# Definition of the probability current

I know that the definition of probability current is given by $$$$J\sim \psi^*\frac{\partial \psi}{\partial x}-\psi \frac{\partial \psi^*}{\partial x}$$$$ However, some papers are using this definition:

$$$$J\sim \psi^*\frac{\partial H}{\partial k_x}\psi$$$$ where $$H$$ is the Hamiltonian of the system and $$k_x$$ is the wavevector. So, how can I understand that these two are equivalent, and is it possible to derive the second one, I mean where does it come from?

update

this question is also related to this one where I think we might have a close answer.

• The fisrt formula applies only to non-relativistic Hamiltonians. The Weyl equation is "relativistic" (although with a different speed of "light"). Commented Mar 7, 2021 at 15:38
• @mikestone, to my knowledge for the relativistic case it is given by $J\sim \psi^*\sigma_x \psi$, where $\psi$ is now a spinor and $\sigma_x$ is the Pauli matrix Commented Mar 7, 2021 at 17:57
• Is not $\partial H/\partial k_x=\sigma_x$ for the Weyl Hamiltonian? I Think that this is what your papers are saying... Commented Mar 7, 2021 at 18:01
• @mikestone, No, it is not as you can see from Eq. 15 in the paper, or do I miss somthing? Commented Mar 7, 2021 at 18:04
• If they were using the usual Weyl hamiltonian it would be $\sigma_x$. In general the current is given by the functional derivative of the Hamiltonian wrt to the gauge field $A_i$, but in a plane-wave Bloch state $A$ appears as $k+A$, so it's also writable as a derivaive wrt $k$. Commented Mar 7, 2021 at 18:11

Ok, I think I get it and would like to share in case any additional input. So in general the probability current can be written as

$$$$J=\frac{-ie\hbar}{2m} [\psi^*(\nabla\psi)-(\nabla\psi^*)\psi]$$$$ knowing that the momentum operator is $$\bf p=-i\hbar\nabla$$, we get

$$$$J=\frac{e}{2m} [\psi^*(\bf p\psi)+(\bf p\psi^*)\psi]=\frac{e}{m}Re[\psi^*(p\psi)]$$$$ Now, we can recall that velocity operator $$\bf{v}$$ can be written with respect to $$\bf p$$ as $${\bf{v}}=\bf{p}/m$$, yeilding

$$$$J=eRe[\psi^*(\bf v\psi)]$$$$

and since $$\bf v\sim\frac{\partial H}{\partial k}$$ we get the desired form of probability current

$$$$J\sim eRe[\psi^*(\bf \frac{\partial H}{\partial k}\psi)]$$$$