View source for Central Potential Scattering and Phase Shifts

{{Quantum Mechanics A}}

We will now discuss scattering from a central potential in a different way.  Recall that the wave function for an incident and scattered wave for a central potential is given by

<math>\psi_k(\mathbf{r})=\psi_k^{(0)}(\mathbf{r})=\psi_k^{(0)}(\mathbf{r})+f_k(\theta)\frac{e^{ikr}}{r},</math>

where <math>\psi_k^{(0)}(\mathbf{r})\!</math> is the incoming wave and <math>f_k(\theta)\!</math> is the scattering amplitude.

To determine <math>f_k(\theta),\!</math> we start with the [[Schrödinger Equation|Schrödinger equation]],

<math> \left( -\frac{\hbar^2 }{2m}\nabla^2+V(r) \right) \psi=\frac{\hbar^2 k^2 }{2m}\psi .</math>
 
As before, this equation may be reduced to an effective one-dimensional equation,
 
<math>\left( -\frac{\hbar^2 }{2m}\frac{d^2 }{dr^2 }+\frac{\hbar^2 l(l+1)}{2mr^2}+V(|\mathbf{r}|) \right) u_l(r) =\frac{\hbar^2 k^2 }{2m}u_l(r),</math>

with the full wave function given by <math>\psi(\mathbf{r})=\frac{u_l(r)}{r}Y_l^m(\theta,\phi).</math>

For a potential with a finite range <math>d,\!</math> we know that, for <math>r \gg d,\!</math> the problem reduces to that of a free particle, and thus

<math> \left( -\frac{\hbar^2 }{2m}\frac{d^2 }{dr^2 }+\frac{\hbar^2 l(l+1)}{2mr^2} \right) u_l(r) =\frac{\hbar^2 k^2 }{2m}u_l(r).</math>

The solution of this equation is a linear combination of the spherical Bessel functions and the spherical Neumann functions,

<math>u_l(r)=A_l j_l(kr) +B_l n_l(kr).\!</math>

When <math>r\to\infty,\!</math> we use the asymptotic approximations of the spherical Bessel functions and the spherical Neumann functions, obtaining

<math>\frac{u(r)}{r}\approx A_l \frac{\sin(kr-l\frac{\pi}{2})}{kr} -B_l \frac{\cos(kr-l\frac{\pi}{2})}{kr}.</math>
 
Let us now define

<math>\frac{B_l }{A_l }=-\tan\delta_l.</math>
 
The angle <math>\delta_l\!</math> is known as the phase shift of the <math>\ l^{\text{th}}</math> wave and it is the phase shift induced by scattering from the potential in the radial part of the wave function. Note that, in the absence of a scattering potential, the boundary condition that the wave function must be finite at the origin causes <math>B_l</math> to vanish for all values of <math>l.</math>  Therefore, the magnitude of <math>B_l</math> compared to <math>A_l</math> is a meausre of the intensity of the scattering.  We may rewrite the above expression as

<math>\frac{u_l(r)}{r}=A_l\frac{\sin(kr-l\frac{\pi}{2} + \delta_l )}{kr}.</math>

Physically, we expect <math>\delta_l < 0\!</math> for repulsive potentials and <math>\delta_l > 0\!</math> for attractive potentials.  Also, if <math>l/k \gg d,\!</math> then the classical impact parameter is much larger than the range of the potential and in this case we expect <math>\delta_l\!</math> to be small.

Because the scattering amplitude has azimuthal symmetry (i.e., it is independent of <math>\phi\!</math>), we can write the full solution of the Schrödinger equation as a superposition of <math>m=0\!</math> spherical harmonics only:

<math>\psi(\mathbf{r})=\sum_{l=0}^{\mathop{ \infty}}a'_l(k)P_l(\cos\theta) \frac{u_l(r)}{r}=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr}.</math>

We now determine the coefficients <math>a_l(k)\!</math> by substituting in the wave function in terms of the scattering amplitude on the left-hand side:

<math>e^{ikr\cos\theta}+f_k(\theta)\frac{e^{ikr}}{r}=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr}</math>

Here, we assume that the incident wave propagates along the <math>z\!</math> direction.  This must hold for large <math>r.\!</math>  We may show that

