Current density operator

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  • Thread starter Paul159
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  • #1
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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 !
 

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  • #2
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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##".
 
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  • #3
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Ok I see thanks !
 
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