).

The correction to energy levels will be given by ()

(Subscript[E, 1])= (1/2)∫(Subscript[ψ, ±]) {((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 2])))(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])] ± ((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 1])))(Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])]}.

We have to add for completeness that eigenfunctions (Subscript[φ, 0]) and (Subscript[φ, nLm]) are not really orthogonal, the first referring to a nucleus with charge Z,the second to a nucleus with charge Z-1, but they are almost orthogonal and this small difference will be neglected in the following.. i.e. we use

(∫_1)(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 1])]=0;   (∫_2)(Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 2])] = 0,

in writing (). Equation () can then be expicitated as

(Subscript[E, 1])= (1/2)(∫_(1,2))(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])]((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 2])))(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])] ± (Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])]((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 2])))(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])] ±

(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 2])]((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 1])))(Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])] + (Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])]((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 1])))(Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 1])] = K ± J

where we have used the facts that integrals are two by two equals, by the exchange  1 ↔ 2, and

K = (∫_(1,2))((|(Subscript[φ, 0])[(Subscript[r, 1])]|)^2)((|(Subscript[φ, nLm])[(Subscript[r, 2])]|)^2)((1/(|(Subscript[r, 1])-(Subscript[r, 2])|))-(1/(Subscript[r, 2])));  J = (∫_(1,2))(Subscript[φ, 0])[(Subscript[r, 1])](Subscript[φ, nLm])[(Subscript[r, 1])](1/(|(Subscript[r, 1])-(Subscript[r, 2])|))(Subscript[φ, 0])[(Subscript[r, 2])](Subscript[φ, nLm])[(Subscript[r, 2])]

In writing the exchange integral J we have used orthogonality.

The energy levels will be given by

E = -((Z^2)/2)- ((((Z-1))^2)/(2(n^2))) + K ± J.

integrals

The angular part of the direct integral canbe easily performed by using multipole expansion

(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))= (∑_(ℓ=0)^∞)(1/(Subscript[r, >]))(((((Subscript[r, <])/(Subscript[r, >]))))^ℓ)(Subscript[P, ℓ])[Cos(γ)];

and the identity relating the relative angle γ with polar angle between the two directions 1 and 2:

(Subscript[P, k])[Cos[γ]] = ((4 π)/(k+1))(∑_(m=-k)^k)(Subscript[Y, km])[(Subscript[Ω, 1])](Subscript[Y, km]^( * ))[(Subscript[Ω, 2])].

The angular part of the first integral, the only non trivial, gives,for a generic term of multipoles expansion

∫(Subscript[dΩ, 1])(Subscript[dΩ, 2]) (Subscript[Y, 00]^2) (Subscript[Y, Lm]^( * ))[(Subscript[Ω, 2])](Subscript[Y, Lm])[(Subscript[Ω, 2])]((4 π)/(k+1))(∑_(m=-k)^k)(Subscript[Y, km])[(Subscript[Ω, 1])](Subscript[Y, km]^( * ))[(Subscript[Ω, 2])].

The integral in (Subscript[Ω, 1]) imply that only k=0 term gives a contribution and the subsequent integral in (Subscript[Ω, 2])is one. We have then

K = ∫(Subscript[dr, 1])(Subscript[dr, 2]) (Subscript[r, 1]^2)(Subscript[r, 2]^2) (((Subscript[R, 1s])[Z;(Subscript[r, 1])])^2) (((Subscript[R, nL])[Z-1;(Subscript[r, 2])])^2)( (1/(Subscript[r, >]))- (1/(Subscript[r, 2]))) ≡ ∫(Subscript[dr, 1])(Subscript[dr, 2]) (Subscript[r, 1]^2)(Subscript[r, 2]^2) (((Subscript[R, 1s])[Z;(Subscript[r, 1])])^2) (((Subscript[R, nL])[Z-1;(Subscript[r, 2])])^2)f[(Subscript[r, 1]),(Subscript[r, 2])];

The kernel vanishes for (Subscript[r, 1])< (Subscript[r, 2])

f[(Subscript[r, 1]),(Subscript[r, 2])] =( (1/(Subscript[r, 1]))- (1/(Subscript[r, 2]))) for (Subscript[r, 1])> (Subscript[r, 2]);  f[(Subscript[r, 1]),(Subscript[r, 2])] = 0 for (Subscript[r, 1])≤ (Subscript[r, 2]).

The (Subscript[r, 1])integral is easily computed

(∫_0^∞)(Subscript[dr, 1]) (Subscript[r, 1]^2) (((Subscript[R, 1s])[Z;(Subscript[r, 1])])^2)f[(Subscript[r, 1]),(Subscript[r, 2])] = (∫_(Subscript[r, 2])^∞)(Subscript[dr, 1]) (Subscript[r, 1]^2) (((Subscript[R, 1s])[Z;(Subscript[r, 1])])^2) ((1/(Subscript[r, 1])) -(1/(Subscript[r, 2]))) = - (Z + (1/(Subscript[r, 2])))Exp[-2 Z (Subscript[r, 2])]

and for K we have

K = -(∫_0^∞)(Subscript[dr, 2]) (Subscript[r, 2]^2)  (((Subscript[R, nL])[Z-1;(Subscript[r, 2])])^2)(Z + (1/(Subscript[r, 2])))Exp[-2 Z (Subscript[r, 2])].

The integral can  be evidently computed for fixed values of n and L, but it is also possible to extract the asymptotic behaviour for large n. The radial wave functions for Z nuclear charge are given by (we use Mathematica notation)

R[n,L,r]=((2 (Z^(3/2)))/(n^2))(Sqrt[(((n-L-1)!)/((n+L)!))])(((((2 Z r)/(n ))))^L)  (e^(-((Z r)/(n ))))  LaguerreL[n-L-1, 2 L + 1,((2 Z r)/(n ))]

To perform the large n limit we will use the folowing asymptotic behaviour of Laguerre polynomials, see notebook [*]:

L[n,α,(x/n)](Underscript[→, n→ ∞]) (n^α) (x^(-α/2))(Subscript[J, α])[2(Sqrt[x])]

Using Stirling approximation for the factorials this gives the following asymptotic form for hydrogen wave functions:

(Subscript[R, as])[n,L,r]≃(((Sqrt[2])  Z BesselJ[1+2 L,2  (Sqrt[2r Z])])/(Sqrt[(n^3) r]))

Inserting in () we have for the asymptotic form of K

K ≃-((((Z-1))^2)/(n^3)) (∫_0^∞)dx (x^2)(Z + (1/x))Exp[-2 Z x] (2/x)(((BesselJ[1+2 L,2  (Sqrt[2x (Z-1)])]))^2).

As

-((((Z-1))^2)/(2(((n+δ))^2))) ≃ -((((Z-1))^2)/(2(n^2))) + ((((Z-1))^2)/(n^3))δ

the integral is just Rydberg defect, (Subscript[δ, c ])in the notation (). The first few values as a function of L are

(Subscript[δ, c])[L] = {0.1684,0.01045,0.000178,... }.

