Problems Chapter 3

Quantum Mechanics
                       K. Konishi, G. Paffuti

Problem

Show that the condition

(Subscript[ψ, +]^′)[0] - (Subscript[ψ, -]^′)[0] = - ((2m g)/(ℏ^2)) ψ[0],

for a potential V[x] = - g δ[x], is compatible with the continuity of the current density.

Solution

The current density is

j =((i ℏ)/(2m))(ψ(∂_x)(SuperStar[ψ]) - (SuperStar[ψ])(∂_x)ψ).

At the discontinuity of the potential, x=0, ψ is continuous while for the derivative of ψ we have

Δ((∂_x)ψ ) = -((2m g)/(ℏ^2)) ψ[0];   Δ((∂_x)(SuperStar[ψ]) ) =- ((2m g)/(ℏ^2)) (SuperStar[ψ])[0];         with   ΔF = (Underscript[Lim, ϵ→0])(F[ϵ] - F[-ϵ]).

For the current :

Δj = ((i ℏ)/(2m))(ψ[0]Δ((∂_x)(SuperStar[ψ]) )-(SuperStar[ψ])[0]Δ((∂_x)ψ ))= - ((i g)/(ℏ^2))(ψ[0](SuperStar[ψ])[0] - (SuperStar[ψ])[0]ψ[0]) = 0.

Problem

A particle moves in a three dimensional well

V[r] = 0;  (0<x<a;  0<y<b; 0<z<c);    V[r] = ∞,   otherwise.

Find the eigenvalues and the eigenstates. Discuss the degeneracy for the low-lying levels, in particular for the cubic box  a=b=c.

Solution

The Hamiltonian is  the sum of three independent one dimensional operators

H = (1/(2m))((Subscript[p, x]^2) + (Subscript[p, y]^2) + (Subscript[p, z]^2))≡ (Subscript[H, x])+(Subscript[H, y])+(Subscript[H, z]).

The operators (Subscript[H, x]). (Subscript[H, y]), (Subscript[H, z]) commute between themselves, then can be simultaneously diagonalized. Each of them has the spectrum of a one dimensional well:

(Subscript[H, x]):  (((ℏ^2)(π^2))/(2m ))((Subscript[n, x]^2)/(a^2)) ; (Subscript[H, y]):  (((ℏ^2)(π^2))/(2m ))((Subscript[n, y]^2)/(b^2)) ;  (Subscript[H, z]):  (((ℏ^2)(π^2))/(2m ))((Subscript[n, z]^2)/(c^2)) ;    (Subscript[n, x]),(Subscript[n, y]), (Subscript[n, z]) = 1,2...

The spectrum of H is

(Subscript[E, (Subscript[n, x])(Subscript[n, y])(Subscript[n, z])])= (((ℏ^2)(π^2))/(2m ))(((Subscript[n, x]^2)/(a^2))+((Subscript[n, y]^2)/(b^2))+((Subscript[n, z]^2)/(c^2))).

The corresponding, normalized, eigenstates are

(Subscript[ψ, (Subscript[n, x])(Subscript[n, y])(Subscript[n, z])])[x,y,z] = (Sqrt[(8/(a b c))]) Sin[((π (Subscript[n, x])x)/a)]Sin[((π (Subscript[n, y])y)/b)]Sin[((π (Subscript[n, z])z)/c)].

For generic (non relative rational) (a^2), (b^2), (c^2) the state are non degenerate. If  the ratios of the edges are rational some levels can be degenerate.

The total number of states up to an energy E is approximately (the integral is for one sector of a sphere, with volume 1/8 of the whole ball):

N[E]=∫d(Subscript[n, x])d(Subscript[n, y])d(Subscript[n, y]) Θ[E- (Subscript[E, (Subscript[n, x])(Subscript[n, y])(Subscript[n, z])])] = (((((2m)/((ℏ^2)(π^2)))))^(3/2))a b c (∫_(x,y,z>0))dx dy dz  Θ[E> (x^2)+(y^2)+(z^2)] =
(((((2m)/((ℏ^2)(π^2)))))^(3/2))a b c (1/8) (4/3)π (E^(3/2)) =((((2m E))^(3/2))/(ℏ^3)) (2/3) (V/(4(π^2))).

V = a b c is the volume. In the problems of chap. 13 it is shown that a more accurate formula is

N[E] = ((((2m E))^(3/2))/(ℏ^3)) (2/3) (V/(4(π^2))) - (((2m)/(ℏ^2)))(S/(16π)) ;

S = 2 (a b + a c + b c) is the surface of the box.

Note on the degeneracy

Let us consider the case a = b = c:

E = (((ℏ^2)(π^2))/(2m (a^2) ))((Subscript[n, x]^2)+(Subscript[n, y]^2)+(Subscript[n, z]^2)).

It is easy to compute energies and degeneracies for the first few levels

While for low energy it is quite clear the appearance of degeneracy 6 due to the permutations of 3 (different) numbers (Subscript[n, x]), (Subscript[n, y]), (Subscript[n, z]) for higher energies things are not so simple. Here is an example of the degeneracy distribution

The total number of states N[E] follow very accurately the prediction (), the number of levels is proportional to E, and very close to 5/6 E/E0:

The problem of levels and of their number is closely related to a classic problem in number theory: find the numbers that can be written as a sum of 3 squares. In number theory usually the value 0 for (Subscript[n, i])is included, and the problem has been solved by Legendre and Gauss. We quote a result, the numbers which cannot be written as a sum of 3 squares have the form (4^a)(8 m + 7).  These are certainly excluded from the spectrum, other numbers, built with the value 0 for (Subscript[n, i])are included in number theory classification and possibly excluded in the spectrum (essentially the number sum of two squares). These are the first numbers of these series

nExcluded= {7,15,23,28,31,39,47,55,60,63,71,79}

confronting with the first E/E0 levels we can check the statement:

E/E0 = {3,6,9,11,12,14,17,18,19,21,22,24,26,27,29,30,33,34,35,36,38,41,42,43,        44,45,46,48,49,50,51,53,54,56,57,59,61,62,65,66,67,68,69,70,72,73,74,75,76,77};

Problem

If the one-dimensional square well has the parameters such that

(Sqrt[((m (a^2) (Subscript[V, 0]))/(2 (ℏ^2)))]) =δ ≪ 1 ,

it has a unique bound state. Compute the energy of this state approximately, in the first non trivial order in δ, as a function of  (Subscript[V, 0]).

