Problems Chapter 4

Quantum Mechanics
 K. Konishi, G. Paffuti

Problem

Two particle of spin 1/2 interact with a Hamiltonian (Subscript[H, 0]) = 1/2 (Subscript[s, 1])·(Subscript[s, 2]).  List all the conserved quantities and write the energy eigenvalues.  An external field B is applied to the system. Let us assume that this field interacts only with the first particle, with V = B·(Subscript[s, 1]). Find the conserved quantities and determine the spectrum of (Subscript[H, 0])+ V.

Solution

Analytical solution

(Subscript[H, 0]) commutes with the 3 components of total spin S = (Subscript[s, 1]) + (Subscript[s, 2]) . The total spin can be 0,1. Writing

(Subscript[s, 1])·(Subscript[s, 2])= (1/2) ( (S^2) - (Subscript[s, 1]^2)- (Subscript[s, 2]^2)) =   (1/2)( S(S+1) - (3/2))

we find the eigenvalues  E = (1/4,-3/4). The degeneracy is (2S+1) i.e. 3 and 1 respectively. A complete set of commuting variables is (S^2) and (Subscript[S, z])where z is an arbitrary axis. The eigenstates of (Subscript[H, 0])denoted with |S,(Subscript[S, z])⟩, are given by

|1,1⟩= u⊗u;   |1,0⟩ = (u⊗d + d⊗u)/(Sqrt[2]); |1,1⟩ = d⊗d ;  |0,0⟩ = (u⊗d - d⊗u)/(Sqrt[2]);

u and d represent up and down states for single particles.

We can choose the direction z along the external field B, in this way the interaction is written as V = B (Subscript[s, 1z]). V no longer commutes with all 3 components of total spin, as (Subscript[s, 1z]) is not rotationally invariant. The surviving symmetry is rotation around the z axis, and correspondingly (Subscript[S, z])is still a good quantum number.

The action of (Subscript[s, 1z]) on the above basis is:

(Subscript[s, 1z])|1,1⟩  = 1/2 |1,1⟩;              (Subscript[s, z])|1,-1⟩ = -1/2 |1,-1⟩;
(Subscript[s, 1z])|1,0⟩  = 1/2 |0,0⟩;              (Subscript[s, 1z])|0,0⟩ =    1/2 |1,0⟩;

The reader can check that the selection rules for the third component of a vector operator are satisfied, i.e.

Δ S = 0,± 1;  0→0 forbidden;  Δ (Subscript[S, z])=0.

The resulting Hamiltonian is

(Subscript[H, 1])=(({{(1/4)+(B/2), 0, 0, 0}, {0, (1/4), 0, (B/2)}, {0, 0, (1/4)-(B/2), 0}, {0, (B/2), 0, -(3/4)}})).

The states 1 and 3 continue to be eigenstates, with eigenvalues 1/4 ± B/2. The submatrix involving states 2 and 4 must be diagonalized giving rise to eigenvalue equation

-(3/16)-((B^2)/4)+(λ/2)+(λ^2)==0 .

Solving this equation and adding the two trivial eigenvalues of (Subscript[H, 1]) (the elements (1,1) and (3,3) of the matrix), the eigenvalues are

(Subscript[λ, 1])= (1/4)+(B/2);       (Subscript[λ, 2])=(1/4)-(B/2);      (Subscript[λ, 3]) = (1/4) (-1-2 (Sqrt[1+(B^2)])) ;      (Subscript[λ, 4])=(1/4) (-1+2 (Sqrt[1+(B^2)])).

For large B

{(Subscript[λ, 1]),(Subscript[λ, 2]),(Subscript[λ, 3]),(Subscript[λ, 4])} → {(B/2),- (B/2),- (B/2),(B/2)};

i.e. two double degenerate levels. In this limit the eigenstates are eigenstates of V, while (Subscript[H, 0]) acts as a small perturbation. Here is a sketch of the behavior of the eigenvalues as a function of B.

Problem

Solution

1

The geometry of the problem is shown in figure below:

Consider for simplicity the double of the spin operator. We are looking for the eigenstate with eigenvalue +1 of

Σ = Cos[θ] (Subscript[σ, z]) + Sin[θ] (Subscript[σ, y]) = (({{Cos[θ], -i Sin[θ]}, {i Sin[θ], -Cos[θ]}})).

