Commutation relation for Hermitian operators

  • #1
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Homework Statement


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.

The Attempt at a Solution



I don't usually post questions without some attempt at an answer but I am at a loss here.
 

Answers and Replies

  • #2
DrClaude
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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.
 
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  • #3
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[itex][\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}][/itex]
How does this help?
 
  • #4
DrClaude
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[itex][\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}][/itex]
How does this help?
You forgot the right-hand-side of the original equality.
 
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  • #5
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[itex][\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}[/itex]
 
  • #6
DrClaude
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[itex][\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}[/itex]
Continue...
 
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  • #7
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[itex]
[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}[/itex]
 
  • #8
DrClaude
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[itex]
[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}[/itex]
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.
$$
\begin{align*}
[\hat{A},\hat{B}]^{\dagger} &=(c\hat{C})^{\dagger} \\
[\hat{B},\hat{A}] &=c^{*}\hat{C}^{\dagger} \\
-[\hat{A},\hat{B}] &=c^{*}\hat{C}^{\dagger}
\end{align*}
$$
 
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  • #9
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Im still having some trouble seeing where to go from here
 
  • #10
DrClaude
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What properties does ##\hat{C}## have? Can you replace the left-hand-side by something else?
 
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  • #11
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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.
 
  • #12
DrClaude
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You got it!
 
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  • #13
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thanks for your patience and for not giving it away too easily
 

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