# Derivation of current density operator using Heisenberg equation of motion [closed]

Electronic (probability) current density is calculated by continuity equation: $$\nabla\cdot \vec J=-\partial_t \rho \tag{1}$$ here $$\vec J$$ is current density operator and $$\rho=e \psi_r^+\psi_r$$ is density. One way to find $$\vec J$$ is to explicitly calculate time derivative of $$\rho$$ and use Schrodinger equation. This method is explained here. This method is easy to understand for me.

I am trying to calculate $$\vec J$$ using Heisenberg equation of motion that is $$\nabla\cdot \vec J = -\frac{i}{\hbar}[H_0,e\psi_{r'}^+\psi_{r'}]\tag{2}$$ I am stuck at calculation of this commutation relation. I take $$H_0$$ as Hamiltonian of free particle: $$H_0 = \frac{\hbar^2}{2m} \int d\vec r \nabla \psi_{r}^+ \nabla \psi_{r} \tag{3}$$ So: $$[H_0,e\psi_{r'}^+\psi_{r'}]= [H_0,\psi_{r'}^+]\psi_{r'}+\psi_{r'}^+[H_0,\psi_{r'}]\tag{4}$$ And $$[H_0,\psi_{r'}]=\frac{\hbar^2}{2m} \int d\vec r [\nabla \psi_{r}^+ \nabla \psi_{r},\psi_{r'}]= \frac{\hbar^2}{2m} \int d\vec r \nabla \psi_{r}^+[ \nabla \psi_{r},\psi_{r'}] +\frac{\hbar^2}{2m} \int d\vec r [\nabla \psi_{r}^+ ,\psi_{r'}]\nabla \psi_{r} \tag{5}$$ how to proceed? how to calculate $$[\nabla \psi_{r}^+ ,\psi_{r'}]$$?

Edit 1:

After a comment, i found that $$[\nabla \psi_{r}^+ ,\psi_{r'}]=\nabla[ \psi_{r}^+ ,\psi_{r'}]$$. After using commutation rules $$[ \psi_{r} ,\psi_{r'}^+]=\delta(r-r')$$, I get: $$\nabla\cdot \vec J = \frac{-i\hbar e}{2m}\int d\vec r \{\psi_r^+ (\nabla_r[\psi_r^+,\psi_{r'}])\nabla \psi_r +\nabla \psi_r^+ (\nabla[\psi_{r},\psi_{r'}^+])\psi_r\}\\ = \frac{-i\hbar e}{2m}\int d\vec r \{\psi_r^+ \nabla_r\delta(r-r')\nabla \psi_r +\nabla \psi_r^+ \nabla\delta(r-r')\psi_r\}\\ = \frac{-i\hbar e}{2m} \{\psi_r^+ \nabla^2 \psi_r +\nabla \psi_r^+ (\nabla)\psi_r\}$$ The very term $$\nabla\psi_r^+ (\nabla)\psi_r$$, it should be $$(\nabla^2\psi_r^+)\psi_r$$. I wonder how the extra $$\nabla$$ moves back?

• Hint $[\nabla_r \psi_r^\dagger, \psi_{r'}] = \nabla_r [\psi_r^\dagger,\psi_{r'}]$ where $\nabla_r$ is the derivative with respect to the $r$ coordinates (and not the $r'$s) Sep 1 at 14:06
• Possible duplicate: physics.stackexchange.com/questions/294837/… Sep 1 at 14:23
• @BySymmetry thank you for quick reply. I have also solved it, except for one little confusion. I have written this is Edit1. Please have a look at it Sep 1 at 14:35

Only you are not doing the math right in your Edit 1: \begin{aligned} \nabla\cdot \vec J =& \frac{-i\hbar e}{2m}\int d\vec r \{\psi_r^+ (\nabla_r[\psi_r^+,\psi_{r'}])\nabla \psi_r +\nabla \psi_r^+ (\nabla[\psi_{r},\psi_{r'}^+])\psi_r\}\\ =& \frac{-i\hbar e}{2m}\int d\vec r \{ \color{red}{ - \psi_r^+ \nabla_r\delta(r-r')\nabla \psi_r} +\nabla \psi_r^+ \nabla\delta(r-r')\psi_r\}\\ =& \frac{-i\hbar e}{2m} \{\nabla_{r'} ( \psi_r^+ \nabla \psi_{r'}) - \nabla ( (\nabla \psi_r^+) \psi_r)\} \\ =& \cdots \end{aligned}