QM: Operator algebra and inner products

[Niles]

Homework Statement

We have a hermitian operator Q and a complex number z. I have an inner product given by:

<f | (zQ) g>,

where f and g are two arbitrary functions. Now taking the hermitian conjugate of a complex number is just taking the complex conjugate of that number. So taking the hermitian conjugate of the above gives me:

<(zQ)^H f | g> = <z^*Q^H f | g> = z<Q^H f | g> = z<Qf | g>

According to my book, this is wrong, because I should end up with z^* in front of the inner product, and not z. Where is my reasoning wrong?

Thanks in advance.Niles.

 

[CompuChip]

It also depends on the book used, some authors prefer to define

\langle f | z g \rangle = z \langle f | g \rangle, <br /> \langle z f | g \rangle = z^* \langle f | g \rangle

while some use

\langle f | z g \rangle = z^* \langle f | g \rangle, <br /> \langle z f | g \rangle = z \langle f | g \rangle

A teacher of mine (mathematician who worked closely with physicists) once said that one is typically used by mathematicians while the other is often used by physicists. You are free to choose which convention you use, as long as you use it consistently.

 

[Niles]

My book uses the first one you mentioned. But I am using the same convention, so this is what troubles me.

Can you spot my error? The place where my book is different is when taking the hermitian conjugate - there the author does not complex conjugate the number z (actually, he does not even assume it is complex.)

I am using Griffiths Introduction to QM.

 

[CompuChip]

I think you are right, in fact you can take out the z right away:
<f | z Q g> = z <f | Q g> = z <Q f | g>
by the property I mentioned and the hermiticity of Q, respectively.

Griffiths Introduction to QM.

I have that book in front of me right now... can you give me a page number?

 

[Niles]

It is problem 3.4, part B on page 110, chapter 3.

But I think you solved my mystery. Because this just tells us that the complex number z has to be a real number, which is the answer to that problem.

Since you are sitting with the book in front of you, I hope it is OK I ask another question on this topic. It is on problem 3.5 part C on the same page. This is my approach:

<f | (QR) g> = <(QR)^Hf | g> = <Q^HR^H f | g>,

which is not what I have to show. But isn't the hermitian conjugate a "linear operator" so to speak? That is what I had in mind when solving it.

Thanks in advance, I really appreciate it.

 

[CompuChip]

For matrices you have the property that
(A B)^T = B^T A^T,
while complex conjugation doesn't change the order:
(A B)* = A* B*

Therefore, the Hermitian conjugate works like the transpose as far as ordering of matrices is concerned:
(A B C ...)^H = ... C^H B^H A^H

You're welcome, glad I could help.