17 Scattering Problems

17.1 Introduction to Scattering Problems

Consider a potential that becomes constant as \(x \to \pm \infty\). For instance, \[V(x) \to \begin{cases} 0 & \quad x \to - \infty \\ V_0 & \quad x \to +\infty \, . \end{cases}\] In a scattering problem, we ask the question: what is the fate of a particle with energy \(E\) incoming from \(x = -\infty\)?

The answer in classical mechanics is simple. Let \(V_{\text{max}}\) denote the maximum value of the potential. Conservation of energy tells us that:

• If \(E<V_{\text{max}}\), the particle will be “reflected" back to \(x = - \infty\) with probability \(1\).

• If \(E>V_{\text{max}}\), the particle will be “transmitted" to \(x = \infty\) with probability \(1\).

The answer in quantum mechanics is much more interesting!

17.2 Scattering Wavepackets

What do we mean by the scattering of particles in quantum mechanics? The proper answer is to consider the scattering of “wavepackets", which you can imagine as Gaussian wave functions.

• For \(t \to - \infty\), the wave function has the form of an incoming wavepacket, \[\psi(x,t) \to \psi_I(x,t) \, .\]

• The incoming wavepacket will then “scatter” from the potential and as \(t\to \infty\), the wave function tends to a sum of reflected and transmitted wavepackets, \[\psi(x,t) \to \psi_R(x,t) + \psi_T(x,t) \, .\]

In the limit \(t \to \infty\), the reflected and transmitted wavepackets are completely separated in space and show no interference. The probability for the particle to be reflected or transmitted is then defined by \[\begin{aligned} R = \lim_{t\to \infty} \int |\psi_R(x,t)|^2 {\rm d}x\,,\\[1ex] T = \lim_{t\to \infty} \int |\psi_T(x,t)|^2 {\rm d}x\, , \end{aligned}\] such that \[R +T = 1\] if the total wave function is correctly normalized.

For a general potential \(V(x)\) and incoming wave function \(\psi_I(x,t)\), computing \(R\) and \(T\) is extremely hard. Fortunately, we do not need to solve this problem in general to answer important questions about scattering experiments.

17.3 Scattering Plane Waves

Scattering experiments typically involve incoming particles with a small energy uncertainty \(\Delta E\). This suggests that instead it is useful to consider stationary wave functions with energy \(E>0\), \[\psi(x,t) = \phi(x)e^{-iEt/\hbar}\,, \qquad \hat H \cdot \phi(x) = E\phi(x) \, .\] It is straightforward to find the form of the Hamiltonian eigenfunction \(\phi(x)\) in the asymptotic regions where the potential is constant: these are plane waves.

• For \(x \to -\infty\), Hamiltonian eigenfunctions obey \[\frac{\partial^2\phi(x)}{\partial x^2} = - k^2 \phi(x)\,, \qquad k = \sqrt{2mE/\hbar^2} \, .\] The general solution is \[\phi(x) = e^{ikx} + r e^{-ikx} \, .\] This is a superposition of incoming and reflected waves. Since the wave functions are not square-normalizable, the overall constant is not physically meaningful: for convenience we have set the coefficient of the incoming wave to \(1\).

• For \(x \to + \infty\), \(V(x) = V_0\). For scattering problems we assume that \(E > V_0\). The Hamiltonian eigenfunctions are solutions to \[\frac{\partial^2\psi(x)}{\partial x^2} = - k'^2 \psi(x)\,, \qquad k' =\sqrt{2m(E-V_0)/\hbar^2} \, .\] We consider the solution \[\psi(x) = t e^{ik'x}\] corresponding to a transmitted wave. There is another solution \(e^{-ik'x}\). which corresponds to a wave incoming from \(x = +\infty\). We discard this solution as we only want waves incoming from \(x = - \infty\).

