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I have a few doubts about the study of the free electron gas model:

  1. We know that the current density expression may be calculated as follows

J(r,t)=$\frac{\hbar}{2mi}(\psi^{*}\nabla\psi -\psi\nabla\psi^{*})$,

where $m$ is the mass of the particle, $t$ is the time, and $\psi^{*}$ the complex conjugate of the wave function.

If the wave function of the free electron gas model is

$\psi(\textbf{r})=\psi_{o}exp[i\textbf{k} .\textbf{r}]$,

where k is the wave-vector, r is the position vector, and $\psi_{o}$ is the wave's amplitude.

How could I calculate J(r,t) from $\psi(\textbf{r})$?

  1. For a real wave function, what would be the result for J(r,t)?

Thanks!

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You do not say what your "doubts" are.

If $\Psi(x)= \sqrt \rho e^{ikx}$ with $\rho=|\psi|$ a real number, you just plug into your formula for $J$ to get $$ J=\rho \hbar \frac k m. $$ Note that the derivatoves of $\rho$ cancel between the two terms in $J$. As $p= \hbar k$ is the momentum $mv$, this is $$ J= \rho v $$ as expected. If $\psi$ is real then $J=0$ everywhere.

What part of this do you find puzzling?

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  • $\begingroup$ Thanks for commenting. I have studied these concepts some years ago, and I was not so sure about how to proceed $\endgroup$ Feb 20 '21 at 13:28
  • $\begingroup$ In your answer, you have restricted your solution to the one-dimensional case. What would be the result in three dimensions? $\endgroup$ Feb 20 '21 at 13:40
  • $\begingroup$ Just the same:If $\psi({\bf x},t)= \sqrt \rho e^{i\theta({\bf x},t)}$ we have ${\bf J}= \rho {\bf v}$ with ${\bf v}= \nabla \theta/m$. $\endgroup$
    – mike stone
    Feb 20 '21 at 13:57
  • $\begingroup$ I see it. However, what is $\theta$ if my wave function is $\psi(\textbf{r})=\psi_{o}exp(i\textbf{k}\cdot\textbf{r})$? $\endgroup$ Feb 20 '21 at 14:08
  • $\begingroup$ $\theta= {\bf k}\cdot {\bf r}$. $\endgroup$
    – mike stone
    Feb 20 '21 at 14:08

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