In this tutorial, we will develop a couple of technical results that will be useful in the development of partial wave theory. The partial wave expansion applies when we are in the intermediate region where we can approximate the potential as \(V \approx 0\), and we then want to connect the solution to the scattering region where the potential is nonzero. This is done by matching boundary conditions.

Let us recall the wave function from the previous section on partial waves, which consists of the incoming and outgoing waves:

\[\begin{equation} \psi = A\sum_{\ell=0}^\infty i^\ell(2\ell+1)(j_\ell(kr) + ika_\ell h^{(1)}_\ell(kr)) P_\ell(\cos\theta) \end{equation}\]

Since each angular momentum \(\ell\) scatters independently, the amplitude of each must be conserved, by conservation of probability. Let’s show that, if we have a wave function written as

\[\begin{equation} \psi = A\sum_{\ell=0}^\infty i^\ell (2\ell+1)(b_\ell h^{(1)}_\ell(kr) + c_\ell h^{(2)}_\ell(kr))P_\ell(\cos\theta), \end{equation}\]

that we require \(|b_{\ell}|^2 = |c_{\ell}|^2\). This is the conservation of probability for an incoming wave function component of angular momentum \(\ell\) scattering into an outgoing wave function component of angular momentum \(\ell\) must conserve the amplitude since each component scatters independently (no mixing between different \(\ell\)).

The continuity equation is

\[\begin{equation} \frac{\partial}{\partial t}|\psi|^2 + \nabla\cdot\mathbf{J} = 0 \end{equation}\]

The time-dependent version of the above wave function is

\[\begin{equation} \psi(r,\theta,t) = \sum_{\ell=0}^\infty i^\ell (2\ell+1)(b_\ell h^{(1)}_\ell(kr) + c_\ell h^{(2)}_\ell(kr)) P_\ell(\cos\theta) e^{-iE_kt/\hbar} \end{equation}\]

because all of the states with the same \(k\) are degenerate. Thus, \(\rho = |\psi|^2\) is time independent so \(\partial_t\rho = 0\). For the probability current part, we have then that

\[\begin{equation} \int \nabla\cdot\mathbf{J} = 0 = \oint \mathbf{J}\cdot \d\mathbf{A} \end{equation}\]

Thus, let us calculate the above integral.

\[\begin{align} \oint \mathbf{J}\cdot \d\mathbf{A} ={}& \frac{i\hbar}{2m}\int_0^\pi \int_0^{2\pi} (\psi \nabla\psi^* - \psi^*\nabla\psi) r^2 \hat{r} \sin\theta \d\varphi\d\theta \\ ={}& \frac{2\pi i\hbar}{2m}\int_0^\pi (\psi \partial_r\psi^* - \text{c.c}) r^2 \sin\theta \d\theta \end{align}\]

where c.c. means complex conjugate. Now, plugging in the wave function, we get

\[\begin{align} \oint \mathbf{J}\cdot \d\mathbf{A} ={}& \frac{\pi i\hbar}{m}\int_0^\pi \left[\sum_{\ell=0}^\infty \sum_{\ell'=0}^\infty i^\ell (2\ell+1) (b_\ell h^{(1)}_\ell(kr) + c_\ell h^{(2)}_\ell(kr))P_\ell(\cos\theta) \right.\\ &\times \left. (-i)^{\ell'} (2\ell'+1)(b_\ell^* \partial_r h^{(2)}_\ell(kr) + c_\ell^* \partial_r h^{(1)}_\ell(kr)) P_{\ell'}(\cos\theta) - \text{c.c}\right] r^2 \sin\theta \d\theta \end{align}\]

