How is the First-Order Perturbation Energy Related to Magnetic Susceptibility?

[schattenjaeger]

Not really a problem, I'm reading these 10 pages from a solid-state book for a professor, and it's a little over my head. Excuse the lack of LATEX but I copied it from another board I posted, and if you can help you likely know the equations involved anyways. I'm not expected to understand it ALL of course(I've had undergrad QM), but I'd like to at least have someone explain this first part to me

...example when talking about magnetic susceptibility of solids where each atom or ion has a closed shell of electrons with high excitation energy, you get diamagnetism with the susceptibility given by chi=-Ze^2*N/(6mc^2)r^2 Z is the number of electrons in the atom, N is number of atoms per unit volume, r^2 is mean square radious of electron charge cloud

that's all good and well, but the mention that the derivation comes from the "first-order perturbation energy <0|e^2/2mc^2*A^2|0> where the 0s denote ground states and A is the magnetic vector potential

now from what little I know of pertubation theory, to get the first order energy correction you do <0|H'|0> where H' is the perturbed Hamiltonian and |0> is the unpertubed ground states. So the equation in the derivation in the preceding paragraph is vaguely of that form, but I don't get how e^2/2mc^2 * A^2 is the perturbed Hamiltonian

 

[Gokul43201]

Please cite the source - which text is this from?

Write the Hamiltonian for a charge in a vector potential (A), where the conjugate momentum is now p' = p-(e/c)A. All the additional terms that arise from the vector potential make up the perturbation.

 

[schattenjaeger]

Text is by Ziemann

Ok, thanks for that, but I don't see how even knowing the first order perturbation energy gets you that expression for susceptibility