# Quick Matrix Element Question using Hermitian Operator

1. Mar 22, 2013

### starryskiesx

Hi there,

This should be very simple...

If I have a state <1|AB|2> where A and B are Hermitian operators, can I rewrite this as <2|BA|1> ?

That would be, taking the complex conjugate of the matrix element and saying that A*=A and B*=B.

Thank you!

2. Mar 22, 2013

### The_Duck

Well, you need to remember to include the complex conjugation:

$\langle 1 | A B | 2 \rangle^* = \langle 2 | B^\dagger A^\dagger | 1 \rangle = \langle 2 | B A | 1 \rangle$

so $\langle 1 | A B | 2 \rangle = \langle 2 | B A | 1 \rangle^*$

3. Mar 22, 2013

### starryskiesx

Great, thank you for the help

4. Mar 22, 2013

### raymo39

not necessarily, you need to also consider that this is only true if A and B are compatible observables, and it is only a matrix element if |1> and |2> are basis vectors of those same compatible observables.

5. Mar 22, 2013

### starryskiesx

So it appears I have a bigger problem!

|1> and |2> are eigenstates of the hamiltonian, my system is that of a harmonic oscillator. What I'm trying to prove is that <1|P|2> = -imw<1|X|2>, starting with the matrix element [P,H] where H is the hamiltonian. I thought I could do this just by swapping eigenstates so now I'm more stuck :)

So far I've tried two methods, one involving writing H = T + V for the harmonic oscillator and finding the commutation relation and the other working with H|1> = E1|1> etc.

Any suggestions?

6. Mar 22, 2013

### Fredrik

Staff Emeritus
What The Duck did holds in general, for arbitrary state vectors |1> and |2>.

7. Mar 22, 2013

### raymo39

I agree, my post was in reply to the original poster. Guess i was late to the party