Central Potential Scattering and Phase Shifts: Difference between revisions

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{{Quantum Mechanics A}}
{{Quantum Mechanics A}}
Recall that for scattering, we have [[Differential cross-section and the Green's function formulation of scattering|Green function]] method
:<math>\psi_k =\psi_k^{(0)} +\int d^3 r' G(\mathbf r,\mathbf r')V(\mathbf r')\psi_k (\mathbf r'),</math>
where the Green's function satisfies that
:<math>(\nabla^2+k^2)G_k(\mathbf{r},\mathbf{r'})=\delta(\mathbf{r}-\mathbf{r'})</math>
and the solution is chosen such that the second term in Eq.(1) corresponds to an outgoing waves.
:<math>G_k(\mathbf{r},\mathbf{r'})=-\frac{1}{4\pi}\frac{e^ { ik| \mathbf{r}-\mathbf{r'}|}}{|\mathbf{r}-\mathbf{r'}|}</math>
and in the asymptotic limit of r goes to infinity, <math>r\to\infty\!</math>,
:<math>\lim_{r \to \infty}k|\mathbf{r}-\mathbf{r'}|= \lim_{r \to \infty} k \sqrt{ r^2 - 2 \mathbf{r} \cdot \mathbf{r}' + r'^2 } \sim kr \left( 1 - \frac{\mathbf{r}\cdot\mathbf{r}'}{r^2} \right) = kr - k\mathbf{\hat{r}} \cdot \mathbf{r}' = kr - \mathbf{k}'\cdot \mathbf{r}',</math>
where <math> \mathbf{k}' = k \mathbf{\hat{r}} </math>.
Thus
:<math>\lim_{r \to \infty} \psi_k(\mathbf{r})=\psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3 r'e^ { -i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})\frac{e^{ikr}}{r}
=\psi_k^{(0)}(\mathbf{r})+f_k(\theta,\phi)\frac{e^{ikr}}{r},</math>
where
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3 r' e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'}),</math>
where <math> \mathbf{k}' = k \mathbf{\hat{r}} </math> and the angles <math>\theta\!</math> and  <math>\phi\!</math> are the angles between <math>\mathbf{\hat{r}}\!</math> (the vector defining the detector)  and <math>\mathbf{k}\!</math>
(the vector defining the in the incoming waves).


For central potentials, i.e. if <math>V(r)=V(|\mathbf{r}|)</math>, then <math>f_k(\theta)\!</math>, i.e. the scattering amplitude does
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
not depend on the azimuthal angle <math>\phi\!</math>.
 
To determine <math>f_k(\theta)\!</math>, we need to find the solution of the [[Schrödinger equation]]:
<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(|\mathbf{r}|) \right) \psi=\frac{\hbar^2 k^2 }{2m}\psi .</math>
<math> \left( -\frac{\hbar^2 }{2m}\nabla^2+V(r) \right) \psi=\frac{\hbar^2 k^2 }{2m}\psi .</math>
   
   
We use [[Spherical Coordinates|spherical coordinates]] and the radial equation is:
As before, this equation may be reduced to an effective one-dimensional equation,
   
   
:<math>\left( -\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial r^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>
<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


For <math>V(r)\!</math> with a finite range <math>d\!</math>, we have shown that for <math>r \gg d\!</math> we have
<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>
:<math> \left( -\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial r^2 }+\frac{\hbar^2 l(l+1)}{2mr^2} \right) u_l(r) =\frac{\hbar^2 k^2 }{2m}u_l(r)</math>


and the solution is a combination of the spherical Bessel functions and the spherical Neumann functions
The solution of this equation is a linear combination of the spherical Bessel functions and the spherical Neumann functions,
:<math>\frac{u(r)}{r}=A_l j_l(kr) +B_l n_l(kr).</math>
When <math>r\!</math> is large enough, we use approximation of the spherical Bessel functions and the spherical Neumann functions.


:<math>\frac{u(r)}{r}\rightarrow A_l \frac{\sin(kr-l\frac{\pi}{2})}{kr} -B_l \frac{\cos(kr-l\frac{\pi}{2})}{kr} .</math>
<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>
   
   
Letting
Let us now define
:<math>\frac{B_l }{A_l }=-\tan\delta_l ,</math>
here the angle <math>\delta_{\ell}\!</math> is called the phase shift of the <math>\ l^{th}</math>wave and it is the difference in phase between radial parts. Note that in the absence of a scattering potential (V=0 everywhere) the boundary condition that the wave function must be finite at the origin causes <math>B_{\ell} </math> to vanish for all values of <math> \ell </math>. Therefore the magnitude of <math>B_{\ell} </math> compared to <math>A_{\ell} </math> is a meausre of the intensity of the scattering. This means that <math>\delta_{\ell}\!</math> is a measure of the amount by which the phase of the radial wave function for angular momentum <math> \ell </math> differs from the no-scattering case. Thus, we can rewrite the above expression (up to a normalization constant) as


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


Physically, we expect <math>\delta_l < 0\!</math> for repulsive potentials and <math>\delta_l > 0\!</math> for attractive potentials.
<math>\frac{u_l(r)}{r}=A_l\frac{\sin(kr-l\frac{\pi}{2} + \delta_l )}{kr}.</math>
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.