We can find eigenvalues and eigenvectors by diagonalizing the Σ, but it is instructive to proceed performing a rotation of reference frame. The rotated frame is obtained by a clockwise rotation of θ around x axis, i.e. (Subscript[R, x])[-θ].In the new frame the state we are looking for is simply ψ' =(({{1}, {0}})). Let us note that it is an eigenstate of (Subscript[σ, z]) in the rotated frame. The state in the first frame is

ψ = (Subscript[R, x]^(-1))[-θ] ψ'= (Subscript[R, x])[θ] ψ' =(({{Cos[θ/2],  i Sin[θ/2]}, {i Sin[θ/2], -Cos[θ/2]}}))(({{1}, {0}})) = (({{Cos[θ/2]}, {i Sin[θ/2]}})).

In fact

Σ·ψ =(({{Cos[θ], -i Sin[θ]}, {i Sin[θ], -Cos[θ]}}))(({{Cos[θ/2]}, {i Sin[θ/2]}})) = (({{Cos[θ/2]}, {i Sin[θ/2]}}))= ψ

It is perhaps useful to clarify a point: we used the relation ψ' = (Subscript[R, x])[-θ]ψ which is the correct one for the components of the wave function.

In terms of kets the transformation law reads (primed kets refer to rotated frame)

U[R]|σ⟩= (Subscript[R, σ'σ]) |σ⟩'  ;     |σ⟩' = (Subscript[R, σ'σ]^( * )) U[R]|σ⟩ .

The state we are looking for is |+⟩' (+ and - refers to indexes 1 and 2). This state, in the original non rotated frame is  (U^(-1))[R]|+⟩' i.e.

(U^(-1))[R](|+⟩)'  = (Subscript[R, 1σ]^( * ))|σ⟩.

The rotation matrix is

(Subscript[R, x])[-θ]=(({{Cos[θ/2],  -i Sin[θ/2]}, {-i Sin[θ/2], -Cos[θ/2]}})),

and it follows that

((U^(-1))[R]|+⟩)'= (Subscript[R, 11]^( * ))|+⟩ + (Subscript[R, 12]^( * ))|-⟩ = Cos[θ/2]|+⟩ + i Sin[θ/2]|-⟩

which coincides with the previous result.

2

The components, as shown in the previous answer, are the conjugate of first row of the rotation matrix which brings the old reference frame to a new one in which z axis is along ζ.  The axes directions transforms as a basis then the i-th axis will have as components, in general, the i-th row of the 3×3 rotation R. In 3 dimensions rotation matrices are real, and  we have, indicating with (Subscript[e, i]) the old basis and with (Subscript[f, i])the new basis:

(Subscript[f, i])= (Subscript[R, ij])(Subscript[e, j])

The third row of the rotation matrix must be:

ζ = {Sin[θ]Cos[φ],Sin[θ]Sin[φ],Cos[θ]};

A generic rotation matrix is given through Euler angles, α, β, γ by

R(α,β,γ) = Rz[γ]Ry[β]Rz[α],

with the convention used in this course. With

Rx[α] = (({{1, 0, 0}, {0, Cos[α], Sin[α]}, {0, -Sin[α], Cos[α]}})); Ry[α]=(({{Cos[α], 0, -Sin[α]}, {0, 1, 0}, {Sin[α], 0, Cos[α]}}));  Rz[α] = (({{Cos[α], Sin[α], 0}, {-Sin[α], Cos[α], 0}, {0, 0, 1}})) .

The definitions are shown in the figure below:

It is apparent that the choice: α =φ, β=θ, γ=0 brings the third axis along the ζ direction, as the reader can check with

ℛ=Rz[0]Ry[θ]Rz[φ] = (({{Cos[θ] Cos[φ], Cos[θ] Sin[φ], -Sin[θ]}, {-Sin[φ], Cos[φ], 0}, {Cos[φ] Sin[θ], Sin[θ] Sin[φ], Cos[θ]}})) .