An analogous computation can be done for exchange integral. After angular integrations we have

J = (2/(2L+1))(∫_0^∞) (Subscript[r, 2]^(L+2))(Subscript[dr, 2])(Subscript[R, 1s])[Z,(Subscript[r, 2])](Subscript[R, nL])[Z-1,(Subscript[r, 2])](∫_(Subscript[r, 2])^∞) (Subscript[r, 1]^(-L+1))(Subscript[dr, 1])(Subscript[R, 1s])[Z,(Subscript[r, 1])](Subscript[R, nL])[Z-1,(Subscript[r, 1])]

The asymptotic behaviour is again estimated with

J≃ ((((Z-1))^2)/(n^3))(4/(2L+1))(∫_0^∞) (Subscript[r, 2]^(L+2))(Subscript[dr, 2])(Subscript[R, 1s])[Z,(Subscript[r, 2])](1/(Sqrt[(Subscript[r, 2])]))J[1+2 L,2  (Sqrt[2(Subscript[r, 2]) (Z-1)])]
(∫_(Subscript[r, 2])^∞) (Subscript[r, 1]^(-L+1))(Subscript[dr, 1])(Subscript[R, 1s])[Z,(Subscript[r, 1])](1/(Sqrt[(Subscript[r, 1])]))J[1+2 L,2  (Sqrt[2(Subscript[r, 1]) (Z-1)])]= ((((Z-1))^2)/(n^3)) (Subscript[δ, A])

The first few values for (Subscript[δ, A])[L]are (for Z = 2):

(Subscript[δ, A])[L]= {0.3833694984731047`,0.03514195468964153`,0.0006497347970519517`}

(Subscript[δ, A])is quite in agreement with experimental data, while (Subscript[δ, c])require second order corrections. The reader can find more details on this argument in the book of Bethe and Salpeter [1], which we have followed quite closely, we have only a small numerical discrepancy on the first value of (Subscript[δ, A]) for helium. In the quoted book the shifts are given in form of infinite series.

References

[1] H.A. Bethe, E.E. Salpeter: Quantum Mechanics of One-And Two-Electron Atoms, Plenum Publishing Corporation; (1977)

Problem

An excited level of an alkaline atom can be described by the wavefunction of the outermost electron in a screened Coulomb potential due to the nucleus and the other electrons (with the net charge |e|) corrected by a small central perturbation V[r]. Show that in this approximation the spectrum of the optical electron can be described by a corrected Rydberg-type formula.

Solution

The spectrum of the radiating (external) electron of an alkaline metal can be described quite accurately by an hydrogen like spectrum:

(Subscript[E, n]) =- (( (Subscript[Z, eff]^2))/(2(((n+δ))^2)))

The parameter δ is called a Rydberg correction, or a Rydberg defect. The meaning of this parametrization is obviously that δ does not depend on the principal quantum number n, while it can depends on orbital momentum L. This parametrization works better for large values of n.

A peripheral electron is submitted, approximatively, to a screened columb potential, as electric field of the nucleus is almost compensated by tightly bound electrons. The idea then is to describe the motion of this electron with a Schrödinger equation for a potential

V[r] =- ((Subscript[Z, eff])/r) + (Subscript[V, 1])[r]

where (Subscript[V, 1])is a small perturbation. The effect of (Subscript[V, 1])is described in first order perturbation theory by

(Subscript[δE, n]) = ⟨n,L|(Subscript[V, 1])|n,L⟩= ∫dr (r^2) (Subscript[V, 1])[r] (((Subscript[R, nL])[r])^2)

For large n the wave function of a coulombic system has a simple limit form, see notebook [*],

(Subscript[R, as])[n,L,r]= ((2 (Subscript[Z, eff]^(3/2)))/(n^(3/2))) ((BesselJ[1+2 L,2  (Sqrt[2(Subscript[Z, eff]) r])] )/( (Sqrt[2(Subscript[Z, eff]) r])))

then

(Subscript[δE, n]) ≃((Subscript[Z, eff]^2)/(n^3)) ∫dr (r^2) (Subscript[V, 1])[r] (2/r)(((Subscript[J, 2L+1])[(Sqrt[2(Subscript[Z, eff]) r])])^2) ≡ ((Subscript[Z, eff]^2)/(n^3))(Subscript[δ, L])

The integral in previous formula is exactly the (perturbative) Rydberg correction, in effect for small δ/n ≪ 1

(Subscript[E, n])=- (( (Subscript[Z, eff]^2))/(2(((n+δ))^2))) ≃ - (( (Subscript[Z, eff]^2))/(2(n^2)))(1- 2(δ/n)) = - (( (Subscript[Z, eff]^2))/(2(n^2))) + (( (Subscript[Z, eff]^2))/(n^3))δ.

As indicated in our formula the parameter can depend on L.

Problem

A free particle moves inside a box of lengths (a,b,c). Discuss how the energy levels change upon varying a,b,c. Show that for isotropic changes, i.e. changes of length unit,  the expression for the mean energy of a stationary state reduces to the equation of state for a nonrelativistic perfect gas.

Solution

The Hamiltonian for a free particle is the sum of three independent Hamiltonians

H = -((ℏ^2)/(2m))(((∂^2)/(∂(x^2)))+((∂^2)/(∂(y^2)))+((∂^2)/(∂(z^2)))).

On a three dimensional box of lengths (a,b,c) also the boundary conditions are written as separated constraints:

ψ[0,y,z]=ψ[a,y,z] = 0;  ψ[x,0,z]=ψ[x,b,z] = 0;ψ[x,y,0]=ψ[x,y,c] = 0;

then the problem is completely separable. Energy eigenvalues are given by

(Subscript[E, (Subscript[n, 1])(Subscript[n, 2])(Subscript[n, 3])])= (((ℏ^2)(π^2))/(2m))(((Subscript[n, 1]^2)/(a^2)) + ((Subscript[n, 2]^2)/(b^2))+((Subscript[n, 3]^2)/(c^2)))

For a small change in the parameters

a→ a(1+(Subscript[δ, a]));     b→ b(1+(Subscript[δ, b]));       c→ c(1+(Subscript[δ, c]));

we have

δ (Subscript[E, (Subscript[n, 1])(Subscript[n, 2])(Subscript[n, 3])])=-2 (((ℏ^2)(π^2))/(2m))((Subscript[δ, a])((Subscript[n, 1]^2)/(a^2)) +(Subscript[δ, b]) ((Subscript[n, 2]^2)/(b^2))+(Subscript[δ, c])((Subscript[n, 3]^2)/(c^2))).

A change of scale would correspond to

(Subscript[δ, a]) = (Subscript[δ, b])=(Subscript[δ, c])=(Subscript[δ, λ]);

and

δ E=-2 (Subscript[δ, λ])(((ℏ^2)(π^2))/(2m))(((Subscript[n, 1]^2)/(a^2)) + ((Subscript[n, 2]^2)/(b^2))+((Subscript[n, 3]^2)/(c^2)))= -2 (Subscript[δ, λ]) E.

Under this transformation the volume of the box changes as (at first order)

δV = a b c (1+(Subscript[δ, a]))(1+(Subscript[δ, b]))(1+(Subscript[δ, c]))≃ a b c (1+(Subscript[δ, a])+(Subscript[δ, b])+(Subscript[δ, c])) → 3(Subscript[δ, λ]) a b c = 3 (Subscript[δ, λ])V.

The energy variation can be written as

δ E = - (2/3) (E/V)δV;    ((δ E)/(δV)) = - (2/3) (E/V).

This is the state equation for a perfect gas as by definition - δE/δV is the pressure, p, and for a perfect gas E = 3/2kT N, then

p = ((N k T)/V).

