Problems Chapter 13

Quantum Mechanics
                       K. Konishi, G. Paffuti

Problem

Write the Green function for a free particle in one dimension by solving the equation (E-H)G=1 in the x representation. Verify the result by writing G as a sum over eigenstates of H.

Solution

x representation

With E = (((ℏ k))^2)/2m the equation for the Green function in one dimension is

((((ℏ^2)(k^2))/(2m))+ ((ℏ^2)/(2m))((d^2)/(dx^2)))G[x-y]= δ[x-y].

The solution with a divergent (i.e. expanding) wave as boundary condition is

G[k;x,y]=G[k;x-y]= (1/(2i k))((2m)/(ℏ^2)) Exp[i k |x-y|];

with

k = i (1/ℏ)(Sqrt[-2m E]);  k>0 for E > 0.

To check the result first we remember that

ϵ[x]≡Sign[x]= {({{+1;, x>0}, {-1;, x<0}})  ;  ϵ'[x] = 2δ[x].

The first derivatives of G are (we omit the k argument as it is kept fixed below)

G'[x]=(1/(2i k))((2m)/(ℏ^2)) i k ϵ[x] (e^(i k |x|));
G''[x]= (1/(2i k))((2m)/(ℏ^2))(- (k^2)((ϵ[x])^2)+ 2i k δ[x])(e^(i k |x|))= - (k^2)G[x]; c.v.d.

k representation

The computation can be performed also in Fourier transform. From eq.() using the known prescription iϵ on poles (with ϵ → 0 )

G[x] = ((2m)/(ℏ^2))∫((dk')/(2π))((e^(i k'x))/((k^2)+iϵ - (k^(′2))))= - ((2m)/(ℏ^2))∫((dk')/(2π))((e^(i k'x))/((k'- (SubPlus[k]))(k'- (SubMinus[k])))).

Where (Subscript[k, ±])are the poles with positive and negative imaginary part,  respectively. The integral can be evaluated by closing the contour in the complex plane with a half-circle at infinity and applying the Cauchy theorem. With x>0 the path must be closed in the half plane Im[k]>0 while for x<0 in the region Im[k]<0. The integral gives immediately (Res is the residue of the integrand)

G[x]=2π i θ[x]Res[(SubPlus[k])] (e^(i (SubPlus[k])x)) - 2π i θ[-x]Res[(SubMinus[k])] (e^(i (SubMinus[k])x)).

Using the values of the poles (remember that ϵ → 0 in all expressions)

(Subscript[k, ±])== ± (Sqrt[(k^2)+iϵ]) = ± (k+iϵ)

we have

Res[(SubPlus[k])]=- ((2m)/(ℏ^2))(1/(2π))(1/((SubPlus[k])-(SubMinus[k])))= - ((2m)/(ℏ^2))(1/(2π))(1/(2k));  Res[(SubMinus[k])]=- ((2m)/(ℏ^2))(1/(2π))(1/((SubMinus[k])-(SubPlus[k])))= + ((2m)/(ℏ^2))(1/(2π))(1/(2k));

and by substitution in () we recover the previous result

G[x] =iθ[x]( - ((2m)/(ℏ^2))(1/(2k)))(e^(i k x))- i θ[-x]((2m)/(ℏ^2))(1/(2k))(e^(-i k x))= ((2m)/(ℏ^2))(1/(2i k))(e^(i k |x|)).

Analyticity

Let us note that the analyticity properties (a cut for E>0) and by definition of the adjoint we must have both

⟨y|(SuperDagger[G])[E]|x⟩= ⟨y|G[(e^(2π i)) E]|x⟩;   ⟨y|(SuperDagger[G])[E]|x⟩= ⟨x|G[E]|y⟩^( * ).

As for E→ (e^(2πi))E one has  k→ - k the above relations imply

(SuperStar[G[k;x,y]]) = G[-k;y,x]

which is indeed verified with our solution ().

Problem

Write the Green function for a free particle in any dimension solving the equation (E-H)G=1 in the x representation.

