I Is the current density operator derived from fundamental considerations?

Paul159
Messages
15
Reaction score
4
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 !
 
Physics news on Phys.org
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##".
 
  • Like
Likes Paul159 and dRic2
Ok I see thanks !
 
From the BCS theory of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly increase the temperature, we must observe a decrease in the actual supercurrent, because the density of electron pairs and total supercurrent momentum decrease. However, this supercurrent...
Back
Top