Equation () is  a direct consequence of scale transformation properties of the model. In a scale transformation x → (1+(Subscript[δ, λ]))x the laplacian changes as

Δ → (((1+(Subscript[δ, λ])))^(-2))Δ ≃ Δ - 2(Subscript[δ, λ]) Δ.

As the Hamiltonian is proportional to the laplacian, at first order in perturbation theory the levels are changed by

δE = - 2(Subscript[δ, λ]) E.

The volume changes as δV = 3 (Subscript[δ, λ])V and expressing (Subscript[δ, λ])as δV/(3 V), equation () follows.

Problem

A free particle moves inside a spherical well of radius R. Discuss how the energy levels change if the sphere is deformed to an ellipsoid of axes (b,b,a), b < a.

Generalize the result to an arbitrary deformation.

Solution

Introduction

This model has been proposed by Migdal to describe levels of deformed nuclei.

A rotation (oblate) ellipsoid is described by the surface

((x^2)/(b^2))+ ((y^2)/(b^2))+ ((z^2)/(a^2)) - 1 = 0

We suppose that this ellipsoid is obtained by a small deformation of a sphere of radius R, without changing the volume.,  The axes and the volume are given by

a = R(1+(Subscript[δ, a]));  b = R(1-(Subscript[δ, b]));   V = (4/3)π (b^2)a≃ (4/3)π (R^3)(1+(Subscript[δ, a])- 2(Subscript[δ, b]))

as the volume must be unchanged we have (Subscript[δ, a])= 2(Subscript[δ, b]).

We have to solve the free particle  Schrödinger equation but with boundary conditions on the surface S givenby equation ():

-((ℏ^2)/(2m)) Δ ψ = E ψ;       ψ[x∈S] = 0.

The equation is simple, while the boundary conditions are a small deformation with respect usual boundary condition on a sphere. We know how to handle small perturbation in an Hamiltonian (perturbation theory), the idea is to trade deformations in boundary conditions for a deformation in the Hamiltonian. Here is very simple, if we change variables with

x = (1+(Subscript[δ, x])) (Subscript[ξ, 1]); y = (1+(Subscript[δ, y])) (Subscript[ξ, 2]); z = (1+(Subscript[δ, z])) (Subscript[ξ, 3]);

we see that if ξ variables lie on a sphere of radius R, then (x,y,z) variables lie on an ellipsoid:

(Subscript[ξ, 1]^2)+(Subscript[ξ, 2]^2)+(Subscript[ξ, 3]^2) = (R^2)         ⇒          ((x^2)/((R^2)(((1+(Subscript[δ, x])))^2))) +  ((y^2)/((R^2)(((1+(Subscript[δ, y])))^2)))+ ((z^2)/((R^2)(((1+(Subscript[δ, z])))^2)))=1.

Under the transformation (), at first order

H = - ((ℏ^2)/(2m))( (Subscript[Δ, ξ]) - 2(Subscript[δ, x])((∂^2)/(∂(Subscript[ξ, 1]^2)))- 2(Subscript[δ, y])((∂^2)/(∂(Subscript[ξ, 2]^2)))- 2(Subscript[δ, z])((∂^2)/(∂(Subscript[ξ, 3]^2)))) ≡ (Subscript[H, 0])[ξ] + V;    V = ((ℏ^2)/m)((Subscript[δ, x])((∂^2)/(∂(Subscript[ξ, 1]^2)))+(Subscript[δ, y])((∂^2)/(∂(Subscript[ξ, 2]^2)))+(Subscript[δ, z])((∂^2)/(∂(Subscript[ξ, 3]^2)))).

The "unperturbed" problem described by (Subscript[H, 0])is a free particle in a spherical well, and we know eigenvalues and eigenfunctions for this problem. The eigenvalues of H are simply computed by first order perturbation theory with V.

In our simple case

(Subscript[δ, x])=(Subscript[δ, y])= -(Subscript[δ, b]);  (Subscript[δ, z])=(Subscript[δ, a])= 2(Subscript[δ, b]).

and

V = ((ℏ^2)/m)(Subscript[δ, b])(-((∂^2)/(∂(Subscript[ξ, 1]^2)))-((∂^2)/(∂(Subscript[ξ, 2]^2)))+ 2  ((∂^2)/(∂(Subscript[ξ, 3]^2))))= - ((ℏ^2)/m)(Subscript[δ, b])( (Subscript[Δ, ξ]) - 3 ((∂^2)/(∂(Subscript[ξ, 3]^2)))).

Matrix elements

The unperturbed Hamiltonian has eigenvalues and eigenfunctions given by

(Subscript[E, k,L])= (((ℏ^2)(k^2))/(2m));            (Subscript[ψ, k,L,m])[r,θ,φ] = (Subscript[R, kL])[r] (Subscript[Y, Lm])[θ,φ];

Where wave number k is fixed by imposing zero boundary conditions on a sphere of radius R to the solution of radial Schrödinger equation

(Subscript[R, kL]^('')) + (2/r)(Subscript[R, kL]^(')) +((k^2)- ((L(L+1))/(r^2)))(Subscript[R, kL])= 0.

The computation of matrix elements of V in equation () is an exercise on Wigner Eckart theorem. In effect V is the zz component of a traceless symmetric tensor, like a quadrupole

(Subscript[Q, ij])= - ((ℏ^2)/m)(Subscript[δ, b])((Subscript[Δ, ξ])(Subscript[δ, ij]) - 3 ((∂^2)/(∂(Subscript[ξ, i])∂(Subscript[ξ, j])))).

By Wigner Eckart theorem

⟨n,L,M| (Subscript[Q, ij])|n,L,M⟩ = C ⟨n,L,M| (Subscript[L, i])(Subscript[L, j])+ (Subscript[L, j])(Subscript[L, i])- (2/3)(L^2) (Subscript[δ, ij])|n,L,M⟩

Angular momentum eigentates |L,m⟩ diagonalize our perturbation and we have

⟨n,L,M| (Subscript[Q, 33])|n,L,M⟩ = C (2 (M^2)- (2/3)L(L+1));       Q ≡⟨n,L,L| (Subscript[Q, 33])|n,L,L⟩ = C (2 (L^2)- (2/3)L(L+1))= C (2/3)L(2L-1)

i.e.

⟨n,L,M| (Subscript[Q, 33])|n,L,M⟩ = ((3 Q)/(L(2L-1)))((M^2)- (1/3)L(L+1)).

which determine how a level is splitted under the perturbation. Q is the marix element on the highest component of the multiplet, M=L. We note that states with ± M are degenerate, and this could be predicted by invariance under y→-y of the perturbation. This transformation change sign to third component on angular momentum, see the analogous situation in Stark effect, problem [].

To compute the effective splitting we have to compute Q. A simple computation, carried below, shows that for a level with unperturbed energy E:

Q = 2 (Subscript[δ, b])E ((2L)/(2L+3))  ⇒  δE = 3(Subscript[δ, b])E (4/((2L+3)(2L-1)))((M^2)- (1/3)L(L+1)).