Solution

We are looking for a (radial symmetric) solution of

(Δ + (k^2))G = ((2m)/(ℏ^2))(δ^((d)))[x].

In dimension d the Laplacian operator has the form

Δ = (1/(r^(d-1)))(∂/(∂r))(r^(d-1))(∂/(∂r)) + angular terms;

eq.() has the form

(1/(r^(d-1)))(∂/(∂r))(r^(d-1))(∂/(∂r))G[r]+(k^2)G[r] =(( 2m)/(ℏ^2))(δ^((d)))[x].

In x≠0 by the change of variables

z = k r;    G = A (z^(1-(d/2)))f[z]

eq.() becomes

f''[z] + (1/z)f'[z] + (1 - (((((d/2)-1))^2)/(z^2)))f[z] = 0.

This is a Bessel equation of order d/2-1. The request of outgoing waves selects as a solution the Hankel function (Subscript[H, (d/2)-1]^((1))):

(Subscript[H, ν]^((1)))[x]= (Subscript[J, ν])[x]+ i (Subscript[Y, ν])[x];

J, Y being Bessel function of first and second kind. The Green function then has the form

G[r] = A (((k r))^(1-(d/2)))(Subscript[H, (d/2)-1]^((1)))[k r].

The small x behavior of the Hankel function is

(Subscript[H, ν]^((1)))[x]∼ i (Subscript[Y, ν])[x] ∼ - i (1/(Sin[π ν]))((((x/2)))^(-ν))(1/(Γ[1-ν]))= -i ((Γ[ν])/π)((((x/2)))^(-ν)),

where we used the identity

Γ[x]Γ[1-x] = (π/(Sin[π x])).

The small r behavior of G follows

G ∼ -i A ((Γ[(d/2)-1])/π)(2^((d/2)-1))(((k r))^(2-d)).

To fix A we use Gauss theorem for a small sphere S of radius R enclosing the origin

(∫_S)(∇^2)(r^(2-d)) = (∫_(∂S)) (∂/(∂r))(r^(2-d)) = (Subscript[Ω, d])(R^(d-1))(2-d)(R^(1-d)) = -(d-2)(Subscript[Ω, d]).

(Subscript[Ω, d]) is the solid angle in d dimensions

(Subscript[Ω, d])= ((2 (π^(d/2)))/(Γ[(d/2)])).

Using the Gauss theorem on the equation () one sees that the singular term in G must be

G ∼(( 2m)/(ℏ^2))(-(1/((d-2)(Subscript[Ω, d]))))(r^(2-d)).

One gets easily

A = (1/i)((2m)/(ℏ^2)) ((k^(d-2))/((π^((d/2)-1))(2^((d/2)+1))));         G[r] = (1/i)((2m)/(ℏ^2)) ((k^(d-2))/((π^((d/2)-1))(2^((d/2)+1)))) (((k r))^(1-(d/2)))(Subscript[H, (d/2)-1]^((1)))[k r].

In particular for d = 3, using

(Subscript[H, (1/2)]^((1)))[x] = - i (Sqrt[(2/(π x))]) (e^(i x));

one recovers the known result

G[r] = -(( m)/(2π (ℏ^2)))((e^(i k r))/r).

Problem

Compute the Green function for a particle in one dimension in a potential V[x] = g δ[x] solving (E-H)G=1 in the x representation and verify the result by considering H eigenstates. Compute the spectral density g[E]. Verify the relation between g[E] and scattering phases by computing the transmission and reflection coefficients.

Solution

The Hamiltonian for the problem is

H = ((p^2)/(2m))+ g δ[x].

The equation for the Green function G is

(((d^2)/(dx^2))+(q^2))G = ((2m)/(ℏ^2))δ(x-y) + ((2m g)/(ℏ^2))δ[x] G[0];      with  E = (((ℏ^2)(q^2))/(2m)).

Note that the equation is not translation invariant. In the free particle case a function  Exp[i q |x-y|] gave rise to a singularity δ[x-y]. The equation () has two δ singularities in x=y and x=0.  It is then natural to look for solutions in the form

G[x,y] = A (e^(i q |x-y|)) + B (e^(i q |x|)).

