# Square of a hermitian operator in matrix form

1. May 23, 2012

### 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?

2. May 23, 2012

### George Jones

Staff Emeritus
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).

3. May 23, 2012

### 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?

4. May 23, 2012

### Fredrik

Staff Emeritus
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.