Proving $\hat{O}\hat{O}^\dagger$ is Hermitian

In summary, the conversation revolved around proving that the operator \hat{O}\hat{O}^{\dagger} is Hermitian, which can be done by showing that it commutes with its Hermitian conjugate. The use of bra-ket notation and the fundamental properties of inner product were discussed, and it was shown that the operator is indeed Hermitian by using integrals and wave functions.
  • #1
latentcorpse
1,444
0
pretty simple question. have to prove [itex]\hat{O} \hat{O}\dagger[/itex] is a Hermitian operator.

i found that

[itex]\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[/itex]

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?
 
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  • #2
do you know how to use bra-ket notation? it is a bit easier then.
 
  • #3
ok.

[itex] \langle \psi|\hat{O} \hat{O}^{\dagger}|\phi \rangle^{\star} = \langle \phi|\hat{O}^{\dagger} \hat{O}| \psi \rangle[/itex]

but then i again encounter the problem of showing that
[itex]\hat{O} \hat{O}^{\dagger}=\hat{O}^{\dagger} \hat{O}[/itex]
?
 
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  • #4
you are not using bra-ket correctly.

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

remeber to use dual-correspondence
 
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  • #5
sorry, i edited the bra-ket. now from the definition of hermition conjugate, O is hermitian iff

[itex]\hat{O}\hat{O}^{\dagger}=\hat{O}^{\dagger}\hat{O}[/itex]
 
  • #6
have you used the dual correspondance and the fundamental property of inner product?
 
  • #7
i don't know what those are. sorry...
 
  • #8
latentcorpse said:
i don't know what those are. sorry...

ok, if you did not know what bra-ket is and how it works you should have said so...
 
  • #9
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
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:
[tex]
\int \psi _1 ^* (\hat{A} \psi _2 ) \, dx = \int (\hat{A}\psi _1 )^* \psi _2 \, dx
[/tex]

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

verify that and continue
 
  • #11
ok. I am getting confused now.
my notes define hermitian conjugate as:

[itex](\int \psi^* \hat{O} \phi dx)^* = \int \phi^* \hat{O}^{\dagger} \psi dx[/itex]

the operator is hermitian if [itex]\hat{O}=\hat{O}^{\dagger}[/itex]

how does that compare to what you have?

is your definition just saying [itex]\hat{A}=\hat{A}^{\dagger}[/itex]

how would you write the first equation you wrote down in bra-ket notation?
 
  • #12
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:
[tex]

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

[/tex]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>
 
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  • #13
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
yes, that is correct done in bra-ket, as you saw, it was really easy.

[tex]

\int \psi _1 ^* (\hat{O}\hat{O}^{\dagger} \psi _2 ) \, dx = \int (\hat{O}^{\dagger}\psi _1 )^* \hat{O}^{\dagger} \psi _2 \, dx
[/tex]

do this one more time.. recall that conjugating reverses the order (AB)* = B*A*
 
  • #15
could we say

[itex](\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[/itex]

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
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.
 

1. What is the definition of a Hermitian operator?

A Hermitian operator is a type of linear operator in quantum mechanics that has a special property called Hermiticity. This means that the operator is equal to its own adjoint or Hermitian conjugate, denoted as $\hat{O}^\dagger$. In other words, the operator and its adjoint produce the same result when acting on a quantum state.

2. What is the importance of proving that $\hat{O}\hat{O}^\dagger$ is Hermitian?

Proving that $\hat{O}\hat{O}^\dagger$ is Hermitian is crucial in quantum mechanics because Hermitian operators represent physical observables. This means that the eigenvalues of a Hermitian operator correspond to the possible outcomes of a measurement of that observable. Therefore, proving Hermiticity ensures that the operator accurately describes a physical quantity.

3. How can you prove that $\hat{O}\hat{O}^\dagger$ is Hermitian?

To prove that $\hat{O}\hat{O}^\dagger$ is Hermitian, you need to show that it is equal to its own adjoint. This can be done by using the definition of Hermitian operators and the properties of adjoints. Specifically, you need to show that $\hat{O}\hat{O}^\dagger = \hat{O}^\dagger \hat{O}$, which can be simplified using the properties of linear operators.

4. What are the consequences if $\hat{O}\hat{O}^\dagger$ is not Hermitian?

If $\hat{O}\hat{O}^\dagger$ is not Hermitian, then it does not accurately represent a physical observable. This means that the eigenvalues of the operator may not correspond to the possible outcomes of a measurement, leading to incorrect predictions and interpretations in quantum mechanics. It is therefore important to always verify that an operator is Hermitian before using it to make predictions about physical systems.

5. Can a non-Hermitian operator have real eigenvalues?

No, a non-Hermitian operator cannot have real eigenvalues. This is because Hermiticity is a necessary condition for an operator to have real eigenvalues. Therefore, if an operator does not satisfy the condition of Hermiticity, it will not have real eigenvalues. This further emphasizes the importance of proving that an operator is Hermitian in quantum mechanics.

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