The derivatives give

(((d^2)/(dx^2))+(q^2))G [x,y] = 2i q A δ[x-y] + 2i q B δ[x].

By substitution in () we see that the equation is satisfied for

A = (m/(i (ℏ^2)q)); B = ((2β A)/(2(i q - β)))(e^(i q |y|));     β ≡ ((m g)/(ℏ^2)).

β plays the role of a momentum scale. The final form of G is

G[x,y]= (m/(i (ℏ^2)q))((e^(i q |x-y|)) + ((e^(i q (|x|+|y|)))/(i (q/β)- 1))).

Analytic properties

Our conventions for the cut in the square root function are

ℏ q = - i (Sqrt[-2m E]); E = ((q^2)/(2m));    physical sheet: Im[q]>0 (first Riemann sheet in E).

The function G is expressed through q then has a cut in the complex plane E. The cut is for Re[E] > 0.

The function G in () has  a pole at q = -iβ.

(Subscript[E, 0]) = (((ℏ^2)(q^2))/(2m))=- (((ℏ^2)(β^2))/(2m)).

For attractive potential near the pole, () becomes

G[x,y]∼ (m/(i (ℏ^2)))((e^(-|β|(|x|+|y|)))/(q+i β))≃ ((|β| (e^(-|β|(|x|+|y|))))/(E-(Subscript[E, 0]))).

The residue at the pole gives the wave function of the bound state (see below).

Spectral density

Spectral density is defined by

g[E] =- (1/π) Im[Tr[G[E]]] ≡ (Subscript[g, 0])[E] + δg[E]

(Subscript[g, 0]) is the density of states for a free particle. The operation Tr means x=y and integration of ().

The integration of the first term in () gives the known result for (Subscript[g, 0]) (see also the text):

(Subscript[g, 0])[E]= V (m/(π (ℏ^2)q))= V ((Sqrt[m])/ℏ)(1/(π(Sqrt[2 E]))).

The second term can be integrated with q→ q+i ϵ (upper side of the cut in the first sheet of E-plane) and then taking the limit ϵ→0:

(∫_(-∞)^(+∞))dx (e^(i 2 (q+iϵ)|x|))= (i/q).

Substitution in δg gives

δg[E] = - (1/π) Im[-(m/( (ℏ^2)(q^2)))(1/(1-i (q/β)))] = (m/(π (ℏ^2)q β))(1/(1+((q^2)/(β^2)))).

Scattering phases and spectral density

In the text it has been shown that the scattering phases δ[E] are related to δg by

δg[E] = (1/π)((d δ[E])/dE).

To compute δ[E] we can consider the regular solution of the equation or compute the transmission and  reflection coefficients.

Regular solutions

In the text it is shown that even and odd solutions have respectively the asymptotic forms

(Subscript[ψ, e])[x] → Cos[q x+(Subscript[δ, e])];       (Subscript[ψ, o])[x]→ Sin[q x+(Subscript[δ, o])].

By imposing the continuity of ψ and the correct discontinuity in x=0, (SubPlus[ψ'[0]])- (SubMinus[ψ'[0]])= (ℏ^2) 2m g one easily obtains

-2q Sin[(Subscript[δ, e])]=2β Cos[(Subscript[δ, e])];    Sin[(Subscript[δ, o])]=0; ⇒   (Subscript[δ, o])= 0;  Tan[(Subscript[δ, e])]=-(β/q).

By a derivative

(Subscript[δg, o])[E] = 0;     (Subscript[δg, e])[E] = (1/π)(m/((ℏ^2)q))((d (Subscript[δ, e]))/dq) =  (m/(π (ℏ^2)q β))(1/(1+((q^2)/(β^2)))),

in agreement with eq.().

Transmission and reflection coefficients

The transmission and reflection coefficients (Subscript[A, T]) and (Subscript[A, R]) are found by solving the Schrödinger equation with the boundary conditions

x<0: (e^(i q x)) + (Subscript[A, R]) (e^(-i q x));    x>0:  (Subscript[A, T])(e^(i q x)).

