Problems Chapter 16

Quantum Mechanics
                      K. Konishi, G. Paffuti

Problem

Write in the Born approximation the amplitude for two identical particles, subject to a spin-independent interaction. Consider in particular the Coulomb interaction.

Solution

Let us consider the system in the center of mass frame. We will consider a central potential.

The problem is formally equivalent to a scattering of a single particle of reduced mass μ = m/2 in a potential V[r]. The amplitude in Born approximation is ( ignoring the problem of the identity of particles )

f[θ]=- (( μ)/(2π (ℏ^2)))⟨p'|V|p⟩ =- (( μ)/(2π (ℏ^2)))∫(d^3)r V[r] (e^((i/ℏ)(p-p')r)).   
 Coulomb:  (Subscript[f, c])[θ]= - (( μ)/(2π (ℏ^2))) (e^2) ((4π (ℏ^2))/(|p-p'((|)^2)))

θ is the scattering angle, p·p'= (p^2)Cos[θ]. We have

|p-p'((|)^2) = 2(p^2)(1-Cos[θ]) = 4 (μ^2)(v^2)((Sin[(θ/2)])^2);    v = velocity in the c.m.

In the Coulomb case

(Subscript[f, c])[θ]= -((e^2)/(2μ (v^2)))(1/((Sin[(θ/2)])^2))= -((e^2)/(m (v^2)))(1/((Sin[(θ/2)])^2)).

Identical particles

Let us now consider identical particles, electrons as an instance. The admitted states satisfying Pauli principle are

(1/(Sqrt[2]))(|p,-p⟩ +|-p,p⟩ );  spin singlet;
(1/(Sqrt[2]))(|p,-p⟩ -|-p,p⟩ );  spin triplet.

We have to compute matrix elements of the Hamiltonian interaction between these states.

For the singlet state

(Subscript[ℱ, s])= - (( μ)/(2π (ℏ^2)))(1/2)2 (⟨p, -p|V|p,-p⟩ +⟨p, -p|V|-p,+p⟩)

The matrix elements depend only on p and θ. In the second matrix element the final momentum is opposite to the first one, then θ → π - θ. For the amplitude we get

(Subscript[ℱ, s])= f[θ] + f[π-θ].

In the same manner for the triplet

(Subscript[ℱ, t])= f[θ] - f[π-θ].

The corresponding differential cross section are

d(Subscript[σ, s])= |(Subscript[ℱ, s])((|)^2)dΩ = |f[θ] + f[π-θ]((|)^2);     d(Subscript[σ, t])= |(Subscript[ℱ, t])((|)^2)dΩ = |f[θ] - f[π-θ]((|)^2) .

For an unpolarized beam, with 3/4 probability to be in a triplet state and 1/4 to be in a singlet state

d(OverBar[σ]) = (3/4)d(Subscript[σ, t]) + (1/4)d(Subscript[σ, s]) .

Problem

Derive the eikonal approximation from the asymptotic form of Legendre polynomials (for large L and small θ):

(Subscript[P, L])[Cos[θ]]∼ (Subscript[J, 0])[L θ].

Solution

The validity of eq.() has been established in the text, see Chapter on WKB approximation.

Let b the impact parameter. The angular momentum is given by L = k b.

At high energy the amplitude is the sum of a very large number of partial waves. Approximating the sum on L with an integral and using ()

f[θ]=(1/(2i k))(∑_L)(2L+1)(Subscript[P, L])[Cos[θ]](Exp[2(Subscript[iδ, L])]-1)≃ (1/(2i k)) (∫_0^∞)k db (2 k b) (Subscript[J, 0])[k b θ](Exp[2i δ]-1)=
-i k (∫_0^∞)b db (Subscript[J, 0])[k b θ]((e^(2i δ))-1)

Let us now consider the phase scattering. For large L the semiclassical approximation works. The semiclassical phase, computed along the classical trajectory, is

φ =(1/ℏ)∫p[x]dx .

