Question about Fermi's gas effective mass.

  • #1
Hi everybody; today I was reading some problems about a metal and the electrons of the conduction band; the man who solved them used the mass of the electron as effective mass (m*).
I don't know why he did that; I have investigated but I don't have fount an explanation.
Can somebody please point me in the right direction to a webpage or book where I can find that?

Thanks for reading.
 

Answers and Replies

  • #2
ZapperZ
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Hi everybody; today I was reading some problems about a metal and the electrons of the conduction band; the man who solved them used the mass of the electron as effective mass (m*).
I don't know why he did that; I have investigated but I don't have fount an explanation.
Can somebody please point me in the right direction to a webpage or book where I can find that?

Thanks for reading.
I've posted in another thread an explanation that you might be able to use:

https://www.physicsforums.com/threads/quasiparticles-of-quantum-mechanics.80512/#post-619236

The whole concept here is to make the problem solvable. Electrons in solids are in a many-body situation. This is often unsolvable as far as understanding their dynamics, etc. So Landau came up with his Fermi Liquid theory in which, under a weak-coupling limit, the many-body problem can be transformed into a many one-body problem. The latter we know how to solve. But a consequence of that is the "renormalization" of the bare particle mass. The bare particle is now a "quasiparticle", having some effective mass that can be derived from the dispersion relation.

So the "electrons" in solids are not the bare electrons that you know and love. Rather, they are electrons "dressed" in a many-body interaction.

Zz.
 
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  • #3
Thank you very much for your anwser. I get it now.
 

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