Computation of the matrix element

The wave function is  ψ=R[r] (Subscript[Y, LL])[θ,φ](we call x,y,z the coordinates to clarify the manipulations). Integrating by parts:

(Q/(Subscript[δ, b])) = - ((ℏ^2)/m)∫(SuperStar[ψ])(Δ - 3 ((∂^2)/(∂(z^2))))ψ = ((ℏ^2)/m)(k^2) -3 ((ℏ^2)/m)∫((∂(SuperStar[ψ]))/(∂z))((∂ψ)/(∂z))≡ ((ℏ^2)/m)(k^2) -3 ((ℏ^2)/m) S

The spherical harmonic (Subscript[Y, LL]) in cartesian coordinates is

(Subscript[Y, LL]) = C ((((x+i y))^L)/(r^L));

then we can write

((∂ψ)/(∂z)) = (z/r)R'(Subscript[Y, LL]) - R L C((((x+i y))^L)/(r^(L+1)))(z/r) = (z/r)( R' - (L/r)R)(Subscript[Y, LL]).

The integral to compute is then transformed in

S = ∫(d^3)x ((Cos[θ])^2)((|(Subscript[Y, LL])|)^2)((( R' - (L/r)R))^2).

The radial integral integratin by parts ( 2 R R'= d/dr ((R^2)))can be transformed as

∫dr (r^2)(((R')^2)-2(L/r)R R'+ ((L^2)/(r^2))(R^2)) = ∫dr ((r^2)((R')^2)+ L (R^2) + (L^2)(R^2)) = ∫dr ((r^2)((R')^2)+ L(L+1)(R^2))

Multiplying by (r^2)R and integrating the radial Schrödinger equation

(1/(r^2))(∂/(∂r))(r^2)(∂/(∂r))R - ((L(L+1))/(r^2))R = - (k^2)R

it follows

∫dr ((r^2)((R')^2)+ L(L+1)(R^2)) = (k^2) ∫dr (r^2)(R^2) = (k^2).

The last integral being just the norm of R. Then

S = (k^2) ∫dΩ ((Cos[θ])^2)((|(Subscript[Y, LL])|)^2).

The simplest way to compute the angular integral is to write the explicit form of the spherical armonic

(Subscript[Y, LL])[θ,φ] = ((((-1))^L)/(2^(L+1)))((e^(i L φ))/(Sqrt[π]))((Sqrt[(2L+1)!])/(L!))((Sin[θ])^L)

it follows

S = (k^2)∫Sin[θ]dθ ((Cos[θ])^2) (1/(2^(2L+1)))(((2L+1)!)/((L!)^2))((Sin[θ])^(2L))=(( (k^2))/(2L+3)).

Substituting in ()

Q =(Subscript[δ, b])( ((ℏ^2)/m)(k^2) -3 ((ℏ^2)/m) (( (k^2))/(2L+3)))= (Subscript[δ, b])2E ( 1 - (3/(2L+3))) = 4 (Subscript[δ, b]) (E/(2L+3)),

the value used in previous section. An alternative way to perform angular integral is to express ((Cos[θ])^2)in terms of spherical harmonics and use integrals for three spherical harmonics, proportional to Clebsch-Gordan coefficients.

Generalization

It is clear that previous calculation can be generalized to an arbitrary change of variables. The idea is to make a transformation x → ξ in such a way that in ξ variables the surface is easily handled, as a sphere. The boundary (Subscript[S, x])will go, in general, in the bondary (Subscript[S, ξ]). Under an infinitesimal transformation

H[x[ξ]] = H[ξ] + δH;  (i.e. H → H + δH)

Let φ[ξ] the zero order eigenfunctions in ξ variables, i.e.

H[ξ]φ[ξ] = (Subscript[E, o]) φ[ξ];   φ[ξ∈ (Subscript[S, ξ])] = 0;

the first order eigenvalues are

(Subscript[E, 0])+ δE = (Subscript[E, 0])+ ∫dξ (SuperStar[φ])[ξ] δH φ[ξ].

A typical case is given by a change of the laplacian, let us consider (all formulas are valid at first order in the small parameter (functions ) (f^i)):

(x^i) = (ξ^i)+ (f^i)[ξ];   (ξ^i)= (x^i) - (f^i)[x];        ((∂(ξ^j))/(∂(x^i)))≃ (Subscript[δ, ij])- ((∂(f^j))/(∂(ξ^i)))

It is easy to show that

(Subscript[Δ, x]) = ((∂^2)/(∂(Subscript[x, i])∂(Subscript[x, i])))= ((Subscript[δ, ji])-((∂(f^j))/(∂(ξ^i))))(∂/(∂(ξ^j)))((Subscript[δ, ki])-((∂(f^k))/(∂(ξ^i))))(∂/(∂(ξ^k))) ≃ (Subscript[Δ, ξ])-((∂(f^j))/(∂(ξ^i)))((∂^2)/(∂(ξ^j)∂(ξ^i)))- (∂/(∂(ξ^i)))(((∂(f^k))/(∂(ξ^i)))(∂/(∂(ξ^k))))= (Subscript[Δ, ξ])-2((∂(f^j))/(∂(ξ^i)))((∂^2)/(∂(ξ^j)∂(ξ^i)))- (((Δf)^i))(∂/(∂(ξ^i)))

from which δH follows.

Equation () can be transformed in such a way to enlight some subtle points. Consider the eigenvalues problem in original x variables

H ψ[x] = E ψ[x]

Once we know E, this is a differential identity, satisfied for every value of x, and independently of the name we use for the variable, of course. With the transformation ()

ψ[x[ξ]]≃ψ[ξ] + (f^k)(∂_k)ψ = ψ[ξ] + δψ.

Using () we have, at first order:

(H[ξ]-E)ψ[ξ] + (δH ψ + H[ξ]δψ- E δψ) = 0

The first term is identically zero, because as already stressed a differential relation do not depend on th ename of variables, ad we have

δH ψ = (E - H[ξ])δψ

the crucial point is that all quantities in this relation have already a factor f exposed, so in other factors we can consider order zero quantities. At zero order E = (Subscript[E, 0])and δψ = δφ, where φ is the zero order solution (). After multiplication by (SuperStar[φ]) and integrating we obtain

∫𝕕ξ (SuperStar[φ]) δH φ = δE = ∫𝕕ξ (SuperStar[φ])( (Subscript[E, 0]) - H[ξ])δφ.

The right hand side is not zero because  H is self adjoint on functions which are zero on (Subscript[S, ξ]) but in  general δφ do not belongs to this space, i.e. δφ ≠ on (Subscript[S, ξ]). Taking the explicit form of the laplacian for H we have (assume φ real for simplicity, and integrate by parts

(∫_ξ) φ Δ δφ = (∫_ξ) φ Δ δφ = - (∫_ξ) ∇φ ∇ δφ = - (∫_ξ) ∇(∇φ  δφ)+ (∫_ξ) (Δ φ)  δφ.

Last term cancel the term  in (Subscript[E, 0])in () while the first can be transformed in a surface integral by use of Gauss theorem, leaving us with:

δE = ((ℏ^2)/(2m))(∫_(Subscript[S, ξ]))n·(δφ ∇φ) = ((ℏ^2)/(2m))(∫_(Subscript[S, ξ]))(Subscript[n, k])( (f^i)((∂_i)φ)((∂_k)φ))

This formula clarify these issues

There is not change in eigenvalues if the variation vanishes on boundaries. This is essential as a change of variables in the interior of the domain is just a relabeling of the points, i.e. a unitary transformation, and eigenvalues do not change for unitary transformations.