In the spinor space the rotation matrices around the principal axes are constructed with the Pauli matrices:

(Subscript[σ, x])= (({{0, 1}, {1, 0}})) ;    (Subscript[σ, y])= (({{0, -i}, {i, 0}}));    (Subscript[σ, z])=(({{1, 0}, {0, -1}}));

(Subscript[R, k])[α] = Exp[i (α/2)(Subscript[σ, k])] = Cos[(α/2)]+i Sin[(α/2)](Subscript[σ, k]).

The matrix ℛ, () is given in spinor space by

R = (Subscript[R, 2])[θ](Subscript[R, 3])[φ]= (({{Cos[(θ/2)] (e^(i φ/2)), Sin[(θ/2)] (e^(-i φ/2))}, {-Sin[(θ/2)] (e^(i φ/2)), Cos[(θ/2)] (e^(-i φ/2))}})).

The spinor we look for is then given by the complex conjugate of the first row of R,

(Subscript[ψ, 1]) = (({{Cos[(θ/2)] (e^(-i φ/2))}, {Sin[(θ/2)] (e^(+i φ/2))}})) .

We can check the result by considering the projection of the spin along ζ and verifying that  ζ.σ (Subscript[ψ, 1]) = (Subscript[ψ, 1]):

ζ.σ = (({{Cos[θ], Sin[θ] (e^(-i φ))}, {Sin[θ] (e^(i φ)), -Cos[θ]}}));      ζ.(Subscript[σψ, 1])=(Subscript[ψ, 1]) .

The second row of R gives the opposite polarized spinor

(Subscript[ψ, 2])= (({{-Sin[(θ/2)] (e^(-i φ/2))}, {Cos[(θ/2)] (e^(+i φ/2))}})) .

The use of γ angle would give an additional arbitrary phase to the spinor, as expected.

Problem

The particle (Λ^0) ,with spin 1/2,  decays  into  p + π in the rest system.  Assume that the (Λ^0) is polarized along z axis. Write the most general amplitude for the process and the angular distribution for the secondary particles. What are the constraints if parity is conserved? Write the expression for the mean value of the longitudinal polarization of the proton (projection of spin along the direction of motion).

Solution

The kinematic is the following:

The problem is invariant under rotations around the z axis (polarization of the Λ particle), so we can put φ=0 for the azimuth angle. Let |Λ> the initial state and S the operator which causes the transition to p, π states, denoted by |n>. We can observe the direction of the proton (surely opposite to that of the π) |θ> and, possibly the spin state (eg. along z) of the proton, σ. The probability amplitude for the event is

A[θ,σ]=<θ,σ| S |Λ> = (∑_n)<θ,σ|n><n|S |Λ> .

One is interested only in the angular momentum variables here. The initial state have J=1/2, (Subscript[J, z])= 1/2, the same must be for |n> states. Quantum number of intermediate states in the sum can be selected as L and σ, L is the angular momentum. As L ⊕ 1/2 must give 1/2 by angular momentum conservation, the only possible states have L=0,1. The two angular momenta must be combined via ClebschGordan coefficients, note that (Subscript[L, z])= 1/2 - σ as (Subscript[J, z])=1/2. The amplitude in θ are clearly (Subscript[δ, σ (Subscript[s, z])]) (Subscript[Y, Lm])[θ,0]. If we call a and b the two matrix elements of S in these two states, the two possible terms in the sum are:

f1[θ,σ] = a C[1,(1/2)-σ;(1/2),σ,(1/2),(1/2)](Subscript[Y, 1,(1/2)-σ])[θ,0];
f2[θ,σ] = b C[0,0;(1/2),σ,(1/2),(1/2)](Subscript[Y, 0,0])[θ,0];

To the second term only sz=1/2 gives a contribution. The total amplitude is

A[θ,σ] = f0[θ,σ]+ f1[θ,σ];

The (relative) probability distribution for the decay (i.e. we do not measure proton spin) is:

P[θ]=(∑_(σ=-1/2)^(+1/2)) |A[θ,σ]((|)^2)= ((|a((|)^2)+|b((|)^2)-2 Re[a (SuperStar[b])] Cos[θ])/(4 π))

Parity

The two states above have opposite parity, as orbital parity is (((-))^L). This means that if both a and b are different from 0 parity is violated (as it is in Nature for this decay). If we look at the angular distribution ()

P[θ]= ((|a((|)^2)+|b((|)^2)-2 Re[a (SuperStar[b])] Cos[θ])/(4 π));

this is clear. If we consider a reflection in the (x,y) plane the spin of Λ remain unchanged (spin is an axial vector), while momentum of the p changes and θ → π - θ. But P[π - θ]≠ P[θ] if ab≠0 so parity must be violated.

