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Ladder operators in Klein -Gordon canonical quantisation

  1. Oct 18, 2015 #1
    The quantum Klein-Gordon field ##\phi({\bf{x}})## and its momentum density ##\pi({\bf{x}})## are given in Fourier space by

    ##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)## and

    ##\pi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} (-i) \sqrt{\frac{ \omega_{{\bf{p}}}}{2}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} - a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)##.

    [These are equations (2.25) and (2.26) from the Peskin and Schroeder.]

    Now, I used the substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained

    ##a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = a^{\dagger}_{{\bf{p}}}##.

    On the other hand, I used the same substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained

    ##a_{{\bf{p}}} = - a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = - a^{\dagger}_{{\bf{p}}}##.

    Can someone explain what's going on?
     
    Last edited: Oct 19, 2015
  2. jcsd
  3. Oct 19, 2015 #2

    vanhees71

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    There is a - missing in the second term of ##\Pi(\vec{p})##. That's seen in the Heisenberg picture and the relation ##\Pi=\dot{\phi}##.
     
  4. Oct 19, 2015 #3
    Sorry, that's my own mistake. Peskin and Schroeder has the minus sign. I've made the edit now.
     
  5. Oct 19, 2015 #4
    Actually, with the minus sign in the second term of ##\pi({\bf{x}})##, I get the inconsistency in the ladder operators.

    Does your second sentence solve the inconsistency or justify the minus sign in the second term?
     
  6. Oct 19, 2015 #5

    vanhees71

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    I don't understand, how you get these equations in the first place. The ##\hat{a}_{\vec{p}}## are an independent set of annihilation operators. There shouldn't be a relation between ##\hat{a}_{\vec{p}}## and ##\hat{a}_{-\vec{p}}^{\dagger}##.
     
  7. Oct 19, 2015 #6
    Well, under the substitution ##{\bf{p}} \rightarrow {\bf{-p}}##,

    ##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)##

    becomes

    ##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{-\bf{p}}}}} \big( a_{-{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{-{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} \big)##,

    and then I compared the coefficients of the exponential ##e^{-i{\bf{p}} \cdot {\bf{x}}}## on both sides to obtain ##a^{\dagger}_{{\bf{p}}} = a_{-{\bf{p}}}## and the coefficients of the exponential ##e^{i{\bf{p}} \cdot {\bf{x}}}## on both sides to obtain ##a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}##.

    But now that you mention it, I'm beginning to think my argument is wrong. The creation and annihilation operators are, in fact, independent of each other, i.e. for instance, ##a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}}## is the same as ##a_{-{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}}## because the integration is over the entire range of momenta from ##-\infty## to ##\infty##.

    Am I correct?
     
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