# I What happens to the eigenvalue if an operator acts on a bra?

1. Mar 23, 2016

### shedrick94

I'm going through a derivation and it shows: (dirac notation)

<una|VA-AV|unb>=(anb-ana)<una|V|unb>

V and A are operators that are hermition and commute with each other and ana and anb are the eigenvalues of the operator A. I imagine it is trivial and possibly doesn't even matter but why does the sign flip when the operator acts on the eigenfunctions.

i.e why is it not <una|VA-AV|unb>=(ana-anb)<una|V|unb>

Last edited: Mar 23, 2016
2. Mar 23, 2016

### blue_leaf77

If V and A commute, then $\langle u_{na}|VA-AV|u_{nb}\rangle=0$.

3. Mar 23, 2016

### shedrick94

Yeh that's part of the proof but that's not what I don't understand. Why do the operations flip the sign of the eigenvalues. I'd have thought <una|VA-AV|unb>=(ana-anb)<una|V|unb>

4. Mar 23, 2016

### blue_leaf77

No, the sign is not relevant. If you are referring to the minus sign in front of $a_{nb}$, that's because of the minus sign in front of $AV$ in $\langle u_{na}|VA-AV|u_{nb}\rangle=0$.

5. Mar 23, 2016

### shedrick94

I don't understand why the signs change at all.

6. Mar 23, 2016

### blue_leaf77

Ah I think I see what you mean, do you mean like this $\langle u_{na}|VA-AV|u_{nb}\rangle= \langle u_{na}|VA|u_{nb}\rangle -\langle u_{na}|AV|u_{nb}\rangle = (a_{nb}-a_{na})\langle u_{na}|V|u_{nb}\rangle$?
Well, I don't know where you get this equation but since V and A commute, you can write $AV-VA$ in place of $VA-AV$, and you will get $(a_{na}-a_{nb})$.

7. Mar 23, 2016

### shedrick94

That's exactly it, thank you :). I didn't realise that you could separate the operators out. It's part of the derivation for finding the energy changes in perturbation theory.

8. Mar 23, 2016

### blue_leaf77

I could because an inner product, $\langle u_1|O|u_2\rangle$, is a linear operation.