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.

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