In summary we have asymptotic plane waves, \[\phi(x) \to \begin{cases} e^{ikx} + r e^{-ikx} & \quad x \to -\infty \\[1ex] t e^{ik'x} & \quad x \to + \infty \end{cases} \, .\]

How do we extract physical information from these wave functions? To answer this question, we consider the probability current. For stationary solutions this takes the form \[\begin{aligned} J & = \frac{\hbar}{2mi}\left( \bar\psi \partial_x \psi - \psi \partial_x \bar\psi \right) \\[1ex] & = \frac{\hbar}{2mi}\left( \bar\phi \partial_x \phi - \phi \partial_x \bar\phi \right)\,, \end{aligned}\] where the phase \(e^{-iEt/\hbar}\) has cancelled out. For a plane wave \(\phi(x) = e^{ikx}\), the probability current is \[J(\phi = e^{ikx}) = \frac{\hbar}{2mi}\left( e^{-ikx} \partial_x e^{ikx} - e^{ikx}\partial_x e^{-ikx} \right) = \frac{\hbar k}{m} \, .\] (once again, this does not violate the statement that the probability current vanishes for square-normalisable stationary solutions because plane waves are not square-normalisable). The probability current in the asymptotic regions is therefore \[J(x) \to \begin{cases} J_I - J_R& \quad x \to -\infty \\[1ex] J_T & \quad x \to + \infty\,, \end{cases}\] where \[J_I := \frac{\hbar k}{m}\,, \quad\qquad J_R := \frac{\hbar k}{m} |r|^2\,, \quad\qquad J_T := \frac{\hbar k'}{m} |t|^2\,,\] are defined such that \(J_I,J_R,J_T>0\). They are the contributions to the probability current in the asymptotic regions from the incident, reflected, and transmitted waves.

There is an immediate constraint on these currents from the continuity equation \(\partial_t P + \partial_xJ = 0\). Since the the wave functions are stationary, \(\partial_t P = 0\), and the probability current is constant in space. We conclude that \[J(-\infty) = J(\infty)\] and therefore \[J_I - J_R = J_T \, .\] This means that if we define \[\begin{aligned} R & := \frac{J_R}{J_I} = |r|^2 \qquad & T & := \frac{J_T}{J_I} = \frac{k'}{k}|t|^2 \end{aligned}\] then \(R+T = 1\).

This suggests we interpret \(R\) and \(T\) as the probability for an incoming particle with definite energy \(E>V_0\) to be reflected and transmitted by the potential. This in fact coincides with a careful analysis of the scattering of wavepackets in the limit \(\Delta E \to 0\) of small uncertainty in energy. We are therefore justified in using the simpler definition in terms of plane waves.

For more information, see section 5.4 of Ramamurti Shankar. Principles of quantum mechanics. Springer Science & Business Media, 2012..

17.4 Example: Finite Step Potential

Let us consider the finite step potential \[V(x) = \begin{cases} 0 & \quad x < 0 \\[1ex] V_0 & \quad x \geq 0 \end{cases}\] where we assume that \(V_0 >0\). This is a simple approximation for a more realistic smooth potential interpolating between \(0\) and \(V_0\). It might represent the boundary between two different materials.

There are two cases to consider: \(E>V_0\) or \(E<V_0\). The “scattering problem” considered above corresponds to \(E>V_0\). However, \(0<E<V_0\) is also extremely interesting and is known as the “tunnelling problem”. We consider it further in the next lecture.

Let us therefore restrict here to \(E>V_0\). In this example, the asymptotic plane wave solutions considered above are valid all they way up to \(x = 0\). In summary, \[\phi(x) = \begin{cases} e^{ikx} + r e^{-ikx} & \quad x < 0 \\[1ex] t e^{ik'x} & \quad x > 0 \end{cases} \, .\] We now need to impose boundary conditions at the discontinuity in the potential at \(x = 0\). Since, the potential remains finite we require that both \(\phi(x)\) and \(\partial_x\phi(x)\) are continuous across \(x = 0\). This requires \[1 + r = t \qquad k (1-r) = k't \, .\] We have two equations for two unknowns \(r,t\) and the solution is unique \[r = \frac{k-k'}{k+k'}\,, \qquad t = \frac{2k}{k+k'} \,,\] and hence \[R = \left( \frac{k-k'}{k+k'} \right)^2\,, \qquad T = \frac{k'}{k}\left( \frac{2k}{k+k'} \right)^2 = \frac{4kk'}{(k+k')^2} \, .\] As a consistency check we see that \(R + T = 1\) as required.