Eigenvalues in particular do not change in infinite domain ( if (f^i)functions decrease toward zero at infinity). This is a minimal rewuirement, as we clearly change variables at our will in the Schrödinger equation in (ℝ^3).

Eigenvaues do change if we modify the boundary and change is do entirely to behaviour of φ (of its derivatives) at the boundaries. This is required also by continuity because we can always imagine to superimpose to our transfomation another one which affects only the interior bringing back x to its original value.

Equation () is the generlization of analogous relations that we will find in some problems of notebook [*], where the motion of a free particle between moving walls will be considered.

The reader can check the equation () with the results of problem [].

Problem

Compute the cross section for photoelectric effect on a hydrogen atom in the ground state, assuming k (Subscript[a, B])≪ 1, where k is the wave number of the photon. Approximate the wave function of the outgoing electron with a plane wave and discuss the validity of the approximations used.

Solution

In the process a light beam of intensity I and frequency ω induce a transition from a bound state, i, and a continuum state, f, of an electron. The quantum numbers of the final state are the momentum p of the particle. The cross section for the process is defined by

dW = N dσ;

N is the photon flux, i.e. the number of photons per second and per unit surface impinging the atom, dW is the probability of transition per unit time.

We will treat the electromagnetic field as a classical field:

A = (Subscript[A, 0])ϵ (e^(i k r))+ c.c.

ϵ is the polarization vector. Fermi golden rule gives the probability of the transition per unit time

dW = ((2π)/ℏ) ((e^2)/((m^2)(c^2)))|(Subscript[A, 0])((|)^2) |⟨f|(e^(i k r))ϵ·p |i⟩(|^2) (Subscript[dΦ, f]) δ((Subscript[E, f])-(Subscript[E, i])-ℏ ω).

i, f, are initial and final state of the electron,  ℏω the photon energy. The strength of the vector potential (Subscript[A, 0])is related to the intensity of incident light by (see text)

(Subscript[A, 0]^2) = ((2π c)/(ω^2)) I = ((2π c)/(ω^2)) ℏ ω N .

The cross section is, from ()

dσ = ((4(π^2)(e^2))/((m^2)c ω))|⟨f|(e^(i k r))ϵ·p |i⟩(|^2) (Subscript[dΦ, f]) δ((Subscript[E, f])-(Subscript[E, i])-ℏ ω).

For long wavelengths, k (Subscript[a, B])≪ 1 we can approximate Exp[i k r ] ≃ 1 (dipole approximation). This approximation works in the X rays regime. For energies ∼ KeV we can approximate the final electron state by a plane wave, a more accurate computations would require the Coulomb wave functions for the emitted electron. In this approximation

(Subscript[ψ, i])[r]= ((2 (a^(-3/2)))/(Sqrt[4π])) (e^(-r/a))  ;   (Subscript[ψ, p])= (e^(i p r/ℏ)) ;             a = Bohr radius.

The matrix element is (in dipole approximation)

(Subscript[V, fi])= ∫(d^3)r (Subscript[ψ, p]^( * ))[r]ϵ(ℏ/i)∇(Subscript[ψ, 1s])[r] = ∫(d^3)r  i ℏ (∇(e^(-i p r)))ϵ (Subscript[ψ, 1s]) = p·ϵ ∫∫(d^3)r  (e^(-i p r)) (Subscript[ψ, 1s]) =
      = p·ϵ (( (a^(-3/2)))/(Sqrt[π])) ((8π/a)/(((((((p/ℏ)))^2)+(1/(a^2))))^2))= p·ϵ  ((8π (a^(3/2)))/((((((((p a)/ℏ)))^2)+1))^2)).

We used the Fourier transform

ℱ[(e^(-μ r))] = -(( d)/(dμ))[((e^(-μ r))/r)] = ((8π μ)/((((k^2)+(μ^2)))^2)).

The phase for a free particle is

(Subscript[dΦ, f]) = (((d^3)p)/(((2π ℏ))^3))= (((p^2)dp dΩ)/(((2π ℏ))^3))= ((m p dE dΩ)/(((2π ℏ))^3)).

Integration of the δ function fix the energy

E = ((p^2)/(2m))= ℏ ω + (Subscript[E, i])= ℏ ω - ((e^2)/(2a))  ≡ ℏ(ω - (Subscript[ω, 0]))⇒ (( (p^2)(a^2))/(ℏ^2)) = ((2m ℏ(ω - (Subscript[ω, 0]))(a^2))/(ℏ^2)) = (ω/(Subscript[ω, 0]))-1;

where

(Subscript[E, i])= - ((e^2)/(2a)) = - ℏ (Subscript[ω, 0]) = energy ground state;     a = ((ℏ^2)/(m (e^2))) = Bohr radius .

Substitution in () gives

dσ = ((32(e^2))/(m ω c)) ((((a p/ℏ))^3)/((((((a p /ℏ))^2)+1))^4))|n· ϵ ((|)^2) dΩ = 64 α (a^2)(((((Subscript[ω, 0])/ω)))^(7/2))(((1- ((Subscript[ω, 0])/ω)))^(3/2))|n· ϵ ((|)^2) dΩ.

n is the direction of the outgoing electron. α = (e^2)/(ℏ c)is the fine structure constant. Using

(1/(4π))∫dΩ (Subscript[n, i])(Subscript[n, j]) = (1/3)(Subscript[δ, ij]) ;

the total cross section is

64 ((4π)/3) α (a^2)(((((Subscript[ω, 0])/ω)))^(7/2))(((1- ((Subscript[ω, 0])/ω)))^(3/2)).

The angular dependence for a polarized light beam can be studied by writing

dσ = σ (3/(4π))|n· ϵ ((|)^2) dΩ.

Let z axis be along the beam direction. n = (Sin[θ]Cos[φ], Sin[θ]Sin[φ], Cos[θ]) the direction of the outgoing electron. For a polarization along x axis:

dσ = σ (3/(4π))((Sin[θ])^2)((Cos[φ])^2) dΩ.

Approximations

We considered the final electron as a free particle, this is correct if its kinetic energy is much higher than Coulomb interaction with nucleus, i.e. for p a/ℏ ≫ 1 of ω/(Subscript[ω, 0])≫ 1. A more accurate computation would require exact Coulombic functions for the final state, we quote the result

64 ((4π)/3) α (a^2) 2π(((((Subscript[ω, 0])/ω)))^4)  ((e^(-4 x ArcCot[x]))/(1- Exp[-2π x]));   x = ℏ/a p.

A comparison of the results is given below, with  (Subscript[σ, 0])=64α (a^2)4π/3:

We see that at high energy (x→0) the two result match while near threshold (x→∞) the cross sections are different.

Problem

Write explicitly the form of the Hamiltonian for matter - radiation interaction taking into account the dipole, quadrupole and magnetic dipole interactions. Discuss the selection rules for each case.

Solution

Let |i⟩and |f⟩the initial and final states of our system. We will consider an atom in the approximation of infinite nuclear mass, then these states depends only on electrons coordinates.

To fix our ideas we will consider the absorption process. In the text it has been shown that the relevant part of electromagnetic interaction for this process is

A = (Subscript[A, 0])ϵ (e^(i k r - i ω t));    ω = (((Subscript[E, f])- (Subscript[E, i]))/ℏ)≡ (Subscript[ω, f])-(Subscript[ω, i]).

