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QFT: Trace Theorem

  1. Mar 27, 2007 #1
    Hi guys,

    Everyone knows that one can calculate the S-Matrix with various tricks. One of them is to use traces to simplify the matrices.
    Can someone tell me or point me to a place where I can find an explanation why I can do this?
    I think I vaguely remember that I have seen once an equation showing why traces can be used, but I can't recall that source (book? lecture notes? websites?)

  2. jcsd
  3. Mar 27, 2007 #2

    Hans de Vries

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    You can find the trace theorem explained in most QFT books, for instance
    in Griffiths chapter 7.7.

    Now how does the Trace come in? First you manipulate the expression
    until [itex]u[/itex] and [itex]{\bar u}[/itex] get back-to-back. This is an inner product:

    [tex]u{\bar u}[/tex]

    Then you replace it with the completeness matrix:

    \sum_{k\ =1,2} u^{(s)}{\bar u}^{(s)}\ =\ (\gamma^\mu p_\mu + mc)

    This is a 4x4 matrix while you need the inner product: [itex]u_0{\bar u_0}+u_1{\bar u_1}+u_2{\bar u_2}+u_3{\bar u_3}[/itex]

    Now, the inner product is the sum of the diagonal elements, and thus: The trace.

    u{\bar u}\ =\ \mbox{Tr}(\ \gamma^\mu p_\mu + mc\ )

    Regards, Hans
    Last edited: Mar 27, 2007
  4. Mar 28, 2007 #3
    It's important to remember just how many different labels and indices various entities in QFT actually having, a great deal of them are supressed and it's easy to forget or get confused between them.

    In a slight elaboration of Hans' explaination, remember that spinors have a spinor index. [tex]\mathbf{u}[/tex] is a spinor with all terms supressed on it. It's a function of momentum, [tex]\mathbf{u}(p)[/tex], it's got spin (which for usual fermions are s = +-1/2) so [tex]\mathbf{u}^{s}(p)[/tex] and then there's the spinor index, which you can think of as the entry in the vector expression for the spinor, [tex]\mathbf{u}_{\alpha}^{s}(p)[/tex].

    An inner product of spinors is [tex]\bar{\mathbf{u}}\mathbf{u}[/tex] (Hans has that the wrong way around), which when you put in all the various indices you have [tex]\bar{\mathbf{u}}_{\alpha}^{r}(p)\mathbf{u}_{\beta}^{s}(q)[/tex].

    You how have an expression which is the inner product of two spinors of different spin (s,r labels), spinor polarisations ([tex]\alpha[/tex], [tex]\beta[/tex]) and momentum (p,q).

    You'll get an equalisation of momentum, so p=q always.

    Summing over spins and averaging means you can make use of the completeness relation Hans mentions, getting rid of spin complications and giving you

    [tex]\sum_{s= \pm \frac{1}{2}} \bar{\mathbf{u}}_{\alpha}^{s}(p)\mathbf{u}_{\beta}^{s}(p) = p_{\mu}(\gamma^{\mu})_{\alpha \beta} + m \mathbb{I}_{\alpha \beta}[/tex]

    From this formate you are able to compute specific reactions which have to take into account polarisation. This would involving saying what polarisation your incoming and outgoing particles are (ie giving values from 0 to 3 to alpha and beta in a specific choice of basis for your spinors and gamma matrices) and then working with the relevent entries within the matrix equation I just gave.

    This is annoying for two reasons. Firstly, many experiments don't care about polarisation so unless it's desperately needed, it's more work than required. If you're only just learning QFT, it's certainly more work than required. Secondly, you have to pick a representation for your gamma matrices and the spinor basis, that meaning you have to give a specific form of your gamma matrices.

    If instead you're only interested in averages and your expriments are blind to polarisation you just sum over them all so [tex]\alpha = \beta[/tex], but for any matrix [tex]M_{\alpha \beta}[/tex] the quantity [tex]M_{\alpha \alpha}[/tex] is the trace. This has the useful property that you don't need to pick a specific form for your spinors and your matrices, you get all the information you need from the gamma's anticommutation relation [tex]\{ \gamma^{\mu} , \gamma^{\nu} \} = g^{\mu \nu}[/tex]. Taking the trace of that immediately gives [tex]\textrm{Tr}(\gamma^{\mu} \gamma^{\nu}) = 4g^{\mu\nu}[/tex] and others follow from that and the anticommuting [tex]\gamma^{5}[/tex] matrix.
    Last edited: Mar 28, 2007
  5. Mar 28, 2007 #4

    Hans de Vries

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    Just a small remark:

    There is the inner product which gives the Lorentz scalar 2m:

    [tex]\bar{\mathbf{u}}\mathbf{u} = 2m[/tex]

    and then there is the inner product the other way round:


    Which, being less sloppy, I should write as the sum of two inner products:

    [tex]\mathbf{u^\uparrow}\bar{\mathbf{u}}^\uparrow +
    \mathbf{u^\downarrow}\bar{\mathbf{u}}^\downarrow\ =
    \mbox{Tr}(\ \gamma^\mu p_\mu + mc\ )[/tex]

    which gives us the Trace.

    Regards, Hans
    Last edited: Mar 28, 2007
  6. Mar 28, 2007 #5
    ^ Too true, my mistake, should have remembered that. Thinking too much about inner products giving scalars and not anything else. Sorry about that.
  7. Mar 30, 2007 #6
    Wow, thanks for the neat explanation, AlphaNumeric and Hans
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