At high energy the trajectory canbe approximated by a straight line parallel to the z axis, with an impact parameter b. Subtracting the free phase we get the scattering phase shift

δ = (1/ℏ)(∫_(-∞)^(+∞))dz (Sqrt[2m(E - V[r])]) - (1/ℏ)(∫_(-∞)^(+∞))dz(Sqrt[2mE ]) ∼ -((Sqrt[2mE ])/ℏ)(1/(2E))(∫_(-∞)^(+∞))dz V[r]= - (m/(2(ℏ^2) k))(∫_(-∞)^(+∞))V[(Sqrt[(z^2)+(b^2)])]dz.

and

f[θ] = -i k (∫_0^∞)b db (Subscript[J, 0])[k b θ]((e^(2i δ))-1).

which coincides with the formula given in the text.

Problem

Compute the scattering amplitude for an impenetrable sphere of radius R in the Kirchhoff approximation.

Solution

This is a standard problem in optics or acoustics. The solution given below is similar to the one presented in ref.[1], except for some adaptations for the Schrödinger equation boundary conditions.

The Green' s formula

The Schrödinger equation reads

Δ ψ[x] + (k^2)ψ[x] = 0,        |x|> R, ψ[x]=0 for |x|=R.

The solution of the scattering problem can be written as

ψ[x] = (Subscript[ψ, i])[x]+ φ[x],

where (Subscript[ψ, i])is the incident wave and φ the scattered wave, which behaves asymptotically as a divergent spherical wave. As (Subscript[ψ, i]) satisfies the free wave equation, also φ is a solution of (), with the boundary condition φ = - (Subscript[ψ, i]) on the sphere.

Let us consider the Green's function

G[x,y]= (1/(|x-y|))Exp[i k|x-y|],

This function is a solution of the free wave equation except at x=y, where it is singular:

(Subscript[Δ, y]) G[x,y] + (k^2)G[x,y] = -4π δ[x-y].

It follows

φ[y](Subscript[Δ, y]) G[x,y] - G[x,y](Subscript[Δ, y])φ[y] = -4π δ[x-y]φ[y].

Integration on the region R external to the sphere gives

(∫_R)φ[y](Subscript[Δ, y]) G[x,y] - G[x,y](Subscript[Δ, y])φ[y] = - 4π φ[x].

The left hand side can be written as

φ[y](Subscript[Δ, y]) G[x,y] - G[x,y](Subscript[Δ, y])φ[y] = (∂/(∂y)){φ[y](∂/(∂y)) G[x,y] - G[x,y](∂/(∂y))φ[y]}.

Using Gauss theorem in () and neglecting contribution at infinity (this is correct as the integrand vanishes more rapidly then 1/(r^2)) we get

4π φ[x]= (∫_sphere){ φ[y](∂/(∂y)) G[x,y] - G[x,y](∂/(∂y))φ[y]}dA.

The additional minus sign in () is due to the fact that Gauss theorem involve the normal direction outside the region R, while in () we used the usual convention considering dA directed toward the outside of the sphere, then inside R.

Formula () is the well known Green's formula, here re-derived just for convenience of the reader.

The Kirchhoff approximation

Let z be the direction  of the incident particle, z=0 the plane through the origin of the sphere and x-z the scattering plane.

The boundary condition ψ = 0 in the surface of the sphere imply φ = - (Subscript[ψ, i]).

The hemisphere at z<0 can be considered the illuminated zone (I) in optics language while the hemisphere for z > 0 is the shadow (S) zone. On I the wave is reflected and approximately we can put there

((∂φ)/(∂n))= ((∂(Subscript[ψ, i]))/(∂n)),

n is the normal direction. This approximation is easily understood by considering a reflection against a plane, let say z=0, the vanishing of ψ and the existence of the reflected wave gives

ψ ≃ (e^(i k z)) + φ =   (e^(i k z)) - (e^(-i k z))  ⇒ (Subscript[(((∂φ)/(∂z))), z=0])= i k = (Subscript[(((∂(Subscript[ψ, i]))/(∂z))), z=0]).

