Calculating Probability Current for Wave Function

[T-7]

Homework Statement

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 :-)

 

[dextercioby]

Apparently A from the general expression you quoted

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

includes the real exponential...

 

[nrqed]

Homework Statement

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.

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

 

[T-7]

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

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.