ϵ is the polarization vector while the field strength (Subscript[A, 0])is related to the intensity I of light beam by

(Subscript[A, 0]^2) = ((2π c)/(ω^2)) I .

In the Coulomb gauge ϵ·k=0 and electric and magnetic field are given by

E = -(1/c)((∂A)/(∂t))= i (ω/c)(Subscript[A, 0])ϵ (e^(i k r - i ω t)) ;   B= ∇⋀A = (Subscript[A, 0])i k⋀ϵ (e^(i k r - i ω t)) .

The multipole expansion is an expansion in powers of (k r). We know that the low orders in this expansion are given by

(Subscript[H, I]) = - d·E - μ·B - (1/6)(Subscript[Q, ij])(∂_i)(Subscript[E, j]).

Fields are computed at r=0, d, μ, (Subscript[Q, ij])are respectively the electric dipole, the magnetic dipole and the quadrupole tensor

d = (∑_a) (Subscript[e, a])(x^((a)));  μ = (∑_a) ((Subscript[e, a])/(2 m c))((Subscript[L, a])+ g (Subscript[S, a])); (Subscript[Q, ij])= (∑_a) (Subscript[e, a])(3 (Subscript[x, i]^((a)))(Subscript[x, j]^((a)))- ((((x^((a)))))^2)(Subscript[δ, ij])).

The sum runs on the electrons. The matrix elements of the interaction take the form

(Subscript[H, d])= -i (Subscript[A, 0])(ω/c) ⟨f|d|i⟩ϵ ;     (Subscript[H, m])=  -i (Subscript[A, 0])(ω/c) ⟨f|μ|i⟩(n⋀ϵ );    (Subscript[H, Q])= (1/6)(Subscript[A, 0])((((ω/c)))^2)⟨f|(Subscript[Q, ij])|i⟩((Subscript[n, i])(Subscript[ϵ, j])).

n = k/|k| is the direction of the light beam. We used k = |k|=ω/c . The reader must note that in this approximation ω∼(Subscript[ω, fi]).

Selection rules

Selection rules follow from parity conservation and Wigner Eckart theorem. Let us consider each term separately.

Electric dipole

Explicitly the only matrix elements different from 0 are

(Subscript[D, z])= ⟨J',(Subscript[J, z]^′)| (Subscript[d, z])|J,(Subscript[J, z])⟩ with ( (Subscript[J, z]^′) = (Subscript[J, z]) );    (Subscript[D, ±]) = ⟨J',(Subscript[J, z]^′)| (Subscript[d, ±])|J,(Subscript[J, z])⟩ ; with ( (Subscript[J, z]^′) = (Subscript[J, z]) ± 1).

Magnetic dipole

Explicitly the only matrix elements different from 0 are

(Subscript[M, z])= ⟨J',(Subscript[J, z]^′)| (Subscript[μ, z])|J,(Subscript[J, z])⟩ with ( (Subscript[J, z]^′) = (Subscript[J, z]) );    (Subscript[M, ±]) = ⟨J',(Subscript[J, z]^′)| (Subscript[μ, ±])|J,(Subscript[J, z])⟩ ; with ( (Subscript[J, z]^′) = (Subscript[J, z]) ± 1).

Electric quadupole

Hamiltonian

It can be useful to check the above expansion starting with the canonical form of the Hamiltonian

H = (1/(2m))(((p - (e/c)A))^2)+ e Φ - g (e/(2m c))S·B .

In Coulomb gauge we can take Φ=0 for the radiation Hamiltonian. A is given by ().

The first order in k gives for matrix elements of the interaction

-(e/(m c))(Subscript[A, 0]) ⟨f|ϵ·p|i⟩.

We used the gauge condition ∇·A = 0 to write A· p + p· A = 2 A· p. Using the equations of motion up to the first order in A:

⟨f|p|i⟩=m ⟨f|((dr)/dt)|i⟩= m (i/ℏ) ⟨f|[H,r]|i⟩= m i ((((Subscript[E, f])-(Subscript[E, i])))/ℏ)⟨f|r|i⟩= i m (Subscript[ω, fi])⟨f|r|i⟩.

Using ω = (Subscript[ω, fi]) one recovers ().

The first order in k for A is

(Subscript[A, i]^((1)))= (Subscript[A, 0])(Subscript[ϵ, i])(i (Subscript[k, j])(Subscript[x, j])) = i (Subscript[A, 0])((1/2)((Subscript[ϵ, i])(Subscript[k, j])+(Subscript[ϵ, j])(Subscript[k, i])) + (1/2)((Subscript[ϵ, i])(Subscript[k, j])-(Subscript[ϵ, j])(Subscript[k, i])))(Subscript[x, j]).

The first term gives, using ϵ·k = 0,

(Subscript[H, Q])= -(( e)/(2m c))i (Subscript[A, 0]) (1/2)((Subscript[ϵ, i])(Subscript[k, j])+(Subscript[ϵ, j])(Subscript[k, i]) -(2/3)(ϵ·k)(Subscript[δ, ij]))⟨f|(Subscript[p, i])(Subscript[x, j])+(Subscript[p, j])(Subscript[x, i])|i⟩.

Using again equations of motion and ω = (Subscript[ω, fi]) we have

(1/m) ⟨f|(Subscript[p, i])(Subscript[x, j])+(Subscript[p, j])(Subscript[x, i])|i⟩ = (1/m)⟨f|(d/dt)((Subscript[x, i])(Subscript[x, j]))|i⟩= i ω ⟨f|((Subscript[x, i])(Subscript[x, j]))|i⟩;

and with some easy algebraic manipulations:

(Subscript[H, Q]) = ((e ω)/(4c))(Subscript[A, 0])((Subscript[ϵ, i])(Subscript[k, j])+(Subscript[ϵ, j])(Subscript[k, i]) -(2/3)(ϵ·k)(Subscript[δ, ij]))⟨f|((Subscript[x, i])(Subscript[x, j]))|i⟩= 
((e ω)/(8c))(Subscript[A, 0])((Subscript[ϵ, i])(Subscript[k, j])+(Subscript[ϵ, j])(Subscript[k, i]) -(2/3)(ϵ·k)(Subscript[δ, ij]))⟨f|(Subscript[x, i])(Subscript[x, j]) + (Subscript[x, j])(Subscript[x, i])- (2/3)(x^2)(Subscript[δ, ij])|i⟩ =
((e ω)/(4c))(Subscript[A, 0])(Subscript[ϵ, i])(Subscript[k, j])⟨f|(Subscript[x, i])(Subscript[x, j]) + (Subscript[x, j])(Subscript[x, i])- (2/3)(x^2)(Subscript[δ, ij])|i⟩ .

which match ().

The antisymmetric term in () gives

-(( e)/(2m c))i (Subscript[A, 0]) ((Subscript[ϵ, i])(Subscript[k, j])-(Subscript[ϵ, j])(Subscript[k, i]) )(Subscript[x, j])(Subscript[p, i])= -(( e)/(2m c))i (Subscript[A, 0]) (Subscript[ϵ, i])(Subscript[k, j])((Subscript[x, j])(Subscript[p, i])- (Subscript[x, i])(Subscript[p, j])) = -(( e)/(2m c))i (Subscript[A, 0]) (ϵ·p k·x -ϵ·x k·p).