If we then use the identity

\[\begin{equation} \int_0^\pi P_\ell(\cos\theta) P_{\ell'}(\cos\theta) \sin\theta \d\theta = \frac{2\delta_{\ell\ell'}}{2\ell+1} \end{equation}\]

then we find

\[\begin{align} \oint \mathbf{J}\cdot \d\mathbf{A} ={}& \frac{\pi i\hbar }{m} \left[\sum_{\ell\ell'} \frac{2\delta_{\ell\ell'}r^2}{2\ell+1} i^\ell (2\ell+1) (b_\ell h^{(1)}_\ell(kr) + c_\ell h^{(2)}_\ell(kr)) (-i)^{\ell'} (2\ell'+1)(b_\ell^* \partial_r h^{(2)}_\ell(kr) + c_\ell^* \partial_r h^{(1)}_\ell(kr)) - \text{c.c}\right] \\ ={}& \frac{2\pi i\hbar r^2}{m} \left[\sum_{\ell} (2\ell+1) (b_\ell h^{(1)}_\ell(kr) + c_\ell h^{(2)}_\ell(kr))(b_\ell^* \partial_r h^{(2)}_\ell(kr) + c_\ell^* \partial_r h^{(1)}_\ell(kr)) - \text{c.c}\right] \\ ={}& \frac{2\pi i\hbar r^2}{m} \left[\sum_{\ell} (2\ell+1) (|b_\ell|^2 h^{(1)}_\ell(kr) \partial_r h^{(2)}_\ell(kr) + c_\ell b_\ell^* h^{(2)}_\ell(kr) \partial_r h^{(2)}_\ell(kr) \right. \\ &\left. + b_\ell c_\ell^* h^{(1)}_\ell(kr) \partial_r h^{(1)}_\ell(kr) + |c_\ell|^2 h^{(2)}_\ell(kr) \partial_r h^{(1)}_\ell(kr))- \text{c.c}\right] \end{align}\]

We see that \(c_\ell b_\ell^* h^{(2)}_\ell \partial_r h^{(2)}_\ell(kr) + b_\ell c_\ell^* h^{(1)}_\ell(kr) \partial_r h^{(1)}_\ell(kr)\) is real, so when we subtract off the complex conjugate we get zero. Thus, we find

\[\begin{align} \oint \mathbf{J}\cdot \d\mathbf{A} ={}& \frac{2\pi i\hbar r^2}{m} \left[\sum_{\ell} (2\ell+1) (|b_\ell|^2 h^{(1)}_\ell(kr) \partial_r h^{(2)}_\ell(kr) + |c_\ell|^2 h^{(2)}_\ell(kr) \partial_r h^{(1)}_\ell(kr))- \text{c.c}\right] \\ ={}& \frac{2\pi i\hbar r^2}{m} \sum_{\ell} (2\ell+1) (|b_\ell|^2 - |c_\ell|^2)(h^{(1)}_\ell(kr) \partial_r h^{(2)}_\ell - h^{(2)}_\ell(kr) \partial_r h^{(1)}_\ell(kr)) \end{align}\]

It is sufficient to note that the term \(h^{(1)}_\ell(kr) \partial_r h^{(2)}_\ell - h^{(2)}_\ell(kr) \partial_r h^{(1)}_\ell(kr) \neq 0\) to conclude that \(|b_\ell|^2 - |c_\ell|^2 = 0\). Note that if we want to be more specific, we can show that

\[\begin{equation} h^{(1)}_\ell(kr) \partial_r h^{(2)}_\ell - h^{(2)}_\ell(kr) \partial_r h^{(1)}_\ell(kr) = -\frac{2i}{kr^2} \end{equation}\]

We can simplify the term with the \(r\) in it because it is the Wronskian, using the familiar theorem from Sturm-Liouville theory: if

\[\begin{equation} (py')' - qy + \lambda \sigma y = 0 \end{equation}\]

Then the Wronskian is given by

\[\begin{equation} W(x) = \frac{c}{p(x)}. \end{equation}\]

If we plug in some specific value of \(r\), then we find the above result (the constant \(c = -2i/k\)). Thus, we have that

\[\begin{align} \oint \mathbf{J}\cdot \d\mathbf{A} ={}& \frac{4\pi\hbar}{m} \sum_{\ell=0}^\infty (2\ell+1) (|b_\ell|^2 - |c_\ell|^2) \end{align}\]

Now, it may be a bit more clear that we should set \(|b_\ell| = |c_\ell|\).