Imposing the continuity of ψ and the discontinuity of  2β ψ[0]on the derivative, it is found

(Subscript[A, T])= ((i q)/(iq-β));   (Subscript[A, R])= (β/(iq-β)).

The potential is even, and in the text it has been shown that in this case one has

(Subscript[δg, e])= - i (m/(2(ℏ^2)q))(1/((Subscript[A, T])+(Subscript[A, R]))) (d/dq)((Subscript[A, T])+(Subscript[A, R]));   (Subscript[δg, o])= - i (m/(2(ℏ^2)q))(1/((Subscript[A, T])-(Subscript[A, R]))) (d/dq)((Subscript[A, T])-(Subscript[A, R])).

By substitution we get

(Subscript[δg, o])[E] = 0;     (Subscript[δg, e])[E] =  (m/(π (ℏ^2)q β))(1/(1+((q^2)/(β^2)))).

in agreement with the previous results.

As expected only the density of even states is changed. For odd states vanishing at x=0 the perturbation g δ[x] has no effects.

Sum of eigenfunctions

One can obtain the result () also by a more complex computation, from the knowledge of the eigenfunctions of the problem.

A set of eigenfunctions for our problem is (see text)

(Subscript[ψ, k]^R)[x] = (1/(Sqrt[2π]))((e^(i k x)) - F[k](e^(i k |x|)));   (Subscript[ψ, k]^L)[x]=(1/(Sqrt[2π]))((e^(-i k x)) - F[k](e^(i k |x|))).

where

F[k] = (1/(1-i (k/β)));          F[-k]= (SuperStar[F])[k];    F+(SuperStar[F])= (2/(1+ ((k^2)/(β^2))))= 2|F((|)^2).

These functions are normalized to δ[k-k']and  k > 0.

(Subscript[E, 0])= - (((ℏ^2)(β^2))/(2m));   (Subscript[ψ, 0])[x] = (Sqrt[|β|]) (e^(-|β||x|)).

The Green function by definition is

G[x,y]= ∑ (((Subscript[ψ, α])[x](Subscript[ψ, α]^( * ))[y])/(E+iϵ - E')) =
                     =((2m)/(ℏ^2))(∫_0^∞)dk  (((Subscript[ψ, k]^R)[x](OverBar[(Subscript[ψ, k]^R)])[y]+ (Subscript[ψ, k]^L)[x](OverBar[(Subscript[ψ, k]^L)])[y])/((q^2)+iϵ - (k^2))) - ((2m |β|)/(ℏ^2))(( (e^(-|β|(|x|+|y|))))/((β^2)+ (q^2)))θ[-β].

The last term gives a contribution only for  g<0 and in this case it reproduces the result ().

In expanding the products in () it is convenient to consider several cases

x, y > 0, or x, y < 0

(Subscript[ψ, k]^R)[x](OverBar[(Subscript[ψ, k]^R)])[y]+ (Subscript[ψ, k]^L)[x](OverBar[(Subscript[ψ, k]^L)])[y]=(1/(2π))((e^(i k(x-y)))+ (e^(-i k (x-y)))- F[k](e^(i k(x+y)))- (SuperStar[F])[k](e^(-ik(x-y)))).

Due to properties () of F[k] this expression is symmetric in k→ - k then the integral in () can be extended to the whole k-axis. The continuum contribution is

(Subscript[G, cont])=-((2m)/(ℏ^2))(∫_0^∞)((dk)/(2π))(((e^(i k(x-y)))/((k^2)-(q^2)+iϵ))- (1/(1-i (k/β)))((e^(i k(x+y)))/((k^2)-(q^2)+iϵ))).

The first integral has poles in k=± q ± i ϵ with residues ±1/2q respectively. The integration path is closed in the region Im[k]>0 or Im[k]<0 if the sign of x-y is ± 1. One has in any case, as the reader can easily check

(∫_0^∞)((dk)/(2π))((e^(i k(x-y)))/((k^2)-(q^2)+iϵ)) = (i/(2q))(e^(i q|x-y|)).

