I started by taking the conjugate transpose of both sides and end up with:
$$(AB – BA)^t = (AB)^t – (BA)^t = A^t B^t – B^t A^t = – ([B,A])^t$$
and this equals $$-iC^t$$ where I'm using $\ {}^t \ $ to symbolize conjugate transpose.
Is this accurate? From here, I'm not really sure where to go to prove $C$ is hermitian.
I am in an introductory Quantum class and a lot of this math is going over my head (I've never really dealt with matrices until now), so detailed explanations would be very helpful. Thanks!
Best Answer
I assume that $A$ and $B$ are hermitian:
$A^\dagger = A, \tag 1$
$B^\dagger = B; \tag 2$
then
$[A, B]^\dagger = (AB - BA)^\dagger = (AB)^\dagger - (BA)^\dagger$ $= B^\dagger A^\dagger - A^\dagger B^\dagger = BA - AB = -[A, B]; \tag 3$
that is, $[A, B]$ is skew-hermitian; thus if we set
$iC^\dagger = -[A, B]^\dagger = [A, B], \tag 4$
then $iC^\dagger$ is also skew. We now observe that an operator $D$ is skew if and only if $iD$ is hermitian, since
$D^\dagger = - D \Longrightarrow (iD)^\dagger =-iD^\dagger = iD, \tag 5$
and
$(iD)^\dagger = iD \Longrightarrow -iD^\dagger = iD \Longrightarrow D^\dagger = -D; \tag 6$
in the light of (5)-(6) we conclude that $C^\dagger$ is hermitian, since
$i(iC^\dagger) = -C^\dagger, \tag 7$
and $-C^\dagger$ is hermitian if and only if the same holds for $C^\dagger$, and hence for $C$ since
$C = (C^\dagger)^\dagger = C^\dagger. \tag 8$