# Commuation Relation

by eNtRopY
Tags: commuation, relation
 P: n/a 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
 P: n/a is the periodic hamiltonian a harmonic oscillator function?
 P: n/a 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
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PF Gold
P: 5,532
Commuation Relation

 Originally posted by eNtRopY Prove that if a periodic Hamiltonian operator commutes with the translation operator then these two operators can have simultaneous eigenstates.
Are you sure that's the problem? I ask because any two operators that commute can have simultaneous eigenstates; the "periodic" condition has nothing to do with it.

Methinks you are supposed to be finding an explicit expression for what exactly those eigenstates are.

In this case, |&theta;>=&Sigma;ein&theta;|n> (n goes from -[oo] to +[oo]) is an eigenstate of &tau;(a) with eigenvalue e-i&theta;.

This is from Sakurai, Modern Quantum Mechanics pp. 261-263.

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