If A and B are hermitian, then i[A,B] is also hermitian

[Shackleford]

(a) I'm not sure what else to do. I don't think I'm properly treating the i.

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http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-11192034.jpg?t=1286843025

[Pengwuino]

You have to take the 'i' as part of what you're looking to see whether or not it's hermitian.

[Shackleford]

You have to take the 'i' as part of what you're looking to see whether or not it's hermitian.

Well, I'm not sure what to do with the 'i." I'm afraid this commutator business is new to me. I'm reviewing my notes, but I'm still wary. Is there a good online resource that would help? Our quantum physics book (Gasiorowicz) isn't that great of a book, but, hey, it's the book MIT is using!

[Pengwuino]

[A,B] is an operator right? So why can't the 'i' in front of the [A,B] be included into the operation? When you do the hermitian conjugate of anything you do the transpose AND the complex conjugate, which will affect the 'i'.

[Shackleford]

[A,B] is an operator right? So why can't the 'i' in front of the [A,B] be included into the operation? When you do the hermitian conjugate of anything you do the transpose AND the complex conjugate, which will affect the 'i'.

I'm not sure I follow you. Any outside sources defining the various operations?

[A,B] = AB - BA

i[A,B] = iAB - BA

[Pengwuino]

I'm not sure I follow you. Any outside sources defining the various operations?

[A,B] = AB - BA

i[A,B] = iAB - BA

Do you know what it means for any quantity to be hermitian? And why did the 'i' only go with the first term out of the commutator?

[Shackleford]

Do you know what it means for any quantity to be hermitian? And why did the 'i' only go with the first term out of the commutator?

An operator that has all real eigenvalues is hermitian. In this case, the i would have to disappear, right?

I don't know. Is this better?

i[A,B] = iAB - iBA

[Pengwuino]

An operator that has all real eigenvalues is hermitian. In this case, the i would have to disappear, right?

I don't know. Is this better?

i[A,B] = iAB - iBA

No, what you are doing has nothing to do with eigenvalues. The quantity i[A,B] is something. It doesn't matter what it is at this point, it is just something. Now, whatever that something is, when you take the hermitian conjugate of it, you complex conjugate everything (which means any 'i's to go '-i') along with taking the transpose (if you think about it as matrices). So you applied that operation correctly, hence the 'dagger's on the A and B operators. However, you did not do anything to the 'i'. The 'i' needs to be conjugated.

At that point, you use the fact that Hermitian operators have the property that A^\dagger = A

[Shackleford]

No, what you are doing has nothing to do with eigenvalues. The quantity i[A,B] is something. It doesn't matter what it is at this point, it is just something. Now, whatever that something is, when you take the hermitian conjugate of it, you complex conjugate everything (which means any 'i's to go '-i') along with taking the transpose (if you think about it as matrices). So you applied that operation correctly, hence the 'dagger's on the A and B operators. However, you did not do anything to the 'i'. The 'i' needs to be conjugated.

At that point, you use the fact that Hermitian operators have the property that A^\dagger = A

Yeah, hermitian operators are self-adjoint. I'm very tired.

I was taking the adjoint of the operator. Is that what you call hermitian conjugate?

Of course, if I make it -'i' the two lines are equal.

[Mindscrape]

Here, if you were to do this with transposes instead, it may look something like this..

i[A,B]=iAB-iBA

(i[A,B])^T=(iAB)^T-(iBA)^T=i^T B^T A^T - i^T A^T B^T

so, what if you include complex conjugation, and the fact that A and B are Hermitian?

[Pengwuino]

Yeah, hermitian operators are self-adjoint. I'm very tired.

I was taking the adjoint of the operator. Is that what you call hermitian conjugate?

Of course, if I make it -'i' the two lines are equal.

Yes, hermitian conjugate, adjoint, same thing for us. So you're taking the adjoint of the entire quantity i[A,B]. So you must take the complex conjugate of the 'i'.

[Shackleford]

Yes, hermitian conjugate, adjoint, same thing for us. So you're taking the adjoint of the entire quantity i[A,B]. So you must take the complex conjugate of the 'i'.

Okay. I guess I didn't know that. That's good to know. Thanks. Anything else I should know you think I don't know? lol.

Let me finish this problem.

[Shackleford]

Here, if you were to do this with transposes instead, it may look something like this..

i[A,B]=iAB-iBA

(i[A,B])^T=(iAB)^T-(iBA)^T=i^T B^T A^T - i^T A^T B^T

so, what if you include complex conjugation, and the fact that A and B are Hermitian?

Ah. I knew the + (adjoint) applied to the whole damn thing. I just didn't do as such because I wasn't sure how to find the adjoint of 'i.'

[Shackleford]

Here, if you were to do this with transposes instead, it may look something like this..

i[A,B]=iAB-iBA

(i[A,B])^T=(iAB)^T-(iBA)^T=i^T B^T A^T - i^T A^T B^T

so, what if you include complex conjugation, and the fact that A and B are Hermitian?

I think I showed it's hermitian by showing they're self-adjoint.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-12094709.jpg?t=1286894968

[Shackleford]

I worked out (15) and (17). However, I'm not exactly sure how to do (16) and (18). Do I simply integrate to get the expectation value? - integrate d<x>/dt and d<p>/dt?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-12182834.jpg?t=1286926348

[Mindscrape]

Yeah, it looks like you've mostly shown that i[A,B] is hermitian given that A and B are also hermitian. You should probably more explicit in what steps do what. I would give you 4/5.

If you got 17 then it should be easy to find the expectation values at t=0. Expectation value for an observable A in a wavefunction [itex]\Psi[/tex] is given by

\langle \Psi | \hat{A} | \Psi \rangle = \int \Psi^\dagger \hat{A} \Psi

[Shackleford]

Yeah, it looks like you've mostly shown that i[A,B] is hermitian given that A and B are also hermitian. You should probably more explicit in what steps do what. I would give you 4/5.

If you got 17 then it should be easy to find the expectation values at t=0. Expectation value for an observable A in a wavefunction [itex]\Psi[/tex] is given by

\langle \Psi | \hat{A} | \Psi \rangle = \int \Psi^\dagger \hat{A} \Psi

(18)

<p>t = e E0 (1/w) sin wt + C

C = <p>0

<x>t = -(1/m) e E0 (1/w^2) cos wt + (1/m) t <p>0 + (1/m) e E0 (1/w^2) + <x>0

[Mindscrape]

That could be right, I haven't checked it myself. You really ought to start a new thread if you want help on a new problem.