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Commutator of [x^2,p^2]

  1. Sep 16, 2010 #1
    I am attempting to calculate the commutator [itex][\hat{X}^2,\hat{P}^2][/itex] where [itex]\hat{X}[/itex] is position and [itex]\hat{P}[/itex] is momentum and am running into the following problem. The calculation goes as follows,

    [tex]
    [\hat{X}^2,\hat{P}^2]=-\left(\underbrace{[\hat{P}^2,\hat{X}]}_{-2i\hbar\hat{P}}\hat{X}+\hat{X}\underbrace{[\hat{P}^2,\hat{X}]}_{-2i\hbar\hat{P}}\right)=2i\hbar\left(\hat{P}\hat{X}+\hat{X}\hat{P}\right)
    [/tex]

    and using that [itex][\hat{X},\hat{P}]=i\hbar[/itex] we find that

    [tex]
    [\hat{X}^2,\hat{P}^2]=2i\hbar\left[\left(\hat{X}\hat{P}-i\hbar\right)+\hat{X}\hat{P}\right]=4i\hbar\hat{X}\hat{P}+2\hbar^2
    [/tex]

    which is wrong because I know from a theorem that if [itex]\hat{A}[/itex] is Hermitian and [itex]\hat{B}[/itex] is Hermitian then [itex][\hat{A},\hat{B}][/itex] is anti-Hermitian, which is definitely not the case here. What am I doing wrong?

    Thanks in advance for any help.
     
  2. jcsd
  3. Sep 16, 2010 #2

    strangerep

    User Avatar
    Science Advisor

    Maybe, in the last term, when checking whether it's anti-Hermitian, are you
    forgetting to swap P and X ?
     
  4. Sep 16, 2010 #3
    Doesn't [itex]2\hbar^2[/itex] destroy the anti-hermicity, since

    [tex]\left(2\hbar^2\right)^{\dagger}\neq -2\hbar^2[/tex]
     
  5. Sep 16, 2010 #4

    strangerep

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    Science Advisor

    [tex]
    (4i\hbar XP + 2\hbar^2)^\dagger ~=~ -4i\hbar PX + 2\hbar^2
    ~=~ -4i\hbar(XP - i\hbar) + 2\hbar^2 ~=~ -4i\hbar XP - 4\hbar^2 + 2\hbar^2
    ~=~ -(4i\hbar XP + 2\hbar^2)
    [/tex]
     
  6. Sep 16, 2010 #5
    Use the rule:

    [tex]
    [\hat{A}^{2}, \hat{B}^{2}] = [\hat{A}^{2}, \hat{B}] \, \hat{B} + \hat{B} \, [\hat{A}^{2} \, \hat{B}]
    [/tex]

    and

    [tex]
    [\hat{A}^{2}, \hat{B}] = \hat{A} \, [\hat{A}, \hat{B}] + [\hat{A}, \hat{B}] \, \hat{A}
    [/tex]
     
  7. Sep 16, 2010 #6
    oh you are right i was forgetting to change the order of X and P. Thanks. Thanks for your comment as well Dickfore I will use that.
     
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