Solution

With the notations of the figure

the (even) solution of Schrödinger equation is

ψ[x]= A Cos[q x]; |x|< a/2 ;          ψ[x] = B Exp[- α |x|] ;

with

ℏ q = (Sqrt[2m((Subscript[V, 0])-ϵ)]) ;   ℏ α = (Sqrt[2m ϵ]) .

Matching the logarithmic derivative at x = a/2  gives

q Tan[q (a/2)]=  α  ⇒  (q^2)(a/2) ≃ α

For small ϵ

((2m (Subscript[V, 0]))/ℏ)(a/2) ≃ (Sqrt[2m ϵ])   ⇒   ϵ = ((m (Subscript[V, 0]^2) (a^2))/(2 (ℏ^2)))  ⇒  E = - ϵ = - ((m (Subscript[V, 0]^2) (a^2))/(2 (ℏ^2)))= - (Subscript[V, 0]) (δ^2).

Problem

Consider the limit a→0, (Subscript[V, 0])→∞ with (Subscript[aV, 0])=f fixed, in the problem of a square well potential. Compare the result with the δ[x] potential.

Solution

The notations are given in the figure below

The potential being even, we can limit the study of solutions to the region x > 0.

The (even, odd) solutions of Schrödinger equation are ( for x > 0 )

ψ[x]= A Cos[q x]; |x|< a/2 ;          ψ[x] = B Exp[- α x] ;

ψ[x]= A Sin[q x]; |x|< a/2 ;          ψ[x] = B Exp[- α x] ;

with

ϵ = |E|;  ℏ q = (Sqrt[2m((Subscript[V, 0])-ϵ)]) ;   ℏ α = (Sqrt[2m ϵ]) .

The match of logarithmic derivative at x = a/2  gives

q Tan[q (a/2)]=  α  ;     q Cot[q (a/2)]=  α  .

In the limit  a→0, (Subscript[V, 0])→∞ with (Subscript[aV, 0])=f fixed,

q → (1/ℏ)(Sqrt[2m (Subscript[V, 0])]) =  (1/ℏ)(Sqrt[2(m/a) f]) → ∞;    q a → 0.

Only the first of the equations () can have a limit

q Tan[q (a/2)] → (((q^2)a)/2)→ ((m f)/(ℏ^2)).

The only eigenvalue is, from ()

E = - ϵ = - (((((m f)/(ℏ^2))))^2)((ℏ^2)/(2m))= -(( m (f^2))/(2 (ℏ^2))).

In a δ potential - g δ[x] there is only one bound state (see text) with energy

(Subscript[E, δ]) = - ((m (g^2))/(2 (ℏ^2)))

The "strength" of the δ potential is defined by

-g = -∫dx V[x]=-∫dx g δ[x].

For a deep well the integration over a region including the well gives

∫d x V[x] = -∫dx (Subscript[V, 0])= - (Subscript[V, 0])a → - f.

We see that the well goes into a δ potential with strength f and its energy eigenvalue () gives the correct result ().

Problem

Find the transmission coefficient for the step potential

V = 0, x<0;   V=(Subscript[V, 0])>0, x≥ 0 ,

for E<(Subscript[V, 0]) and E>(Subscript[V, 0]).

Solution

For  E < (Subscript[V, 0]) there is not transmitted wave as for x > 0 the solution is exponentially depressed.

For E > (Subscript[V, 0]) the solution giving a transmitted wave is

ψ[x]= (e^(i k x)) + A (e^(-i k x)) ;   x < 0;             ψ[x] = B (e^(i q x));  x > 0 .

Where

ℏ k = (Sqrt[2 m E]) ;     ℏ q = (Sqrt[2 m (E - (Subscript[V, 0]))]) .

Matching ψ'/ψ at x = 0 gives

k ((1-A)/(1+A)) = q  ⇒ A = ((k-q)/(k+q)) .

The reflection and transmission coefficients, are fixed by the ratio of the currents

R = |A((|)^2)  ; T = (q/k)|B((|)^2)= 1 - R.

We have :

R = (((((k-q)/(k+q))))^2) ;  T = ((4 k q)/(((k+q))^2)).

To check this result we compute B from the continuity of ψ at x=0:

B = 1 + A = ((2 k)/(k+q))  ;     ⇒     T=  (q/k)((|B((|)^2))/(|A((|)^2))) = ((4 k q)/(((k+q))^2));  c.v.d.

Problem

Discuss the qualitative features of the low-lying energy levels in a deep double well potential, by using the symmetry argument (parity) and the non degeneration theorem.

Solution

The situation is outlined below:

In a first approximation we have two separate wells, exactly degenerate. This cannot represents the true ground state of the model as we know that the ground state is non degenerate and even. A good candidate for an approximate wave function is a superposition of the two separate localized solutions describing the particle in each well:

ψ = (((Subscript[ψ, L])[x] + (Subscript[ψ, R])[x])/(Sqrt[2])).

Parity invariance impose (Subscript[ψ, L])[-x]= (Subscript[ψ, R])[x] (a better way to say this is that parity invariance allow to choose the phases in such a way that (Subscript[ψ, L])[-x]= (Subscript[ψ, R])[x] ).

