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Zwiebach quick calculation 10.6

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data
    I am somewhat new to creation and annhilation operators, but I can reduce the problem to showing that:

    [tex] a_{p_1} ^{\dag} a_{p_2} ^{\dag} = a_{p_1} ^{\dag} + a_{p_2} ^{\dag} [/tex]

    [tex] a_{p_1} ^{\dag} [/tex] is the creation operator for a particle with momentum p1 (if I understand it correctly)

    Can someone help prove that?
    2. Relevant equations

    3. The attempt at a solution
    Last edited: Oct 8, 2007
  2. jcsd
  3. Oct 8, 2007 #2
    This can't be right since if it were, then:
    [tex]a_{p_1}^{\dag} a_{p_2} ^{\dag} a_{p_3} ^{\dag} = a_{p_1}^{\dag}(a_{p_2}^{\dag} + a_{p_3}^{\dag}) = a_{p_1}^{\dag}a_{p_2}^{\dag} + a_{p_1}^{\dag}a_{p_3}^{\dag} = a_{p_1}^{\dag} + a_{p_2}^{\dag} + a_{p_1}^{\dag} + a_{p_3}^{\dag}[/tex]
    In order to do the calculation, apply H and P as defined in equations (10.60) and (10.61) directly to the state in (10.65). Use induction on k.
  4. Oct 8, 2007 #3
    That's weird though. How can it not be true if

    \sum_{\vec{j}} \vec{j} a_j ^{\dag} a_j a_{p1}^{\dag} a_{p2}^{\dag}
    \sum_{\vec{j}} \vec{j} a_j ^{\dag} a_j a_{p1}^{\dag}
    \sum_{\vec{j}} \vec{j} a_j ^{\dag} a_j a_{p2}^{\dag}

  5. Oct 8, 2007 #4
    In general, for operators, AB = AC does not imply B = C even when A is non-zero.
  6. Oct 8, 2007 #5
    Yes. I was wrong.

    But then somehow you need to split up that long product of [tex] a_{p_k}^{\dag}[/tex] operators. You can commute them by 10.59, but they doesn't really help. Is there another way to split it into two terms?
  7. Oct 8, 2007 #6
    The formula [A,BC] = B[A,C] + [A.B]C will do nicely.
  8. Oct 10, 2007 #7
    Ah, I see. Thanks.

    For QC 10.7, I am trying to use

    [tex] \sum_{\vec{p}} a_p^{\dag} a_p | \Omega > = \sum_{\vec{p}} (a_p a_p^{\dag} -1) | \Omega >[/tex]

    So, there should probably be a Kronecker delta that has an index that gets summed over or something but I am not really sure where this comes in?

    Just to be sure, I want to show that [tex] \sum_{\vec{p}} a_p^{\dag} a_p | \Omega > = n | \Omega > [/tex].

    where n is the number of particles in the state omega, right?

    By the way, I should probably not be using omega because we are using that for the vacuum state and we want an arbitrary state here, right?
    Last edited: Oct 10, 2007
  9. Oct 10, 2007 #8
    QC 10.7 is done exactly the same way that QC 10.6 is done.
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