#1
May1003, 06:18 PM

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 


#2
May1203, 12:29 PM

P: n/a

is the periodic hamiltonian a harmonic oscillator function?



#3
May1203, 01:15 PM

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 



#4
May1403, 03:50 PM

Emeritus
Sci Advisor
PF Gold
P: 5,540

Commuation RelationMethinks you are supposed to be finding an explicit expression for what exactly those eigenstates are. In this case, θ>=Σe^{inθ}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. 261263. 


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