Now, what are the consequences of this condition on the following wave function:

\[\begin{equation} \psi = A\sum_{\ell=0}^\infty i^\ell (2\ell+1)(j_\ell(kr) + ika_\ell h^{(1)}_\ell(kr))P_\ell(\cos\theta)? \end{equation}\]

Let’s show that that we have partial wave phase shifts \(\delta_\ell\), and find their relationship to \(a_\ell\).

If we return to our original wave function, we have that

\[\begin{equation} \psi = A\sum_{\ell=0}^\infty i^\ell(2\ell+1)(j_\ell(kr) + ika_\ell h^{(1)}_\ell(kr)) P_\ell(\cos\theta) \end{equation}\]

and since \(j_\ell(kr) = \frac{1}{2}(h^{(1)}_\ell(kr) + h^{(2)}_\ell(kr))\), then we get

\[\begin{equation} \psi = A\sum_{\ell=0}^\infty i^\ell(2\ell+1)\left[\left(\frac{1}{2} + ika_\ell \right)h^{(1)}_\ell(kr) + \frac{1}{2}h^{(2)}_\ell(kr)\right] P_\ell(\cos\theta) \end{equation}\]

Since we have that the magnitudes of the coefficients must be equal, they must differ only by a phase. Therefore, we set

\[\begin{equation} \frac{1}{2} + ika_\ell = \frac{1}{2}e^{2i\delta_\ell},\qquad a_\ell = \frac{1}{2ik}(e^{2i\delta_\ell} - 1) = \frac{1}{k}e^{i\delta_\ell}\sin(\delta_\ell) \end{equation}\]

and the solution is then written as

\[\begin{equation} \psi = A\sum_{\ell=0}^\infty i^\ell\frac{(2\ell+1)}{2}\left[e^{2i\delta} h^{(1)}_\ell(kr) + h^{(2)}_\ell(kr)\right] P_\ell(\cos\theta) \end{equation}\]

so the outgoing wave is phase-shifted by \(2\delta\) in comparison to the incoming wave. The 2 in the exponent is a mere convention; we think of the incoming wave as being shifted by \(\delta\), and then shifted again on the way out.

Next, let’s relate the scattering phase shifts can be related to the scattering cross section. The scattering cross section can be rewritten as

\[\begin{align} \sigma ={}& 4\pi \sum_{\ell =0 }^\infty (2\ell+1) |a_\ell|^2 \\ ={}& 4\pi \sum_{\ell =0 }^\infty (2\ell+1) \bigg|\frac{1}{k}e^{i\delta_\ell}\sin\delta_\ell\bigg|^2 \\ ={}& \frac{4\pi}{k^2} \sum_{\ell =0 }^\infty (2\ell+1)\sin^2\delta_\ell \end{align}\]

Thus, if we have the partial wave phase shifts, we can calculate the scattering cross section.