The physical mechanism which allow the superposition is the tunneling between the two wells. As an effective Hamiltonian in the basis (Subscript[ψ, L]), (Subscript[ψ, R]) one can consider

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

ϵ is proportional to the tunnel probability. The ground state of H is (). The reader can verify that H commute with

P = (({{0, 1}, {1, 0}}))

which exchange (Subscript[ψ, L]), (Subscript[ψ, R]) and acts as parity operator in this system. The state  () is even.

Problem

A particle of mass m is confined in a potential well V[x], with

V[x] ={ ({{0;,      0≤x≤ a}, {∞;,                   x<0 and x>a}})

The particle is in in the ground state.

Solution

1

The force can be found equating the work done in an adiabatic expansion to the energy variation of the system:

F δa = - δE; F =- (( ∂E)/(∂a)).

For the ground state

E = (((π^2)(ℏ^2))/(2m (a^2)));           F = (((π^2)(ℏ^2))/(m (a^3))).

2

In a sudden expansion the state is frozen, i.e. remains, in the short interval of the expansion,

(Subscript[ψ, 0])[x] = (Sqrt[(2/a)])Sin[π (x/a)]θ[x]θ[a-x]

After the expansion energy eigenstates are given by

(Subscript[φ, n])[x] = (Sqrt[(2/(2a))])Sin[n π (x/(2 a))]

The requested probabilities are

(Subscript[P, n]) =((|(∫_0^a)dx (Subscript[ψ, 0])[x](Subscript[φ, n])[x]|)^2) = ({{1/2;,      n=2}, { 0      ;,       n = 2L; L > 1}, {(32/(π^2))(1/(((4-(n^2)))^2)), n = 2L+1.}})

It is easy to check (try for example with Mathematica  ) that (∑_n)(Subscript[P, n])= 1.

3

In an adabatic expansion the state "follows" the expansion and continues to be the ground state of the expanding well. The total energy change is then

ΔE = (((π^2)(ℏ^2))/(2m (a^2)))-  (((π^2)(ℏ^2))/(2m 4 (a^2)))= (3/8)(((π^2)(ℏ^2))/(m (a^2))).

As expected from the definition of force, equation (),

ΔE = (∫_a^(2a))F[a]da .

Problem

A particle of mass m is subject to the one dimensional potential

V[x] ={ ({{-g δ[x];,      x≤ a}, {∞;,       x≥a}})

Solution

1

The potential is sketched in the following figure

For E < 0 the general form of the solution of the Schrödinger equation is

ψ[x] ={ ({{A Exp[α x],,       x<0}, {B Exp[α x]+C Exp[-α x],,      x>0}})

With a δ potential in (Subscript[x, 0]), ψ must be continuous and the discontinuity of its first derivative is fixed by Schrödinger equation:

ψ'[(SuperPlus[(Subscript[x, 0])])]- ψ'[(SuperMinus[(Subscript[x, 0])])] = ((2m)/(ℏ^2))  (Underscript[Lim, ε→0])(∫_((Subscript[x, 0])-ε)^((Subscript[x, 0])+ε))V[x]ψ[x]𝕕x.

In the case at hand the constraint on ψ and its derivative are

A = B + C;     α(B-C) - α A = - ((2m)/(ℏ^2))g A.

i.e.

B =( 1-((m g )/(α (ℏ^2))));  C = ((m g )/(α (ℏ^2))) A .

At point x = a the function must vanish

ψ[a] = B Exp[α a]+C Exp[-α a] = 0

Using eq ()

1- (e^(-2 a α)) = ((α (ℏ^2))/(g m))

From the plots of the functions

(Subscript[f, 1])[x]=1 - Exp[-2 a x];  (Subscript[f, 2])[x]= x (( (ℏ^2))/(g m))

it is evident that the possibility on a non trivial solution depends on derivative at the origin, i.e. we have an intercept if

(( (ℏ^2))/(g m)) ≤ 2 a,   ⇒    g ≥ (( (ℏ^2))/(2 m a)).

Only in this range we can have a bound state.

2

As a → ∞ condition () gives

α = ((g m)/(ℏ^2))

which is the known result for a simple δ - potential.

3

There is no degeneracy in the continuum levels. To show this let us write the form of a stationary wave function for positive energy. k is a positive number and E = (ℏ^2)(k^2)/2m

ψ ={ ({{A (e^(i k x))+ B (e^(-i k x)),, x<0}, {C (e^(i k x))+ D (e^(-i k x)),,       0<x<a}, {0,, x≥ a}})

The condition ψ[a] = 0 gives a relation between C and D. Continuity and gap in first derivative, eq.(), give two more condition. Three equations for 4 coefficients fix the solution up to an overall normalization factor.

Problem

A particle of mass m is subjected to the one dimensional potential

V[x] = ({{∞;,      x≤ -a}, {g δ[x];;,       -a≤x≤2a}, {∞, x>2a}}).

We want to study the first two energy levels. Let us call (SubMinus[ψ])[x],(SubPlus[ψ])[x]the wave functions  for negative and positive x respectively.

Solution

1

The vanishing of ψ at the external boundaries imply

(SubMinus[ψ])[x] = A Sin[k( x + a)] ;       (SubPlus[ψ])[x] = B Sin[k( x -2 a)]

where E = (ℏ^2) (k^2) /2m. At x = 0 we have to match the functions and the derivative's gap

(SubPlus[ψ])[0]=(SubMinus[ψ])[0]       ⇒      A Sin[k a]=- B Sin[2k a];

(Subscript[ψ, +]^('))[0]-(Subscript[ψ, -]^('))[0]= ((2m g)/(ℏ^2))ψ[0]     ⇒    k(B Cos[2k a] - A Cos[k a]) = ((2m g)/(ℏ^2))A Sin[k a].

From first conditions it follows

Sin[k a](A + 2 B Cos[k a]) = 0.

The first possibility is Sin[k a] = 0. In this case

k = ((π n)/a);       (Subscript[E, n]) = ((ℏ^2)/(2m))(((π^2) (n^2))/(a^2));  n = 0,1,2 ...