<math>e^{ikr\cos\theta}=\sum_{l=0}^{\mathop{ \infty}}(2l+1)i^l P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2} ) }{kr},</math>

so that

<math>\sum_{l=0}^{\mathop{ \infty}}(2l+1)i^l P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2} ) }{kr}+f_k(\theta)\frac{e^{ikr}}{r}=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr} .</math>

Using the fact that

<math>\sin (x)=\frac{e^{ix}-e^{-ix}}{2i},\!</math>

we may rewrite this as
 
<math> 
\begin{align}
\sum_{l=0}^{\mathop{ \infty}}(2l+1)i^l P_l(\cos\theta) \frac{1}{kr} \frac{1}{2i} \left( e^{i\left( kr-l\frac{\pi}{2} \right)} - e^{-i\left( kr-l\frac{\pi}{2} \right)} \right) +f_k(\theta)\frac{e^{ikr}}{r} \\
= \sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{1}{kr} \frac{1}{2i} \left( e^{i\left(kr-l\frac{\pi}{2}+\delta_l \right)} - e^{-i\left(kr-l\frac{\pi}{2}+\delta_l \right)}\right).
\end{align}
</math>

By matching the coefficients of <math>e^{-ikr}\!</math>, we get

<math>a_l(k)=(2l+1)i^le^{i\delta_l},</math>

and doing the same for <math>e^{ikr}\!</math> yields

<math>f_k(\theta)=\frac{1}{k}\sum_{l=0}^{\infty}(2l+1)e^{i\delta_l }\sin\delta_lP_l(\cos\theta) .</math>

Note that <math>\delta_l\!</math> is a function of <math>k\!</math> and therefore a function of the incident energy. If <math>\delta_l(k)\!</math> is known, then we can reconstruct the entire scattering amplitude and consequently the differential cross section. The phase shifts themselves must be determined by solving the [[Schrödinger Equation|Schrödinger equation]].

The differential scattering cross section is

<math>\frac{d\sigma}{d\Omega}=|f_k(\theta) |^2=\frac{1}{k^2 }\left|\sum_{l=0}^{\infty}(2l+1)e^{i\delta_l }\sin\delta_lP_l(\cos\theta) \right|^2.</math>

By integrating <math>\frac{d\sigma}{d\Omega}\!</math> over the solid angle <math> \Omega, \!</math> we obtain the total scattering cross section:

<math>
\begin{align}
\sigma_{tot} &= \int \frac{d\sigma}{d\Omega}\,d\Omega \\
&= \sum_{l=0}^{\infty}\sum_{l'=0}^{\infty}(2l+1)(2l'+1)e^{i\delta_l }e^{-i\delta_{l'}}\sin\delta_l\sin\delta_{l'} \int_{0}^{2\pi} d\phi\,\int_{0}^{\pi}d\theta\,\sin\theta P_l(\cos\theta)P_{l'}(\cos\theta) \\
&= \frac{4\pi}{k^2 }\sum_{l=0}^{\mathop{ \infty}}(2l+1)\sin^2\delta_l
\end{align}
</math>

The last equality follows from the orthogonality of the Legendre polynomials,

<math>\int_{-1}^{1}dx\,P_l(x) P_{l'}(x)=\frac{2}{(2l+1)}\delta_{ll'}.</math>
 
Finally, note that since <math>P_l(1) = 1\!</math> for all <math>l,\!</math> we obtain

<math>f_k(0) = \frac{1}{k}\sum_{l=0}^{\infty} \left(2l+1\right)e^{i\delta_l}\sin\delta_lP_l(1) = \frac{1}{k}\sum_{l=0}^{\infty} \left(2l+1\right)e^{i\delta_l}\sin\delta_l.
</math>

If we take the imaginary part of the scattering amplitude,then

<math>\Im m[f_k(0)] = \frac{1}{k} \sum_{l=0}^{\infty} \left(2l+1\right) \sin^2 \delta_l = \frac{k}{4\pi} \sigma_{tot}.</math>

Therefore,

<math>\sigma_{tot}=\frac{4\pi}{k}\Im mf(0).</math>

This relationship is known as the optical theorem.  The optical theorem is a general law of wave scattering theory that relates the forward scattering amplitude to the total cross section of the scattering. It was originally discovered independently by Sellmeier and Lord Rayleigh in 1871.

