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Deduction of the action of this unitary on the wavefunction

  1. Jul 20, 2015 #1

    julian

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    I have operators satisfying ##[\hat{Z} , \hat{E}] = i \hbar##. The operators ##\hat{Z}## and ##\hat{E}## are taken to be Hermitian. You consider the unitary operator

    ##U_\lambda = \exp (i \lambda \hat{Z} / \hbar)##.

    I have proved that

    ##U_\lambda \hat{E} U_\lambda^\dagger = \hat{E} - \lambda \quad Eq.1##

    I know how to apply Eq.1 to any function of ##\hat{E}##, ##f (\hat{E})##, which may be Taylor expanded. Moreover consider the application of the transformation to some matrix elements

    ##\int_0^\infty \psi^* (E) f (\hat{E}) \phi (E) dE = \int_0^\infty \psi^* U^\dagger U f (\hat{E}) U^\dagger U \phi dE = \int \psi^* U^\dagger f (\hat{E} - \lambda) U \phi dE##

    I'm told that you can deduce from this that (and similarly for ##\psi^*##):

    ##U \phi (E) = \phi (E - \lambda)##.

    I not sure I understand this. What am I missing?

    Thanks.
     
    Last edited: Jul 20, 2015
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  3. Jul 20, 2015 #2

    blue_leaf77

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    The integration should run from ##-\infty## to ##\infty##. Then
    $$\int_{-\infty}^\infty \psi^* (E) f (E) \phi (E) dE = \int_{-\infty}^\infty \psi^* (E-\lambda) f (\hat{E-\lambda}) \phi (E-\lambda) dE$$
    Now compare this with ##\int_{-\infty}^\infty (U\psi)^\dagger f (E - \lambda) U \phi dE##
     
  4. Jul 21, 2015 #3

    julian

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    Actually ##E## is the Energy and is bounded from below, I've taken the lower bound to be ##E=0##, so that there are no energy-eigenstates for ##E < 0##. I probably should have mentioned that.

    Did a major edit here.
     
    Last edited: Jul 21, 2015
  5. Jul 21, 2015 #4

    blue_leaf77

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    Are you sure ##\hat{E}## is energy and should be bound to zero? What if ##E-\lambda## is negative, ##\phi(E-\lambda)## will then have no defined state?
    Actually ##U## operator is called the translation operator, you may remember that there is a pair of well-known operators which satisfy the same commutator relation as our problem here, which are the momentum and position operators. Besides how come you explicitly write that the eigenvalues of ##\hat{E}## is continuous? Perhaps it will help if you tell us in which context/discussion you encountered this problem.
     
    Last edited: Jul 21, 2015
  6. Jul 21, 2015 #5

    julian

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    It is energy - I'm looking into the problems that arise when you attempt defining a putative Hermitian operator conjugate to the Hamiltonian operator (so a putative time operator. I'm aware that time is not a dynamical quantity but I'm ignoring that for the moment). In non-relativistic problems the energy has a lower bound and we may as well take it be 0 as we are allowed to.

    By the way I'm not worrying too much about some of the mathematical subtleties right now - just trying to get a first handle on it.

    The contradiction I'm leading up to is applying ##U## to an energy eigenstate, ##\phi (E) = \delta (E-E_0)##, which only has an amplitude at some particular energy ##E_0##, with ##\lambda## is sufficiently big to displace ##E_0## below ##E=0## where, by assumption, there are no energy eigenstates, proving no Hermitian ##\hat{Z}## exists.

    In relativistic quantum mechanics there is the gap ##-mc^2 < E < mc^2##, so we can apply similar arguments there.
     
    Last edited: Jul 21, 2015
  7. Jul 21, 2015 #6

    julian

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    Come to think of it, my old prof's notes I'm looking at might have a mistake...when you do ##\phi(E) \mapsto \phi (E - \lambda)## what this does is moves the function to the right along the ##E-##axis by ##\lambda##.

    If I want to shift the energy eigenstate ##\phi (E) = \delta (E-E_0)## to the left along the ##E-##axis (creating an energy-eigenstate with negative energy and hence a contradiction) dont I need to do ##\phi (E) \mapsto \phi (E+\lambda)##?

    Obviously ##\phi (E+\lambda) = \delta (E+\lambda-E_0)## only has an amplitude at ##E = E_0 - \lambda##!
     
    Last edited: Jul 21, 2015
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