# How do I determine if it's a Hermitian Operator or not

1. Jan 25, 2006

### MixilPlixit

First post so please go easy on me, here goes:

I have looked over the basic definition of what is a Hermitian operator such as: <f|Qf> = <Qf|f> but I still am unclear what to do with this definition if I am asked prove whether or not i(d/dx) or (d^2)/(dx^2) for example are Hermitian operators. I am using Griffiths' "Intro to Quantum Mechanics" and it really seems like he skips a lot of steps. Steps I need to make sure I understand. Can someone help enlighten me.

2. Jan 26, 2006

### Meir Achuz

You have to put the operator in the integral \int\psi*[id/dx]\psi dx.
Then integrate by parts, using the BC at the endpoints. If you get the same result after integratilng by part, the operator is hermitian.

3. Jan 26, 2006

### inha

This is exactly how it's done in Griffiths too. Maybe Mixil had some specific part of this in mind?

4. Jan 26, 2006

### MixilPlixit

I think I am a little hazy on integration by parts, so that apeears to be where my problem lies. Can someone recommend a good refresher on the matter?

5. Jan 26, 2006

### ghosts_cloak

6. Jan 26, 2006

### MixilPlixit

OK, that somewhat helps but if I have $$\int \frac{d^2}{d \phi^2}d \phi$$ how do I evaluate that? It's been a while since calc obviously. It's the 2nd derivative thing that's throwing me. Hopefully I am not tiring you guys, cause I am certainly getting bleary. Thanks for your help this far. I am thankful for your continued assistance.

7. Jan 26, 2006

### George Jones

Staff Emeritus
You can't evaluate that. On what is $\frac{d^2}{d \phi^2}$ operating?

Regards,
George

8. Jan 26, 2006

### MixilPlixit

I am given $$Q = \frac{d^2}{d \phi^2}$$ And am asked to prove if eigenvalues are real. In order to do that I need to determine if Q is in fact a Hermitian operator. So I am at: $$\int f*(\frac{d^2}{d \phi^2})g d\phi$$ This is integrated over 0 to 2 pi.

Last edited: Jan 26, 2006
9. Jan 26, 2006

### George Jones

Staff Emeritus
You need to use integration by parts twice. For the first integration by parts, it may be helpful to think of

$$\left( \frac{d^2}{d \phi^2} \right)g$$

as

$$\frac{dg'}{d \phi}$$

where

$$g' = \frac{dg}{d \phi}.$$

Regards,
George

10. Jan 26, 2006

### MixilPlixit

For the first integration then would $$v = \frac{dg}{d \phi}$$ and $$du = (\frac{df}{d\phi})^*$$?

 Well I think I solved it, but it feels wrong. I got $$-\int f^*(\frac{d^2}{d \phi^2}) g d\phi$$ It the initial integral but minus. Can that be right?

Last edited: Jan 26, 2006
11. Jan 27, 2006

### dextercioby

Assuming the operator is densly defined, can u at least identify the operator's adjoint...? Next thing to worry you is to find whether your operator is bounded or not...

Daniel.