Commuation Relation


by eNtRopY
Tags: commuation, relation
eNtRopY
#1
May10-03, 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
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jonnylane
#2
May12-03, 12:29 PM
P: n/a
is the periodic hamiltonian a harmonic oscillator function?
eNtRopY
#3
May12-03, 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

Tom Mattson
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#4
May14-03, 03:50 PM
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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, |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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