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Commutator calculus

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data
    |phi (n)> being eigen states of hermitian operator H ( H could be for example the hamiltonian
    of anyone physical system ). The states |phi (n)> form an orthonormal discrete basis.
    The operator U(m,n) is defined by:

    U(m,n)= |phi(m)><phi(n)|
    Calculate the commutator:
    [H,U(m,n)]

    ( this is part of the first problem in Cohen, Tannoudji, Diu, Laloe textbook in quantum mechanics.)

    3. The attempt at a solution[itex]\[/itex]

    [HU-UH] (ψ) = H|phi(m)><phi (n)|ψ> - |phi (m)><phi(n)| H| ψ>

    = <phi(n)|ψ> H |phi (m>) - |phi (m><phi(n)| <ψ | H

    and then ? i did not find symbol phi.
     
  2. jcsd
  3. Feb 28, 2014 #2

    hilbert2

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    Note that the operator ##H## can be written as ##H=\sum\limits_{k}E_{k}\left|\phi(k)\right>\left<\phi(k)\right|##, where the ##E_{k}## are its eigenvalues. Also, note that the vectors ##\left|\phi(k)\right>## form an orthonormal set.
     
  4. Feb 28, 2014 #3

    ChrisVer

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    you should use an eigenstate of the Hamiltonian instead of [itex]\psi[/itex] (if you ask how you can do that, you can expand [itex]\psi[/itex] as a superposition of the Hamiltonian's eigenstates)
    Then in general, following correct paths you will reach the desired result/
     
  5. Mar 1, 2014 #4
    many thanks for help. I find a commutator value depending upon the system energy :

    commutator = E(m) U(m,n) if E= E(m) and - E(n) U(m,n) if E=E(n).

    in other cases it's zero.
    Is this correct ?
     
  6. Mar 1, 2014 #5

    hilbert2

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    I got something like [H,U(m,n)] = (E(m)-E(n))U(m,n). The commutator does not depend on what state the quantum system is in.
     
  7. Mar 1, 2014 #6

    vela

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    You can say that ##\hat{H}\lvert \phi_m \rangle \langle \phi_n \vert \psi \rangle = \langle \phi_n \vert \psi \rangle \hat{H}\lvert \phi_m \rangle## because ##\langle \phi_n \vert \psi \rangle## is a number, though it doesn't really help you in this case. What you can't do is say ##\lvert \phi_m \rangle \langle \phi_n \lvert \hat{H} \rvert \psi \rangle## equals ##\lvert \phi_m \rangle \langle \phi_n \lvert \langle \psi \rvert \hat{H}## because ##\hat{H}\lvert \psi \rangle## and ##\langle \psi \rvert \hat{H}## aren't the same. One's a bra; the other, a ket. You need to be a bit more precise with your notation, otherwise you're invariably going to make errors.

    You have, so far,
    \begin{align*}
    [\hat{H},\hat{U}] &= \hat{H}\hat{U} - \hat{U}\hat{H} \\
    &= \hat{H}\lvert \phi_m\rangle\langle\phi_n\rvert - \lvert \phi_m\rangle\langle\phi_n\rvert\hat{H}
    \end{align*} Now in the first term, apply ##\hat{H}## to the ket ##\lvert \phi_m \rangle##. What do you get? Similarly, in the second term, what do you get when ##\hat{H}## acts on the bra ##\langle \phi_n \rvert##?
     
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