Now since we are seeking the scattering amplitude with azimuthal symmetry, we can write
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.
the solution of the Schrödinger equation as a superposition of <math>m=0\!</math> spherical harmonics
 
only:
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=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{u_l(r) }{r},</math>
 
:<math>\psi=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr} ,</math>
<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>
where the Legendre polynomials are
 
:<math>P_l(x)= \frac{1}{2^ll!}\frac{d^l }{dx^l }(x^2-1 )^l,</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>P_0(x)=1;P_1(x)=x;P_2(x)=\frac{1}{2}(3x^2 -1); \dots .</math>
 
Let us fix the coeffcients <math>a_l(k)\!</math> by
<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>
:<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>
 
which must hold at large <math>r\!</math> and where we chose the coordinates by letting the incident waves
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
propagate along z-direction. Note that (due to an entirely separate argument):
 
:<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>
<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
 
:<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>
so that
We fix the coefficients <math>a_l(k)\!</math> by matching the incoming spherical waves on both sides of the above equation.
 
Note that this does not involve <math>f_k(\theta)\!</math> since the scattering amplitude controls the outgoing spherical waves.
<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>


Since <math>\sin (x)=\frac{e^{ix}-e^{-ix}}{2i}\!</math>,
we may rewrite this as
:<math>  
<math>  
\begin{align}
\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}}(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) .
= \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}
\end{align}
</math>
</math>


By matching the coefficients of <math>e^{-ikr}\!</math>, we get
By matching the coefficients of <math>e^{-ikr}\!</math>, we get
:<math>a_l(k)=(2l+1)i^le^{i\delta_l}</math>
and for the coefficients of <math>e^{ikr}\!</math>, we get
:<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 we can reconstruct the entire scattering amplitude and consequently the differential
cross section. The phase shifts must be determined from the solution of the Schrödinger equation.


<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
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>\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>
:<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}
\begin{align}
\sigma_{tot} &= \int \frac{d\sigma}{d\Omega} d\Omega \\
\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) \\
&= \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,
&= \frac{4\pi}{k^2 }\sum_{l=0}^{\mathop{ \infty}}(2l+1)\sin^2\delta_l
\end{align}
\end{align}
</math>
</math>
which follows from the orthogonality of the Legendre polynomials


:<math>\int_{-1}^{1}dxP_l(x) P_{l'}(x)=\frac{2}{(2l+1)}\delta_{ll'}.</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 have
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>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>
</math>
If we take the imaginary part of the scattering amplitude,then
If we take the imaginary part of the scattering amplitude,then
:<math>\Im mf_k(0) = \frac{1}{k} \sum_{l=0}^{\infty} \left(2l+1\right) \sin^2 \delta_l = \frac{k}{4\pi} \sigma_{tot}.  
 
</math>
<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, we have
 
:<math>\sigma_{tot}=\frac{4\pi}{k}\Im mf(0).</math>
Therefore,
This relationship is known as the optical theorem.
 
The optical theorem is a general law of wave scattering theory. You can see that it 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.
<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:
Referring back to the formula for the scattering amplitude, one more important quantity can be discussed:
Line 111: Line 115:
<math>S_l(k)=e^{2i\delta_l(k)}</math>
<math>S_l(k)=e^{2i\delta_l(k)}</math>


This quantity, for now referred to as the partial scattering due to various angular momenta, 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>.
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>:
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}{f_l(r)}\frac{d f_l(r)}{dr}\right)_{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>f_l(r)</math> beyond the region of scattering, the following relation is found:
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(k) = 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>
<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 with algebraic manipulation:
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>
<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>, that is to say <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, therefore in general the real phase angles <math>\xi_l\!</math> (the ''hard sphere phase shifts'') are expressed such that:
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>
<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.
Note that these phase shifts are present for any potential, not just that of a hard sphere.


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


Consider a beam of point particles of mass m scattering from a finite spherical attractive well of depth <math> V_0 \!</math> and radius <math> a \!</math>
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>
<math>
V(r) = \begin{cases}  
V(r) = \begin{cases}  
-V_0,  & r < a \\
-V_0,  & r < a \\
Line 142: Line 146:
</math>
</math>


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


:<math>
<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>
\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)R_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.


