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If the external potential $V$ in the time-dependent Schroedinger equation doesn't depend on time, then we can separate the wavefunction as spatial part and time part.

$$ \Psi(x,t)=\psi(x) \theta(t) $$

$\psi(x)$ is the solution of well-known time-independent Schroedinger equation, and $\theta(t) = e^{-Et/\hbar}$. Then the probability density is time-independent, or stationary:

$$ \rho=|\Psi(x,t)|^{2} = |\psi(x)|^{2} $$

Thus, $ \frac{d}{dt} \int \rho d \tau =0 $.

However, introduce an imaginary potential $V(\vec{x})=V_{1}(\vec{x})+i V_{2}(\vec{x})$, where $V_{1}$ and $V_{2}$ are real function.

Then the time-dependent Schroedinger equation is

$$ i \hbar \frac{\partial}{\partial t}\Psi(\vec{x},t)= -\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(\vec{x},t)+ \left\{V_{1}(\vec{x})+i V_{2}(\vec{x}) \right\}\Psi(x,t) $$

Let's calculate $ \frac{d}{dt} \int \rho d \tau $ at first.

$$ \begin{align*} \frac{d}{dt} \int \rho d \tau &= \frac{d}{dt} \int \Psi^{\ast}\Psi d \tau = \int \frac{\partial}{\partial t}(\Psi^{\ast} \Psi)d \tau \\ &=\int \left (\Psi^{\ast} \frac{\partial}{\partial t} \Psi + \Psi \frac{\partial}{\partial t} \Psi^{\ast} \right)d \tau\\ &= \int \frac{1}{2} \mathrm{Re} \left(\Psi^{\ast} \frac{\partial}{\partial t}\Psi \right) d \tau \end{align*} $$

From the given Schroedinger equation,

$$ \frac{1}{2} \mathrm{Re} \left(\Psi^{\ast} \frac{\partial}{\partial t} \Psi \right) =\frac{V_{2}(\vec{x})}{2 \hbar} \Psi^{\ast}\Psi $$

Thus,

$$ \frac{d}{dt} \int \rho d \tau = \frac{V_{2}(\vec{x})}{2 \hbar} \neq 0 $$

$V_{2}$ is generally not zero, meaning that the density isn't stationary.

This would violate the statement that the wavefunction is stationary if the external potential doesn't depend on time.

There no problem if the potential is real, but I guess imaginary potential has a kind of physical meaning. How can I think about this?

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  • $\begingroup$ You seem to have implicitly assumed that the wave function is normalizable. This might not necessarily be true when you are dealing with such a potential. $\endgroup$ – Aritra Feb 1 '17 at 9:25
  • $\begingroup$ @Arita Then the last integration part among equations is a possible danger. Is there any example? $\endgroup$ – Patche Feb 1 '17 at 9:57
  • $\begingroup$ expanded below as an answer. Hope this helps! $\endgroup$ – Aritra Feb 1 '17 at 10:35
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The answer in short is that when you are dealing with a complex potential, the direct consequence is that you are dealing with complex energy eigenvalues & hence a non-hermitian Hamiltonian.Therefore, for a complex potential, the eigenstates of H are not stationary anymore and that itself negates the initial assumption in the question.

To expand on this, let me reproduce the calculations from "Complex potential well" -- Max Lewandowski, Universität Postdam, 2011 [ http://users.physik.fu-berlin.de/~pelster/Projects/lewandowski.pdf ]

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