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Summing spinor products

  1. Nov 29, 2008 #1

    CompuChip

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    Hi,

    I doubted whether to post this in the homework section, but decided not to because a) it's only a very small part and b) I think my question is important in general.

    Looking at page 107 of Zee's Introduction to QFT I am trying to get from (16) to (17). For this I need to evaluate several "inner" products like
    [tex]\sum_{\alpha} (u^\dagger)^\alpha(p, s) u_\alpha(p, s')[/tex]
    and
    [tex]\sum_{\alpha} (v^\dagger)^\alpha(p, s) u_\alpha(-p, s')[/tex]
    where - using standard notation, I think, u and v are the amplitudes of plane wave solutions [itex]u(p, s) e^{-i \vec p \cdot \vec x}[/itex], [itex]v(p, s) e^{i \vec p \cdot \vec x}[/itex] of the Dirac equation. Question: How do I do that? :smile:

    It's also somewhat related to page 105 of Zee, where he calculates
    [tex]\sum_s u_\alpha(p, s) \overline u_\beta(p, s) = \frac12 (\gamma^0 + 1)_{\alpha\beta}[/tex]
    (and similarly for u -> v) in the rest frame and then without any explanation jumps to the general (8) and (9),
    [tex]\sum_s w_\alpha(p, s) \overline w_\beta(p, s) = \left( \frac{\gamma^\mu p_\mu \pm m}{2m} \right)_{\alpha\beta}[/tex]
    (with the plus sign for w = u and minus for w = v).
    Apart from the fact I don't really see how he does that, I also don't think it will work for my question above, as [itex]u^\dagger u[/itex] is not an invariant, so we cannot first calculate it in the rest frame [plus I am not summing over s].
     
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  3. Nov 29, 2008 #2

    StatusX

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    For your second question, the idea is that you can say a lot about a quantity if you know how it transforms. For the case of:

    [tex]
    \sum_s u_\alpha(p, s) \overline u_\beta(p, s) = \frac12 (\gamma^0 + 1)_{\alpha\beta}
    [/tex]

    you know the effect of a Lorentz transformation on the LHS: just apply the spinor rotation matrices to each spinor. This has the effect of conjugating the LHS by the spinor rotation matrices (ie, [itex]u \rightarrow S u[/itex] , [itex]\overline u \rightarrow \overline u S^{-1}[/itex], so LHS -> [itex] S [/itex] LHS [itex]S^{-1}[/itex]). Then the RHS must transform the same way. Since you know it in the rest frame, you can get it in any other frame by applying to it the transformation that takes the rest frame to the frame in question.

    But there's an easier way. You know how [itex]\gamma^\mu[/itex] transforms under such a conjugation: it's four vector index gets rotated. Then for the expression to be Lorentz invariant, it must be the case that we're contracting with [itex]\gamma^\mu[/itex] a four vector which only has a time component in the rest frame. The only possibility is the four momentum, and picking factors so our equation reduces to the above in the rest frame, we get the general expression you wrote down.

    For the first question, the easiest thing is probably to evaluate it in the rest frame, where you have an explicit form for u and v. Then the quantity has no spinor indices (they are all contracted), so actually it is a spacetime scalar and so this gives you the general answer in any frame.
     
  4. Nov 30, 2008 #3

    CompuChip

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    Thanks a lot for the elaborate answer on the second part, I think I got it now.

    Actually, I thought that we preferred working with [itex]\overline u u \propto u^\dagger \gamma^0 u[/itex] (proportionality factors of - and i ignored :smile:) precisely because [itex]u^\dagger u[/itex] itself is not Lorentz invariant. So I would expect something like: [itex]\overline u \gamma^\mu u[/itex] transforms as a vector, and [itex]u^\dagger u[/itex] is (related to -- upto factors of -1, i and [itex](\gamma^0)^2) = \pm 1[/itex] --) the [itex]\mu = 0[/itex] component of it.