Referring back to the formula for the scattering amplitude, one more important quantity can be discussed:

<math>S_l(k)=e^{2i\delta_l(k)}</math>

This quantity, for now referred to as the partial scattering for angular momentum <math>l\!,</math> is the ratio of the coefficients of the outgoing and incoming waves for a wave scattered on a potential of finite range <math> a.\!</math>

These ratios can simplify the problem of evaluating the continuity of the waveform at the boundary <math> r = a.\!</math>  In general, if the interior wave function is known to be smoothly continuous across the boundary at <math> r = a,\!</math> then the phase shifts can be expressed in terms of the logarithmic derivatives evaluated at the boundary <math> r = a:\!</math>

<math>\beta_l=\left(\frac{a}{u_l(r)}\frac{d u_l(r)}{dr}\right)_{r=a}</math>

Using the above equations for the form of <math>u_l(r)</math> beyond the region of scattering, the following relation is found:

<math>\beta_l = ka\frac{j'_l(ka)\cos(\delta_l) - n'_l(ka)\sin(\delta_l)}{j_l(ka)\cos(\delta_l) - n_l(ka)\sin(\delta_l)}</math>

Thus, after some algebra,

<math>S_l = e^{2i\delta_l}= -\frac{j_l(ka) -in_l(ka)}{j_l(ka) + in_l(ka)}  \frac{\beta_l - ka\frac{j'_l(ka) - in'_l(ka)}{j_l(ka)-in_l(ka)}}{\beta_l - ka\frac{j'_l(ka)+in'_l(ka)}{j_l(ka)+in_l(ka)}}.</math>

Note that, if <math>\beta_l \to \infty,\!</math> which corresponds to <math>f_l(a)=0,\!</math> then only the first portion of this expression survives.  This is a special quantity corresponding to hard sphere scattering; we may define the phase angles <math>\xi_l,\!</math> known as the hard sphere phase shifts, as

<math>e^{2i\xi_l} = -\frac{j_l(ka)-in_l(ka)}{j_l(ka)+in_l(ka)}.</math>

Note that these phase shifts are present for any potential, not just that of a hard sphere.

== Scattering by Square Well Potential and Hard Sphere ==

Consider a beam of point particles of mass <math>m\!</math> scattering from a finite spherical attractive well of depth <math> V_0 \!</math> and radius <math> a, \!</math>

<math>
V(r) = \begin{cases} 
-V_0,  & r < a \\
0 , & r > a
\end{cases}
</math>

The effective [[Schrödinger Equation|Schrödinger equation]] for <math>r<a \!</math> is 

<math>\frac{d^2 R_l}{dr^2} + \frac{2}{r} \frac{dR_l}{dr} -\frac{l(l+1)}{r^2}R_l + \frac{2m}{\hbar^2}(E+V_0)u_l = 0.</math>

Its solution is

<math>u_{l}(r) = A_l j_l(\kappa r),\!</math>

where <math>\kappa^2 = \frac{2m}{\hbar^2}(E+ V_0).\!</math>

In the region <math>r>a,\!</math> <math>u_l\!</math> is

<math>u_l(r)= B j_l (kr)+C n_l(kr).\!</math>
 
Here, <math>A,\!</math> <math>B,\!</math> and <math>C\!</math> are arbitrary constants and <math>k^2 =  \frac{2m}{\hbar^2}E.\!</math>

For large <math> r,\!</math>

<math>u_l(r)\approx\frac{\sin{(kr-l\pi/2 + \delta_l)}}{kr},</math>

where

<math> \frac{C}{B}= -\tan{\delta_l}.</math>

We now apply the boundary conditions at <math>r=a,\!</math> which are continuity of <math>u_l\!</math> and of its logarithmic derivative.  We obtain

<math>
-\frac{C}{B}= \tan{\delta_l}=\frac{kj_l'(ka)j_l(\kappa a)-\kappa j_l(ka) j_l'(\kappa a)}{kn_l'(ka)j_l(\kappa a)-\kappa n_l(ka) j_l'(\kappa a)}.
</math>

Let us now consider two limiting cases:

'''(a)''' <math>ka \ll l \!</math> and <math>\kappa a \ll l:\!</math>  In this case, we find that, with some simplification, <math>\tan{\delta_l}\sim k^{2l+1} \sim E^{l+1/2}.\! </math>  This behavior is a result of the centrifugal barrier that keeps waves of energy far below the barrier from feeling the effect of the potential.