This term is the free propagator. The second term has an additional pole in k=-i β which contributes only for x+y <0 if  β>0, otherwise it gives a contribution also for x+y>0. The second contribution to the continuum part of G is then

β>0:     -i (1/(1-i (q/β))) (e^(i q (x+y)))
β<0:   -i (1/(1-i (q/β))) (e^(i q (x+y))) - |β|((e^(-|β|(x+y)))/((β^2)+ (q^2)))

The whole expression for the continuum part becomes

(Subscript[G, cont])=(m/(i (ℏ^2)q))((e^(i q |x-y|)) + ((e^(i q (|x|+|y|)))/(i (q/β)- 1)))+ ((2m |β|)/(ℏ^2))(( (e^(-|β|(|x|+|y|))))/((β^2)+ (q^2)))θ[-β]

For x, y < 0 the computation is identical. By summing the possible contribution of the bound state we cancel the last term above and get

G[x,y]=(m/(i (ℏ^2)q))((e^(i q |x-y|)) + ((e^(i q (|x|+|y|)))/(i (q/β)- 1)))

x,>0>y > 0, or  y>0>x

We leave to the reader the exercise to show that by using the relations () the same expression holds in this region, reproducing ().

Problem

Compute the Green function for a particle in one dimension in a potential V[x] = g δ[x-a] + g δ[x+a] solving (E-H)G=1 in the x representation. Verify the analyticity properties in the complex E plane and compute the parameters for resonant states. Compare the results with the Gamow Siegert approach. Compute the spectral density g[E] from G and verify the result by using the scattering phases.

Solution

The Hamiltonian of our problem is

H = ((p^2)/(2m))+ g δ[x-a] + g δ[x+a]

The equation for the Green function reads

(((d^2)/(dx^2))+(k^2))G = ((2m)/(ℏ^2))δ(x-y) + ((2m g)/(ℏ^2))(δ[x-a] G[a]+δ[x+a] G[-a]).

We write

E = (((ℏ^2)(k^2))/(2m)) ;   β ≡ ((m g)/(ℏ^2)) .

We look for a solution in the form

G[x,y] = A (e^(i k |x-y|)) + (Subscript[B, 1]) (e^(i k |x+a|))+ (Subscript[B, 2]) (e^(i k |x-a|)).

Using

(∂_x)|x|= ϵ[x];    (∂_x)ϵ[x] = 2δ[x];

substitution in () gives

A = (m/(i (ℏ^2)k));
(Subscript[B, 1])= -i β A (e^(-2i k a)) ((((e^(i k|y+a|))(β-i k) - β (e^(2ik a))(e^(i k |y-a|))))/((2β Sin[k a]+k (e^(-i k a)))(2β Cos[k a]-i k (e^(-i k a)))));
(Subscript[B, 2])= -i β A (e^(-2i k a)) ((((e^(i k|y-a|))(β-i k) - β (e^(2ik a))(e^(i k |y+a|))))/((2β Sin[k a]+k (e^(-i k a)))(2β Cos[k a]-i k (e^(-i k a)))));

Analytic properties

The propagator has singularities in the k-plane at the origin, and at zeros in the denominators in (), poles in E.

In the E-plane there is a cut on the real positive axis and poles at

2β Cos[k a] - i k (e^(-i k a))= 0   ⇒ (2β - i k)Cos[k a] - k Sin[k a] = 0;
2β Sin[k a] +  k (e^(-i k a))  = 0   ⇒ (2β - i k)Sin[k a] + k Cos[k a] = 0.

With k a = x+i y the two equations for the poles can be written as

{ ({{y=-β a (1+(e^(-2y))Cos[2x]}, {x= β a Sin[2x](e^(-2y))          }}) ; { ({{y=β a ((e^(-2y))Cos[2x]-1)}, { x= - β a Sin[2x](e^(-2y))        }}).

