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.