# Probability current with complex potential

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.3, Exercise 5.3.1.

Shakar’s derivation of the probability current in 3-d is similar to the one we reviewed earlier, so we don’t need to repeat it here. We can, however, look at a slight variant where the potential has a constant imaginary part, so that

$\displaystyle V\left(\mathbf{r}\right)=V_{r}\left(\mathbf{r}\right)-iV_{i} \ \ \ \ \ (1)$

where ${V_{r}\left(\mathbf{r}\right)}$ is a real function of position and ${V_{i}}$ is a real constant. A Hamiltonian containing such a complex potential is not Hermitian.

To see what effect this has on the total probability of finding a particle in all space, we can repeat the derivation of the probability current. From the Schrödinger equation and its complex conjugate, we have

 $\displaystyle i\hbar\frac{\partial\psi}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V_{r}\psi-iV_{i}\psi\ \ \ \ \ (2)$ $\displaystyle -i\hbar\frac{\partial\psi^*}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi^*+V_{r}\psi^*+iV_{i}\psi^* \ \ \ \ \ (3)$

Multiply the first equation by ${\psi^*}$ and the second by ${\psi}$ and subtract to get

$\displaystyle i\hbar\frac{\partial}{\partial t}\left(\psi\psi^*\right)=-\frac{\hbar^{2}}{2m}\left(\psi^*\nabla^{2}\psi-\psi\nabla^{2}\psi^*\right)-2iV_{i}\psi\psi^* \ \ \ \ \ (4)$

As in the case with a real potential, the first term on the RHS can be written as the divergence of a vector:

 $\displaystyle \mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}(\Psi^*\nabla\Psi-\Psi\nabla\Psi^*)\ \ \ \ \ (5)$ $\displaystyle \nabla\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(\psi^*\nabla^{2}\psi-\psi\nabla^{2}\psi^*\right)\ \ \ \ \ (6)$ $\displaystyle \frac{\partial}{\partial t}\left(\psi\psi^*\right)$ $\displaystyle =$ $\displaystyle -\nabla\cdot\mathbf{J}-\frac{2V_{i}}{\hbar}\psi\psi^* \ \ \ \ \ (7)$

If we define the total probability of finding the particle anywhere in space as

$\displaystyle P\equiv\int\psi^*\psi d^{3}\mathbf{r} \ \ \ \ \ (8)$

then we can integrate 4 over all space and use Gauss’s theorem to convert the volume integral of a divergence into a surface integral:

 $\displaystyle \frac{\partial}{\partial t}\left(\int\psi\psi^*d^{3}\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle -\int\nabla\cdot\mathbf{J}d^{3}\mathbf{r}-\frac{2V_{i}}{\hbar}\int\psi\psi^*d^{3}\mathbf{r}\ \ \ \ \ (9)$ $\displaystyle \frac{\partial P}{\partial t}$ $\displaystyle =$ $\displaystyle -\int_{S}\mathbf{J}\cdot d\mathbf{a}-\frac{2V_{i}}{\hbar}P \ \ \ \ \ (10)$

We make the usual assumption that the probability current ${\mathbf{J}}$ tends to zero at infinity fast enough for the first integral on the RHS to be zero, and we get

$\displaystyle \frac{\partial P}{\partial t}=-\frac{2V_{i}}{\hbar}P \ \ \ \ \ (11)$

This has the solution

$\displaystyle P\left(t\right)=P\left(0\right)e^{-2V_{i}t/\hbar} \ \ \ \ \ (12)$

That is, the probability of the particle existing decays exponentially. Although Shankar says that such a potential can be used to model a system where particles are absorbed, it’s not clear how realistic it is since the Hamiltonian isn’t hermitian, so technically the energies in such a system are not observables.