Commutation relation for Hermitian operators

[jimmycricket]

Homework Statement

The Hermitian operators $\hat{A},\hat{B},\hat{C}$ satisfy the commutation relation$[\hat{A},\hat{B}]=c\hat{C}$.
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.

 

[DrClaude]

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.

 

[jimmycricket]

$[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]$
How does this help?

 

[DrClaude]

$[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]$
How does this help?

You forgot the right-hand-side of the original equality.

 

[jimmycricket]

$[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}$

 

[DrClaude]

$[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}$

Continue...

 

[jimmycricket]

$[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}$

 

[DrClaude]

$[\hat{A},\hat{B}]^{\dagger}=[\hat{B},\hat{A}]=-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}$

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.
$$
\begin{align*}
[\hat{A},\hat{B}]^{\dagger} &=(c\hat{C})^{\dagger} \\
[\hat{B},\hat{A}] &=c^{*}\hat{C}^{\dagger} \\
-[\hat{A},\hat{B}] &=c^{*}\hat{C}^{\dagger}
\end{align*}
$$

 

[jimmycricket]

Im still having some trouble seeing where to go from here

 

[DrClaude]

What properties does $\hat{C}$ have? Can you replace the left-hand-side by something else?

 

[jimmycricket]

Ahh ok I think I am there $-[\hat{A},\hat{B}]=(c\hat{C})^{\dagger}=c^{*}\hat{C}^{\dagger}=c^*\hat{C}$ (since C is hermitian)
So we now have $-[\hat{A},\hat{B}]=-c\hat{C}=c^*\hat{C}$
$$\Longrightarrow \frac{c^*}{c}=-1$$
Only pure imaginary numbers satisfy this last condition.

 

[DrClaude]

You got it!

 

[jimmycricket]

thanks for your patience and for not giving it away too easily