Hence, the solution for above differential equation is  
'''(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.
:<math> {R_{l}} = A_l j_l(\kappa r) \!</math>  
where <math>\kappa^2 = \frac{2m}{\hbar^2}(E+ V_0)\!</math>.


Another solution for the region <math>r>a \!</math> is
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>\ {R_l}= B j_l (kr)+C n_l(kr)</math>
<math> \tan{\delta_l} \approx \frac{\gamma (ka)^{2l+1}}{E-E_R},</math>
where <math> A,B,\!</math> and <math> C \!</math> are arbitrary constants and <math>k^2 =  \frac{2m}{\hbar^2}E \!</math>.


Now, for large <math> r \!</math>,
so that the partial wave cross section is
:<math> R_l \approx\frac{\sin{(kr-l\pi/2 + \delta_l)}}{kr}</math>


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


We'll basically apply 2 boundary conditions now, continuity of <math>R_l\!</math> and matching the logarithmic derivatives at <math>r=a\!</math>, as a result we obtain,


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


Lets discuss some limiting cases:
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>


(a) Consider the case when <math>ka \ll l \!</math> and <math>\kappa a \ll l\!</math>, then with some simplification we get <math>\tan{\delta_l}\approx k^{2l+1} \approx E^{l+1/2}\! </math>.
For <math> r\rightarrow\infty</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.
:<math>
u_{l}(r) \longrightarrow \frac{A_{l}}{kr\cos\delta_{l}}\sin\left(kr-l\frac{\pi}{2}+\delta_{l}\right)
</math>


(b) When the phase shift is <math>\pi/2\!</math> then the partial wave cross section  <math>\sigma_l(k)= \frac{4\pi (2l+1)}{k^2}\sin^2{\delta_l}\!</math> is maximum. Then, we have a resonant scattering.
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:


From(a), we see that the phase shift is small for <math> ka \!</math> small. But, when <math> ka \!</math> changes and passes the resonant condition, the phase shift rises rapidly and has a sharp peak at resonant energy <math>E_R\!</math>. This can be represented as
:<math> f_{k}(\theta)\approx -\frac{1}{k}e^{-ikd} P_{0}(\cos\theta) \sin kd    </math>
:<math> \tan{\delta_l} \approx \frac{\gamma (ka)^{2l+1}}{E-E_R}</math>.


So, partial wave cross section is  
Therefore, the total 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>,
:<math> \sigma= \int d\Omega |f_{k}(\theta)|^{2} = 4\pi d </math>
which is the ''Breit-Wigner formula'' for resonant cross section


==Problems==
==Problems==
[[Phy5645/scatt1|Problem 1]]<br/>
 
[[Worked Problem for Scattering on a Delta-Shell Potential|Problem 2]]
'''(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]]

Latest revision as of 13:50, 18 January 2014

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_k(\mathbf{r})=\psi_k^{(0)}(\mathbf{r})=\psi_k^{(0)}(\mathbf{r})+f_k(\theta)\frac{e^{ikr}}{r},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_k^{(0)}(\mathbf{r})\!} is the incoming wave and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_k(\theta)\!} is the scattering amplitude.

To determine Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_k(\theta),\!} we start with the Schrödinger equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( -\frac{\hbar^2 }{2m}\nabla^2+V(r) \right) \psi=\frac{\hbar^2 k^2 }{2m}\psi .}

As before, this equation may be reduced to an effective one-dimensional equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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),}

with the full wave function given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\mathbf{r})=\frac{u_l(r)}{r}Y_l^m(\theta,\phi).}

For a potential with a finite range Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d,\!} we know that, for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \gg d,\!} the problem reduces to that of a free particle, and thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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).}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_l(r)=A_l j_l(kr) +B_l n_l(kr).\!}

When Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\to\infty,\!} we use the asymptotic approximations of the spherical Bessel functions and the spherical Neumann functions, obtaining

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}.}

Let us now define

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{B_l }{A_l }=-\tan\delta_l.}

The angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_l\!} is known as the phase shift of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l^{\text{th}}} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_l} to vanish for all values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l.} Therefore, the magnitude of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_l} compared to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_l} is a meausre of the intensity of the scattering. We may rewrite the above expression as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{u_l(r)}{r}=A_l\frac{\sin(kr-l\frac{\pi}{2} + \delta_l )}{kr}.}

Physically, we expect Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_l < 0\!} for repulsive potentials and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_l > 0\!} for attractive potentials. Also, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l/k \gg d,\!} then the classical impact parameter is much larger than the range of the potential and in this case we expect Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_l\!} to be small.