Longitudinal polarization of the proton

In the previous problem it has been shown that the spinor with positive and negative spin projection along an axis with polar angles ( θ, φ=0) are respectively:

(SubPlus[ψ]) = (({{Cos[(θ/2)] (e^(-i φ/2))}, {Sin[(θ/2)] (e^(+i φ/2))}})) ;     (SubMinus[ψ]) = (({{-Sin[(θ/2)] (e^(-i φ/2))}, {Cos[(θ/2)] (e^(+i φ/2))}})).

The corresponding amplitudes will be:

(Subscript[A, ±])=∑ A[θ,σ](Subscript[ψ, ±])[σ];

i.e.

(SubPlus[A])= (((-a+b) Cos[(θ/2)])/(2 (Sqrt[π]))) ;    (SubMinus[A])= -(((a+b) Sin[(θ/2)])/(2 (Sqrt[π]))) .

We see that for θ = 0 (SubMinus[A]) = 0, while for θ=π, (SubPlus[A]) = 0. This is due to the conservation of angular momentum: for motions along z axis the orbital momentum (Subscript[L, z])vanishes, and the spin of the proton must be in positive z direction, i.e. with positive longitudinal polarization for θ=0 and negative longitudinal polarization for θ = π. the sum over the final polarizations gives correctly the same result as before:

P[θ] = |(SubPlus[A])((|)^2)+ |(SubMinus[A])((|)^2) = ((|a((|)^2)+|b((|)^2)-2 Re[a (SuperStar[b])] Cos[θ])/(4 π)) .

The mean value of the polarization will be given by the difference of counting with polarization + and - over the total events:

(SubPlus[N])= ∫ |(SubPlus[A])((|)^2) Sin[θ]dθ ;      (SubMinus[N])= ∫ |(SubMinus[A])((|)^2) Sin[θ]dθ ;

mean Polarization = (((SubPlus[N]) - (SubMinus[N]))/((SubPlus[N]) + (SubMinus[N]))) .

Again this measurement checks parity conservation: mean longitudinal polarization is the mean value of a pseudoscalar, < s·p >, so must be zero if parity is conserved.

Problem

A particle of spin S interacts with an external uniform magnetic field B,

H = - μ·B = - β (Subscript[S, z]).

At time t = 0 the spin is oriented along n = (Sin[θ]Cos[φ],Sin[θ]Sin[φ], Cos[θ]) , i.e.  S·n |0⟩ = S |0⟩.

Solution

1

Let us first consider the eigenstate of S·n for a spin 1/2 particle:

|n⟩ = (({{(e^(-i φ/2))Cos[(θ/2)]}, {(e^(i φ/2))Sin[(θ/2)]}})).

To verify that | n⟩ is an eigenvector the reader can easily check that  n·σ | n⟩ = | n⟩.

For the general case with spin S in the direction n, we consider as suggested the equivalent system of N = 2S spin 1/2 particles, with parallel spin:

|0⟩ = (Subscript[(({{(e^(-i φ/2))Cos[(θ/2)]}, {(e^(i φ/2))Sin[(θ/2)]}})), 1]) (Subscript[(({{(e^(-i φ/2))Cos[(θ/2)]}, {(e^(i φ/2))Sin[(θ/2)]}})), 2]) ... (Subscript[(({{(e^(-i φ/2))Cos[(θ/2)]}, {(e^(i φ/2))Sin[(θ/2)]}})), N])

The spin operator is

S = ∑ (Subscript[s, i]) .

The mean value of (Subscript[S, x]) is

⟨(Subscript[S, x])⟩ = N ⟨n|(Subscript[s, 1x])|n⟩= (N/2)((e^(i φ))Cos[(θ/2)]Sin[(θ/2)] +(e^(-i φ))Cos[(θ/2)]Sin[(θ/2)] ) = S Cos[φ]Sin[θ].