    Anyway, in the rest frame, using the notation from Zee (pp. 104,5), with u(p, +) = (1, 0, 0, 0) and u(p, -) = (0, 1, 0, 0):
    [tex]\sum_{\alpha} u^\dagger_\alpha(p, s) u_\alpha(p, s') = \delta_{ss'}[/tex]
    However, looking at the answer I am supposed to get in my calculation, I would expect something like
    [tex]-i \frac{\vec p^2 + m^2}{m} \delta_{ss'}[/tex]
    I don't know which factors should come from the contraction and which should just appear elsewhere in my calculation though. But I have
    [tex]i m \int \frac{d^3p}{\sqrt{p^2 + m^2}} \sum_{s, s'} b^\dagger(p, s) b(p, s') u^\dagger_\alpha(p, s) u_\alpha(p, s')[/tex]
    and this should ultimately give
    [tex]\int d^3p \sqrt{\vec p^2 + m^2} \sum_s b^\dagger(p, s) b(p, s)[/tex].

    Should I write the result of the summation as something like [itex]\delta_{ss'} \frac{p^\mu p_\mu}{m^2} = \delta_{ss'} \frac{\vec p^2 + m^2}{m^2}[/itex], as this reduces to [itex]\delta_{ss'}[/itex] in the rest frame and is Lorentz invariant? (Just guessing now, I don't really see yet why one would want to do that while [itex]\delta_{ss'}[/itex] seems perfectly all right already).

    Am I making any sense?
     
  5. Nov 30, 2008 #4

    CompuChip

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    The factor i which I mentioned does not, as I suspected, come from the contraction but it drops out against some other i in my calculation (luckily).

    In some obscure lecture notes I found the idea of writing down u, which satisfies [itex](\not p - m) u = 0[/itex], in the rest frame and act on it by [itex](\not p + m)[/itex]. Then because [itex](\not p - m)(\not p + m) = 0[/itex] the result will be a solution. Following this procedure, I have found
    [tex]u(p, +) = (E + m, 0, p^3, p^1 + i p^2)[/tex] and [tex]u(p, -) = (0, m + E, p^1 - i p^2, -p^3)[/tex].
    Then if I do the contraction, I get
    [tex]u(p, s) u(p, s') = \delta_{ss'}( E(p) / m )[/tex]
    which gives me the correct answer. Does that mean that my ad-hoc argument from the previous post was right?
     
    Last edited: Nov 30, 2008
  6. Nov 30, 2008 #5

    StatusX

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    Sorry, I didn't read it carefully and assumed it was [itex]\overline u[/itex] instead of [itex]u^\dagger[/itex]. You're correct in pointing out this implies the quantity transforms as the time component of a four vector. If you've found it to be 1 in the rest frame (ignoring the s indices), then the general answer is simply [itex] p^0/m [/itex], ie, the only (relevant) four vector time component which is 1 in the rest frame. Using the mass shell condition, this can be rewritten as [itex] \sqrt{\vec p^2 + m^2}/m [/itex], which seems to be what you found in your last post.

    This has different units than the answer you found, which means it should be easy to find the mistake. I'm guessing there's different conventions for u(p,s) being used. What are the units of u(p,s)?


    [itex]p^\mu p_\mu = m^2[/itex] in every frame, that's why it's a scalar. You're forgetting gamma factors. [itex]\vec p^2 + m^2[/itex] would not work because it's not a scalar.
     
  7. Dec 1, 2008 #6

    CompuChip

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    That, too, was a source of confusion. I was following Zee, who only wrote down an explicit form u = (1, 0, 0, 0) and (0, 1, 0, 0) in the rest frame, and then babbled something about normalization conditions without being explicit. I found out that to recover the results he writes down later on, you have to normalize by a factor of 1/sqrt(2 m (m + p^0)) = [1/E] [IIRC, don't have the precise solutions at hand now].

    So by explicitly doing the calculation (as I explained earlier, using that [itex](\not p + m)(\not p - m) = 0[/itex] we can act, for example, with [itex](\not p - m)[/itex] on the rest frame form to get a solution u of [itex](\not p + m) u = 0[/itex]), I managed to get the right answer. I think I also get the idea of the "smart guess" approach (examine how it transforms and construct something which transforms in the same way and gives the right result in the rest frame) although the practical implementation is still giving me some difficulties. I guess it just takes some practice (and in the beginning, a lot of calculations to check the result) :smile:

    Thank you very much for your help StatusX.
     
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