Square of a hermitian operator in matrix form

[Arijun]

If we have a hermitian operator Q and we know it's matrix representation [Q], does that mean that [Q²] = [Q]²?
For example, I'm pretty sure that's the case for p² for a harmonic oscillator. We have p=ic(a₊-a_-) and so
p²=c²(a₊-a_-)(-a₊+a_-)*=c²(a₊-a_-)(a₊-a_-)=p p
Which tells us that [p²]=[p]². But could we do that with any hermitian operator?

 

[George Jones]

Yes. In general, if $C = AB$, then $\left[ C \right] = \left[ AB \right] = \left[ A \right] \left[ B \right]$ for any $A$ and $B$ (not just for hermitian operators).

 

[Arijun]

But if it's not hermitian, then the operator may not commute with its Hermitian transpose (right?), so how would we know whether [C²]=[C] [C]* or [C]* [C], and generally wouldn't there be some commutator relationship in there?

 

[Fredrik]

C² means CC (the composition of C with C), not C*C or CC*.

The result [A_°B]=[A] follows immediately from the definition of matrix multiplication. Actually, I think matrix multiplication is defined the way it is precisely to ensure that this result holds.

$$
\begin{align}
[A\circ B]_{ij} &=((A\circ B)e_j)_i=(A(Be_j))_i =(A(Be_j)_k e_k)_i =(A_{kj} e_k)_i\\
&=(Ae_k)_i _{kj}=[A]_{ij}_{kj} = ([A])_{ij}
\end{align}
$$

If you're having difficulties following this, read the post I linked to in the quote below. Ask if it's still not clear.

The relationship between linear operators and matrices is explained e.g. in post #3 in this thread. (Ignore the quote and the stuff below it).