Example: Consider the problem from last week, with the 3D infinite spherical barrier. In this case, the potential was $$\begin{equation} V(r) = \begin{cases} \infty & r < a \\ 0 & r \geq a \end{cases} \end{equation}$$ so the wave function for \(r < a\) was zero. For this problem
  1. Compute the partial wave phase shifts \(\delta_\ell\)
  2. Compute the partial wave shift for the case \(ka \ll 1\)
Solution:
  1. The partial wave shifts can be gotten from the partial wave amplitudes, which we calculated last time to be $$\begin{equation} a_\ell = \frac{ij_\ell(ka)}{kh^{(1)}_\ell(ka)} \end{equation}$$ Then, the partial wave phase shifts are $$\begin{align} e^{2i\delta_\ell} ={}& 1 + 2ika_\ell \\ ={}& 1 - \frac{2j_\ell(ka)}{h^{(1)}_\ell(ka)} \\ ={}& \frac{h^{(1)}_\ell(ka) - 2j_\ell(ka)}{h^{(1)}_\ell(ka)} \\ ={}& \frac{j_\ell(ka) + iy_\ell(ka) - 2j_\ell(ka)}{h^{(1)}_\ell(ka)} \\ ={}& -\frac{h^{(2)}_\ell(ka)}{h^{(1)}_\ell(ka)} \end{align}$$ Thus, we have that $$\begin{equation} 2\delta_\ell = \pi - 2\Arg(h^{(1)}_\ell(ka)) \end{equation}$$ so the phase shift is $$\begin{equation} \delta_\ell = \frac{\pi}{2} - \Arg h^{(1)}_\ell(kr) = \frac{\pi}{2} - \arctan \left(\frac{y_\ell(ka)}{j_\ell(ka)}\right) \end{equation}$$
  2. The phase shift is given by plugging in the low-energy forms $$\begin{equation} j_\ell(z) \sim \frac{z^\ell 2^\ell \ell!}{(2\ell)!},\qquad y_\ell(z) \sim -\frac{(2\ell+1)!}{z^{\ell+1}2^\ell \ell!} \end{equation}$$ $$\begin{align} \delta_\ell ={}& \frac{\pi}{2} - \arctan\left(-\frac{(2\ell+1)!(2\ell)! }{(ka)^{2\ell+1}2^{2\ell} \ell!^2}\right) \\ ={}&\frac{\pi}{2} + \arctan\left(\frac{(2\ell+1)!(2\ell)! }{(ka)^{2\ell+1}2^{2\ell} \ell!^2}\right) \\ ={}&\pi - \arctan\left(\frac{(ka)^{2\ell+1}2^{2\ell} \ell!^2}{(2\ell+1)!(2\ell)! }\right) \\ ={}&\pi - \frac{(ka)^{2\ell+1}2^{2\ell} \ell!^2}{(2\ell+1)!(2\ell)!} \end{align}$$ Where we have taken the first term in the Taylor series for \((ka)\ll 1\) in the last step. This gives us \(\delta_\ell = \pi - (ka)^{2\ell+1}2^{2\ell} \ell!^2/(2\ell+1)!(2\ell)!\).
Example: Let us now consider a 3D spherical well potential $$\begin{equation} V(r) = \begin{cases} - V_0 & r < a \\ 0 & r \geq a \end{cases} \end{equation}$$
  1. Compute the partial wave phase shifts for each \(\ell\).
  2. Compute the phase shift \(\delta_0\) for \(\ell = 0\).
  3. In the low-energy limit \(ka\ll 1\), compute the scattering cross section. Do this by taking only the \(\ell=0\) term.
Solution:
