# Probability current

1. Oct 11, 2007

### T-7

1. The problem statement, all variables and given/known data

I have a wave function,

$$\psi(x)=Ae^{ik_{0}x-ax^{2}}$$

All I am required to do is calc. the prob. current. This I have done using the usual

$$j = \frac{\hbar}{m}.|A|^{2}.\frac{\partial \varphi}{\partial z}$$

where $$\varphi}$$ is the imaginary component of the exponent of e.

It comes (of course) to $$j = \frac{\hbar}{m}.|A|^{2}.k_{0}$$.

But, doing it the long way (I just thought I would), with

$$j = \frac{\hbar}{2mi}.(\psi*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi*}{\partial x})$$

I end up with the same result, but $$e^{-2ax^{2}}$$ attached. Since taking $$\psi*$$ only changes the sign of the imaginary component, this seems inevitable... anyone know what I'm doing wrong?

Cheers :-)

2. Oct 11, 2007

### dextercioby

Apparently A from the general expression you quoted

$$j=\frac{\hbar}{m}|A|^{2}\frac{\partial \phi}{\partial x}$$

includes the real exponential...

3. Oct 11, 2007

### nrqed

???? Where did you see this equation? It does not look right to me. are you sure this is not only for plane waves??

4. Oct 14, 2007

### T-7

I'm thinking the simpler equation can only apply where the exponent of e in the wave function is wholly imaginary.

Using the equation

$$j = \frac{\hbar}{2mi}\left( \psi^{*} \frac{\partial\psi}{\partial x} - \psi \frac{\partial\psi^{*}}{\partial x} \right)$$

I obtain

$$\frac{\hbar}{m}|A|^{2}e^{-2ax^{2}} k_{0}$$

I presume that's correct. In which case, I don't think the $$e^{-2ax^{2}}$$ should be absorbed into A. The probability current is a function of x.

Hmmm.

Last edited: Oct 14, 2007