5.3. The optical theorem

We conclude this chapter by deriving an important theorem which results from the conservation of matter. This is the optical theorem, which also exists in classical optics (there it derives from total energy conservation). To study the optical theorem, we should therefore study the conservation of matter in quantum mechanics. For the Schrödinger equation we can derive a material conservation law a follows. Suppose we have a volume $V$ in three-dimensional space which is bounded by a surface $\Gamma$. We calculate the rate of change of the amount of material in that space. That amount of material is given by $$Q = \int_V \left| \psi \right|^2 d^3r.$$ We then find for the rate of change (the dot indicates derivative with respect to time): $$\dot{Q} = \int_V \left\langle \dot{\psi} | \psi \right\rangle d^3r + \int_V \left\langle\psi | \dot{\psi} \right\rangle d^3r.$$ The time-derivatives of the wave function are given by the time-dependent Schrödinger equation: $${\rm i}\hslash \frac{\partial}{\partial t} \psi({\bf r}, t) = \hat{H} \psi({\bf r}, t)$$ and its hermitian conjugate $$-{\rm i}\hslash \frac{\partial}{\partial t} \psi^*({\bf r}, t) = \hat{H} \psi^*({\bf r}, t).$$ Taking for the Hamiltonian the form $-\hbar^2/(2m) \nabla^2 + V({\bf r})$, we obtain $$\dot{Q} = -\frac{\hslash}{2{\rm i}m} \int_V \left[ \psi^*({\bf r}) \nabla^2 \psi({\bf r}) - \psi({\bf r}) \nabla^2 \psi^*({\bf r}) \right] d^3r$$ (note that the terms containing $V$ cancel). Using Green's second identity, the volume integral can be transformed into a surface integral: $$\dot{Q} = -\frac{\hbar}{2{\rm i}m} \int_\Gamma \left[ \psi^*({\bf r}) \nabla \psi({\bf r}) - \psi({\bf r}) \nabla \psi^*({\bf r}) \right] d{\bf a}.$$ Here, $d{\bf a}$ is a vector, the norm of which is that of a small surface segment, and directed along the outward surface normal. The change in material can in this case only be caused by flow of material through the boundary. In particular, we have $$\dot{Q} = -\int {\bf j}\cdot d{\bf a}.$$ so that we see that the expression for the particle flux is $${\bf j}= \frac{\hbar}{2{\rm i}m}\left[ \psi^*({\bf r}) \nabla \psi({\bf r}) - \psi({\bf r}) \nabla \psi^*({\bf r}) \right]$$

Note that using the diverence theorem, the equation relating the material change to the surface integral can be related to the continuity equation: $$\dot{Q} = \int_V \dot{\rho} d^3 r = -\int {\bf j}\cdot d{\bf a}= - \int_V \nabla \cdot {\bf j}d^3r.$$ As the volume is arbitrary, this equation can only hold when $$\dot{\rho} + \nabla \cdot {\bf j}= 0.$$

Let's now turn again to the scattering problem, where we are dealing with an incoming and an outgoing wave: $$\psi = \phi_{\text{in}} + \phi_{\text{out}}.$$ The total current in this wave is then given as $${\bf j}= \frac{\hslash}{2{\rm i}m} \left[ \left( \phi^*_{\text{in}} + \phi^*_{\text{out}}\right) \nabla \left( \phi_{\text{in}} + \phi_{\text{out}}\right) - \left( \phi_{\text{in}} + \phi_{\text{out}}\right) \nabla \left( \phi^*_{\text{in}} + \phi^*_{\text{out}}\right) \right].$$ As this expression consists of a complex number minus its complex conjugate, we can write this in the form $${\bf j}= \frac{\hslash}{m} {\text{Im}} \left[ \left( \phi^*_{\text{in}} + \phi^*_{\text{out}}\right) \nabla \left( \phi_{\text{in}} + \phi_{\text{out}}\right)\right].$$ This can be rewritten as $${\bf j}= {\bf j}_{\text{in}} + {\bf j}_{\text{out}} + \frac{\hslash}{m} {\text{Im}} \left( \phi^*_{\text{in}}\nabla \phi_{\text{out}} + \phi^*_{\text{out}}\nabla \phi_{\text{in}} \right) = {\bf j}_{\text{in}} + {\bf j}_{\text{out}} + \frac{\hslash}{m} {\text{Im}} \left( \phi^*_{\text{in}}\nabla \phi_{\text{out}} - \phi_{\text{out}}\nabla \phi^*_{\text{in}} \right).$$

