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Invariance of energy under change of origin

  1. Nov 28, 2012 #1
    Suppose you have a particle in one dimension in an energy eigenstate, i.e. Hψ(x)=Eψ(x) for some E. For an observer B in a coordinate frame with the origin translated some distance K to the right, the wavefunction of the particle looks like ψ'(x) = ψ(x+K).

    Surely, we expect the energy that B measures to be the same as you measure, so Hψ'(x) = Eψ'(x) or in other words Hψ(x+K) = Eψ(x+K). But how can we prove this?
     
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  3. Nov 28, 2012 #2

    Bill_K

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    <E> = ∫ψ*(x) H(x) ψ(x) dx = ∫ψ*(x') H(x') ψ(x') dx' where x' = x + k.
     
  4. Nov 28, 2012 #3
    If you have a potential, it will produce some non physical factor in psi, which gets killed off by it's conjugate when we turn psi into a probability density function
     
  5. Nov 28, 2012 #4
    I'm not sure I understand this notation. The hamiltonian H is an operator, i.e. it's a functional that takes a function f(x) to some other function f'(x). I can see how it can be dependent on time like H(t) - that just means that the way it takes f(x) to f'(x) changes with time. But what does H(x) mean?

    Also, even if you've proved the above result for <E>, that just says that the energy expectation values of ψ(x) and ψ'(x) are equal. Does that necessarily prove that ψ'(x) is an energy eigenvector with energy E?
     
  6. Nov 28, 2012 #5

    George Jones

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    I am too busy to do the calculation, but I have a few thoughts.

    Doesn't the Hamiltonian also have to be "shifted"? For example, consider a harmonic oscillator.

    The chain rule might help.

    Your example is a particular case of something much more general.
     
  7. Nov 28, 2012 #6

    tom.stoer

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    usually

    [tex]H(x) = -\partial_x^2 + V(x)[/tex]

    If you translate x to x+k in the wave function, then you have two options

    1.)
    [tex]x \to x+k[/tex]
    [tex]dx \to dx[/tex]
    [tex]\partial_x \to \partial_x[/tex]
    [tex]\psi(x) \to \psi(x+k)[/tex]
    [tex]V(x) \to V(x+k)[/tex]
    Then obviously this is a symmetry of the system and nothing changes

    2.)
    [tex]\psi(x) \to \psi(x+k)[/tex]
    [tex]V(x) \to V(x)[/tex]
    Then usually this is not a symmetry of the system; it corresponds to
    [tex]\psi(x) \to \psi(x)[/tex]
    [tex]V(x) \to V(x-k)[/tex]
    and that means that you translate the oparticle w/o translating the potential, so the energy will change (only in the trivial case V(x) = V = const this is a symmetry)
     
  8. Nov 29, 2012 #7
    that is the nice way of doing it.
     
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