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AspiringResearcher
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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.
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