In the shadow zone in a first approximation ψ = 0 then

((∂φ)/(∂n))= -((∂(Subscript[ψ, i]))/(∂n)),

i.e. in the shadow zone both φ and its normal derivative are (approximately)  opposite to the incident wave.

The scattered wave ()  can then be written as a sum of the integrals over the two zones

φ[x]=-(1/(4π)) (∫_I){(Subscript[ψ, i])[y](∂/(∂y)) G[x,y] + G[x,y](∂/(∂y))(Subscript[ψ, i])[y]}dA - (1/(4π)) (∫_S){(Subscript[ψ, i])[y](∂/(∂y)) G[x,y] - G[x,y](∂/(∂y))(Subscript[ψ, i])[y]}dA

In the wave region,  |x| ≫ R we can approximate the Green's function with

G[x,y]≃ (1/r)Exp[i k r]Exp[-ik n·y].

with

r = |x|;  n = x/r = (Sin[θ],0,Cos[θ]).

The shadow integral

Let us first consider the integral over the hemisphere S. As |x|>R inside the sphere

(∂/(∂y)){(Subscript[ψ, i])[y](∂/(∂y)) G[x,y] - G[x,y](∂/(∂y))(Subscript[ψ, i])[y]} ≡(∂/(∂y))H = 0,

then the flux of the vector  H  on a closed surface is zero, this means that the flux through S is equal to the flux through the disk D, intersection of the plane z=0 with the sphere, easier to compute.

With

x = r(Sin[θ],0,Cos[θ]);   y = (ξ,η,ζ);    (ξ,η) = b(Cos[(Subscript[φ, 0])],Sin[(Subscript[φ, 0])]);

k = k(0,0,1);            h = normal to D (z axis)= (0,0,1);

we have, from () and using that the normal to D is directed along z:

h·(∂/(∂y)) G[x,y]=-i (1/r)(e^(ikr))(e^(-i k n·y ))k n·h =-i (1/r)(e^(ikr))(e^(-i k b Sin[θ]Cos[(Subscript[φ, 0])] ))k Cos[θ]    
 h·(∂/(∂y))(Subscript[ψ, i])[y]= i h·k (e^(i k· y)) = i k (e^(i k ζ))  → (on S)-> i k

Then the integral on S is

φ[x]=+ (1/(4π)) (∫_0^R)b db(∫_0^(2π))d(Subscript[φ, 0])  (1/r)(e^(ikr)) (e^(-i k b Sin[θ]Cos[(Subscript[φ, 0])] ))i k (1+Cos[θ]) =
           ((i k)/(4π))(1+Cos[θ])(1/r)(e^(ikr)) (∫_0^R)b db(∫_0^(2π))d(Subscript[φ, 0]) (e^(-i k b Sin[θ]Cos[(Subscript[φ, 0])] ))=
 ((i k)/(4π))(1+Cos[θ])(1/r)(e^(ikr))(∫_0^R)b db 2π (Subscript[J, 0])[k b Sin[θ]] = ((i k)/(4π))(1+Cos[θ])(1/r)(e^(ikr)) 2π  (R/(k Sin[θ]))(Subscript[J, 1])[k R Sin[θ]].

Then the contribution to scattering amplitude from the shadow zone is

(Subscript[f, S]) = ((i )/2)R(((1+Cos[θ]))/(Sin[θ]))(Subscript[J, 1])[k R Sin[θ]].

The illuminated zone

Here the integral has to be done directly on the hemisphere. The normal is

h = (Sin[(Subscript[θ, 0])]Cos[(Subscript[φ, 0])],Sin[(Subscript[θ, 0])]Sin[(Subscript[φ, 0])],Cos[(Subscript[θ, 0])]).