  1. The Schrödinger equation is given by $$\begin{equation} -\frac{\hbar^2}{2mr^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial R}{\partial r}\right) + V_\text{eff}(r)R = ER \end{equation}$$ where $$\begin{equation} V_\text{eff}(r) = V(r) + \frac{\hbar^2\ell(\ell+1)}{2mr^2} \end{equation}$$ In the region \(r > a\), we have that the wave function is the typical form $$\begin{equation} \psi_\text{out} = A\sum_{\ell=0}^\infty i^\ell \frac{(2\ell+1)}{2}(h^{(1)}_\ell(kr) e^{2i\delta_\ell} + h^{(2)}_\ell(kr))P_\ell(\cos\theta) \end{equation}$$ and in the region \(r < a\), we have $$\begin{equation} \psi_\text{in} = B\sum_{\ell=0}^\infty i^\ell (2\ell+1) c_\ell j_\ell(k_0 r) P_\ell(\cos\theta) \end{equation}$$ Note that the \(i^\ell(2\ell+1)\) is not strictly necessary, but we include it for convenience. Then, we have that the wave function should be continuous at \(r = a\), and its derivative should also be continuous there. Thus, we have that the continuity condition is $$\begin{equation} \frac{A}{2}[h^{(1)}_\ell(ka) e^{2i\delta_\ell} + h^{(2)}_\ell(ka)] = Bc_\ell j_\ell(k_0 a) \end{equation}$$ and the $$\begin{equation} \frac{Ak}{2}[h^{(1)\prime}_\ell(ka) e^{2i\delta_\ell} + h^{(2)\prime}_\ell(ka)] = Bk_0c_\ell j_\ell'(k_0 a) \end{equation}$$ If we solve for the phase shift by dividing these two equations, we find that $$\begin{equation} \frac{k[h^{(1)\prime}_\ell(ka) e^{2i\delta_\ell} + h^{(2)\prime}_\ell(ka)]}{h^{(1)}_\ell(ka) e^{2i\delta_\ell} + h^{(2)}_\ell(ka)} = k_0 \frac{j'_\ell(k_0a)}{j_\ell(k_0a)} \end{equation}$$ Then, we can rearrange a little to get $$\begin{align} kj_\ell(k_0a)[h^{(1)\prime}_\ell(ka) e^{2i\delta_\ell} + h^{(2)\prime}_\ell(ka)] ={}& k_0 j'_\ell(k_0a)[h^{(1)}_\ell(ka) e^{2i\delta_\ell} + h^{(2)}_\ell(ka)] \\ kj_\ell(k_0a)h^{(1)\prime}_\ell(ka) e^{2i\delta_\ell} + kj_\ell(k_0a) h^{(2)\prime}_\ell(ka) ={}& k_0 j'_\ell(k_0a)h^{(1)}_\ell(ka) e^{2i\delta_\ell} + k_0 j'_\ell(k_0a) h^{(2)}_\ell(ka) \end{align}$$ Then rearranging, we find $$\begin{equation} e^{2i\delta_\ell}[kj_\ell(k_0a)h^{(1)\prime}_\ell(ka) - k_0 j'_\ell(k_0a)h^{(1)}_\ell(ka) ] = k_0 j'_\ell(k_0a) h^{(2)}_\ell(ka) - kj_\ell(k_0a) h^{(2)\prime}_\ell(ka) \end{equation}$$ and finally solving for $$\begin{equation} e^{2i\delta_\ell} = \frac{k_0 j'_\ell(k_0a) h^{(2)}_\ell(ka) - kj_\ell(k_0a) h^{(2)\prime}_\ell(ka)}{kj_\ell(k_0a)h^{(1)\prime}_\ell(ka) - k_0 j'_\ell(k_0a)h^{(1)}_\ell(ka) } \end{equation}$$ As a sanity check, we can let \(V_0\to-\infty\), which means that, defining \(\kappa = \sqrt{2m(-E-V_0)}/\hbar\), the Bessel functions get replaced by modified Bessel functions so $$\begin{align} e^{2i\delta_\ell} ={}& \frac{k_0 i'_\ell(\kappa_0a) h^{(2)}_\ell(ka) - ki_\ell(\kappa a) h^{(2)\prime}_\ell(ka)}{ki_\ell(\kappa_0a)h^{(1)\prime}_\ell(ka) - k_0 i'_\ell(\kappa a)h^{(1)}_\ell(ka) } \\ ={}& 1 + \frac{\kappa i'_\ell(\kappa a) h^{(2)}_\ell(ka) - ki_\ell(\kappa a) h^{(2)\prime}_\ell(ka)}{ki_\ell(\kappa a)h^{(1)\prime}_\ell(ka) - \kappa i'_\ell(\kappa a)h^{(1)}_\ell(ka) } - 1 \\ ={}& 1 + \frac{\kappa i'_\ell(\kappa a) h^{(2)}_\ell(ka) - ki_\ell(\kappa a) h^{(2)\prime}_\ell(ka) - ki_\ell(\kappa a)h^{(1)\prime}_\ell(ka) + \kappa i'_\ell(\kappa a)h^{(1)}_\ell(ka) }{ki_\ell(\kappa a)h^{(1)\prime}_\ell(ka) - \kappa i'_\ell(\kappa a)h^{(1)}_\ell(ka) } \\ ={}& 1 + \frac{2\kappa