A and B coefficients are determined, up to an overall normalization by ()

B Cos[2π n] - A Cos[π n]=0,     ⇒   B = (((-1))^n)A.

As Sin[k a] = 0 in this class of solutions ψ[0] = 0.

2

The second possibility allowed by eq.() is

A = - 2 B Cos[k a].

Substitution in () gives

k Cos[2 k a] + 2 k ((Cos[k a])^2) = - ((4m g)/(ℏ^2))Cos[k a]Sin[k a],

i.e. with ξ = 2 k a

ξ (2 Cos[ξ] + 1) = - Λ Sin[ξ];     Λ = ((4m g a)/(ℏ^2)) > 0 .

We can plot the two curves

(Subscript[f, 1])[ξ] = ξ (2 Cos[ξ] + 1);   (Subscript[f, 2])[ξ] = - Λ Sin[ξ],

for different values of  Λ. The graph below show curves for Λ = 0.5, 2, 5.

One easily convince oneself that the first intersection, giving the first state in this group, corresponds to a value of ξ = 2ka lying in the region

((2π)/3)< ξ < π ;   (π/(3a)) < k < (π/(2a)).

Comparison with () show that ground state of the system is the first state of this second group, while n=1 state in () is the first excited state.

In the two limits  g → 0 and g → ∞ also Λ → 0, ∞.  from the graph it is clear that as Λ → 0 the intersection tends to the first zero of the curve (Subscript[f, 1])[ξ], i.e. ξ = 2π/3. while for Λ → ∞ the intersection approach the zero of the second curve, Sin[ξ], i.e. ξ = π. In  the two limits then

as g→ 0:    k → (π/(3a));    as g→ ∞ :  k → (π/(2a)).

Problem

A particle of mass m and energy E < 0  moves in a one-dimensional potential

V[x] = f δ[x] - g δ[x-a];   g > 0.

Solution

1

For a single δ potential the solution for E < 0  has the form (Θ is the Heavside theta function, or step function, in Mathematica is UnitStep[x] ):

ψ[x] = A  (e^( β( x-a))) Θ[a-x] + B (e^(-β (x-a)))Θ[x-a];    E = - (((β^2)(ℏ^2))/(2m))

For a δ potential matching conditions require continuity of ψ and assigned gap for derivatives :

ψ[(SuperPlus[a])]= ψ[(SuperMinus[a])];               ψ'[(SuperPlus[a])]- ψ'[(SuperMinus[a])] = ((2m)/(ℏ^2))  (Underscript[Lim, ε→0])(∫_(a-ε)^(a+ε))V[x]ψ[x]𝕕x.

In our case

A = B;   -β B - β A = ((2m)/(ℏ^2))(- g A)    ⇒            β =(( m g)/(ℏ^2)).

2

Negative energy, normalizable, solutions must decrease at x → ± ∞. The most general form of this solution is, with above notation () for E:

ψ[x] = (Subscript[A, 1])(e^(β x))Θ[-x] + ( (Subscript[A, 2])(e^(β x))+ (Subscript[B, 2])(e^(-β x)))Θ[x]Θ[a-x] + (Subscript[B, 3])(e^(-β x))Θ[x-a].

We have two set of equations of type (), i.e.  four matching conditions, two for functions:

(Subscript[A, 1])= (Subscript[A, 2])+ (Subscript[B, 2]);         (Subscript[A, 2])(e^(β a))+ (Subscript[B, 2])(e^(-β a)) = (Subscript[B, 3])(e^(-β a));

and two for derivatives

β(- (Subscript[A, 1]) + ((Subscript[A, 2])- (Subscript[B, 2]))) = ((2 m f)/(ℏ^2)) (Subscript[A, 1]) ;    -β ((Subscript[B, 3])(e^(-β a))+ (Subscript[A, 2])(e^(β a))- (Subscript[B, 2])(e^(-β a))) = - ((2 m g)/(ℏ^2))(Subscript[B, 3])(e^(-β a)).

3

We have a system of four homogenous equations in four unknown quantities (Subscript[A, 1,])(Subscript[A, 2]),(Subscript[B, 2]),(Subscript[B, 3]). The system has a nontrivial solution only if the determinant of the coefficients is zero, and this will determine one (or more) values for β, i.e. energies for the bound states.

Instead of writing te determinant let us proceed by elimination of variables. let us solve for (Subscript[A, 2])and (Subscript[B, 2]). From ()

(Subscript[A, 2])= (1+(( m f)/(β (ℏ^2))))(Subscript[A, 1]);   (Subscript[B, 2])= - ((m f)/(β (ℏ^2)))(Subscript[A, 1]);

while from ()

(Subscript[A, 2])(e^(β a))= ((m g)/(β (ℏ^2)))(Subscript[B, 3])(e^(-β a));   (Subscript[B, 2])(e^(-β a)) =  (1-(( m g)/(β (ℏ^2))))(Subscript[B, 3])(e^(-β a));

Eliinating the unknown (Subscript[A, i])and (Subscript[B, i])we arrive at

(e^(2 a β)) (g m-β (ℏ^2)) (f m+β (ℏ^2)) = f g (m^2).

With

x =(( β (ℏ^2))/(m g));     A = ((2 m g a)/(ℏ^2));

the previous condition take the form

(x+(f/g))(x-1) = - (( f)/g) Exp[- A x]

4

We look for solutions with x > 0.

For f = 0 we recover x=1, which is the solution ().

For f = -g, i.e. a symmetric δ potential, the condition becomes

1 - (( m g )/(β (ℏ^2))) = ± (( m g )/(β (ℏ^2)))Exp[-β a].

which has been found in [ ** ].

