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I Simultaneous eigenfunctions

  1. Mar 29, 2017 #1

    dyn

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    If time evolution of a general ket is given by | Ψ > = e-iHt/ħ | Ψ (0) > where H is the Hamiltonian. If i have a eigenbasis consisting of 2 bases |a> and |b> of a general Hermitian operator A and i write e-iHt/ ħ |a> = e-iEat/ ħ |a> and e-iHt/ħ |b> = e-iEbt/ ħ |b> ; does this mean that operator A and the Hamiltonian share a set of eigenfunctions ? ie they commute ?

    And how would the time evolution of a ket be written if its operator did not commute with the Hamiltonian ?
     
    Last edited: Mar 29, 2017
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  3. Mar 29, 2017 #2

    blue_leaf77

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    If ##E_a \neq E_b## then ##A## and ##H## share the same eigenvectors. If on the other hand ##E_a = E_b##, any linear combination of ##|a\rangle## and ##|b\rangle## is an eigenvector of ##H## but obviously they cannot be an eigenvector of ##A## as well since ##a\neq b##.
    Expand the eigenvector of ##A## into the eigenvectors of ##H##.
     
  4. Mar 30, 2017 #3

    dyn

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    If A is a general operator and Ea ≠ Eb then why do A and H share the same eigenvectors ?
     
  5. Mar 30, 2017 #4

    blue_leaf77

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    If ##[A,H]=0## and the spectrum of ##H## is non-degenerate, then for any pair of eigenvectors ##|a\rangle## and ##|b\rangle## of ##H##
    $$
    \langle a|[A,H]|b\rangle = 0 \\
    (E_a-E_b) \langle a|A|b\rangle = 0
    $$
    This means when ##|a\rangle \neq |b\rangle## thus ##E_a -E_b \neq 0##, the scalar ##\langle a|A|b\rangle = 0##. That is, the matrix of ##A## is diagonal with respect to the basis of the eigenstates of ##H## and the two operators have the same set of eigenvectors.
     
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