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Commuation Relation |
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| May10-03, 06:18 PM | #1 |
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Commuation Relation
Prove that if a periodic Hamiltonian operator commutes with the translation operator then these two operators can have simultaneous eigenstates.
{{ H(x) = H(x+a) } & { T(a)f(x) = f(x+a) } & { [T(a),H(x)] = 0 } } --> {{[uni][psi] [subset] complex functions, there exists [psi] [subset] complex functions such that { H(x)[psi](x) = E[psi](x) } & { T(a)[psi](x) = c(a)[psi](x) } } Thanks dudes. eNtRopY |
| May12-03, 12:29 PM | #2 |
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is the periodic hamiltonian a harmonic oscillator function?
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| May12-03, 01:15 PM | #3 |
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No it's independent of time. I mean that it is periodic in space... like a Bloch function.
V(x+a) = V(x) Actually, I believe I figured out the answer, but if you guys want to post your answer that's cool too. eNtRopY |
| May14-03, 03:50 PM | #4 |
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Commuation RelationMethinks you are supposed to be finding an explicit expression for what exactly those eigenstates are. In this case, |θ>=Σeinθ|n> (n goes from -[oo] to +[oo]) is an eigenstate of τ(a) with eigenvalue e-iθ. This is from Sakurai, Modern Quantum Mechanics pp. 261-263. |
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