For small |f| the solution is unique :

x ≃ 1 - (f/g)(e^(-A)) ;    β = ((m g)/(ℏ^2))(1-(f/g)(e^(-A)));    E≃ - ((m (g^2))/(ℏ^2))+ ((m f g)/(ℏ^2))(e^(-((2 m g a)/(ℏ^2)))).

For the general case let us look at graphs of the to sides of equation ():

It is apparent that if derivative in x = 0 of the first curve is bigger than derivative of the second curve then there is no solution

(f/g)- 1 > (f/g)A

This happens for A < 1 and suficiently large f:

f > (g/(1-A)) = (g/(1-((2 m g a)/(ℏ^2)) )).

Problem

A particle of mass m moves in the one dimensional potential ((Subscript[V, 0])> 0)

V[x] = ({{∞;,      x< 0}, {0;,       0≤x≤a}, {(Subscript[V, 0]);,      x>a}}).

Solution

1

({{The potential is sketched
in the figure. The region
 x<0 is forbidden.

,                 }})

The solution for a bound state has the form

ψ[x] ={ ({{A Sin[k x],    , 0≤ x ≤ a}, {B Exp[-β x],  , x> a}})   ;        with      k = (Sqrt[((2m E)/(ℏ^2))]);  β = (Sqrt[((2m ((Subscript[V, 0])-E))/(ℏ^2))]) .

Continuity of ψ'/ψ at x = a gives the constraint

β = - k Cot[ k a].

Defining ξ = ka; η = βa the eigenvalue can be found by the intersection of the two curves

η = - ξ Cot[ξ];    (ξ^2)+ (η^2) = ((2m (Subscript[V, 0])(a^2))/(ℏ^2))

As is evident from the graph the condition to have only one bound state is

(π/2) < (Sqrt[((2m (Subscript[V, 0])(a^2))/(ℏ^2))]) < (3/2) π.

2

For large (Subscript[V, 0]) the wave function for x>a canbe neglected and we have

ψ[x]≃ (Sqrt[(2/a)])Sin[((π x)/a)];         k = (π/a);  E = (((ℏ^2)(π^2))/(2m (a^2)))

3

If the barrier behind x = 2 a is dropped we have the following potential

A particle which at t = 0 is confined in the region 0 ≤ x ≤ a can tunnel below the barrier and escape. The penetration factor for a rectangular barrier has been computed in the text:

D = ((4 (k^2) (β^2))/(4 (k^2) (β^2)+((((k^2)+(β^2)))^2) ((Sinh[β a])^2)))  ≃ ((16(k^2))/(β^2))Exp[- 2 β a];    k≃ (π/a);  β≃ (Sqrt[((2m (Subscript[V, 0]))/(ℏ^2))]) .

The number of hits per second against the barrier at x=a for a particle of velocity v is v/(2 a) the the probability of decay per unit time is

W = (dP/dt) = (v/(2a))D ≃ ((ℏ k)/m)(1/(2a))((16(k^2))/(β^2))Exp[- 2 β a].

The number of hits per second can also be computed from the current inside the interval [0, a]. The wave function is stationary and superposition of two opposite traveling waves:

ψ[x]= (Sqrt[(2/a)]) (((e^(i k x))- (e^(-i k x)))/(2i)) = (SubPlus[ψ]) + (SubMinus[ψ])

The current relative to right movers, the one which hits the barrier, is the result we used in our previous formula:

j = (ℏ/(2m i))( (Subscript[ψ, +]^( * )) ((d (SubPlus[ψ]))/dx)- ((d (Subscript[ψ, +]^( * )))/dx) (SubPlus[ψ]))= (ℏ/(2m i))(1/(2a))2i k = ((ℏ k)/(2 a m)).

Problem

Consider the space divided in two regions separate by a plane. The potential energy being 0 and -(Subscript[U, 0]) in the two half spaces. A plane wave impinge on the plane with an incidence angle α. Describe the reflection and the transmission (refracted wave) of the wave. Write and verify the condition for conservation of  the number of particles.

Solution

The Schrödinger in the two region is

Δψ + (k^2)ψ = 0;    (k^2) = 2 m E/(ℏ^2)
Δψ + (q^2)ψ = 0;    (q^2)= (k^2) + ((2m (Subscript[U, 0]))/(ℏ^2)) ≡ (k^2)+ (β^2);

With the notations shown in the figure we look for a solution in the form

y> 0:  ψ = Exp[i (Subscript[k, 1])r] + (Subscript[A, R]) Exp[i (Subscript[k, 2])r];  (Subscript[k, 1])= k(Sin[α],-Cos[α]); (Subscript[k, 2])= k(Sin[α'],Cos[α']);
y < 0: ψ = (Subscript[A, T]) Exp[i q r];    q = q(Sin[β],-Cos[β]).

Continuity for ψ,  for the separation plane y=0 give

Exp[i k Sin[α] x]+ (Subscript[A, R])Exp[i k Sin[α'] x]= (Subscript[A, T]) Exp[i q Sin[β] x]

As the relation must hold for all x the phases must be equal:

k Sin[α]= k Sin[α'] = q Sin[β]  ⇒    α'= α;   (q/k) = ((Sin[α'])/(Sin[β])).

With q/k = n we recognize the usual Snell reflection and refraction laws:

α'= α;    n =  ((Sin[α])/(Sin[β])).

As the phases are equal from eq.() and continuity of ψ it follows

1 + (Subscript[A, R])= (Subscript[A, T]).

The continuity of (∂_x)ψ,  using the equality of phases and () , gives

k Sin[α] + k (Subscript[A, R]) Sin[α] =q (Subscript[A, T]) Sin[β]

which is automatically satisfied with  () and (): this is obvious because we imposed () for all x.

The continuity of (∂_y)ψ gives

- k Cos[α] + k (Subscript[A, R]) Cos[α] = -q (Subscript[A, T]) Cos[β] ⇒ Cos[α](1-(Subscript[A, R])) = n (Subscript[A, T])Cos[β].

