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Show that Linear Combination is not Hermitian

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    Hermitian if: A*=A

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

    3. 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!
     
  2. jcsd
  3. Jun 8, 2012 #2
    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, [itex]\psi, \phi[/itex]. An operator [itex]\hat S[/itex] is Hermitian if

    [itex]\langle \psi | \hat S | \phi \rangle = \langle \phi | \hat S | \psi \rangle[/itex]

    Knowing that [itex]\hat A, \hat B[/itex] are Hermitian, plug in [itex]\hat A \pm i \hat B[/itex] for [itex]\hat S[/itex] above.
     
  4. Jun 8, 2012 #3
    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!!!
     
  5. Jun 8, 2012 #4
    Let me clarify a bit, as I wrote something that was a bit misleading. For any operator [itex]\hat C[/itex],

    [tex]\langle \psi | \hat C | \phi \rangle = \langle \phi | \hat C^* | \psi \rangle[/tex]

    And only for a Hermitian operator is [itex]\hat C^* = \hat C[/itex].

    But really, this problem is done when you realize that, if [itex]\hat C = \hat A \pm i \hat B[/itex], then [itex]\hat C^* = \hat A \mp i \hat B[/itex].
     
  6. Jun 9, 2012 #5
    This is going to sound idiotic... but what does the upside down +/- sign denote?
     
  7. Jun 9, 2012 #6

    George Jones

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    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 [itex]\hat C = \hat A + i \hat B[/itex], then [itex]\hat C^* = \hat A - i \hat B[/itex].
     
  8. Jun 9, 2012 #7
    AHhhh! so it's just a thing I have to remember. Thank you!
     
  9. Jun 9, 2012 #8
    That Ahh, isn't screaming in fear... it is in understanding in...understanding or (whatever word I was actually looking for)
     
  10. Jun 9, 2012 #9
    Could you show how you do this part please. I'm assuming you integrate ∫ψ(A-iB)ψ*
    but im not sure where to begin on this and I feel I'm missing something very basic
     
  11. Jun 10, 2012 #10
    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.
     
  12. Jun 10, 2012 #11
    ah. I was assuming A and B to be functions, and that was making things alot more complex.
     
  13. Jun 10, 2012 #12
    A and B are Operators. You cant 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 dont 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.
     
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