Show that Linear Combination is not Hermitian

[Ryomega]

Homework Statement

Show that linear combinations A-iB and A+iB are not hermitian if A and B (B≠0) are Hermitian operators

Homework Equations

Hermitian if: A*=A

Hermitian if: < A l C l B > = < B l C l A >

The Attempt at a Solution

So I've seen this question everywhere but not the solution to it.
I get that the solution isn't (A+iB)* = (A*+i*B*) = (A*-iB*) (since i*=-i)
So that's not helping me prove all its non-hermitianess, but it doesn't seem right since if I changed the order of the Hermitian:

< +iB l C l A > ≠ < A l C* l -iB >

Is that where I should be going with this? Or am I completely going wrong?
I know it's against the rules, but could someone show me the solution? I've been stuck on this for an entire day now and I'm fed up.

Thanks a lot!

 

[Muphrid]

Your brackets don't look like the way I was taught Dirac notation. The operators go in the middle, not on the left or on the right.

Consider two states, \psi, \phi. An operator \hat S is Hermitian if

\langle \psi | \hat S | \phi \rangle = \langle \phi | \hat S | \psi \rangle

Knowing that \hat A, \hat B are Hermitian, plug in \hat A \pm i \hat B for \hat S above.

 

[Ryomega]

Hi, sorry I'm fairly new to this notation thing. After reading it a thousand times I think I got it. Rest assured, I think I have learned Dirac notation as you have.

And thank you for the help. Integrating A - iB and A + iB makes them not equal to each other. Which now makes sense.

Thanks a lot!

 

[Muphrid]

Let me clarify a bit, as I wrote something that was a bit misleading. For any operator \hat C,

\langle \psi | \hat C | \phi \rangle = \langle \phi | \hat C^* | \psi \rangle

And only for a Hermitian operator is \hat C^* = \hat C.

But really, this problem is done when you realize that, if \hat C = \hat A \pm i \hat B, then \hat C^* = \hat A \mp i \hat B.

 

[Ryomega]

This is going to sound idiotic... but what does the upside down +/- sign denote?

 

[George Jones]

\hat C = \hat A \pm i \hat B, then \hat C^* = \hat A \mp i \hat B.

This is going to sound idiotic... but what does the upside down +/- sign denote?

The top plus on the left goes with the top minus on the right; the bottom minus on the left goes with the bottom plus on the right. For example, when \hat C = \hat A + i \hat B, then \hat C^* = \hat A - i \hat B.

 

[Ryomega]

AHhhh! so it's just a thing I have to remember. Thank you!

 

[Ryomega]

That Ahh, isn't screaming in fear... it is in understanding in...understanding or (whatever word I was actually looking for)

 

[drsauce]

Integrating A - iB and A + iB makes them not equal to each other. Which now makes sense.

Could you show how you do this part please. I'm assuming you integrate ∫ψ(A-iB)ψ*
but I am not sure where to begin on this and I feel I'm missing something very basic

 

[Ryomega]

A and iB are simply constants.
Integrating a constant just gives itself back. So you get A-iB and A+iB.
By definition A*= A if they are Hermitian.
Since they don't equal each other, they are not Hermitian.

 

[drsauce]

ah. I was assuming A and B to be functions, and that was making things a lot more complex.

 

[Wayn3]

A and iB are simply constants.
Integrating a constant just gives itself back. So you get A-iB and A+iB.
By definition A*= A if they are Hermitian.
Since they don't equal each other, they are not Hermitian.

A and B are Operators. You can't just integrate over them. You should really pick up the book "Modern Quantum Mechanics" by Sakurai and read the first 30-50 pages.

There are two ways to introduce quantum mechanics. You either take the wave mechanics route, which leads to a very flawed understanding of quantum mechanics, or you learn it the way dirac introduced it.

Now, I don't know whether you americans really use the "*" to indicate that you mean the Operator which is dual to non-starred operator. In my opinion, that would be very misleading. We use a different symbol.

If you want to know what C* really is, then you first need to study these "states" a bit.

If you have ket (state) |a>, then that is a vector on some complex vector space.
The bra <a| is the bra dual to |a>, that means, its a vector on the complex vector space which is dual to the vector space of which |a> is element. These are properties introduced in linear algebra which you need to know in order to understand quantum mechanics.

If you progress to operators, you need to understand that C operator only acts on ket |a>, not on bra <a|. Only the operator dual to C - C* - can act on <a|.

Sakurai explains this in great detail.

In order to show your task, i would work from the fact that an operator is defined by the way it acts on a state. Thus, if you want to prove hermiticity, or the lack of, you should take a look at:

<a|C|a>

and show that it is (not) equal to

(<a|C|a>)*


Keep in mind that <a|b> = <b|a>* and that you can act as if C|a> was just a ket.