Solving ()and () for (Subscript[A, R])and (Subscript[A, T]) gives

(Subscript[A, R])= ((Cos[α]-n Cos[β])/(Cos[α]+n Cos[β])) ;  (Subscript[A, T])= ((2 Cos[α])/(Cos[α]+n Cos[β]));

the Fresnel reflection laws adapted to this case.

Let us note that it is not true the usual one dimensional relation 1 = (Subscript[A, R]^2) + n (Subscript[A, T]^2). The unitarity instead imply

k Cos[α] = k (Subscript[A, R]^2) Cos[α]+ q (Subscript[A, T]^2)Cos[β] ⇒  Cos[α] = (Subscript[A, R]^2) Cos[α]+ n (Subscript[A, T]^2)Cos[β]

which is indeed satisfied by ().

In fact, consider an area Sin the separation plane. The number of the incident particle per second is

(Subscript[j, inc]) S Cos[α]= k S Cos[α].

The number of particles reflected and refracted per second

(Subscript[j, refl])S Cos[α] = k  (Subscript[A, R]^2)S Cos[α] ;    (Subscript[j, T])S Cos[β] = q  (Subscript[A, T]^2)S Cos[β].

The conservation of particle's number gives ().

Problem

A particle of mass m moves in one dimension and is subject to a harmonic  force - k x. At time t=0 the center of the force is suddenly shifted by (Subscript[x, 0]). The particle is assumed to be in the ground state for t<0.

Solution

For t<0 and t>0 the Hamiltonian is

(Subscript[H, 0])= ((p^2)/(2m)) + (1/2)m (ω^2)(x^2)  ;   (Subscript[H, 1]) = ((p^2)/(2m)) + (1/2)m (ω^2)(((x-(Subscript[x, 0])))^2) ;   ω = (Sqrt[k/m]) .

At negative times the system is in the state (Subscript[ψ, 0])of  (Subscript[H, 0]). Just after the translation the state is unchanged.

1

The mean value of (Subscript[H, 1])do not change for t>0, it is then sufficient to do the computation in the limit t→(SuperPlus[0]).

⟨(Subscript[H, 1])⟩ = ⟨(Subscript[ψ, 0])|(Subscript[H, 0]) +(1/2)m (ω^2)(Subscript[x, 0]^2)- m (ω^2)x (Subscript[x, 0])|(Subscript[ψ, 0])⟩ = (1/2)ℏ ω + (1/2)m (ω^2)(Subscript[x, 0]^2) .

We used ⟨(Subscript[ψ, 0])|x|(Subscript[ψ, 0])⟩ = 0.

2

Let (Subscript[φ, n]) the eigenstates of (Subscript[H, 1.])We have for the initial state

ψ[x,0]=(Subscript[ψ, 0])[x]= (∑_(n=0)^∞) (Subscript[a, n]) (Subscript[φ, n])[x].

The time evolution is

ψ[x,t] =  (∑_(n=0)^∞) (Subscript[a, n]) (Subscript[φ, n])[x] (e^(-i (Subscript[E, n])t/ℏ)).

The requested probabilities are time independent:

(Subscript[P, n])[t] = |(Subscript[a, n])(e^(-i (Subscript[E, n])t/ℏ))((|)^2)=(Subscript[a, n]^2) .

and can be computed on the state at t→(SuperPlus[0]).

The wave functions required to answer the question are

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

(Subscript[φ, 0])[x] = (((((m ω)/(π ℏ))))^(1/4)) Exp[- ((m ω)/(2ℏ)) (((x-(Subscript[x, 0])))^2)]; (Subscript[φ, 1])[x] = (((((m ω)/(π ℏ))))^(1/4)) (1/(Sqrt[2]))2 (Sqrt[((m ω)/ℏ)]) (x-(Subscript[x, 0]))Exp[- ((m ω)/(2ℏ)) (((x-(Subscript[x, 0])))^2)].

We have

(Subscript[a, n]) = (∫_(-∞)^(+∞))dx (Subscript[ψ, 0])[x] (Subscript[φ, n])[x],

and an elementary integral gives

(Subscript[P, 0])= |(Subscript[a, 0])((|)^2)= (e^(-((m ((Subscript[x, 0])^2) ω)/(2 ℏ))))   ;  (Subscript[P, 1])= |(Subscript[a, 1])((|)^2)= ((m ω)/(2ℏ))(Subscript[x, 0]^2)(e^(-((m ((Subscript[x, 0])^2) ω)/(2 ℏ))))  .

It is interesting to perform the computation using creation and annihilation operators.

Let  a, (SuperDagger[a])the standard operators for (Subscript[H, 0]) and b, (SuperDagger[b])the corresponding operators for (Subscript[H, 1])

a = (Sqrt[((m ω)/(2ℏ))]) x + i (1/(Sqrt[2m ω ℏ]))p;   (SuperDagger[a]) = (Sqrt[((m ω)/(2ℏ))]) x - i (1/(Sqrt[2m ω ℏ]))p;

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

Clearly

b = a - C;  (SuperDagger[b])= (SuperDagger[a]) - C ;   C = (Sqrt[((m ω)/(2ℏ))]) (Subscript[x, 0]) .

The expansion () reads

(Subscript[|0⟩, 0]) = (∑_n) (Subscript[a, n]) |n⟩_1;     where   b|n⟩_1= (Sqrt[n])|n-1⟩_1 .

The condition for (Subscript[|0⟩, 0]) to be the ground state of (Subscript[H, 0])is

(Subscript[a|0⟩, 0])= 0 ⇒ (b + C)(∑_n) (Subscript[a, n]) |n⟩_1 = 0 .

Using the definition of b we get the recurrence relation

(Subscript[a, n])= (((-C))/(Sqrt[n]))(Subscript[a, n-1]) = ... ((((-C))^n)/(Sqrt[n!]))(Subscript[a, 0]) .