Let us consider the attractive and the repulsive case.

y = |β|a (1+(e^(-2y)));   y = |β|a (1-(e^(-2y))).

It is apparent that y < 0 for all β. In conclusion

The spectral density

The spectral density is defined by

g[E] =- (1/π) Im[Tr[G[E]]] ≡ (Subscript[g, 0])[E] + δg[E]

(Subscript[g, 0]) is the density of states for a free particle. The operation Tr means x=y and integration in (). The first term gives the free particle contribution, see also the previous problem. V is the volume:

(Subscript[g, 0])[E]= V (m/(π (ℏ^2)q))= V ((Sqrt[m])/ℏ)(1/(π(Sqrt[2 E]))).

The second term can be integrated by writing k→k+iϵ (upper side of the cut) and then taking the limit ϵ→ 0. Using

(Subscript[I, 1])= (Underscript[Lim, ϵ→0])(∫_(-∞)^(+∞))dx (e^(i 2 (k+i ϵ)|x±a|))= (i/k);

(Subscript[I, 2])= (Underscript[Lim, ϵ→0])(∫_(-∞)^(+∞))dx (e^(i (k+i ϵ)(|x+a|+|x-a|)))= (e^(2i k a))((i+2 k a)/k).

one gets easily

Tr[δG[E]]= ((2m β)/(i (ℏ^2)(k^2)))(((i(β-i k) - (e^(4ia k))(2 a k + i)β))/((e^(4i a k))(β^2)-(((β - i k))^2)))

and finally after some simplifications

δg[E]=-(1/π)Im[Tr[δG[E]]]=
((2m β)/(π (ℏ^2)k))(((-2 a (β^3)+(β^2)- (β+2a((k^2)-(β^2)))Cos[4 a k]β - k (4a β-1)Sin[4a k]β + (k^2)))/((k^4)+2(β^2)(k^2)+4(β^3)k Sin[4k a]+2(β^4)+2(β^2)((k^2)-(β^2))Cos[4 a k]))

A typical behavior of δg with peaks corresponding to resonant states is shown in the figure below.

Scattering phases

In the text it has been shown that the scattering phases δ[E] are related to δg by

δg[E] = (1/π)((d δ[E])/dE).

To compute δ[E] we can consider the regular solutions of the equation.

In the text it is shown that even and odd solutions have respectively the asymptotic form

(Subscript[ψ, e])[x] → Cos[q x+(Subscript[δ, e])];       (Subscript[ψ, o])[x]→ Sin[q x+(Subscript[δ, o])].

It is sufficient to consider the positive values of x. In the two regions 0<x<a and x>a the two solutions have the form

(Subscript[ψ, e])={ ({{A Cos[k x]}, {Cos[k x + (Subscript[δ, e])]}})  ;   (Subscript[ψ, o])={ ({{A Sin[k x]}, {Sin[k x + (Subscript[δ, o])]}});

By imposing the continuity of ψ at x=a and the correct discontinuity 2β ψ[a] for the derivative one easily obtains

(Subscript[δ, e])= - a k + ArcTan[((-2β + k Tan[a k])/k)];  (Subscript[δ, o])= - a k + ArcTan[(k/(2β + k Cot[a k]))].

The spectral density is written as

δg[E] = (1/π)((d (Subscript[δ, e])[E])/dE)+ (1/π)((d (Subscript[δ, o])[E])/dE)=
  =((2m β)/(π (ℏ^2)k))(((-2 a β + 2a k Tan[a k]+1)/(((Tan[a k])^2)(k^2)+ (k^2)-4β k Tan[a k]+4 (β^2)))- ((2 a β + 2a k Cot[a k]-1)/(((Cot[a k])^2)(k^2)+ (k^2)+4β k Cot[a k]+4 (β^2)))).

After some (boring) simplifications this expression is shown to be equal to equal to ().

Problem

Compute the eigenvalues for a free particle in a volume bounded by a parallelepiped with edges (a,b,c) and verify the form of the Weyl expansion.

