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A question regarding a Hamiltonian.

  1. Oct 13, 2011 #1

    MathematicalPhysicist

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  2. jcsd
  3. Oct 13, 2011 #2

    dextercioby

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    In the phase space (x,p), consider the transformations/rotations

    [tex] 2p'=p+x [/tex]

    [tex] 2x'=p-x [/tex]

    What is H(x(x',p'),p(x',p')) equal to ?
     
  4. Oct 13, 2011 #3

    MathematicalPhysicist

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    But aren't x and p not commutable? ([tex][x,p]=i\hbar[/tex]).

    I mean [tex]p'^2-x'^2=1/4 (p^2+x^2+px+xp- p^2-x^2 +px+xp)=1/2 \{x,p\}[/tex]
     
  5. Oct 13, 2011 #4

    Bill_K

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    When he writes H = xp, he means a classical Hamiltonian. In QM it's not Hermitian until you symmetrize it.

    "... its quantum counterpart (obtained by symmetrization)..."
     
  6. Oct 13, 2011 #5

    MathematicalPhysicist

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    Ok, thanks.
    So in QM we would take (xp+px)/2.
     
  7. Oct 13, 2011 #6

    dextercioby

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    Actually [itex]\frac{1}{2}\left(\bar{\displaystyle{\hat{x}\hat{p}+\hat{p}\hat{x}}}\right) [/itex] (the bar should extend on both terms in the bracket), but for practical purposes, the operator without the bar is enough.
     
    Last edited: Oct 13, 2011
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