From the normalization condition

1 = (∑_n)| (Subscript[a, n])((|)^2)= (Subscript[a, 0]^2)(∑_n) (((C^2)^n)/(n!)) = (Subscript[a, 0]^2) Exp[(C^2)]     ⇒       (Subscript[a, 0]^2) = Exp[-(C^2)].

The probability to find the n - th state has a Poisson distribution

(Subscript[P, n])= | (Subscript[a, n])((|)^2) = (((C^2)^n)/(n!)) Exp[-(C^2)] .

The result for n = 0, 1 coincides with the previous one.

3

As the function do not change at t=0 the operators x and p do not have discontinuities. The same is true for (OverDot[x])= p/m . The derivative of p instead is discontinuous as

(dp/dt)= -(∂/(∂x))V[x]

and V changes discontinuously.

t < 0:   (∂/(∂x))V[x] = m (ω^2) x    ;    t > 0:   (∂/(∂x))V[x] = m (ω^2) (x - (Subscript[x, 0]))    ;

and the discontinuity is

Δ ⟨(dp/dt)⟩ = m (ω^2)  (Subscript[x, 0]).

Problem

A schematic neutron interferometer is shown in the figure

The Hamiltonian describing neutron motion is

H = (((Subscript[p, x]^2)+(Subscript[p, z]^2))/(2M)) + M g z

M ≃ 1.7 (10^(-24))gr is the neutron mass, g≃980 cm/(sec^2). z is the vertical direction, x the horizontal direction, the incident direction of the beam.

Suppose that at t=0 the plane BCED is vertical, while ABD is on the x axis.

Take v=3 (10^5)cm/sec for the velocity of the incident neutron. Neglect the small distortion in the paths due to gravitational field.

Solution

1

λ = (h/p)= ((2π ℏ)/p).

2

The wave function acquires a different phase in the two paths ABCEF and ABDEF not for the lengths of the paths( equal) but for the variation of the wavelength. In effect due to gravitational potential V[z] = M g z, in the horizontal paths CE and BD the wave number k = 2π/λ is given by

(Subscript[k, CE]) = ((Sqrt[2M(E-M g (Subscript[h, 0]))])/ℏ) ;   (Subscript[k, BD]) = ((Sqrt[2M E])/ℏ) = k.

Their difference is

Δk = k - k (((1-((M g (Subscript[h, 0]))/E)))^(1/2))≃ (((M^2)g (Subscript[h, 0]))/(k (ℏ^2))).

The induced phase shift is

ΔΦ = L Δk ≃ (((M^2)g (Subscript[h, 0])L λ)/(2π (ℏ^2))).

The maximal phase difference is for

ΔΦ = 2π n = (((M^2)g (Subscript[h, 0])L λ)/(2π (ℏ^2))).

3

As φ varies the height of CE path changes form (Subscript[h, 0])to 0. Using the preceding formula the number of times the two paths have maximal interference (i.e. maximal intensity in F) is

𝒩 = (((M^2)g (Subscript[h, 0])L λ)/((((2π))^2) (ℏ^2))) = ((Mg ((Subscript[h, 0])L))/(2π ℏ v))≃ 8.8.

The interference has been measured experimentally, see R. Colella, A. Overhauser, S. Werner: Phys. Rev. Lett. 34, 1472 , (1975).

Problem

An excited nucleus (with excitation energy G) decays into the ground state by emitting a photon.

H = ((p^2)/(2M)) + (1/2)M (ω^2)(r^2).

This analysis shows that a fraction of absorption and re - emission of a gamma ray from a nucleus occurs in a recoil - free manner, when the nucleus is bound in a crystal solid. This phenomenon is known as the Moessbauer effect.

Solution

1

Energy conservation gives

(Subscript[E, γ])= G .

2

If the photon carries a momentum p = ℏ k, momentum conservation imply for the nucleus a momentum -p = - ℏ k. The kinetic energy of the nucleus is

(Subscript[E, K])= (((ℏ^2)(k^2))/(2M)).

Energy conservation requires (as (Subscript[E, γ])= p c):

k ℏ c + (((ℏ^2)(k^2))/(2M)) = G  ;   or      (Subscript[E, γ])+ ((Subscript[E, γ]^2)/(2M (c^2)))= G.

Solving iteratively for (Subscript[E, γ]) we find

(Subscript[E, γ])≃ G - ((G^2)/(2M (c^2)))+...

3

By hypothesis the oscillator center does not acquire momentum in the emission process (has an infinite mass). The energy of the photon can be affected by an excitation of the oscillator level,  0 → N, however. Energy conservation gives

G = (Subscript[E, γ]) + (Subscript[E, f])-(Subscript[E, i]) ;   (Subscript[E, γ])= G - ℏ ω N,   N = 0,1,2...

N is related to the quantum numbers of the oscillator after the photon emission by

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

4

The operator Exp[i k z] shifts the momentum of the nucleus by (0,0, -ℏ k), this is the effect of the momentum conservation law. Then

(Subscript[P, N])= |⟨(Subscript[ψ, f])|(e^(i k z))|(Subscript[ψ, 0])⟩((|)^2)= |⟨N|(e^(i k z))|0⟩((|)^2) ;    k ℏ c = G - ℏ ω N .

The reader can easily convince himself that the normalization is consistent, i.e. the sum of probabilities is unity, due to completeness of harmonic oscillator eigenstates. The second expression in () refers to states of the one dimensional oscillator

H = ((p^2)/(2M))+ ((M (ω^2))/2)(z^2) ,

the only one entering in the matrix elements of an operator which depends only on z. To compute the matrix element let us write

⟨N| = (1/(Sqrt[N!]))⟨0| (a^N) ;    Exp[i k z] = Exp[i C( a + (SuperDagger[a]))];      C = k (Sqrt[(ℏ/(2M ω))]) .