The reflection and transmission probabilities as a function of the dimensionless ratio \(E/V_0\) are shown above. There are two important limits:

• As \(E \to \infty\) or \(V_0 \to 0\) we find \(R \to 0\) and \(T \to 1\): the potential step effectively disappears and the incoming particle is transmitted with probability \(1\). Note that this coincides with the classical expectation for particles with \(E>V_0\).

• As \(E \to V_0\) from above we find \(R \to 1\) and \(T \to 0\): the incoming particle is reflected with probability \(1\). This coincides with the classical expectation for particles with \(E\leq V_0\).

As mentioned above, an important question is what happens in quantum mechanics when \(E<V_0\). We will address this problem in the next lecture!

17.5 Problems

1. Scattering off a finite step potential:

   Consider the potential \[V(x) = \begin{cases} 0 & x \leq 0 \\ V_0 & x > 0 \end{cases} \, .\]

   1. What are the boundary conditions on the wavefunction at \(x=0\)?

   2. Construct Hamiltonian eigenfunctions appropriate for incoming particles of energy \(E>V_0\) sent from \(x = -\infty\).

   3. Compute the reflection and transmission probabilities \(R\) and \(T\) and sketch them as a function of \(E/V_0\).

   4. Check that \(R+T=1\). Explain!

   5. Discuss the behaviour of \(R\), \(T\) in the limits \(E \gg V_0\) and \(E \to V_0\).

   Solution ▶ See lecture 17.

2. Scattering off a delta function potential:

   Consider the potential well \[V(x) = - \alpha \delta(x)\] where \(\alpha >0\).

   1. Show that a Hamiltonian eigenfunction with energy \(E>0\) obeys \[\phi''(x) = -\frac{2m}{\hbar^2}\left(E+\alpha \delta(x) \right) \phi(x) \, .\]

   2. The wavefunction \(\psi(x)\) is continuous at \(x = 0\). By integrating part \((a)\) over an interval \((\epsilon,-\epsilon)\) show that the derivative of the wavefunction has a discontinuity at \(x = 0\), \[\lim_{\epsilon \to 0} \left( \phi'(\epsilon) - \phi'(-\epsilon) \right) = - \frac{2m\alpha}{\hbar^2}\phi(0) \, .\]

   3. Find the constant \(k\) in terms of \(m\), \(\hbar\), \(E\) such that \[\phi(x) = \begin{cases} e^{ikx} + r e^{-ikx} & \text{if} \quad x<0 \\ t e^{ikx} & \text{if} \quad x>0 \end{cases}\] is a Hamiltonian eigenfunction with \(E>0\) for \(x \neq 0\). What is the physical interpretation of the three terms?

   4. Impose the boundary conditions from part (a) and show that \[R = |r|^2 = \frac{1}{1+2\hbar^2E/m\alpha^2} \qquad T = |t|^2 = \frac{2\hbar^2E/m\alpha^2}{1+2\hbar^2E/m\alpha^2} \, .\] What is the physical interpretation of \(R\), \(T\) and why does \(R+T = 1\)?

