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

• Shackleford
In summary, you need to take the 'i' as part of what you're looking to see whether or not it's hermitian. You have to take the 'i' as part of what you're looking to see whether or not it's hermitian. When you do the hermitian conjugate of anything you do the transpose AND the complex conjugate, which will affect the 'i'.
Shackleford
(a) I'm not sure what else to do. I don't think I'm properly treating the i.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-11192023.jpg?t=1286843024

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

Pengwuino said:
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!

[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'.

Pengwuino said:
[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

Shackleford said:
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?

Pengwuino said:
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

Shackleford said:
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 $$A^\dagger = A$$

Pengwuino said:
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.

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?

Shackleford said:
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'.

Pengwuino said:
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.

Mindscrape said:
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.'

Mindscrape said:
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

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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?

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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

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

Mindscrape said:
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

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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.

## 1. What does it mean for a matrix to be hermitian?

A hermitian matrix is a square matrix that is equal to its own conjugate transpose. This means that the matrix is symmetric across its main diagonal and all of its elements along the diagonal are real numbers.

## 2. How do you determine if a matrix is hermitian?

To determine if a matrix is hermitian, you calculate its conjugate transpose by taking the complex conjugate of each element and then taking the transpose of the resulting matrix. If the resulting matrix is equal to the original matrix, then it is hermitian.

## 3. What is the significance of proving that i[A,B] is hermitian?

If i[A,B] is hermitian, it means that the commutator of two hermitian matrices is also hermitian. This is important in quantum mechanics, as hermitian matrices represent observables and the commutator represents the uncertainty between two observables. It also allows for the use of the spectral theorem, which states that all hermitian matrices can be diagonalized by a unitary matrix.

## 4. What is the relationship between the commutator and hermitian matrices?

The commutator of two matrices is defined as [A,B] = AB - BA. In the case of hermitian matrices, the commutator is equal to i times the anti-commutator of the two matrices, i.e. [A,B] = i{A,B}. This relationship is important in proving that i[A,B] is hermitian.

## 5. Can i[A,B] ever not be hermitian?

No, i[A,B] is always hermitian as long as A and B are hermitian. This is because the commutator of two hermitian matrices is always anti-hermitian, and multiplying by i will result in a hermitian matrix.

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