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Homework Help: Hamiltonian and symmetries

  1. Jun 5, 2010 #1
    [QM] Hamiltonian and symmetries

    1. The problem statement, all variables and given/known data
    Let there be the hamiltonian:
    [tex]H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2(x^2+y^2+z^2)+kxyz+\frac{k^2}{\hbar \omega}x^2y^2z^2[/tex]
    Find the expectation value of the three components of [tex]\vec r[/tex] in the ground state using ONLY the symmetry properties of the hamiltonian.

    2. Relevant equations



    3. The attempt at a solution
    I define this parity:
    [tex]\Pi_{xy}: x\rightarrow-x\ \ \ \ y\rightarrow-y[/tex]
    Then the hamiltonian commutes with this parity: [tex][H, \Pi_{xy}]=0[/tex]
    The ground state is not degenerate, so it has a definite parity with respect to [tex]\Pi_{xy}[/tex]:
    [tex]<gs|x|gs>=<gs|\Pi_{xy}\Pi_{xy}x\Pi_{xy}\Pi_{xy}|gs>=-<gs|\Pi_{xy}x\Pi_{xy}|gs>=-<gs|x|gs>[/tex]
    So <gs|x|gs>=0;
    Is it right?
     
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 5, 2010 #2

    kuruman

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    Re: [QM] Hamiltonian and symmetries

    That is not the definition of the parity operator I am familiar with. What happened to
    [tex]z\rightarrow-z?[/tex]
    Only because you defined "this parity" so that it commutes.
     
    Last edited: Jun 5, 2010
  4. Jun 6, 2010 #3
    Yes, I know this is not the usual parity... but I don't really know any other way to solve this problem
     
  5. Jun 6, 2010 #4

    kuruman

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    As I mentioned earlier, your Hamiltonian is invariant under cyclic permutations. In other words, the system cannot distinguish x from y from z (it doesn't know the alphabet). What do you think that implies about the expectation values <x>, <y> and <z>?
     
  6. Jun 7, 2010 #5
    That they are all the same?
     
  7. Jun 7, 2010 #6

    kuruman

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    Correct. Since they are all the same, calculating one of them will give you the others. So what do you think that value can be and why?
     
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