Using the identity

(Subscript[ϵ, iαβ])(Subscript[ϵ, iμν])= (Subscript[δ, αμ])(Subscript[δ, βν])- (Subscript[δ, βμ])(Subscript[δ, αν]) ;

the product can be written as

(ϵ·p k·x -ϵ·x k·p)= ϵ⋀k ·(p⋀x) = k⋀ϵ·L.

The formal manipulations are allowed as only x and p commuting components are involved.

The interaction matrix element takes the form

- i (e/(2m c)) (Subscript[A, 0])  k⋀ϵ·L.

The term proportional to the spin is

- i (e/(2m c))g (Subscript[A, 0])  k⋀ϵ·S.

Their sum reproduces ().

Problem

An atomic system is subjected to an external oscillating electric field. Compute the induced dipole. (Dispersion law).

Solution

Let

E[t] = (Subscript[E, 0])Cos[ω t] = (1/2)(Subscript[E, 0])((e^(-i ω t))+ (e^(i ω t))) ,

the external electric field. We want to compute the mean value of the dipole operator at first order in perturbation theory on a stationary state perturbed with ().

At zero order parity conservation imply

⟨i|d|i⟩ = 0 .

Let us work in the interaction representation. At first order in perturbation theory the evolution of the initial state of the system, |i⟩,  has a correction (see text)

|(Subscript[ψ, i]^((1)))⟩= (∑_k) (Subscript[a, k])[t]|k⟩,

where

(Subscript[a, k])[t] =- (1/ℏ)( (Subscript[F, ki]) (((e^(i((Subscript[ω, ki])-ω)t))-1)/((Subscript[ω, ki])-ω))+ (Subscript[F, ki]^†) (((e^(i((Subscript[ω, ki])+ω)t))-1)/((Subscript[ω, ki])+ω)));         F = -(1/2)d·(Subscript[E, 0]) .

The mean value of the dipole operator at first order is

(Subscript[𝒟, i])[t] = ⟨i|(Subscript[d, i]^I)|(Subscript[ψ, i]^((1)))⟩ + ⟨(Subscript[ψ, i]^((1)))|(Subscript[d, i])|i⟩ = (∑_k) ((Subscript[((Subscript[d, i]^I)), ik])(Subscript[a, k])[t] + (Subscript[a, k]^( * ))[t](Subscript[((Subscript[d, i]^I)), ki])).

In interaction representation

(Subscript[((Subscript[d, i]^I)), ik]) = (e^(i (Subscript[ω, ik])t)) (Subscript[d, ik])= (e^(-i (Subscript[ω, ki])t)) (Subscript[d, ik]).

Steady behavior of 𝒟

Let us consider the terms time dependent in (), i.e. let us forget for the moment the -1 in the numerators. The exponential in (Subscript[ω, ki]) in () cancels the corresponding term in () and we have for 𝒟:

(Subscript[𝒟, i])[t] = (1/(2ℏ))(Subscript[E, oj]) ( ((⟨i|(Subscript[d, i])|k⟩⟨k|(Subscript[d, j])|i⟩)/((Subscript[ω, ki])-ω))(e^(-i ω t))+ ((⟨i|(Subscript[d, i])|k⟩⟨k|(Subscript[d, j])|i⟩)/((Subscript[ω, ki])+ω))(e^(i ω t))) +
                 (1/(2ℏ))(Subscript[E, oj]) ( ((⟨i|(Subscript[d, j])|k⟩⟨k|(Subscript[d, i])|i⟩)/((Subscript[ω, ki])-ω))(e^(-i ω t))+ ((⟨i|(Subscript[d, j])|k⟩⟨k|(Subscript[d, i])|i⟩)/((Subscript[ω, ki])+ω))(e^(i ω t))) = (Subscript[α, ij]) (Subscript[E, j])[t];

with

(Subscript[α, ij])=  (1/ℏ) ( ((⟨i|(Subscript[d, i])|k⟩⟨k|(Subscript[d, j])|i⟩)/((Subscript[ω, ki])-ω))+ ((⟨i|(Subscript[d, j])|k⟩⟨k|(Subscript[d, i])|i⟩)/((Subscript[ω, ki])+ω))).

(Subscript[α, ij]) is the polarization tensor and () describes the induce time dependent dipole of the system. The form () is the dispersion law, due the well known relation between polarization and refractive index. Formula () is clearly inconsistent at resonance frequencies. We will see in chap.(13) that near a resonance the effective evolution for a state is described by the substitution

(Subscript[E, k])→ (Subscript[E, k])- (i/2) ℏ (Subscript[γ, k]) ;

where γ is the line-width. Assuming for definiteness that |i⟩is the fundamental state, with (Subscript[γ, i])=0, our formulas change with (Subscript[ω, ki])→ (Subscript[ω, ki])-(Subscript[iγ, k])/2. The cancellation of time dependence in the exponentials between (Subscript[a, k]) factors and (d^I)matrix elements is always at work and the polarization tensor becomes

(Subscript[α, ij])≃  (1/ℏ) ( ((⟨i|(Subscript[d, i])|k⟩⟨k|(Subscript[d, j])|i⟩)/((Subscript[ω, ki])-ω - i ((Subscript[γ, k])/2)))+ ((⟨i|(Subscript[d, j])|k⟩⟨k|(Subscript[d, i])|i⟩)/((Subscript[ω, ki])+ω- i ((Subscript[γ, k])/2)))).

This line-width effect do not cancel in the factors -1 in the numerators of () and this is why these factors do not contribute to the steady behavior of induced dipole. The approximation woks for times t≫ 1/γ, t≫ 1/ω, otherwise the transient terms must be taken into account.

As ω → 0 () reduces to usual static polarizability.

Dispersion and resonances

Let us consider for simplicity a state invariant under rotations. We can write

(Subscript[α, ij])= α (Subscript[δ, ij]);     α =(1/3) (1/ℏ) ( ((⟨i|d|k⟩⟨k|d|i⟩)/((Subscript[ω, ki])-ω - i ((Subscript[γ, k])/2)))+ ((⟨i|d|k⟩⟨k|d|i⟩)/((Subscript[ω, ki])+ω- i ((Subscript[γ, k])/2)))).

Near a resonance

α ≃ (1/3) (1/ℏ)  ((⟨i|d|k⟩⟨k|d|i⟩)/((Subscript[ω, ki])-ω - i ((Subscript[γ, k])/2)));

and we have the typical dispersive structure

Re[α] = (1/(3ℏ))|⟨i|d|k⟩((|)^2) (((Subscript[ω, ki])-ω)/(((((Subscript[ω, ki])-ω))^2)+ ((Subscript[γ, k]^2)/4))) ;    Im[α]=(1/(3ℏ))|⟨i|d|k⟩((|)^2) (((Subscript[γ, k])/2)/(((((Subscript[ω, ki])-ω))^2)+ ((Subscript[γ, k]^2)/4))) .

These properties are related to refraction index n by

(n^2) = ϵ = 1+4π ρ α ;

where ρ is the number of particle per unit volume.

Problem

Consider the electron - electron interactions as a small perturbation in a helium atom. The unperturbed Hamiltonian (Subscript[H, 0]) describes two independent electrons in a Coulombic potential. Write at zeroth order in the perturbation the ionization energy of the system. Let us suppose that the helium is in a state 2(s^2) (both electrons in 2s level). Show that the electron-electron interactions can induce a decay of the system and write the probability per unit time of this process (autoionization).

Solution

In this problem we will use atomic units m=1, e=1, ℏ=1.

Neglecting the electron-electron interaction the ground state of helium atom has an energy

(Subscript[E, 1])= - 2 (Z^2) (1/(2(n^2))) (Underscript[→, n = 1])  - (Z^2).

The continuum spectrum begin when the energy of one of the electrons become positive, then the ionization limit is

(Subscript[E, c])= -(1/2)(Z^2).

Let us now consider in this approximation the 2(s^2)state of the helium atom. Its energy is

(Subscript[E, 1])= - 2 (Z^2) (1/(2(n^2))) (Underscript[→, n = 2])  -((Z^2)/4) > (Subscript[E, c]).

This bound state is embedded in the continuum spectrum and switching on the electrostatic electron-electron interaction will give rise to a level disintegration, or in physical terms

He[2(s^2)] → (SuperStar[He])[1s]+ (SuperMinus[e]).

We want to compute the width for this process in perturbation theory. The unperturbed Hamiltonian is

(Subscript[H, 0])= ((Subscript[p, 1]^2)/(2m))+ ((Subscript[p, 2]^2)/(2m)) - ((Ze^2)/(Subscript[r, 1]))- ((Ze^2)/(Subscript[r, 2]));

the perturbation being

(Subscript[H, I])= + ((e^2)/(|(Subscript[r, 1])-(Subscript[r, 2])|)).

The initial state wave function is

(Subscript[Ψ, i]) = (Subscript[ψ, 2s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 2])]|S⟩.

| S⟩ is the singlet spin state, (Subscript[ψ, 2s])the hydrogenoid wave function. The interaction commutes with spin, then also the final state is a singlet

(Subscript[Ψ, f])= (1/(Sqrt[2]))((Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, k])[(Subscript[r, 2])] + (Subscript[ψ, 1s])[(Subscript[r, 2])](Subscript[ψ, k])[(Subscript[r, 1])])|S⟩.

The singlet spin state is antisymmetric, then Pauli principle imply a symmetric two-electron orbital wave function. (Subscript[ψ, k]) is the coulombic wave function for the continuum spectrum. With the usual normalization for single particle wave functions

⟨(Subscript[ψ, k'])|(Subscript[ψ, k])⟩ = (((2π))^3)(δ^3)(k-k'),

the states () are normalized with

⟨Ψ[k]|Ψ[k']⟩ = (((2π))^3)(δ^3)(k-k').

The Fermi golden rule gives for the transition probability per unit time (in atomic units):

dW=2π |⟨(Subscript[Ψ, f])|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|(Subscript[Ψ, i])⟩((|)^2) (((d^3)k)/(((2π))^3))δ((Subscript[E, 2(s^2)]) - (Subscript[E, 1s])- ((k^2)/2)) = 
      = |⟨(Subscript[Ψ, f])|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|(Subscript[Ψ, i])⟩((|)^2)(q/(4(π^2)))d Ω

We have performed the integral over k and used

q = (Sqrt[2((Subscript[E, 2(s^2)]) - (Subscript[E, 1s]))]).

The matrix element can be written as

A = ⟨(Subscript[Ψ, f])|(1/(|(Subscript[r, 1])-(Subscript[r, 2])|))|(Subscript[Ψ, i])⟩ = (2/(Sqrt[2]))∫(d^3)(Subscript[r, 1])(d^3)(Subscript[r, 2])(Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 1])](1/(|(Subscript[r, 1])-(Subscript[r, 2])|))(SuperStar[ψ])[q][(Subscript[r, 2])](Subscript[ψ, 2s])[(Subscript[r, 2])] .

We will approximate continuum states by plane waves

A≃ (2/(Sqrt[2]))∫(d^3)(Subscript[r, 1])(d^3)(Subscript[r, 2])(Subscript[ψ, 1s])[(Subscript[r, 1])](Subscript[ψ, 2s])[(Subscript[r, 1])](1/(|(Subscript[r, 1])-(Subscript[r, 2])|))(e^(-i q r))[(Subscript[r, 2])](Subscript[ψ, 2s])[(Subscript[r, 2])] .

The integrals can be easily computed with Fourier transforms. From

(Subscript[ψ, 1s])= (2/(Sqrt[4π]))(Z^(3/2))(e^(-Z r));  (Subscript[ψ, 2s]) = (1/(Sqrt[4π]))(1/(2(Sqrt[2])))(Z^(3/2))(2-Z r)(e^(- Z r/2)) ;

A = (Sqrt[2]) ((Z^(9/2))/(32(π^(3/2))))∫(((d^3)k)/(((2π))^3))(Subscript[ℱ, k])[(e^(-3 Z r/2))(2-Z r)](Subscript[ℱ, k])[(e^(-i q x))(2-Z x)(e^(-Z x/2))* (1/(|x|))].

ℱ stands for Fourier transform, and * denotes convolution product. Using the elementary results

ℱ[(e^(-μ r))] = ((8π μ)/((((k^2)+(μ^2)))^2)) ; ℱ[r (e^(-μ r))] =-  (d/(dμ)) ℱ[(e^(-μ r))]=((8π (3(μ^2)-(k^2)))/((((k^2)+(μ^2)))^3)) ; ℱ[(1/(|x|))] = ((4π)/(k^2)) ;

one has

(Subscript[ℱ, k])[(e^(-3 Z r/2))(2-Z r)] = ((32π Z (k^2))/((((k^2)+(((((3Z)/2)))^2)))^3));  (Subscript[ℱ, k])[(e^(-i q x))(2-Z x)(e^(-Z x/2))] = ((4π Z((Z^2) -4 (((k+q))^2)))/((((((k+q))^2)+((((Z/2)))^2)))^3));

and

A = 16 (Sqrt[2]) (π^(3/2)) (Z^(13/2))∫(((d^3)k)/(((2π))^3))(1/(k^2))(( (k^2))/((((k^2)+(((((3Z)/2)))^2)))^3))(( ((Z^2) -4 (((k+q))^2)))/((((((k+q))^2)+((((Z/2)))^2)))^3))=
4 (Sqrt[(2/π)])(Z^(13/2)) ∫(k^2)dk Sin[θ]dθ (( 1)/((((k^2)+(((((3Z)/2)))^2)))^3))(((Z^2)-4( (k^2)+(q^2)+2k q Cos[θ]))/((((k^2)+(q^2)+2k q Cos[θ]+((((Z/2)))^2)))^3)) = 
16 (Sqrt[(2/π)])(Z^5) (∫_0^∞)dk (( (k^2))/((((k^2)+(((((3Z)/2)))^2)))^3))((128 (-16 ((((k^2)-(q^2)))^2)+(Z^4)))/(((16 ((((k^2)-(q^2)))^2)+8 ((k^2)+(q^2)) (Z^2)+(Z^4)))^2)) ≃ 0.075 .

From ()

W = |A((|)^2)(q/π) ≃ 2.5 (10^(-3)) a.u. = 0.07 eV.

More accurate computations [1], using variational principle, give W∼ 4·(10^(-3)) a.u. , of the same order of magnitude.

References

[1] M. Draeger, G. Handke, W. Ihra, and H. Friedrich,  Phys. Rev. A 50, 3793 (1994); A.Bürgers, D. Wintgen, J.M. Rost:  J. Phys.  B 28, 3163, (1995).