'''(b)''' When the phase shift is <math>\pi/2,\!</math> the partial wave cross section <math>\sigma_l(k)= \frac{4\pi (2l+1)}{k^2}\sin^2{\delta_l}\!</math> is maximized.  This is known as a resonant scattering.

From '''(a)''', we see that the phase shift is small for <math> ka \!</math> small.  However, when <math> ka \!</math> changes and passes through the resonance condition, the phase shift rises rapidly and has a sharp peak at resonant energy <math>E_R\!</math>. This can be represented as

<math> \tan{\delta_l} \approx \frac{\gamma (ka)^{2l+1}}{E-E_R},</math>

so that the partial wave cross section is

<math>\sigma_l= \frac{4\pi(2l+1)}{k^2}\frac{[\gamma (ka)^{2l+1}]^2}{(E-E_R)^2 +[\gamma (ka)^{2l+1}]^2 }.</math>
 
This is the Breit-Wigner formula for a resonant cross section.


Let us also consider a hard sphere, given by the potential,
:<math>
V(\mathbf{r})=
\begin{cases}
\infty & r \leq d \\
0 & r>0
\end{cases}
</math>

For scattering state
:<math>u_{l}(r)=
\begin{cases}
A_{l}j_{l}(kr)+B_{l}n_{l}(kr) & r\geq d \\
0 & r \leq d
\end{cases}
</math>

For <math> r\rightarrow\infty</math>,
:<math>
u_{l}(r) \longrightarrow \frac{A_{l}}{kr\cos\delta_{l}}\sin\left(kr-l\frac{\pi}{2}+\delta_{l}\right)
</math>

Matching continuity boundary condition at <math>r=d \!</math>, we get,
:<math>\frac{B_{l}}{A_{l}}=-\frac{j_{l}(kd)}{n_{l}(kd)}=-\tan\delta_{l}</math>
so the scattering phase shift of the <math>l^{th} \!</math> wave is:
:<math>
\delta_{l}=\tan^{-1}\frac{j_{l}(kd)}{n_{l}(kd)}
</math>
For <math>kd \ll 1  </math>,
:<math>j_{l}\approx \frac{x^{l}}{(2l+1)!!}</math>
:<math>n_{l}\approx \frac{(2l-1)!!}{x^{l+1}}</math>

so
:<math> \delta_{l}\approx\tan^{-1}\left(-\frac{(kd)^{2l+1}}{2l+1}\right)  \approx -\frac{(kd)^{2l+1}}{2l+1}  </math>
The <math>l=0\!</math> term dominates in the scattering process, so the scattering amplitude and the cross section are:

:<math> f_{k}(\theta)\approx  -\frac{1}{k}e^{-ikd} P_{0}(\cos\theta) \sin kd    </math>

Therefore, the total cross section is
:<math> \sigma= \int d\Omega |f_{k}(\theta)|^{2} = 4\pi d </math>

==Problems==

'''(1)''' Consider the scattering of a particle from a real spherically symmetric potential. If <math>\frac{d\sigma (\theta) }{d\Omega }</math> is the differential cross section and <math>\sigma</math> is the total cross section, then show that
 
<math>\sigma \leq \frac{4\pi}{k}\sqrt{\frac{d\sigma (0) }{d\Omega }}</math>

for a general central potential using the partial-wave expansion of the scattering amplitude and the cross section.

[[Phy5645/Cross Section Relation|Solution]]

'''(2)''' Consider an attractive delta shell potential (<math>\lambda > 0\!</math>) of the form,

<math>V(\textbf{r})=-\frac{\hbar^2 \lambda}{2m} \delta(r-a).</math>

'''(a)''' Derive the equation for the phase shift caused by this potential for arbitrary angular momentum.

'''(b)''' Obtain the expression for the <math>s\!</math> wave phase shift.

'''(c)''' Obtain the <math>s\!</math> wave scattering amplitude.

[[Worked Problem for Scattering on a Delta-Shell Potential|Solution]]