# Hermitian Conjugate

1. Mar 10, 2009

### latentcorpse

pretty simple question. have to prove $\hat{O} \hat{O}\dagger$ is a Hermitian operator.

i found that

$\left( \int \int \int \psi^{\star}(\vec{r}) \hat{O} \hat{O}^{\dagger} \phi(\vec{r}) d \tau \right)^{\star} = \int \int \int \phi^{\star}(\vec{r}) \hat{O}^{\dagger} \hat{O} \phi(\vec{r}) d \tau$

so all we need to get the result is to establish if the operator commutes with it's hermitian conjugate. i'm guessing it does but don't know why - can someone explain?

2. Mar 11, 2009

### malawi_glenn

do you know how to use bra-ket notation? it is a bit easier then.

3. Mar 11, 2009

### latentcorpse

ok.

$\langle \psi|\hat{O} \hat{O}^{\dagger}|\phi \rangle^{\star} = \langle \phi|\hat{O}^{\dagger} \hat{O}| \psi \rangle$

but then i again encounter the problem of showing that
$\hat{O} \hat{O}^{\dagger}=\hat{O}^{\dagger} \hat{O}$
????

Last edited: Mar 11, 2009
4. Mar 11, 2009

### malawi_glenn

you are not using bra-ket correctly.

an operator is hermitian if <b|R|a> = <a|R|b>*

remeber to use dual-correspondence

Last edited: Mar 11, 2009
5. Mar 11, 2009

### latentcorpse

sorry, i edited the bra-ket. now from the definition of hermition conjugate, O is hermitian iff

$\hat{O}\hat{O}^{\dagger}=\hat{O}^{\dagger}\hat{O}$

6. Mar 11, 2009

### malawi_glenn

have you used the dual correspondance and the fundamental property of inner product?

7. Mar 11, 2009

### latentcorpse

i don't know what those are. sorry...

8. Mar 11, 2009

### malawi_glenn

ok, if you did not know what bra-ket is and how it works you should have said so....

9. Mar 11, 2009

### latentcorpse

yeah, well we use bra-kets in our course so i guess i'm supposed to know it. i just don't recognise the terms "dual correspondence" and "fundamental property of inner product", perhaps i do know them, just not by those names.

i tried googling them but to no avail...

10. Mar 11, 2009

### malawi_glenn

here is a good introduction:

http://www.physics.unlv.edu/~bernard/phy721_99/tex_notes/node6.html

Now using integrals and wave functions:

Operator A is hermitian if:
$$\int \psi _1 ^* (\hat{A} \psi _2 ) \, dx = \int (\hat{A}\psi _1 )^* \psi _2 \, dx$$

so we get
$$\int \psi _1 ^* (\hat{O}\hat{O}^{\dagger} \psi _2 ) \, dx = \int (\hat{O}^{\dagger}\psi _1 )^* \hat{O}^{\dagger} \psi _2 \, dx$$

verify that and continue

11. Mar 11, 2009

### latentcorpse

ok. im getting confused now.
my notes define hermitian conjugate as:

$(\int \psi^* \hat{O} \phi dx)^* = \int \phi^* \hat{O}^{\dagger} \psi dx$

the operator is hermitian if $\hat{O}=\hat{O}^{\dagger}$

how does that compare to what you have?

is your definition just saying $\hat{A}=\hat{A}^{\dagger}$

how would you write the first equation you wrote down in bra-ket notation?

12. Mar 11, 2009

### malawi_glenn

https://www.physicsforums.com/attachment.php?attachmentid=13008&d=1205092708 [Broken]

page 53.

eq. 4.58. since expectation value is real. (4.57) so the opreator O there is hermitian if O dagger = O.

That is to say:
$$\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx$$

If you want to do in bra-ket:

<b|O O^dagger |a> = <a|O O^dagger |b>* ??

consinder bra <b|O
DUAL is
O dagger |b>

Consider ket O^dagger |a>
DUAL is
<a| O

now use <r|t>* = <t|r>

Last edited by a moderator: May 4, 2017
13. Mar 11, 2009

### latentcorpse

all we need to do is put together the two duals you derived into a bra-ket to give

<a| O O(dagger) |b>

so we can now say

<a| O O(dagger) |b>* = <b| O O(dagger) |a> which implies a hermitian operator

is that ok?

out of interest, for getting it to work in terms of integrals,

i took the conjugate of the integral you wrote at the bottom left of post 10 but then got lost?

14. Mar 11, 2009

### malawi_glenn

yes, that is correct done in bra-ket, as you saw, it was really easy.

$$\int \psi _1 ^* (\hat{O}\hat{O}^{\dagger} \psi _2 ) \, dx = \int (\hat{O}^{\dagger}\psi _1 )^* \hat{O}^{\dagger} \psi _2 \, dx$$

do this one more time.. recall that conjugating reverses the order (AB)* = B*A*

15. Mar 11, 2009

### latentcorpse

could we say

$(\int \psi^* \hat{O} \hat{O}^{\dagger} \phi dx)^*=(\int (\hat{O}^{\dagger} \psi)^* \hat{O}^{\dagger} \phi dx)^* = \int (\phi \hat{O}^{\dagger})^* \hat{O}^{\dagger} \psi dx = \int \phi^* \hat{O} \hat{O}^{\dagger} \psi dx$

so operator is hermitian?

when doing it in bra-ket notation, do you have to evaluate the bra and the ket dual seperately like you did or is it possible to do it all in one line?

16. Mar 11, 2009

### malawi_glenn

you can of course say that the dual to O Odagger |a> is <a| O O dagger, then it is even more simple.

Yes, that is one other way that you can show it by using integrals and wave functions. Then you have been true to the definition in your course.