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Group symmetry of Hamiltonian

  1. Dec 17, 2013 #1
    ##H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2x^2##
    Parity
    ##Px=-x##
    end ##e## neutral are group of symmetry of Hamiltonian.
    ## PH=H##
    ##eH=H##
    so I said it is group of symmetry because don't change Hamiltonian? And ##e## and ##P## form a group under multiplication. Is there any way to right some representation of this group? Thx for the answer.
     
  2. jcsd
  3. Dec 17, 2013 #2

    ChrisVer

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    what is [itex]e[/itex]?
    Your representation depends on the space it acts. For example the parity when acting on the 4 Lorentz vectors in Minkowski space can be represented by diag[1,-1,-1,-1].
    On three dim space, it is still diag[-1,-1,-1] so I guess the parity for that is equivalent to [itex]Z_{2}[/itex] so that's why in the Harmonic oscillator you give, your Hamiltonian eigenstates to that of the parity, are the same and they are conserved (being either cosine or sinus )
    I am not sure, but I also somehow think it corresponds to the reflections (although it's not the group itself).

    So to write a representation, you have to define the vector space the representation acts on.
     
  4. Dec 17, 2013 #3
    ##e## is neutral.
    ##e^2=e##
    This iz ##Z_2## group.
    Hamiltonian eigenstates are Hermite functions.
    So could I say that ##H## is invariant under ##Z_2##?
     
  5. Dec 17, 2013 #4

    ChrisVer

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    the identity ok.
    Hermite funcs are funcs of fixed parity as well...
    I'm not sure if you can, personally I'd prefer saying parity...
     
  6. Dec 18, 2013 #5
    I'm just confused what I have now when I know that ##\{e,P\}## is symmetry group of Hamiltonian? I know that ##PH=HP=H##. So ##[P,H]=0##. Right?
     
  7. Dec 19, 2013 #6
    Note that PH (or HP) is not H! Your mistake is that you are acting P on (just) H, but the way you should interpret "PH" is as follows: for any function f, PH is defined as (PH)(f) := P(Hf). And note that for general functions f, P(Hf) won't be the same as Hf! Hence PH is not H.

    What is true, however, is indeed [P,H] = 0. Why? Well, again that statement is equivalent to "for any f, [P,H]f = 0", and this is true, because
    [itex]PH f(x) = P \left[ \left( \alpha \frac{\mathrm d^2}{\mathrm d x^2} + \beta x^2 \right) f(x)\right] = \left( \alpha \frac{\mathrm d^2}{\mathrm d (-x)^2} + \beta (-x)^2 \right) f(-x) = \left( \alpha \frac{\mathrm d^2}{\mathrm d x^2} + \beta x^2 \right) f(-x) = \left( \alpha \frac{\mathrm d^2}{\mathrm d x^2} + \beta x^2 \right) P f(x) = HP f(x) [/itex]

    The value in knowing [P,H] = 0, is that now you know the eigenstates of H are eigenstates of P. Why? Well suppose f is an eigenstate of H, i.e. Hf = Ef.
    Then the claim is [itex]Pf = \pm f[/itex]. To see this, note that [itex]E Pf = P E f = P Hf = HP f[/itex], but this means Pf is an eigenstate of H with eigenvalue E. If we assume there is only one for every energy, then Pf has to be proportional to f! (and it's clear any eigenvalue of P has to square to 1)
     
    Last edited: Dec 19, 2013
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