# Proof help - quantum operators

1. Sep 22, 2009

### burningbend

If A and B are hermitian operators, then AB is hermitian only if the commutator=0.

basically i need to prove that, but i don't really know where to start ofther than the general <f|AB|g> = <g|AB|f>*

obv physics math is not my strong point. thanks :)

2. Sep 22, 2009

### burningbend

i understand the proof of a single operator and the expansion of the relation to integral notation, but i don't understand the significance of the multiplication of the operators and why there is the condition there.

3. Sep 22, 2009

### burningbend

how's this?

<f|AB|g>=<g|AB|f>*

left=($$\sum$$<g|A|m><m|B|f>)*
left=$$\sum$$<m|A*|g><f|B*|m>
left=$$\sum$$<f|B*|m><m|A*|g>
left=<f|B*A*|g>
since B and A are Hermitian,
<f|AB|g> = <f|BA|g>
so AB-BA=0

good enough?

Last edited: Sep 22, 2009
4. Sep 23, 2009

### jambaugh

Begin with the definition of "Hermitian operator" i.e. it is equal to its adjoint, then apply it to AB and see what is necessary to make it Hermitian.

I don't think you need to invoke the "bra", "ket" products.

5. Sep 23, 2009

### burningbend

so,

AB=(AB)*
AB=B*A*
B and A are Hermitian, so,
AB=BA
AB-BA=0

and that's it?

i haven't had linear algebra (lol) so the (AB)*=B*A* is a little confusing to me, which i kinda figured out based on using the brackets and the completeness relation. so this would be it?

6. Sep 23, 2009

### jambaugh

That's all there is to it!