Solution

In three dimensions Weyl's formula gives for the asymptotic number of the eigenvalues of a free particle

(dN/dE)= g[E]=(1/(4(π^2)))(((((2m)/(ℏ^2))))^(3/2)) V (Sqrt[E]) - (S/(16π))(((2m)/(ℏ^2))) + (C/(12(π^2)))(((((2m)/(ℏ^2))))^(1/2))(1/(Sqrt[E])) +...

V is the volume, S the surface and C the mean curvature. The last term changes if cusps are present, so we shall not consider it.

The definition of g[E] is

g[E]=∑ δ[E-(Subscript[E, i])].

It is convenient to work with the Laplace transformation of g (see text)

Z[β] = (∫_0^∞)Exp[-β E]g[E] dE = ∑ Exp[-β (Subscript[E, i])].

Z is the statistical partition function of the system, and the high energy behavior of g[E] is related to the small β behavior of Z[β].

The eigenstates can be classified by three integers (Subscript[n, 1]),(Subscript[n, 2]),(Subscript[n, 3]) and the eigenvalues of H are

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

Using the definition of multiplication of two series it follows from () :

Z[β]=(∑_((Subscript[n, 1]),(Subscript[n, 2]),(Subscript[n, 3]))) Exp[-β (Subscript[E, (Subscript[n, 1]),(Subscript[n, 2]),(Subscript[n, 3])])] = (Subscript[Z, 1])[a,β](Subscript[Z, 1])[b,β](Subscript[Z, 1])[c,β];

where

(Subscript[Z, 1])[a,β]= (∑_1^∞) Exp[- β ((ℏ^2)/(2m))(((π^2)(n^2))/(a^2))]= (∑_0^∞) Exp[- β ((ℏ^2)/(2m))(((π^2)(n^2))/(a^2))] - 1 .

The sum in () can be estimated with the Euler-McLaurin formula:

(∑_0^n)f[n]= (∫_0^∞)f[x]dx + (1/2)f[0] + ... 𝒪(f')

In our case f[x]∼ Exp[-β (x^2)] and the correction terms are depressed by powers of β. A trivial integration gives

(Subscript[Z, 1])[a,β]≃(∫_0^∞)dn Exp[- β ((ℏ^2)/(2m))(((π^2)(n^2))/(a^2))] - (1/2) = (((((2m)/(ℏ^2))))^(1/2)) (a/(π (β^(1/2))))((Sqrt[π])/2) - (1/2) .

Inserting this result in () we get, at first two orders at low β:

Z[β]≃(((((2m)/(ℏ^2))))^(3/2)) ((a b c)/(((2(Sqrt[π])))^3))(β^(-3/2)) - (1/2)(((2m)/(ℏ^2)))(1/(((2(Sqrt[π])))^2))(a b + a c+ b c)(β^(-1)).

As

V = a b c;   S = 2 (a b + a c+ b c);

we have

Z[β]≃(((((2m)/(ℏ^2))))^(3/2)) (V/(((2(Sqrt[π])))^3))(β^(-3/2)) - (1/2)(((2m)/(ℏ^2)))(1/(((2(Sqrt[π])))^2))(S/2)(β^(-1)).

To get g[E] from Z[β] we have to take the inverse Laplace transform of this result. This is easily done by noticing that

(Subscript[ℒ, β])[(E^(α-1))]=∫dE (e^(-β E))(E^(α-1))=(β^(-α))Γ[α]  ⇒    (Subscript[ℒ, E]^(-1))[(β^(-μ))] = ((E^(μ-1))/(Γ[μ])).

In particular

(Subscript[ℒ, E]^(-1))[(β^(-3/2))] = ((E^(1/2))/(Γ[(3/2)]))= (2/(Sqrt[π]))(Sqrt[E]) ;              (Subscript[ℒ, E]^(-1))[(β^(-1))]= 1.

Taking the inverse Laplace transform in ()

g[E]≃(((((2m)/(ℏ^2))))^(3/2))  (E^(1/2)) (V/(4(π^2))) -  (((2m)/(ℏ^2)))(S/(16π)) + ...

in agreement with Weyl's formula.