On I, with γ the angle between x and h:

h·(∂/(∂y)) G[x,y]=-i (1/r)(e^(ikr))(e^(-i k n·y ))k n·h =-i (1/r)(e^(ikr))(e^(-i k R Cos[γ] ))k Cos[γ];

h·(∂/(∂y))(Subscript[ψ, i])[y]= i h·k (e^(i k· y)) = i k Cos[(Subscript[θ, 0])] (e^(i k R Cos[(Subscript[θ, 0])])) ;
Cos[γ]= Cos[θ]Cos[(Subscript[θ, 0])] + Sin[θ]Sin[(Subscript[θ, 0])]Cos[φ]

φ[x]=-(1/(4π)) (∫_I){(Subscript[ψ, i])[y](∂/(∂y)) G[x,y] + G[x,y](∂/(∂y))(Subscript[ψ, i])[y]}dA - (1/(4π)) (∫_S){(Subscript[ψ, i])[y](∂/(∂y)) G[x,y] - G[x,y](∂/(∂y))(Subscript[ψ, i])[y]}dA

The integral over I is

φ[x]=-((R^2)/(4π)) (∫_(π/2)^π)Sin[(Subscript[θ, 0])]d(Subscript[θ, 0])(∫_0^(2π))d(Subscript[φ, 0]) ((e^(ikr))/r) Exp[i k R Cos[(Subscript[θ, 0])]-i k R Cos[γ]] i k (Cos[(Subscript[θ, 0])]-Cos[γ])

The integral can be estimated by saddle point method. In (Subscript[φ, 0]) the saddle point is clearly (Subscript[φ, 0]) = 0. In (Subscript[θ, 0]) we have the condition of stationary phase (with (Subscript[φ, 0])= 0)

0=(∂/(∂(Subscript[θ, 0])))(Cos[(Subscript[θ, 0])] - Cos[γ])= - Sin[(Subscript[θ, 0])] + Cos[θ]Sin[(Subscript[θ, 0])]-Sin[θ]Cos[(Subscript[θ, 0])]= -Sin[(Subscript[θ, 0])]+Sin[(Subscript[θ, 0])-θ],

with solution

(Subscript[θ, 0])= (π/2)+(θ/2).

The phase at the stationary point is

i k R (Subscript[(Cos[(Subscript[θ, 0])]- Cos[γ]), (Subscript[θ, 0])=(π/2)+(θ/2)]) = -2i k R Sin[(θ/2)];

The quadratic fluctuations are, with (Subscript[θ, 0])== (π/2)+(θ/2) + u

i k R (Cos[(Subscript[θ, 0])]- Cos[γ])≃ -2i k R Sin[(θ/2)]Cos[u] - i k R Cos[(θ/2)] Sin[θ](((-1)/2)(φ^2) )≃
i k R(Sin[(θ/2)](u^2) + (1/2)(φ^2) Cos[(θ/2)] Sin[θ])

Using the Fresnel integrals

(∫_(-∞)^∞)Exp[i α (x^2)]d x = ((Sqrt[π])/(Sqrt[-i α])),

the integral over quadratic fluctuations gives

(1/(-i))(1/kR)(π/(Sqrt[(1/2)Sin[(θ/2)]Cos[(θ/2)] Sin[θ]]))= i(2/(k R))(π/(Sin[θ])).

Finally the saddle point value of the non exponential factors is

Sin[(Subscript[θ, 0])](Cos[(Subscript[θ, 0])]-Cos[γ])-> Cos[(θ/2)](-2 Sin[(θ/2)])= - Sin[θ].

Then from () the contribution of the illuminated zone to the scattering amplitude is

(Subscript[f, I])= -((R^2)/(4π))(i k )(- Sin[θ])(2/(k R))((i π)/(Sin[θ]))Exp[-2i k R Sin[(θ/2)]] = - (R/2)Exp[-2i k R Sin[(θ/2)]]

The full scattering amplitude is then

f[θ]=- (R/2)Exp[-2i k R Sin[(θ/2)]]+ ((i )/2)R (((1+Cos[θ]))/(Sin[θ]))(Subscript[J, 1])[k R Sin[θ]] .

The first term, with constant amplitude, represents the reflected wave.

References