1

I often see two definition of current in the book and literature, and I am a little bit confused.

  1. The current density

    $$\textbf{J}_1(\textbf{r})=\frac{-ie\hbar}{2m_e}\sum\limits_{n\textbf{k}}\{\psi^*_{n\textbf{k}}(\textbf{r})\nabla\psi_{n\textbf{k}}(\textbf{r})-[\nabla\psi^*_{n\textbf{k}}(\textbf{r})]\psi_{n\textbf{k}}(\textbf{r})\}$$

  2. The current

    $$\textbf{J}_2=e\sum\limits_{n\textbf{k}}\langle \psi_{n\textbf{k}}|\frac{1}{m_e}\hat{\textbf{p}}|\psi_{n\textbf{k}}\rangle , \quad \text{where} \quad \hat{\textbf{p}}=-i\hbar\nabla . $$

What's the difference between these two expressions? Could $\int\textbf{J}_1(\textbf{r})d\textbf{r}$ lead to $\textbf{J}_2$?

Plus: $\textbf{J}_2$ could be written as $\textbf{J}_2=-\frac{ie\hbar}{m_e}\sum\limits_{n\textbf{k}}\int\psi^*_{n\textbf{k}}(\textbf{r})\nabla\psi_{n\textbf{k}}(\textbf{r}) d\textbf{r}$

Memories
  • 127

2 Answers2

0

The first expression is the correct expression namely the Noether current. It guarantees a real valued current. In many cases, such as this one, the two terms are equal (or opposite if you don't count the minus sign) and then the second expression is also correct.

my2cts
  • 27,443
0

I think $\nabla$ operator is an anti-symmetric operator, i.e.,

$\nabla^\dagger=-\nabla$

because of the fact that $-i\hbar\nabla=(-i\hbar\nabla)^\dagger=i\hbar\nabla^\dagger$

where $\nabla^\dagger$ means it act on the wavefunction to the left-side of it.

so

$\frac{-i\hbar e}{2m_e}\int d\textbf{r} [\psi^*_{n\textbf{k}}\nabla\psi_{n\textbf{k}}-(\nabla\psi^*_{n\textbf{k}})\psi_{n\textbf{k}}]=\frac{-i\hbar e}{2m_e}\langle \psi_{n\textbf{k}}|\nabla-\nabla^\dagger|\psi_{n\textbf{k}}\rangle=\frac{-i\hbar e}{2m_e}\langle \psi_{n\textbf{k}}|2\nabla|\psi_{n\textbf{k}}\rangle=\frac{e}{m_e}\langle \psi_{n\textbf{k}}|\hat{\textbf{p}}|\psi_{n\textbf{k}}\rangle$

I feel this answe is not perfect, maybe you could make it perfect

Memories
  • 127