The number of states is

N[E] = (∫_0^E)g[x]dx = (((((2m)/(ℏ^2))))^(3/2))  (2/3)(E^(3/2)) (V/(4(π^2)))-  (((2m)/(ℏ^2)))(S/(16π))E.

Problem

Show that in the pole approximation for a decay state χ for all t

(Subscript[P, χχ])[t] + (∑_s)(Subscript[P, sχ])[t] = 1.

Solution

We use the notation of the text. The survival probability is

(Subscript[P, χχ])[t] = Exp[-Γ t/ℏ];

where the line width Γ is given by

Γ = (∑_s)((|⟨s|R[(Subscript[E, R])]|χ⟩|)^2) 2π δ[(Subscript[E, R])-(Subscript[E, s])].

(Subscript[E, R]) is the resonance energy and R[E] is the decay matrix.

The states are classified by energy and some other quantum numbers, α and

(∑_s) ≡ (∑_α)∫g[α,E]dE

where g[α,E] is the density of states in dE around E. Then Γ is written explicitly as

Γ = (∑_α)((|⟨α,(Subscript[E, R])|R[(Subscript[E, R])]|χ⟩|)^2)  2π g[α,(Subscript[E, R])].

In the text it is shown that

(Subscript[P, sχ])[t]= ((|⟨s|R[E]|χ⟩|)^2)(((1+ (e^(-Γt/ℏ))-2(e^(-Γt/2ℏ))Cos[((Subscript[E, R])-E)t/ℏ])/((((E-(Subscript[E, R])))^2)+((Γ^2)/4)))).

This expression has poles in E = (Subscript[E, R]) ± i Γ/2. In the pole approximation we can write (with Γ≪ E )

(∑_s)(Subscript[P, sχ])[t]= (∑_α)g[α,(Subscript[E, R])]((|⟨α,(Subscript[E, R])|R[(Subscript[E, R])]|χ⟩|)^2)  ∫dE (((1+ (e^(-Γt/ℏ))-2 (e^(-Γt/2ℏ))Cos[((Subscript[E, R])-E)t/ℏ])/((((E-(Subscript[E, R])))^2)+((Γ^2)/4)))).

The first two terms in the integral can be computed with Jordan lemma as integrals in the complex E-plane closing the contour at infinity

∫dE (((1+ (e^(-Γt/ℏ)))/((((E-(Subscript[E, R])))^2)+((Γ^2)/4))))= (1+ (e^(-Γt/ℏ)))2π i (1/(2i (Γ/2)))= ((2π)/Γ)(1+ (e^(-Γt/ℏ))).

The third integral can be written

∫dE ((2Cos[((Subscript[E, R])-E)t/ℏ])/((((E-(Subscript[E, R])))^2)+((Γ^2)/4)))= ∫dE (1/((((E-(Subscript[E, R])))^2)+((Γ^2)/4)))((e^(i(E-(Subscript[E, R]))t/ℏ))+(e^(-i(E-(Subscript[E, R]))t/ℏ))).

In applying Jordan lemma for t>0 the first integral must be closed in the upper half plane (Im[E]>0) while the second in the lower half plane, then

∫dE ((2Cos[((Subscript[E, R])-E)t/ℏ])/((((E-(Subscript[E, R])))^2)+((Γ^2)/4)))= 2π i((( (e^(-Γt/2ℏ)))/(2i (Γ/2)))) - 2πi ((( (e^(-Γt/2ℏ)))/(-2i (Γ/2))))= - ((4π)/Γ) (e^(-Γt/2ℏ))

Inserting () and () in eq.() we have, using ()

(∑_s)(Subscript[P, sχ])[t]= (∑_α)g[α,(Subscript[E, R])]((|⟨α,(Subscript[E, R])|R[(Subscript[E, R])]|χ⟩|)^2)  ((2π)/Γ) (1+ (e^(-Γt/ℏ)) - 2  (e^(-Γt/ℏ))) = (1-(e^(-Γt/ℏ))).

This and () verify eq.().