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Zwiebach pages 174-175

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data
    I tried to verify the expression for q1(t) in equation 10.52 and found that it was only true if
    adjoint (a_p) = a_p and adjoint(a_{-p}) = a_{-p}. Am I missing something?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 6, 2007 #2
    Equations (10.52) have nothing to do with [itex]a_p[/itex]. For the first of these two equations just add equation (10.39) page 173 to its adjoint. For the second one, add equation (10.41) to its adjoint.
  4. Oct 6, 2007 #3
    Yes. You could do that.

    But also, the equation for q1 in 10.52 says that one half of a(t) plus its adjoint must be equal to the real part of a(t). If you use the expression in 10.49 for a(t) and its adjoint that is when I run into problems.
  5. Oct 7, 2007 #4
    Now I see what you mean. You are treating the first term on the r.h.s of the first of equations (10.49) as if it were the operator [itex]q_1(t)[/itex]. but it is not.
  6. Oct 8, 2007 #5
    No. I am treating the real part of the expression for a(t) in 10.49 as if it were the operator for [itex]q_1(t)[/itex]. I rewrote everything in terms of sines and cosines and found the real part.
  7. Oct 8, 2007 #6
    Sorry, that was an assumption on my part. Can you show us what you did?
  8. Oct 8, 2007 #7
    I'll show it step by step.
    Do you agree that

    [tex]q_1(t) = cos(E_p t)a_{p} + cos(E_p t) a_{-p}^{\dag} [/tex]
    Last edited: Oct 8, 2007
  9. Oct 8, 2007 #8
    It doesn't look too good. The l.h.s. is hermitian, but the r.h.s. isn't. How did you derive it?
  10. Oct 8, 2007 #9
    It is just the real part of a(t) from equation 10.49 . Just use Euler's formula.
  11. Oct 8, 2007 #10
    I see now. What is the real part of [itex]ie^{-iE_pt}[/itex]?
  12. Oct 10, 2007 #11
    Do you mean "what is the real part of [itex]e^{-iE_pt}[/itex]?"

    Then it would be cos(E_p t).
  13. Oct 10, 2007 #12


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    I guess by "real part" you mean that you dropped all the terms with a factor of i?
    You can't do that because "a" itself (and a^dagger) must be viewed as an imaginary quantity!! You must keep *all* the terms including those with factors of i!
    The way you can show that the result is real is to show that q_1^dagger gives back q_1.

    When I write down q_1, it has a total of 8 terms. But one can see right away that q_1^dagger =q_1 as is obvious from 10.52.

    Hope this makes sense
  14. Oct 11, 2007 #13
    No, I meant it as I wrote it. This is a trivial multiplication after using the Euler formula. Get this right and nrqed's post will be clear to you.
  15. Oct 11, 2007 #14
    Yes, that's what I did.

    I am kind of confused about why you wrote "imaginary" and not "complex" and why jimmmysnyder replaced the a_p with an i.

    Moreover, why shouldn't the creation and annhilation operators both be real since you cannot create an imaginary particle or destroy a particle and get an imaginary result? Maybe I just don't understand creation and annhilation operators.
  16. Oct 11, 2007 #15


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    I don't think that Jimmy was replacing a_p with an i. I think he was just checking something about how you picked the real and imaginary parts. I am not too sure what he meant to explain next.
    The fact that the operators are not "real" has nothing to do with the partilces they create or annihilate.

    I guess we have to step back a second. The first thing is that since we are talking about operators and not numbers, we should not really talk about whether they are real or imaginary. We should talk instead about whether they are hermitian or not. If an operator is hermitian, then A^dagger= A. An operator which is associated to an observable (something that can be measured) must be hermitian. This is the case for q and p. On the other hand, "a" and "a^dagger" are not observables (in the sense that their eigenvalues are not things that are measured). So they do not need to be hermitian...and indeed they are *not* hermitian.
  17. Oct 11, 2007 #16
    I'm sorry for the confusion and I promise to clear it up as soon as you tell me the answer.
  18. Oct 11, 2007 #17
    Reread the paragraph below equation (10.44) on page 173.
  19. Oct 11, 2007 #18
    OK. So in conclusion you cannot find the real part of an expression with non-Hermitian operators just by taking out the factors of i.
  20. Oct 11, 2007 #19


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    One does not even talk about "real" and "imaginary" part of an operator. One talks about the hermitian and non-hermitian componeents.

    for any operator A, one may write

    [tex] A = \frac{1}{2} ( A+ A^\dagger) + \frac{1}{2} (A - A^\dagger) [/tex]

    The first term is hermitian by construction (for any operator A) and the second term is not hermitian by construction.
    If the operator A is hermitian, the second term will be zero, obviously.
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