Using the identities

a (e^(i C(a + (SuperDagger[a]))))|0⟩=[ a, (e^(i C(a + (SuperDagger[a]))))]|0⟩ = i C (e^(i C(a + (SuperDagger[a]))))|0⟩;
(a^2) (e^(i C(a + (SuperDagger[a]))))|0⟩=[ (a^2), (e^(i C(a + (SuperDagger[a]))))]|0⟩ = (((i C))^2) (e^(i C(a + (SuperDagger[a]))))|0⟩;....

one has

⟨N|(e^(i k z))|0⟩ = ((((i C))^N)/(Sqrt[N!]))⟨0| (e^(i C(a + (SuperDagger[a]))))|0⟩ = ((((i C))^N)/(Sqrt[N!])) (e^(-((C^2)/2))) .

In the last result we used Baker-Haussdorff formula, valid for [X,Y] a c-number :

(e^X)(e^Y) = (e^(X+Y+(1/2)[X,Y])).

The requested probability is thus a Poisson distribution,

(Subscript[P, N])=(( (C^(2N)))/(N!))Exp[- (C^2)].

Problem

An electron (mass m and charge -e) moves in the x-y plane, subject to a magnetic field directed along z: B = (0, 0, B). In the following the electron spin is neglected. The Hamiltonian is

H = ((((p+ (e/c)A))^2)/(2m)) ;

A is the electromagnetic vector potential.

(Laudau level: see Section 14.3 of the text for further discussions).

Solution

1

The Hamiltonian is

H = ((((p+ (e/c)A))^2)/(2m)) = (((((Subscript[p, x])- ((e B)/c)y))^2)/(2m))+ ((Subscript[p, y]^2)/(2m)).

H commutes with (Subscript[p, x]), then we can look for H eigenstates in the form

Ψ[x,y] = (e^(i p x/ℏ))ψ[y].

Substitution into the Schrödinger equation gives for ψ[y]

(((Subscript[p, y]^2)/(2m)) + (((( ((e B)/c)y - p))^2)/(2m)))ψ[y] = E ψ[y].

This is the Schrödinger equation for a linear harmonic oscillator with center in

(Subscript[y, 0])= ((p c)/(e B)) ,

and frequency equal to the Larmor frequency

ω = ((e B)/(m c)).

The eigenvalues are

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

and are independent on p. This means that each level has an infinite degeneracy, labeled by p, with -∞ < p < ∞.

These energy levels are known as Landau levels.  The present discussion in fact overlaps with the one in Section 14.3 of the main Text.

We remember from classical physics that the Larmor frequency corresponds to the angular frequency of an electron moving around a circle in a uniform magnetic field, with (the charge is - e):

((d p)/dt)=- (e/c)v⋀ B .

2

In this gauge the Hamiltonian takes the form

H = ((((p+ (e/c)A))^2)/(2m)) = (((((Subscript[p, x])- ((e B)/(2c))y))^2)/(2m))+ (((((Subscript[p, y])+((e B)/(2c))x))^2)/(2m)).

To simplify the formulas we will put in the following ℏ = m = eB/c = 1. In these new units

H =  (1/2)(((-i (∂_x)- (y/2)))^2)+ (1/2)(((-i (∂_y)+ (x/2)))^2).

An elegant method of solution is to introduce complex variables

z = x+i y;   (OverBar[z]) = x-i y .

The commutation relations are

[z, (∂_z)] = [(OverBar[z]),(∂_(OverBar[z]))] = -1;   [(OverBar[z]),(∂_z)] = [z,(∂_(OverBar[z]))] = 0 .

We can also define the analogous of annihilation and creation operators

a = (Sqrt[2])((∂/(∂z))+ ((OverBar[z])/4));    (SuperDagger[a])= (Sqrt[2])(-(∂/(∂(OverBar[z])))+ (z/4));    with       [a,(SuperDagger[a])] = 1 .

An elementary computation gives for H

H = -2 (∂/(∂z))(∂/(∂(OverBar[z])))+ (1/8)(OverBar[z]) z + (OverBar[z]) (∂/(∂z)) - z (∂/(∂(OverBar[z]))) = (1/2)((SuperDagger[a])a + a (SuperDagger[a])) = (SuperDagger[a])a + (1/2) .

The spectrum is again the Landau levels ().

To compute the degeneracy let us note that the Schrödinger equation for the ground state

a (Subscript[ψ, 0])= 0  ⇒ ( (∂/(∂z))+ ((OverBar[z])/4))(Subscript[ψ, 0]) = 0 ;

has infinite solutions

(Subscript[ψ, 0,m])= (Subscript[C, m]) ((OverBar[z])^m) (e^(-(( 1)/4)(OverVector[z]) z)) = (Subscript[C, m]) (((x-i y))^m) Exp[-(1/4)((x^2)+ (y^2))];   m = 0,1,2...

All solutions have (Subscript[E, 0])= ℏ ω / 2, an are labeled by the eigenvalues of angular momentum ((Subscript[L, 3])):

L = -i (x (∂_y) - y (∂_x)).

L commutes with H and  L (Subscript[ψ, 0,m])= m (Subscript[ψ, 0,m]) .

Excited state can be constructed by acting with (SuperDagger[a])on each function (Subscript[ψ, 0,m]) .

(Subscript[ψ, N,m])= (1/(Sqrt[N!]))((((SuperDagger[a])))^N) (Subscript[ψ, 0,m]) .

Remarks

((((((Subscript[p, x])- ((e B)/c)y))^2)/(2m))+ ((Subscript[p, y]^2)/(2m)))ψ[x,y]= E ψ[x,y] ;

φ[x,y] = Exp[- (ω/(2ℏ))x y]ψ[x,y],

((((((Subscript[p, x])- ((e B)/(2c))y))^2)/(2m))+ (((((Subscript[p, y])+((e B)/(2c))x))^2)/(2m)))φ[x,y] = E φ[x,y].