   5. Sketch \(R\), \(T\) as a function of energy \(E>0\) and explain intuitively their behaviour as \(E \to \infty\).

   Solution ▶

   1. A Hamiltonian eigenfunction in the potential \(V(x) = - \alpha \delta(x)\) obeys \[-\frac{\hbar^2}{2m} \phi''(x) - \alpha \delta(x) \phi(x) = E\phi(x) \, .\] Rearranging this equation, \[\phi''(x) = -\frac{2m}{\hbar^2}\left(E+\alpha \delta(x) \right) \phi(x) \, .\]

   2. Integrating this equation over an interval \((\epsilon,-\epsilon)\), \[\int^\epsilon_{-\epsilon} \phi''(x) dx = -\frac{2m}{\hbar^2} \int^\epsilon_{-\epsilon} (E + \alpha \delta(x))\phi(x) \, .\] Using the fundamental theorem of calculus and the defining property of the delta-function \(\delta(x)\), we find \[\phi'(\epsilon) - \phi'(-\epsilon) = -\frac{2m \alpha}{\hbar^2} \phi(0) -\frac{2m E}{\hbar^2} \int^\epsilon_{-\epsilon} \phi(x) dx \, .\] In the limit \(\epsilon \to 0\) the second term on the right tends to zero, leaving \[\lim_{\epsilon \to 0} \left( \phi'(\epsilon) - \phi'(-\epsilon) \right) = - \frac{2m\alpha}{\hbar^2}\phi(0) \, .\]

   3. A by now standard computation shows that \(k = \sqrt{2mE/\hbar^2}\). The components \(e^{ikx}\), \(r e^{-ikx}\), \(t e^{ikx}\) correspond to the incident, reflected and transmitted waves.

   4. Imposing the boundary conditions from part (b), we find \[1+r = t \qquad ik(1-r) - i k t = \frac{2m\alpha}{\hbar^2} t \, .\] The solution for \(r\), \(t\) is \[r = - \frac{im\alpha}{\hbar^2k+im\alpha} \qquad t = - \frac{\hbar^2k}{\hbar^2k+im\alpha} \, .\] We therefore find \[\begin{aligned} \nonumber |r|^2 & =\frac{m^2\alpha^2}{k^2\hbar^4+m^2\alpha^2} = \frac{1}{1+2\hbar^2E/m\alpha^2} \\ |t|^2 & =\frac{k^2\hbar^4}{k^2\hbar^4+m^2\alpha^2} = \frac{2\hbar^2E/m\alpha^2}{1+2\hbar^2E/m\alpha^2} \, . \end{aligned}\] The quantities \(R = |r|^2\), \(T=|t|^2\) are the probabilities for a particle with energy \(E\) to be reflected transmitted. The relation \(R+T=1\) ensures that probabilities sum to \(1\).

   5. The solutions look like the following.

      

      As \(E \to \infty\), \(R \to 0\) and \(T \to 1\). As the energy of the particle increases so does the probability of transmission.

3. Critical scattering off a potential barrier:

   Consider the potential barrier with \(V_0>0\), \[V(x) = \begin{cases} 0 & \quad x \leq 0 \\ V_0 & \quad 0<x<L \\ 0 & \quad x \geq L \end{cases} \, .\]

   1. Show that the wavefunction \[\phi(x) = \begin{cases} e^{ikx}+ re^{-ikx} & x<0 \\ A+ B x & 0<x<L \\ t e^{ikx} & x>L \end{cases}\] is a Hamiltonian eigenfunction with energy \(E = V_0\) where \[k = \sqrt{2mV_0 / \hbar^2}\, .\]

   2. What are the boundary conditions at \(x = 0\) and \(x = L\)?

   3. Imposing the boundary conditions, eliminate \(A\), \(B\) and show that \[|r|^2 = \frac{k^2L^2}{k^2L^2+4} \qquad |t|^2 = \frac{4}{k^2L^2+4} \, .\]

   4. Show that \(|r|^2+|t|^2 = 1\) and discuss the physical significance of \(|r|^2, |t|^2\).

   5. Sketch \(|r|^2\), \(|t|^2\) as a function of the dimensionless ratio \(\gamma = kL\) and explain the behaviour as \(\gamma \to 0\) and \(\gamma \to \infty\).