i'_\ell(\kappa a) j_\ell(ka) - 2ki_\ell(\kappa a) j'_\ell(ka)}{ki_\ell(\kappa a)h^{(1)\prime}_\ell(ka) - \kappa i'_\ell(\kappa a)h^{(1)}_\ell(ka) } \end{align}$$ Since the asymptotic form is \(i_\ell(z) \sim e^z/2z\), we have that \(i'_\ell(z) \sim e^2/2z\) and thus $$\begin{align} e^{2i\delta_\ell} ={}& 1 + \frac{2\kappa i'_\ell(\kappa a) j_\ell(ka) - 2ki_\ell(\kappa a) j'_\ell(ka)}{ki_\ell(\kappa a)h^{(1)\prime}_\ell(ka) - \kappa i'_\ell(\kappa a)h^{(1)}_\ell(ka)} \\ ={}& 1 + 2\frac{\kappa j_\ell(ka) - k j'_\ell(ka)}{kh^{(1)\prime}_\ell(ka) - \kappa h^{(1)}_\ell(ka)} \\ ={}& 1 - 2\frac{\kappa j_\ell(ka)}{\kappa h^{(1)}_\ell(ka)} \\ ={}& 1 - \frac{2j_\ell(ka)}{h^{(1)}_\ell(ka)} \end{align}$$ which agrees for \(a_\ell = ij_\ell(ka)/kh^{(1)}_\ell(ka)\). Thus we have confidence that our answer is correct.
  2. For the result from the previous part $$\begin{align} e^{2i\delta_\ell} ={}& -\frac{kj_\ell(k_0a) h^{(2)\prime}_\ell(ka) - k_0 j'_\ell(k_0a) h^{(2)}_\ell(ka)}{kj_\ell(k_0a)h^{(1)\prime}_\ell(ka) - k_0 j'_\ell(k_0a)h^{(1)}_\ell(ka)} \end{align}$$ we can observe that the denominator is the complex conjugate of the numerator. This makes sense because the magnitudes of the numerator and denominator must be equal! Letting $$\begin{equation} z = kj_\ell(k_0a)h^{(1)\prime}_\ell(ka) - k_0 j'_\ell(k_0a)h^{(1)}_\ell(ka) = |z|e^{i\varphi} \end{equation}$$ where \(\varphi = \Arg(z)\). Then, we have that $$\begin{equation} e^{2i\delta_\ell} = e^{i\pi} \frac{z^*}{z} = e^{i\pi} e^{-2i\varphi} \end{equation}$$ so that $$\begin{align} \delta_\ell ={}& \frac{\pi}{2} - \varphi \\ ={}& \frac{\pi}{2} - \Arg[kj_\ell(k_0a)h^{(1)\prime}_\ell(ka) - k_0 j'_\ell(k_0a)h^{(1)}_\ell(ka)] \end{align}$$ Now, let us consider the case \(\ell=0\). Then, we plug in the functions $$\begin{equation} j_0(z) = \frac{\sin z}{z},\qquad h^{(1)}_0(z) = -i \frac{e^{iz}}{z} \end{equation}$$ This gives us $$\begin{align} \delta_\ell ={}& \frac{\pi}{2} - \Arg\left[k \frac{\sin(k_0 a)}{k_0 a}\left(i\frac{e^{ika}}{(ka)^2} + \frac{e^{ika}}{ka}\right) + ik_0\left(-\frac{\sin(k_0 a)}{(k_0a)^2} + \frac{\cos(k_0 a)}{k_0 a}\right) \frac{e^{ika}}{ka}\right] \\ ={}& \frac{\pi}{2} - \Arg\left[k \sin(k_0 a)\left(i\frac{1}{(ka)} + 1\right) + ik_0\left(-\frac{\sin(k_0 a)}{(k_0a)} + \cos(k_0 a)\right) \right] - ka \\ ={}& \frac{\pi}{2} - \Arg\left[\sin(k_0 a)\left(i\frac{1}{a} + k\right) + i\left(-\frac{\sin(k_0 a)}{a} + k_0\cos(k_0 a)\right) \right] - ka \\ ={}& \frac{\pi}{2} - \Arg\left[\sin(k_0 a)k + ik_0\cos(k_0 a)\right] - ka \\ ={}& \frac{\pi}{2} - \Arg\left[ik_0\left(\cos(k_0 a) - i\frac{k}{k_0}\sin(k_0 a)\right)\right] - ka \\ ={}& - \Arg\left[\left(\cos(k_0 a) - i\frac{k}{k_0}\sin(k_0 a)\right)\right] - ka \\ ={}& - \arctan\left(-\frac{k}{k_0}\tan(k_0 a)\right) - ka \\ ={}& \arctan\left(\frac{k}{k_0}\tan(k_0 a)\right) - ka \end{align}$$
