Solving Schrodinger's Equation with a weak Imaginary Potential

[AspiringResearcher]

Homework Statement

A particle of energy $E$ moves in one dimension in a constant imaginary potential $-iV$ where $V << E$.

a) Find the particle's wavefunction $\Psi(x,t)$ approximating to leading non-vanishing order in the small quantity $\frac{V}{E} << 1$.

b) Calculate the probability current density of the wavefunction you just calculated, and interpret your result briefly: what is the effect of the small imaginary potential?

Homework Equations

$$-\frac{\hbar^2}{2m}\psi'' -iV\psi = E\psi \text{ (time-independent schrodinger equation for energy E)}$$
$$j = Re(\bar{\Psi}\frac{\hbar}{im}\frac{d\Psi}{dx})\text{ (probability current density)}$$

First order approximations for $x << 1$:

• $\sin(x) \approx 1$
• $\cos(x) \approx 1 - \frac{x^2}{2}$
• $e^x \approx 1 + x$

The Attempt at a Solution

I attempted this problem in a few different ways and it was never obvious to me how the Taylor approximation would work out.

Since the particle is traveling with a definite energy E, the use of the time-independent Schrodinger equation is justified. The time dependence is easy to calculate.

The first approach I used was to solve the time-independent Schrodinger equation with the standard method for second-order homogeneous linear differential equations with constant coefficients. I modified this approach by assuming that the coefficient $r$ in $\psi = e^{rx}$ would be complex, setting $r = a + bi$ for simplicity.

Assume $\psi(x) = e^{(a+bi)x}$.

The characteristic equation is then
$$-\frac{\hbar^2}{2m}(a+bi)^2 = E + iV$$
which leads to the equations
$$a^2 -b^2 = \frac{-2mE}{\hbar^2} \text{ for the real part}$$
$$2abi = -\frac{2miV}{\hbar^2} \text{ for the imaginary part}$$
Therefore, $$ab = -\frac{mV}{\hbar^2} \implies b = -\frac{mV}{a\hbar^2}$$
$$\implies a^2 - b^2 = a^2 - \frac{m^2V^2}{a^2\hbar^4} = -\frac{2mE}{\hbar^2}$$
$$\implies a^4 + \frac{2mE}{\hbar^2}a^2 - \frac{m^2V^2}{\hbar^4} = 0$$
Using the quadratic formula,
$$a^2 = \frac{-\frac{2mE}{\hbar^2}\pm \sqrt{\frac{4m^2E^2}{\hbar^4}+\frac{4m^2V^2}{\hbar^4}}}{2}$$
Since $a$ must be a real number, the solution with the subtraction will not work, so we have
$$a^2 = -\frac{mE}{\hbar^2} + \frac{m}{\hbar^2}\sqrt{E^2+V^2} = \frac{m}{\hbar^2}(\sqrt{E^2+V^2}-E)$$
Now, taking the square root, I assume that $a$ will be negative or else $\psi$ will be non-normalizable and will blow up, so we have
$$a = -\frac{\sqrt{m}}{\hbar}\sqrt{\sqrt{E^2+V^2}-E}$$
Substituting this value for $a$ into the formula for $b$, we obtain
$$b = \frac{\sqrt{m}V}{\hbar\sqrt{\sqrt{E^2+V^2}-E}}$$

I do not know how to proceed from here on out. Clearly the $\frac{\sqrt{m}}{\hbar}$ term can be factored out in the exponential, but I cannot see how Taylor approximation for $e^{(a+bi)x}$ would lead to vanishing terms.

In my other attempt, I tried a solution $\psi(x) = f(x)\phi(x)$ for some function $f(x)$, where $\phi$ is just the wavefunction for a free particle with definite energy E. However, this was not enlightening.

EDIT: I posted this in the "Advanced Physics Homework" section because I did not feel it was appropriate for the introductory homework section. This is from an upper-division undergraduate QM class.

 

[mfb]

You might have neglected some higher order terms already by using the time-independent Schroedinger equation. You can neglect more to simplify the expression for a and b.

What is $\sqrt{E^2+V^2}$ approximately for V<<E?