   Solution ▶

   1. For \(x<0\) and \(x<L\) the Hamiltonian operator is \(\hat H = \frac{\hat p^2}{2m}\) and therefore \[\hat H \cdot e^{\pm i kx} = -\frac{\hbar^2}{2m} \partial_x^2 e^{\pm ikx} = \frac{\hbar^2k^2}{2m} e^{\pm ikx} = V_0 e^{\pm ikx} \, ,\] since \(k = \sqrt{2mV_0/\hbar^2}\). For \(0\leq x \leq L\) the Hamiltonian operator is \(\hat H=\frac{\hat p^2}{2m}+V_0\) and therefore \[\hat H \cdot (A+Bx) = V_0(A+Bx)\, .\] In all regions we therefore have Hamitlonian eigenfunctions with energy \(E=V_0\).

   2. Since the potential has a finite discontinuities at \(x = 0\) and \(x = L\), we require that the wavefunction \(\psi(x)\) and its first derivative \(\psi'(x)\) are continuous at \(x=0\) and \(x=L\).

   3. The boundary condition at \(x=0\) is \[1+r = A \qquad ik(1-r) = B \, .\] The boundary condition at \(x = L\) is \[A+BL = te^{ikL} \qquad B = ik t e^{iKL} \, .\] There are four linear equations for four unkowns \(r\), \(A\), \(B\), \(t\). Eliminating \(A\) an \(B\), the solution for \(r\) and \(t\) is \[r = \frac{k L}{k L+2 i} \qquad t= \frac{2 i e^{-i k L}}{k L+2 i}\] and therefore \[|r|^2 = \frac{k^2 L^2}{k^2 L^2+4} \qquad |t|^2 = \frac{4}{ k^2 L^2+4} \, .\]

   4. It is clear that \(|r|^2+|t|^2=1\). The wavefunction describes the reflection and transmission of particles incoming from the left with definite energy \(E = V_0\). The quantities \(|r|^2\), \(|t|^2\) are probabilities for an incoming particle to be reflected, transmitted. The relation \(|r|^2 +|t|^2 =1\) is the statement that the total probability is \(1\).
      

   5. In terms of the dimensionless ration \(\gamma = kL\), \[|r|^2 = \frac{\gamma^2}{\gamma^2+4} \qquad |t|^2 = \frac{4}{ \gamma^2+4} \, .\] The sketch should shown \(|r|^2\) starting at \(0\) when \(\gamma=0\), increasing monotonically for \(0<\gamma<\infty\), and asymptote to \(1\) as \(\gamma\to \infty\).

      

      1. The limit \(\gamma \to 0\) corresponds to \(L<<\frac{1}{k}\), where \(|r|^2\to 0\), \(|t|^2\to1\): tunnelling is inevitable. If the barrier length is much less than the incoming wavelength \(( \propto \frac{1}{k})\), the particle does not resolve the barrier and so the probability of transmission is high.

      2. The limit \(\gamma \to \infty\) corresponds to \(L>> \frac{1}{k}\), where \(|r|^2\to1\) and \(|t|^2\to0\): reflection is inevitable. If the barrier length is much greater than the incoming wavelength \(( \propto \frac{1}{k})\), the probability density decays to near zero inside the barrier and the probability of transmission is high.

4. Scattering off a finite potential barrier:

   Consider particles with definite energy \(E>V_0>0\) incoming from the left on the potential barrier \[V(x) = \begin{cases} V_0 & \text{if} \quad 0<x<L \\ 0 & \text{otherwise} \end{cases}\, .\]

   1. Write down appropriate Hamiltonian eigenfunctions in the regions \(x<0\), \(0<x<L\), and \(x>L\).

   2. Impose the boundary conditions at \(x=0\) and \(x=L\) and hence compute the transmission and reflection coefficients \(R\), \(T\).

   3. Verify that \(R+T=1\).

   4. How do \(R\), \(T\) behave when \(E \to V_0^+\) and \(E \to \infty\)? Explain!

   5. What happens to \(R\), \(T\) when \[E = \frac{\hbar^2}{2m}\left(\frac{n\pi}{L} \right)^2\] for \(n\in \mathbb{Z}_{>0}\) such that \(E>V_0\)? Why is this happening?

