Undergrad Is the current density operator derived from fundamental considerations?

[Paul159]

Hello,

I found this article. In equation (1) the authors wrote that the current operator is given by : $- \frac{\delta H}{\delta A}$.
I just would like to know if this relation is a just definition or if it can be derived from more fundamentals considerations ?

Thanks !

 

[Fred Wright]

Quoting from "Many-body quantum theory in condensed matter physics" by Bruus and Flensberg:
"In analytical mechanics $\vec A$ enters through the Lagrangian: $L=\frac{1}{2}mv^2-V+q \vec v \cdot \vec A$ since by the Euler-Lagrange equations yields the Lorentz force. But $\vec p=\frac{\partial L}{\partial \vec v}=m\vec v + q\vec A$, and via a Legendre transform we get $H(r,p)=\vec p \cdot \vec v - L(r,v)=\frac{1}{2}mv^2+V=\frac{1}{2m}(\vec p - q\vec A)^2 +V$. Considering infinitesimal variation $\delta \vec A$ we get $\delta H = H(\vec A +\delta \vec A)-H(\vec A)=-q\vec v \cdot \delta \vec A$".

 

[Paul159]

Ok I see thanks !