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Proving [itex][a_k^\dagger, a_q^\dagger]=0[/itex]

  1. Nov 25, 2014 #1
    I am trying to prove the commutation relations between the creation and annihilation operators in field theory. I was already able to show that [itex][a_k, a_q^\dagger]=i\delta(k-q)[/itex]. I want to show that [itex][a_k^\dagger, a_q^\dagger]=0,[/itex] but I am stuck on the last step. I think my confusion has to do with interpreting the delta function and I was hoping someone can help me out. So far I have shown that

    [tex]a_k^\dagger = \int \! d^3\!x \: e^{-ikx}\left[ \sqrt{\frac{E_k}{2}} \phi(x) - i \sqrt{\frac{1}{2E_k}} \pi(x) \right],[/tex]

    where [itex]\phi(x)[/itex] and [itex]\pi(x)[/itex] are the field and conjugate momentum operators respectively. Based on this I was able to show that

    [tex][a_k^\dagger, a_q^\dagger]=\int \! d^3y \: d^3x \: e^{-i(kx+qy)} \left[ -i \sqrt{\frac{E_k}{E_q}} [\phi(x), \pi(y)] + i \sqrt{\frac{E_q}{E_k}} [\phi(y),\pi(x)] \right].[/tex]

    Using the commutation relation [itex][\phi(x), \pi(y)]=i \delta^3(x-y)[/itex] this becomes

    [tex][a_k^\dagger, a_q^\dagger]=\int \! d^3y \: e^{-i (k+q) y} \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]

    And I believe the integral over y gives us a delta function. If I have done everything correctly then this should be

    [tex][a_k^\dagger, a_q^\dagger]=\delta^3(k+q) \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]

    Since I already know the commutation relations I know that the final expression should equal zero for any arbitrary value of [itex]k[/itex] and [itex]q[/itex]. But my final expression doesn't really say that for [itex]k=-q[/itex]. I don't really understand where this discrepancy is coming from, though. Is there some way to interpret this delta function that I'm not understanding or have I made a computational error somewhere? Thanks
     
  2. jcsd
  3. Nov 25, 2014 #2

    Demystifier

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    Actually it does, because the square bracket vanishes in that case and ##\delta(0)\cdot 0 =0##.
     
    Last edited: Nov 25, 2014
  4. Nov 25, 2014 #3
    I don't really understand why. I guess the [itex]\infty \cdot 0 = 0[/itex] is what's throwing me off. Is there a way to put this on firmer mathematical footing?
     
  5. Nov 25, 2014 #4

    Matterwave

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    Just take it a step back to the integral. The integral of 0 is 0, even if the integral is over all of space. :)
     
  6. Nov 26, 2014 #5

    strangerep

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    Hint: what's the formula for ##E_k## ... ? ;)
     
  7. Nov 26, 2014 #6

    bhobba

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    You need to study distribution theory - get the following book - you wont regret it:
    https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    The Dirac Delta function only has a meaning in integrals. When integrated it gives the value of the function before it at zero. In your case that's zero, so, when integrated gives zero. Hence, on a distribution basis is the same as zero.

    Its a bit strange on first acquaintance, but once you get used to it its easy - which is why I really do suggest getting the book I mentioned - it will make everything clear - and much much more besides eg Fourier transforms become a doodle.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  8. Nov 26, 2014 #7

    Demystifier

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    The ##\delta##-function has a mathematical meaning only inside an integral:
    $$\int dx\, \delta(x) f(x) =f(0)$$
    which is ##0## if ##f(0)=0##.

    Another, more heuristic way to think of a product ##\delta(k-q) f(k,q)## is to replace the Dirac ##\delta(k-q)## with the Kronecker ##\delta_{kq}##. In this way you can put
    $$\delta_{kq}f(k,q)=f(k,k)=0$$
    without doing the integral/sum.
     
  9. Dec 8, 2014 #8
    Thank you for the replies. I will check out that book, Bill!
     
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