   6. Sketch \(R\), \(T\) as a function of energy.

   Hint (b): after imposing boundary conditions, you should find four linear equations for four unknowns. At this stage you may use a computer to do the linear algebra.

   Solution ▶

   1. The hamiltonian eigenfunctions appropriate for the scattering of particles incoming from \(x = -\infty\) with energy \(E\) are \[\phi(x) = \begin{cases} e^{ik'x} + r e^{-ik'x} & \quad x<0 \\ Ae^{ikx} + B e^{-ikx} & 0<x<L \\ t e^{ik'x} & x>L \end{cases} \, ,\] where \[\begin{aligned} k = \sqrt{2mE/\hbar^2} \\ \nonumber k' =\sqrt{2m(E-V_0)/\hbar^2} \end{aligned}\] and \(r\), \(t\), \(A\), \(B\) are constant we need to determine. For scattering problems, the wavefunctions are not normalizable and so the overall constant is not physically meaningful. We have used this freedom to set the coefficient of the incoming wave to \(1\).

   2. We now impose \(\phi(x)\) and \(\partial_x\phi(x)\) are continuous at \(x=0\) and \(x=L\). The boundary conditions at \(x = 0\) give \[\begin{aligned} 1+r & = A+B \\ \nonumber k'(1-r) & = k(A-B) \end{aligned}\] while those at \(x=L\) give \[\begin{aligned} Ae^{ikL} + Be^{-ikL} & = t e^{ik'L} \\ \nonumber k(Ae^{ikL} - Be^{-ikL}) & = k't e^{ik'L} \, . \end{aligned}\] This gives four linear equations for the four unknowns \(r\), \(t\), \(A\), \(B\). These are just simultaneous linear equations and the solution can of course be found by hand with some patience. Instead, you can use a computer. In any case, the solution for \(r\) and \(t\) is \[\begin{aligned} r & = \frac{(k'^2-k^2) \sin (k L)}{\left(k'^2+k^2\right) \sin (k L)+2 i k' k \cos (k L)} \\ \nonumber t & = \frac{2 i k'k e^{-i k' L}}{\left(k'^2+k^2\right) \sin (k L)+2 i k' k \cos (k L)} \, . \end{aligned}\] The reflection and transmission probabilities are \[\begin{aligned} R & = |r|^2 = \frac{\left(k'^2-k^2\right)^2 \sin ^2(k L)}{\left(k'^2+k^2\right)^2 \sin ^2(k L)+4 k'^2 k^2 \cos ^2(k L)} \\ \nonumber T & = |t|^2 =\frac{4 k'^2 k^2}{\left(k'^2+k^2\right)^2 \sin ^2(k L)+4 k'^2 k^2 \cos ^2(k L)} \, . \end{aligned}\] It is straightforward to see that \(R +T=1\) using \(\sin^2z +\cos^2z =1\).
      

   3. 1. The limit \(E \to V_0\) corresponds to \(k' \to 0\). In this limit we find \(R \to 1\) and \(T \to 0\). We recover the classical expectation for \(E\leq V_0\).

      2. The limit \(E \to \infty\) corresponds to \(k,k'\to \infty\) with \(k/k' = 1\). We find \(R \to 0\) and \(T\to 1\). The potential well has effectively vanishes and the particle is transmitted with probability \(1\). We recover the classical expectation for \(E>V_0\).

   4. This corresponds to \(kL = n\pi\) with \(n \in \mathbb{Z}_{>0}\). At these points \(R=0\) and \(T=1\). They correspond to ‘transmission resonances’ where the particle is transmitted with probability \(1\). Roughly, at these points there is destructive interference between the waves reflected from the boundaries at \(x = 0\) and \(x = L\).

   5. The reflection / transmission coefficients looks like: