Proving Bloch's Theorem


by naele
Tags: bloch, proving, theorem
naele
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#1
Feb26-12, 02:06 PM
P: 202
One of the more common ways of showing that a Hamiltonian with periodic potential commutes with the translation operator is to write the following (like Ashcroft and Mermin p. 133)

[tex]
T(R)H(r)\psi(r)=H(r+R)\psi(r+R)=H(r)T(R)\psi(r)
[/tex]

I suspect this might be a dumb question, but what allows us to write [itex]T(R)H(r)\psi(r)=H(r+R)\psi(r+R)[/itex], that is why is the translation operator acting on both the Hamiltonian and the wave, and not just on the Hamiltonian?
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Colen
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May2-12, 03:14 AM
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I think its because the potential is periodic then the Hamiltonian is too: H(x)=H(x+a), you can then sub this in directly and the translation operator now just acts on psi
Hurkyl
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May2-12, 04:03 AM
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Quote Quote by naele View Post
I suspect this might be a dumb question, but what allows us to write [itex]T(R)H(r)\psi(r)=H(r+R)\psi(r+R)[/itex]
Because that is the definition of how the space translation operator acts on a ket.

It may help to write [itex]\theta(r) = H(r) \psi(r)[/itex]. [itex]\theta(r)[/itex] is a ket. What [itex]T(R) \theta(r)[/itex]....


It may help more to consider more traditional function notation for what I believe is being written:
[tex] (T(R) H \psi)(r) = (H \psi)(r + R).[/tex]


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