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Commuation Relation

  1. May 10, 2003 #1
    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
     
    Last edited by a moderator: May 10, 2003
  2. jcsd
  3. May 12, 2003 #2
    is the periodic hamiltonian a harmonic oscillator function?
     
  4. May 12, 2003 #3
    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
     
  5. May 14, 2003 #4

    Tom Mattson

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    Gold Member

    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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