# Derivation of the current density for the free electron gas

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!

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?

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