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Fermion Anticommutation Relations (nightmare!)

  1. Sep 6, 2012 #1
    Hi. I've been thinking about this proof for over a day now and have reached the point where I can't come up with any new approaches!

    I'm trying to prove equation (5.15) in these notes:
    http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf


    Just above eqn (5.15) we are told that the proof should be exactly the same as the proof of (5.5) which is done on p107.

    I completely understand the proof on p107 for the commutation relations.
    When trying to prove the anticommutation relations, the only difference is going to be a minus sign on the second term.

    In other words, the proof is the same as on p107 except we have

    [tex] \{ \psi( \vec{x}), \psi^\dagger \vec{y} \} = \displaystyle\sum_{s} \in \frac{d^3p}{(2 \pi)^3} \frac{1}{2E_{\vec{p}}} \left( u^s( \vec{p} \bar{u}^s(\vec{p}) \gamma^0 e^{i \vec{p} \cdot ( \vec{x}-\vec{y})} + v^s(\vec{p}) \bar{v}^s(\vec{p}) \gamma^0 e^{-i \vec{p} \cdot (\vec{x} - \vec{y})} \right) [/tex]

    This means that if we follow through the next few steps on p107 we arrive at

    [tex] \int \frac{d^3p}{(2 \pi)^3} \frac{1}{E_{\vec{p}}} ( p_i \gamma^i + m ) \gamma^0 e^{i \vec{p} \cdot ( \vec{x} - \vec{y} )}[/tex]

    and as far as I can tell there is no way to make that into a delta function!!!!!


    ANY HELP IS GREATLY APPRECIATED!!!
     
  2. jcsd
  3. Sep 6, 2012 #2

    Bill_K

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    Do you really trust this reference?? Because other places give different results!

    For a scalar field the commutator is usually quoted as [..x, y are 4-vectors..]

    [φ(x), φ*(y)] = iΔ(x - y)

    where Δ is the "invariant delta function", defined by

    iΔ(x) = (2π)-3∫d3k/2ω [e-ik·x - eik·x]

    Δ(x) vanishes for equal times, and its time derivative at equal times is a 3-dimensional δ function: (∂Δ(x)/∂x0)|(x0 = 0) = - δ3(x). Notice it is not true that [φ(x), φ*(y)] at equal times is a δ-function.. you must include the time derivative.

    The corresponding relation for Dirac spinor fields is

    {ψ(x), ψ†(y)} = -iS(x - y)

    where

    S(x - y) = (iγμμ + m)Δ(x - y)
     
  4. Sep 7, 2012 #3
    I think relations (5.15) are correct (although I'm not sure for the (2π)[itex]^{3}[/itex] that probably depends on the normalization of u and v spinors). I think you can extract an expression for c and b operators using the spinor scalar product. For example, if I'm not wrong, the following equation should be true:
    $$b^{r}_{\vec{p}}=\int{d^3x U^{\dagger}(r\vec{p})\psi(x)}$$
    where:

    $$U(r\vec{p})=\frac{1}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_{\vec{p}}}}u^s(\vec{p})e^{-ipx}$$

    When you have extract both b and c you can anticommutate them and using the anticommutation relations on ψ and ψ[itex]^{+}[/itex] you should reach your goal.

    I hope I'm not wrong, as I'm using my old notes and there should be something different from yours but I think the main idea should be ok.
     
  5. Sep 7, 2012 #4

    Bill_K

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    Tong's proof is trying to go the other way. He assumes (5.15) and tries to prove (5.14). But as I indicated above, (5.14) is wrong. {ψ(x), ψ†(y)} is not equal to a delta function.
     
  6. Sep 7, 2012 #5
    Maybe I'm missing something. Is ψ the fermionic field? If so, I'm quite sure that 5.14 are correct. I think you got wrong when you said that:

    $$[\phi(x),\phi^\dagger(y)]=i\Delta(x-y)$$

    as the correct relations, for scalar fields, should be:

    $$[\phi(x),\dot{\phi}^\dagger(y)]=i\delta^3(\vec{x}-\vec{y})$$
    and
    $$[\phi(x),\phi(y)]=i\Delta(x-y)$$.

    And the same thing is valid for fermionic fields. You can see for example Mandl-Shaw "Quantum Field Theory" equations 3.25 and 3.42.
    I hope I'm not wrong, if so I'm sorry :biggrin:
     
  7. Sep 7, 2012 #6

    Bill_K

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    Einj, The commutation relations for the Klein-Gordon field are slightly different depending on whether the field is real or complex.

    For a real field, [φ(x), φ(y)] = -iΔ(x - y) (Sorry, I had the sign wrong.)

    Whereas for a complex scalar field, [φ(x), φ*(y)] = -iΔ(x - y)

    Some references call the invariant function D(x -y) instead of Δ(x - y).

    Note that your first two equations are equivalent. You can get the second one by taking the time derivative wrt y of the first one, since the equal-time time derivative of Δ(x - y) is δ3(x - y).

    No, for a Dirac field ψ(x) there's an additional factor: S(x - y) = (iγμμ + m)Δ(x - y). The expression derived by the OP has this factor in it.
     
  8. Sep 7, 2012 #7
    Ok, I knew that (and of course I said the same thing because S is quite the equivalent of Δ but for fermionic field). However, what should be the fermonic equivalent of my second equation?
     
  9. Sep 7, 2012 #8

    Bill_K

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    One big difference between S and Δ: S is a 4 x 4 matrix!

    To get a delta function out of Δ, you need a time derivative. But in the Dirac case, the time derivative is already there, thanks to the (iγμμ + m) in front. At equal times, Δ(x - y) is zero, so all the terms in the definition of S drop out except for the time derivative. You get

    {ψ(x), ψ†(y)}|(at x0 = y0) = - γ00Δ(x - y) = γ0 δ3(x - y)

    (To leave no doubt about the notation, ψ† is what appears in the bilinear covariants, e.g. jμ = eψ†γμψ.) And so, if you'd rather, using ψ† = ψ*γ0,

    {ψ(x), ψ*(y)}|(at x0 = y0) = δ3(x - y)

    Maybe this is what Tong had in mind when he wrote (5.14).
     
  10. Sep 7, 2012 #9
    Ok, understood. Sorry for the misunderstanding.
     
  11. Sep 8, 2012 #10
    Last edited: Sep 8, 2012
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