Show eigenvalues of hermitian operator are real

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http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-20165642.jpg?t=1287612122 [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-20165727.jpg?t=1287612136 [Broken]

Thanks.
 
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Those look fine. In the first one you've assumed orthonormality. You could have also noted that
[tex]\langle \psi_m | A | \psi _n \rangle = a_n \langle \psi_m | \psi_n \rangle[/tex]
[tex]\langle \psi_m | A^\dagger | \psi _n \rangle = a^*_m \langle \psi_m | \psi_n \rangle[/tex]
which when subtracted leads to
[tex](a_n-a_m^*)\langle \psi_m | \psi_n \rangle =0 [/tex]
This shows that the conjugate eigenvalue must actually be real, and that the states must be orthonormal.

For the second one I don't know what equations the problem tells you to use, so I'm not sure if it's okay or not.
 
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Those look fine. In the first one you've assumed orthonormality. You could have also noted that
[tex]\langle \psi_m | A | \psi _n \rangle = a_n \langle \psi_m | \psi_n \rangle[/tex]
[tex]\langle \psi_m | A^\dagger | \psi _n \rangle = a^*_m \langle \psi_m | \psi_n \rangle[/tex]
which when subtracted leads to
[tex](a_n-a_m^*)\langle \psi_m | \psi_n \rangle =0 [/tex]
This shows that the conjugate eigenvalue must actually be real, and that the states must be orthonormal.

For the second one I don't know what equations the problem tells you to use, so I'm not sure if it's okay or not.
I fumbled through the first part. Your method is more concise, though. However, I'm not sure why am is conjugated. From the definition, when you operate the hermitian operator [tex]A^\dagger [/tex] on the left, it simply goes back to A in the Dirac notation. But in the integral form, I can see why the conjugation is there. In Dirac notation, is it there because the bra is the conjugate of the ket and so the conjugation of the operator is implied?

Here are the equations. I didn't really use them all that much. Should I try to do so?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-21082445.jpg?t=1287667815 [Broken]
 
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Right, the ket conjugates the eigenvalue by definition of the notation.

For the second question you assumed that A and B commute, which is not true in general. Try a different approach, try just using the dirac notation to get where you want to go. Start with

[tex]\langle \phi | (AB)^\dagger | \psi \rangle = \langle \psi | A B | \phi \rangle^*[/tex]

and see where it takes you. You could even use integral form if you really wanted, but I, personally, like the dirac notation because it's concise.
 
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Right, the ket conjugates the eigenvalue by definition of the notation.

For the second question you assumed that A and B commute, which is not true in general. Try a different approach, try just using the dirac notation to get where you want to go. Start with

[tex]\langle \phi | (AB)^\dagger | \psi \rangle = \langle \psi | A B | \phi \rangle^*[/tex]

and see where it takes you. You could even use integral form if you really wanted, but I, personally, like the dirac notation because it's concise.
Okay. Let me try that. If you want, check my other thread, too. That's the next problem I'm going to work on.

https://www.physicsforums.com/showthread.php?t=440132
 
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Right, the ket conjugates the eigenvalue by definition of the notation.

For the second question you assumed that A and B commute, which is not true in general. Try a different approach, try just using the dirac notation to get where you want to go. Start with

[tex]\langle \phi | (AB)^\dagger | \psi \rangle = \langle \psi | A B | \phi \rangle^*[/tex]

and see where it takes you. You could even use integral form if you really wanted, but I, personally, like the dirac notation because it's concise.
http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-21221122.jpg?t=1287717351 [Broken]
 
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Um, yeah, it looks like you're there.
 
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Yeah, I think I might have done an illegal assumption or operation. I wasn't 100% satisfied with what I had earlier. I don't remember what I put, but it should have been something simple like this

[tex]
\langle \phi | (AB)^\dagger | \psi \rangle = \langle (AB) \phi | \psi \rangle = \langle B \phi | A^\dagger \psi \rangle = \langle \phi | B^\dagger A^\dagger \psi \rangle
[/tex]
Step 1 uses changing between bras and kets, step 2 uses the adjoint to move B phi, and step 3 uses bra ket equivalence again.
 
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When I looked back at your solution it wasn't a whole lot different from mine, so I didn't think I needed to reiterate. I don't remember what I put, but it should have been something like this

[tex]
\langle \phi | (AB)^\dagger | \psi \rangle = \langle (AB) \phi | \psi \rangle = \langle B \phi | A^\dagger \psi \rangle = \langle \phi | B^\dagger A^\dagger \psi \rangle
[/tex]
Oh, okay. Thanks for the help so far. I'm trying to do my best to do the homework problems and understand everything well enough so I can do better on the next exam.

Could you help me on the next problem or two? I don't know why I'm having trouble. It looks like a fairly straightforward problem. :frown:

https://www.physicsforums.com/showthread.php?t=440132
 
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Well, I'm reviewing my notes for the exam on Wednesday. I got full-credit for this problem. However, I don't understand how the last line is valid. The psi and phi should be reversed since I'm taking the hermitian conjugate of a conjugate.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-21221122.jpg?t=1287717351 [Broken]
 
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Actually, now that I look back on it, yeah, all your work would have done would have been to take you in a circle back to [itex](AB)^\dagger[/itex]. If you understand the solution I put up, go with this solution since it's pretty concise and really just uses bra-ket formalism. Maybe both the grader and I just got lazy before and didn't fully check your work.
 
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Actually, now that I look back on it, yeah, all your work would have done would have been to take you in a circle back to [itex](AB)^\dagger[/itex]. If you understand the solution I put up, go with this solution since it's pretty concise and really just uses bra-ket formalism. Maybe both the grader and I just got lazy before and didn't fully check your work.
Yeah, I understand what you did. The book actually proves it like that, too. I overlooked that part when doing the homework.

I'm not actually sure how to interpret what I wrote since there are a few different ways to write out this stuff.

Would it be <phi /(AB)^+/psi>^+ = <psi /(AB)/phi> ?
 
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Yeah, that's what you would get.
 

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