# Working with operators

1. Oct 15, 2011

### atomicpedals

So I'm doing some proofs on a Saturday night... working on proving that (AB+BA) is self-adjoint, that is (AB+BA)=(AB+BA)* (using a * instead dagger).

What I want to know is if the following is true:

(AB+BA)*=B*A*+A*B* ?

2. Oct 15, 2011

### capandbells

That's correct.

3. Oct 15, 2011

### atomicpedals

Can I express it as two integrals?

$\int\psi(AB)\psi*dx+\int\psi(BA)\psi*dx$

4. Oct 16, 2011

### dextercioby

I think what you're trying to prove doesn't work in the general case, for A and B unbounded, but only for both of them bounded or at least A. If one of these restraining conditions is met, then

$$(AB+BA)^{\dagger} = (AB)^{\dagger} + (BA)^{\dagger} = B^{\dagger} A^{\dagger} + A^{\dagger}B^{\dagger} = AB+BA$$

for A and B self-adjoint.

For the general case, you can only show that AB+BA is symmetric for A and B s-adj.

5. Oct 16, 2011

### atomicpedals

I think I have to assume that A and B are Hermitian. After thinking about it showing that (AB+BA) is also Hermitian should also effectively demonstrate they're self-adjoint. In the way I've been wanting to go about it (which you outlined perfectly), I don't think I have a mathematical justification to state that

$$B^{\dagger} A^{\dagger} + A^{\dagger}B^{\dagger} = AB+BA$$

6. Oct 16, 2011

### atomicpedals

Ok, does this follow at all?

$$(AB+BA)^{\dagger} = (AB)^{\dagger} + (BA)^{\dagger} = B^{\dagger} A^{\dagger} + A^{\dagger}B^{\dagger} = \int \psi * (AB)^{\dagger} \psi dr + \int \psi * (BA)^{\dagger} \psi dr = \int (B\psi) * A^{\dagger} \psi dr +\int (A\psi) * B^{\dagger} \psi dr = AB + BA$$

7. Oct 16, 2011

### vela

Staff Emeritus
There's no need to bring integrals into this. If an operator A is self-adjoint, you have $A=A^\dagger$. Use that fact to replace all the adjoints in $B^\dagger A^\dagger + A^\dagger B^\dagger$. What do you end up with?

8. Oct 16, 2011

### atomicpedals

I end up with (AB+BA), however I'm being pedagogically required to use the integral definition of an adjoint hence the invocation of integrals (mercifully it doesn't have to be an epsilon-delta style proof).

Last edited: Oct 16, 2011
9. Oct 16, 2011

### vela

Staff Emeritus
Well, it's not correct to write something like
$$\hat{A} = \int \psi^* \hat{A} \psi \,d\vec{r}$$The two sides of the equations are different types of objects. In particular, the righthand side is just a number.

Also, have you established the property $(A+B)^\dagger = A^\dagger+B^\dagger$ already? How about $(AB)^\dagger = B^\dagger A^\dagger$?

How are you justifying going from $\int \psi^* (AB)^\dagger \psi\,d\vec{r}$ to $\int (B\psi)^* A^\dagger \psi\,d\vec{r}$?

10. Oct 16, 2011

### atomicpedals

I have established

$(AB)^\dagger = B^\dagger A^\dagger$

as $(AB)^\dagger = \int \psi^* (AB)^\dagger \psi dr = \int (AB \psi)^* \psi dr = \int (B \psi)^* A^\dagger \psi dr = \int \psi^* B^\dagger A^\dagger \psi dr \Rightarrow B^\dagger A^\dagger$

No good?

Last edited: Oct 16, 2011
11. Oct 16, 2011

### vela

Staff Emeritus
That's fine. It's just hard knowing which properties you can use and which you can't yet since you're being asked to justify some pretty basic things.

So if you can show that you can distribute the dagger through a sum, you're essentially done. Have you done that?

12. Oct 16, 2011

### atomicpedals

Yeah, my prof is very much of the school of thought "If you want to use it, prove it."

I guess I'm not totally sure how I would show the distribution of the dagger across the sum using the integral formalism I'm being asked to adopt.

13. Oct 16, 2011

### vela

Staff Emeritus
Start the same way and use $(A+B)\psi = A\psi+B\psi$ and the linearity of integration.

14. Oct 16, 2011

### atomicpedals

Ah, OK. Now I get it. Thanks for the help!