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Commutation relation for Hermitian operators

  1. Nov 27, 2014 #1
    1. The problem statement, all variables and given/known data
    The Hermitian operators [itex]\hat{A},\hat{B},\hat{C}[/itex] satisfy the commutation relation[itex][\hat{A},\hat{B}]=c\hat{C}[/itex].
    Show that c is a purely imaginary number.

    3. The attempt at a solution

    I don't usually post questions without some attempt at an answer but I am at a loss here.
     
  2. jcsd
  3. Nov 27, 2014 #2

    DrClaude

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    Staff: Mentor

    What happens if you take the Hermitian conjugate of that equality?

    P.-S.: Don't forget that the homework template has three sections. You should've written down the relevant equations.
     
  4. Nov 27, 2014 #3
    [itex][\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}][/itex]
    How does this help?
     
  5. Nov 27, 2014 #4

    DrClaude

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    You forgot the right-hand-side of the original equality.
     
  6. Nov 27, 2014 #5
    [itex][\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}[/itex]
     
  7. Nov 27, 2014 #6

    DrClaude

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    Staff: Mentor

    Continue...
     
  8. Nov 27, 2014 #7
    [itex]
    [\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}[/itex]
     
  9. Nov 27, 2014 #8

    DrClaude

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    You're almost there. To help you, I'll rewrite what you have there so that it may be more obvious what the last steps are.
    $$
    \begin{align*}
    [\hat{A},\hat{B}]^{\dagger} &=(c\hat{C})^{\dagger} \\
    [\hat{B},\hat{A}] &=c^{*}\hat{C}^{\dagger} \\
    -[\hat{A},\hat{B}] &=c^{*}\hat{C}^{\dagger}
    \end{align*}
    $$
     
  10. Nov 27, 2014 #9
    Im still having some trouble seeing where to go from here
     
  11. Nov 27, 2014 #10

    DrClaude

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    What properties does ##\hat{C}## have? Can you replace the left-hand-side by something else?
     
  12. Nov 27, 2014 #11
    Ahh ok I think Im there [itex]-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}=c^*\hat{C}[/itex] (since C is hermitian)
    So we now have [itex]-[\hat{A},\hat{B}]=-c\hat{C}=c^*\hat{C}[/itex]
    [tex]\Longrightarrow \frac{c^*}{c}=-1[/tex]
    Only pure imaginary numbers satisfy this last condition.
     
  13. Nov 28, 2014 #12

    DrClaude

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    Staff: Mentor

    You got it!
     
  14. Nov 28, 2014 #13
    thanks for your patience and for not giving it away too easily
     
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