  3. The phase shift from the previous section is given by $$\begin{align} \delta_0 = \arctan\left(\frac{k}{k_0}\tan(k_0 a)\right) - ka \end{align}$$ and in the low-energy limit, we can rewrite this as $$\begin{align} \delta_0 ={}& \arctan\left(\frac{ka}{k_0a}\tan(k_0 a)\right) - ka \\ ={}& \frac{ka}{k_0a}\tan(k_0 a) - ka \\ ={}& ka\left(\frac{\tan(k_0 a)}{k_0a} - 1\right) \end{align}$$ The first term (\(s\)-wave term) in the scattering cross section series is then given by $$\begin{align} \sigma ={}& \frac{4\pi}{k^2}\sin^2\delta_0 \\ ={}& \frac{4\pi}{k^2}\sin^2\left[ka\left(\frac{\tan(k_0 a)}{k_0a} - 1\right)\right] \\ ={}& \frac{4\pi}{k^2}\left[ka\left(\frac{\tan(k_0 a)}{k_0a} - 1\right)\right]^2 \\ ={}& 4\pi a^2 \left(\frac{\tan(k_0 a)}{k_0a} - 1\right)^2 \end{align}$$ For the case \(k_0 a\to 0\), the scattering cross section goes to zero which makes sense because $$\begin{equation} k_0 = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \end{equation}$$ and we already assumed \(E\approx 0\), so \(k_0 = \sqrt{2mV_0}/\hbar\), and as \(k_0\to 0\), this occurs when \(V_0\to 0\), so there is no potential at all. If instead \(k_0 a = n\pi\), then (returning to the previous formula) we have \(\delta_0 \sim 0\) as well, and the scattering cross section goes to zero, even when the potential is nonzero. This is called the Ramsauer-Townsend effect. It makes sense because, setting \(k_0 = 2\pi/\lambda\), we have that \(\lambda = 2a/n\), and the wave length is commensurate with the spherical well. On the other hand, if \(k_0 a\to \pi/2\), then the scattering cross section diverges! This makes sense because when \(k_0a = \pi/2\), the first bound state appears (at zero energy!). Then, \(\delta_0\to 1\) and \(\sigma = 4\pi/k^2\), and \(k\) is very large! The incoming particle resonates strongly with the bound state, causing the cross section to be very large. Note that the origin of the bound state can be found by looking at $$\begin{equation} \kappa = -k_0\cot(k_0 a) \end{equation}$$ where \(\kappa = \sqrt{-2mE}/\hbar\), and \(k_0 = \sqrt{2m(E+V_0)}/\hbar\). We can rewrite this equation as (defining \(w_0 = \sqrt{2mV_0}a/\hbar\) and \(w = k_0 a\)) $$\begin{align} -\cot(k_0 a) ={}& \frac{\kappa}{k_0} \\ -\cot(w) ={}& \frac{\kappa a}{w} \\ ={}& \frac{1}{w}\frac{a\sqrt{-2mEa^2}}{\hbar} \\ ={}& \frac{1}{w}\frac{a\sqrt{-2m(E+V_0) + 2mV_0}}{\hbar} \\ ={}& \frac{1}{w}\sqrt{-a^2\frac{2m(E+V_0)}{\hbar^2} + \frac{2mV_0a^2}{\hbar^2}} \\ ={}& \frac{1}{w}\sqrt{-w^2 + w_0^2} \\ ={}& \sqrt{\frac{w_0^2}{w^2} - 1} \end{align}$$ The first solution occurs at \(w = \pi/2\) when \(w_0 = \pi/2\). If \(w_0 < \pi/2\), there are no solutions. This is when we get the energy resonance! Since \(w = k_0 a = \pi/2\), this corresponds to the behaviour we found above. Note that the peak of the cross section \(\sigma\) is analogous to a peak in the transmission coefficient \(T\). This is the definition of a resonance.