Because the scattering amplitude has azimuthal symmetry (i.e., it is independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} ), we can write the full solution of the Schrödinger equation as a superposition of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=0\!} spherical harmonics only:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}.}

We now determine the coefficients Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_l(k)\!} by substituting in the wave function in terms of the scattering amplitude on the left-hand side:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}}

Here, we assume that the incident wave propagates along the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\!} direction. This must hold for large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r.\!} We may show that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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},}

so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} .}

Using the fact that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin (x)=\frac{e^{ix}-e^{-ix}}{2i},\!}

we may rewrite this as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} }

By matching the coefficients of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-ikr}\!} , we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_l(k)=(2l+1)i^le^{i\delta_l},}

and doing the same for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{ikr}\!} yields

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_k(\theta)=\frac{1}{k}\sum_{l=0}^{\infty}(2l+1)e^{i\delta_l }\sin\delta_lP_l(\cos\theta) .}

Note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_l\!} is a function of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\!} and therefore a function of the incident energy. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_l(k)\!} 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.

The differential scattering cross section is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.}

By integrating Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\sigma}{d\Omega}\!} over the solid angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega, \!} we obtain the total scattering cross section:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-1}^{1}dx\,P_l(x) P_{l'}(x)=\frac{2}{(2l+1)}\delta_{ll'}.}

Finally, note that since for all we obtain

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}.}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_{tot}=\frac{4\pi}{k}\Im mf(0).}

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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_l(k)=e^{2i\delta_l(k)}}

This quantity, for now referred to as the partial scattering for angular momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l\!,} is the ratio of the coefficients of the outgoing and incoming waves for a wave scattered on a potential of finite range Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a.\!}

These ratios can simplify the problem of evaluating the continuity of the waveform at the boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = a.\!} In general, if the interior wave function is known to be smoothly continuous across the boundary at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = a,\!} then the phase shifts can be expressed in terms of the logarithmic derivatives evaluated at the boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = a:\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_l=\left(\frac{a}{u_l(r)}\frac{d u_l(r)}{dr}\right)_{r=a}}

Using the above equations for the form of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_l(r)} beyond the region of scattering, the following relation is found:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}}

Thus, after some algebra,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)}}.}

Note that, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_l \to \infty,\!} which corresponds to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_l(a)=0,\!} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi_l,\!} known as the hard sphere phase shifts, as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{2i\xi_l} = -\frac{j_l(ka)-in_l(ka)}{j_l(ka)+in_l(ka)}.}

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\!} scattering from a finite spherical attractive well of depth Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0 \!} and radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, \!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r) = \begin{cases} -V_0, & r < a \\ 0 , & r > a \end{cases} }

The effective Schrödinger equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r<a \!} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.}

Its solution is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{l}(r) = A_l j_l(\kappa r),\!}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa^2 = \frac{2m}{\hbar^2}(E+ V_0).\!}

In the region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>a,\!} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_l\!} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_l(r)= B j_l (kr)+C n_l(kr).\!}

Here, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A,\!} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B,\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} are arbitrary constants and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k^2 = \frac{2m}{\hbar^2}E.\!}

For large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r,\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_l(r)\approx\frac{\sin{(kr-l\pi/2 + \delta_l)}}{kr},}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{C}{B}= -\tan{\delta_l}.}

We now apply the boundary conditions at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=a,\!} which are continuity of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_l\!} and of its logarithmic derivative. We obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\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)}. }

Let us now consider two limiting cases:

(a) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ka \ll l \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa a \ll l:\!} In this case, we find that, with some simplification, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan{\delta_l}\sim k^{2l+1} \sim E^{l+1/2}.\! } 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi/2,\!} the partial wave cross section Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_l(k)= \frac{4\pi (2l+1)}{k^2}\sin^2{\delta_l}\!} is maximized. This is known as a resonant scattering.

From (a), we see that the phase shift is small for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ka \!} small. However, when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ka \!} changes and passes through the resonance condition, the phase shift rises rapidly and has a sharp peak at resonant energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_R\!} . This can be represented as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan{\delta_l} \approx \frac{\gamma (ka)^{2l+1}}{E-E_R},}

so that the partial wave cross section is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_l= \frac{4\pi(2l+1)}{k^2}\frac{[\gamma (ka)^{2l+1}]^2}{(E-E_R)^2 +[\gamma (ka)^{2l+1}]^2 }.}

This is the Breit-Wigner formula for a resonant cross section.


Let us also consider a hard sphere, given by the potential,

For scattering state

For ,

Matching continuity boundary condition at , we get,

so the scattering phase shift of the wave is:

For ,

so

The term dominates in the scattering process, so the scattering amplitude and the cross section are:

Therefore, the total cross section is

Problems

(1) Consider the scattering of a particle from a real spherically symmetric potential. If is the differential cross section and is the total cross section, then show that

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

Solution

(2) Consider an attractive delta shell potential () of the form,

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

(b) Obtain the expression for the wave phase shift.

(c) Obtain the wave scattering amplitude.

Solution

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