In a scattering problem we look at the scattered flux generated by a stationary flux of incoming particles. In the stationary limit, there is no generation or absorption of new matter: the particles flowing into some sphere centred around the scattering potential should also come out at the same rate. Therefore we must have: $$\int_{\text{S}} {\bf j}\cdot d{\bf a}= 0,$$ as this expression calculates the total matter flux through the sphere's surface.

As we have written the total flux as a sum over an outgoing, an ingoing and a mixing term, we can also divide the total flux through the surface up into these three contributions. The flux of the incoming wave turns out to be zero! This is because the beam described by $\exp({\rm i}kz)$ gives an incoming flux on one side of the sphere, and an equal outgoing flux on the other side. The outgoing flux through the sphere is given by $$\int_{\text{S}} {\bf j}_{\text{out}} \cdot d{\bf a}= \int \left| f(\theta) \right|^2 \frac{\hslash k}{m} \frac{1}{r^2} 2\pi r^2 \sin\theta d\theta = \frac{\hslash k}{m} \sigma_{\text{tot}}.$$ Therefore, we obtain $$-k \sigma_{\text{tot}} = {\text{Im}} \int \left( \phi^*_{\text{in}} \nabla \phi_{\text{out}} - \phi_{\text{out}}\nabla \phi^*_{\text{in}} \right) \cdot d{\bf a}.$$ Using Green's theorem again, the expression on the right hand side can be reworked: $$-k \sigma_{\text{tot}} = {\text{Im}} \int \left( \phi^*_{\text{in}} \nabla^2 \phi_{\text{out}} - \phi_{\text{out}}\nabla^2 \phi^*_{\text{in}} \right) d^3r.$$ Writing $\psi = \phi_{\text{in}} + \phi_{\text{out}}$, this can be rewritten as $$-k \sigma_{\text{tot}} = {\text{Im}} \int_V \left[ \phi^*_{\text{in}} \nabla^2 \left( \psi - \phi_{\text{in}}\right) - \left( \psi - \phi_{\text{in}}\right)\nabla^2 \phi^*_{\text{in}} \right] d^3r = {\text{Im}} \int_V \left[ \phi^*_{\text{in}} \nabla^2 \left( \psi \right) - \psi\nabla^2 \left( \phi^*_{\text{in}} \right)\right] d^3r. \tag{5.16}\label{Eq:OptTheor}$$ Now we use that the full wave function satisfies the full Schrödinger equation  $$\nabla^2 \psi = -k^2 \psi + \frac{2m V({\bf r}) }{\hslash^2} \psi,$$ and the incoming wave satisfies the Schrödinger equation with potential 0: $$\nabla^2 \phi_{\text{in}} = - k^2 \phi_{\text{in}}.$$ Putting this into \eqref{Eq:OptTheor} gives: $$-k \sigma_{\text{tot}} = {\text{Im}} \int_V \left( - \phi^*_{\text{in}} k^2 \psi + \phi^*_{\text{in}} \frac{2m V(r)}{\hslash^2} \psi + \psi k^2 \phi^*_{\text{in}}\right) d^3r.$$ The first and the third term in the integral cancel and we have $$-k \sigma_{\text{tot}} = {\text{Im}} \int_V \phi^*_{\text{in}} \frac{2m V(r)}{\hslash^2} \psi \; d^3 r.$$ The last term is recognised as the exact scattering amplitude for $\theta=0$ (up to a negative pre-factor) -- see Eq. (5.15).

Therefore, we find the optical theorem: $$\sigma_{\text{tot}} = {\text{Im}} \frac{4\pi f(\theta=0)}{k}.$$ If the wave is scattered, we see an attenuation in the forward direction compared to the case where the incoming particles would not be scattered. The forward scattering amplitude $f(0)$ is therefore related to the scattering of the particles (i.e. $\sigma_{\text{tot}}$). It is important to realise that the optical theorem holds exactly; in the first Born approximation, the theorem does not hold.