Likewise

⟨(Subscript[S, y])⟩ =  S Cos[φ]Sin[θ] ;  ⟨(Subscript[S, z])⟩ =  S Cos[θ] .

2

We can write the state () as

|0⟩ =((e^(-i φ/2))Cos[(θ/2)](Subscript[α, 1])+(e^(i φ/2))Sin[(θ/2)](Subscript[β, 1]))  ... ((e^(-i φ/2))Cos[(θ/2)](Subscript[α, N])+(e^(i φ/2))Sin[(θ/2)](Subscript[β, N]))= (∑_(k=-N)^N) (Subscript[c, k])|S,(Subscript[S, z])= (k/2)⟩.

where α, β are the up and down spinors. The vector |S,(Subscript[S, z])= (k/2)⟩is the symmetric combination

|S,(Subscript[S, z])= (k/2)⟩= (((({{N}, {ℓ}})))^(-1/2))∑ (Subscript[α, 1])(Subscript[α, 2])... (Subscript[β, ℓ+1])...(Subscript[β, N])

of terms with ℓ spins up and N-ℓ spins down, We must have

(ℓ/2) - ((N-ℓ)/2) =(( 2ℓ - N)/2) = (k/2)      ⇒      k = 2 ℓ - N;   ℓ =(( N+k)/2) .

The coefficients (Subscript[c, k])are

(Subscript[c, k])= (((({{N}, {ℓ}})))^(1/2)) ((((e^(-i φ/2))Cos[(θ/2)]))^ℓ) ((((e^(i φ/2))Sin[(θ/2)]))^(N-ℓ)).

The requested probability is then

(Subscript[P, k])= (({{N}, {ℓ}}))((Cos[(θ/2)])^(2ℓ)) ((Sin[(θ/2)])^(2(N-ℓ)));   ℓ = ((N+k)/2).

The reader can easily check that the  probabilities sum up to 1.

3

The time evolution of the operators is given by

(Subscript[S, x])[t] = (e^(i H t/ℏ)) (Subscript[S, x])(e^(-i H t/ℏ)) =  (e^(-i β (Subscript[S, z]) t/ℏ)) (Subscript[S, x])(e^(i (Subscript[βS, z]) t/ℏ)) ; etc.

In particular (Subscript[S, z])[t]=(Subscript[S, z])[0]=(Subscript[S, z]). The time evolution equations are

i (d/dt)(Subscript[S, x])[t] = (1/ℏ)[(Subscript[S, x]),H]= -(β/ℏ)[(Subscript[S, x]),(Subscript[S, z])]= i β (Subscript[S, y]);
i (d/dt)(Subscript[S, y])[t]= (1/ℏ)[(Subscript[S, y]),H]= -(β/ℏ)[(Subscript[S, y]),(Subscript[S, z])]= -i β (Subscript[S, x]);   
i (d/dt)(Subscript[S, z])[t]= 0.

From the first two equations

(Subscript[(Overscript[S, ..]), x]) = - (β^2)(Subscript[S, x]) ; (Subscript[(Overscript[S, ..]), y]) = - (β^2)(Subscript[S, y]) .

Solving and imposing the initial condition one easily get

(Subscript[S, x])[t] = (Subscript[S, x])Cos[β t] + (Subscript[S, y])Sin[β t];   (Subscript[S, y])[t] = (Subscript[S, y])Cos[β t] - (Subscript[S, x])Sin[β t].

For the average values at time t

⟨t|(Subscript[S, x])|t⟩= ⟨0|(Subscript[S, x])[t]|0⟩= S (Cos[φ]Sin[θ]Cos[β t] + Sin[φ]Sin[θ]Sin[β t])= S Cos[φ-β t]Sin[θ];

⟨t|(Subscript[S, y])|t⟩= ⟨0|(Subscript[S, y])[t]|0⟩= S (Sin[φ]Sin[θ]Cos[β t] - Cos[φ]Sin[θ]Sin[β t])= S Sin[φ-β t]Sin[θ].

Problem

A particle of mass m, spin 1/2 and charge e is bound in a three dimensional harmonic potential. A weak uniform magnetic field is applied to the system. The Hamiltonian is

H = (1/(2m))(((p-(e/c)A))^2)+(1/2)m (ω^2)(r^2)- μ·B ;  μ = g (Subscript[μ, B]) s = (g/2)(Subscript[μ, B])σ.        (Subscript[μ, B])= ((e ℏ)/(2mc)).

Solution

1

The Hamiltonian for B = 0 is the sum of three one dimensional oscillators, so the energy levels are

(Subscript[E, N])= ℏ ω (N+(3/2)) ;   N = 0,1,2...

We have an orbital degeneracy equal to the number of ways to get N as the sum of three numbers ( (Subscript[n, 1]), (Subscript[n, 2]), (Subscript[n, 3])), corresponding to the quantum numbers of the three independent one dimensional oscillators:

N = 0, deg. = 1;  N = 1, deg. = 2, N = 3, deg. = 6, N = 4, deg. = 12 ...  In general   deg[N] = (N+1)(N+2). The eigenfunctions have the form

(Subscript[ψ, N])[x,y,z] =C (Subscript[H, (Subscript[n, 1])])[(Sqrt[((m ω)/ℏ)]) x] (Subscript[H, (Subscript[n, 2])])[(Sqrt[((m ω)/ℏ)])y] (Subscript[H, (Subscript[n, 3])])[(Sqrt[((m ω)/ℏ)])z] Exp[- ((m ω)/(2ℏ))(r^2)],

Where (Subscript[H, i])are Hermite polynomials. The parity is (((-1))^N), product of the parity of single oscillators.  Each eigenfunction is a product of a polynomial of degree N times a rotation invariant exponential, then the possible angular momenta are

L = N,N-2,....0 (1).

Even / Odd sequence of L for even/odd parity levels.

Each level has an additional doubling in the degeneracy due to spin.

The first levels are

2

In weak field approximation

H = (1/(2m))(p^2)+(1/2)m (ω^2)(r^2)- ((e B ℏ)/(2m c))(Subscript[L, z]) - ((g B (Subscript[μ, B]))/2)(Subscript[σ, z]) =  (1/(2m))(p^2)+(1/2)m (ω^2)(r^2) - B (Subscript[μ, B])((Subscript[L, z])+ (g/2) (Subscript[σ, z])).

Operators (L^2), (Subscript[L, z]), (s^2), (Subscript[s, z]) commute with H.

3

Eigenstates of H can be classified by N, (Subscript[L, z]), (Subscript[s, z]). The eigenvalues are

(Subscript[E, N,(Subscript[L, z]),(Subscript[s, z])])= ℏ ω (N+(3/2)) - (Subscript[μ, B])B((Subscript[L, z])+ g (Subscript[s, z])).

Problem

A constant electric field directed along z axis acts on a particle of mass m and charge q.

⟨r[0]⟩ = 0;   ⟨p[0]⟩ = (Subscript[p, 0])(OverHat[x]) ;  ⟨L[t]⟩ = 0.

Solution

1

The Hamiltonian is

H = ((p^2)/(2m)) - q (Subscript[E, 0]) z

The equations of motion are

(OverDot[p]) = (i/ℏ)[H,p]= q (Subscript[E, 0])(OverHat[z]) ;   (OverDot[r]) = (i/ℏ)[H,r]= (p/m) ;  ⇒ (Overscript[r, ..]) = ((q (Subscript[E, 0]))/m) (OverHat[z])    (constant acceleration).

Their solution is identical to the classical solution

r[t]=r[0] + (t/m)p[0]+(1/2)((q (Subscript[E, 0]))/m)(t^2) (OverHat[z]) ;              p[t]=p[0]+ q (Subscript[E, 0])t  (OverHat[z]) .

2

By definition

L[t] = r[t]⋀ p[t].

From the solutions () it follows that

(Subscript[L, z])[t] = (Subscript[L, z])[0];
(Subscript[L, x])[t] = (Subscript[L, x])[0] + q (Subscript[E, 0])t y[0] + (1/2)((q (Subscript[E, 0]))/m)(t^2) (Subscript[p, y])[0];
(Subscript[L, y])[t] = (Subscript[L, y])[0] - q (Subscript[E, 0])t x[0] - (1/2)((q (Subscript[E, 0]))/m)(t^2) (Subscript[p, x])[0].

The equations of motion for L are

(OverDot[L]) = (i/ℏ)[H,L].

or

(Subscript[(OverDot[L]), z])[t] = 0;      (Subscript[(OverDot[L]), x])[t] = q (Subscript[E, 0])y[t];       (Subscript[(OverDot[L]), y])[t] = - q (Subscript[E, 0])x[t].

(Subscript[L, z]) is a constant of motion, in agreement with ().  The equations for (Subscript[L, x]) and (Subscript[L, y])can be integrated  using () and reproduce ():

(Subscript[L, x])[t] = (Subscript[L, x])[0]+ q (Subscript[E, 0])(∫_0^t)dt y[t]= (Subscript[L, x])[0] + q (Subscript[E, 0])t y[0] + (1/2)((q (Subscript[E, 0]))/m)(t^2) (Subscript[p, y])[0];
(Subscript[L, y])[t] = (Subscript[L, y])[0]- q (Subscript[E, 0])(∫_0^t)dt x[t]= (Subscript[L, y])[0] - q (Subscript[E, 0])t x[0] - (1/2)((q (Subscript[E, 0]))/m)(t^2) (Subscript[p, x])[0].

3

With the initial conditions given in the text the mean value of () give immediately:

⟨(Subscript[L, x])[t]⟩ = 0;    ⟨(Subscript[L, z])[t]⟩ = 0;    ⟨(Subscript[L, y])[t]⟩ = - (1/2)((q (Subscript[E, 0]))/m)(t^2) (Subscript[p, 0]) .

Problem

A particle of mass m and spin 1/2 is described by a one-dimensional Hamiltonian

H = ((p^2)/(2m)) + V[x];      V[x] = (((Subscript[σ, 3])+1)/4)m (ω^2)(x^2) .

ψ[0,x] = (Subscript[ψ, 0])[x] ((|+⟩-|-⟩)/(Sqrt[2]));

|+⟩,|-⟩ up, down eigenstates of (Subscript[s, z]) ; (Subscript[ψ, 0])[x] = (((((m ω)/(π ℏ))))^(1/4)) Exp[- ((m ω)/(2ℏ))(x^2)] = ground-state wave function for the oscillator (freq. ω ).

Solution

1

H commutes with (Subscript[s, z]) so we can classify the states with the eigenvalues ±1/2 of this operator. For up states (eigenvalues +1/2, or +1 for (Subscript[σ, 3])) H is the Hamiltonian (Subscript[H, ω ])of an harmonic oscillator, with eigenvalues and eigenfunctions

(Subscript[E, n]) = ℏ ω (n+(1/2)), n=0,1,...;   (Subscript[ψ, n])[x]= (Subscript[C, n]) (Subscript[H, n])[(Sqrt[((m ω)/ℏ)])x]Exp[- ((m ω)/(2ℏ))(x^2)].

For down states (eigenvalues -1/2, or -1 for (Subscript[σ, 3])) , the system describes a free particle of an Hamiltonian (H^((free))) with eigenvalues and eigenfunctions

(Subscript[E, p])= ((p^2)/(2m)),  -∞ < p < ∞;    (Subscript[ψ, p])[x] = (1/(Sqrt[2π ℏ]))Exp[i p x].

2

The state is not an eigenfunction of H as it is not an eigenfunction of (Subscript[s, z]). The mean value of H on this state is

⟨ψ|H|ψ⟩= (1/2)⟨(Subscript[ψ, 0])|(Subscript[H, ω])|(Subscript[ψ, 0])⟩ + (1/2)⟨(Subscript[ψ, 0])|(H^((free)))|(Subscript[ψ, 0])⟩ = (1/4)ℏ ω + (1/2)⟨(Subscript[ψ, 0])|((p^2)/(2m))|(Subscript[ψ, 0])⟩=(1/4)ℏ ω + (1/2)((ℏ ω)/4)= (3/8)ℏ ω.

3

The spin up part of the state evolve as an eigenstate of H:

(Subscript[ψ, 0])[x]|+⟩→ (e^(-i (Subscript[E, 0])t/ℏ))  (Subscript[ψ, 0])[x]|+⟩ ;   (Subscript[E, 0])=(( ℏ ω)/2) .

For the down part of the state we can write

(Subscript[ψ, 0])[x]|-⟩= (∫_(-∞)^(+∞))dk a[k](e^(i k x)) |-⟩→  (∫_(-∞)^(+∞))dk a[k](e^(-i (((k^2)ℏ t)/(2m))))(e^(i k x)) |-⟩ .

Using the general formula

(∫_(-∞)^(+∞))dz Exp[-A (z^2)+ B z] = (∫_(-∞)^(+∞))dz Exp[-A (((z- (B/(2A))))^2)+ ((B^2)/(4A))] = (Sqrt[(π/A)]) Exp[((B^2)/(4A))];   Re[A]>0 .

for

(Subscript[ψ, 0])[x] = (((((m ω)/(π ℏ))))^(1/4)) Exp[- ((m ω)/(2ℏ))(x^2)] ≡ N Exp[- A (x^2)]

one easily gets

a[k] = (N/(2π))(Sqrt[(π/A)]) Exp[-(( (k^2))/(2A))].

The free part of the wave function at x = 0 and time t is then

(∫_(-∞)^(+∞))dk a[k](e^(-i (((k^2)ℏ t)/(2m)))) = (N/(2π))(Sqrt[(π/A)])(∫_(-∞)^(+∞))dk (e^(-(( (k^2))/(2A))))(e^(-i (((k^2)ℏ t)/(2m)))) = (N/(Sqrt[1+i ω t])).

The state is

ψ[x=0,t]= (N/(Sqrt[2]))((e^(-i ω t/2))|+⟩ - (1/(Sqrt[1+i ω t]))|-⟩).

To compute the probabilities for measuring up and down components it is convenient to normalize () in spin space:

|ψ⟩ =(Sqrt[((Sqrt[ 1+(ω^2)(t^2)])/(1+(Sqrt[ 1+(ω^2)(t^2)])))])  ((e^(-i ω t/2))|+⟩ - (1/(Sqrt[1+i ω t]))|-⟩).

The requested probability is then

(SubMinus[P])=|⟨-|ψ⟩((|)^2) = (1/(1+(Sqrt[ 1+(ω^2)(t^2)]))) .

Problem

Let us consider the Hamiltonian of Hydrogen atom, together with the spin-orbit interaction term for the electron and the spin-spin interaction between electron and proton (hyperfine interaction):

(Subscript[H, so])= A L·(Subscript[s, e]) ;         (Subscript[H, ss])= B (Subscript[s, p])·(Subscript[s, e]) .

A and B are constants.

(L^2), (J^2), (F^2), (Subscript[F, z]);    J = L + (Subscript[s, e]) ; F = J + (Subscript[s, p]) .

Solution

1

The energy levels are

(Subscript[E, n])=- (1/(2 (n^2))) ((m (e^4))/(ℏ^2)).

Taking into account the spin variables the total degeneracy is 4 (n^2), a factor (n^2) is the Coulomb degeneracy, a factor 2 is due to electron spin and another factor of 2 is due to nuclear spin.

2

Within the degenerate subspace of fixed n, L we can write

(Subscript[H, so])= A L·(Subscript[s, e]) =(A/2)[(J^2)- (L^2)-(Subscript[s, e]^2)] = (A/2)[ J(J+1) - L(L+1) - (3/4)].

Taking into account the rules for angular momentum sum we can compute energies. The results for energies and degeneracies are summarized in the following table.

3

In a similar way we can write

H = (Subscript[H, 0])+ A L·(Subscript[s, e]) + B (Subscript[s, p])·(Subscript[s, e]) = (Subscript[H, 0])+ (A/2)[(J^2)- (L^2)-(Subscript[s, e]^2)]+ (B/2)[(S^2)-(3/4)-(3/4)].

The states |L,J,F,(Subscript[F, z])⟩of the previous table are not, in general, eigenstates of H, as (J^2) do not commutes with (S^2), this happens only when the values of L and F determine uniquely the S value. We are using the notation

S = (Subscript[s, e])+ (Subscript[s, p]) ; F = L + S.

One gets easily the following table