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

  1. The current density


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

  • $\begingroup$ Hint: $|\psi_{nk}\rangle=\int \langle\mathbf{r}|\psi_{nk}\rangle |\mathbf{r}\rangle d\mathbf{r}=\int \psi_{nk}(\mathbf{r})|\mathbf{r}\rangle d\mathbf{r}$. $\endgroup$ Jan 24, 2019 at 14:19

2 Answers 2


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.

  • $\begingroup$ This doesn't make sense. First of all, setting the two terms in the first expression equal will just give you zero. Second of all, the momentum operator measures the momentum observable, and so its expectation value must be real. Since the second expression is just a scaled sum of expectation values, the second expression is also guaranteed to be real. $\endgroup$ Jan 24, 2019 at 18:47
  • $\begingroup$ I found this in Sakurai's book, Eq. (2.4.17). It shows $\int\textbf{J}_1(\textbf{r})d\textbf{r}$can lead to $\textbf{J}_2$, but without details $\endgroup$
    – Memories
    Jan 24, 2019 at 22:04
  • $\begingroup$ @probably_someone your first argument is based on incorrect reading. Your second argument is only true for momentum eigen functions. $\endgroup$
    – my2cts
    Jan 25, 2019 at 5:20
  • $\begingroup$ @my2cts The second argument is true in general. See e.g. physics.stackexchange.com/questions/221027/… $\endgroup$ Jan 25, 2019 at 12:41

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


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.


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

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


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