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Qft question (Srednicki)

  1. May 20, 2008 #1
    I have been working through Srednicki this summer to teach myself qft, and all too often I've gotten stuck on a small point and ended up spending a great deal of time clearing it up by myself. While this is probably an important part of the learning process, I am progressing a bit too slowly, so I thought I would post some of the questions I have here if anyone is willing to help me out. Please forgive me in advance if any of the questions are overly simple or stupid.

    To start with, I'm stuck in section 39 on working out the anti-commutation relations for the creation and annihilation operators. For example, take the calculation at the top of pg 246 (eq 39.16)

    $\{b_s(\textbf{p}),b_{s'}^{\dag}(\textbf{p}')\} = \int d^3xd^3ye^{-ipx+ip'y}\{\overline{u}_s(\textbf{p})\gamma^0\Psi(x),\overline{\Psi}(y)\gamma^0u_{s'}(\textbf{p}')\}$\\
    $=\int d^3xd^3y e^{-ipx+ip'y}\overline{u}_s(\textbf{p})\gamma^0\{\Psi(x),\overline{\Psi}(y)\}\gamma^0u_{s'}(\textbf{p}')$

    Srednicki starts with the second line, but how does that follow from the first line that I have written?
  2. jcsd
  3. May 20, 2008 #2
    I was almost done my first quantum field theory course before I remembered that spinors have indices, and that those indices can be put to good use. The eigenfunctions [tex]u(\textbf{p})[/tex] are not quantum operators, or anything fancy. They're just complex spinors. So write
    [tex]\bar\Psi \gamma^0 u[/tex] as [tex]\psi^*_i \gamma^0_{ij} u_j[/tex] (summation implied).

    Now the fields themselves obey the canonical anticommutation relation
    [tex]\left\{ \Psi^*_i(x) \gamma^0_{ij}, \Psi_k(y) \right\} = \delta_{jk} \delta(x-y)[/tex].

    This should make things clear.
  4. May 20, 2008 #3
    Indeed it does, thanks!

    Though I think there is a missing (or extra, if you like) [tex]\gamma^0[/tex] in your third line. Since [tex]\overline{\Psi}=\Psi^{\dag}\gamma^0[/tex], should we not have [tex]\Psi^*_i \gamma^0_{ij} \gamma^0_{jk} u_k[/tex]?
  5. May 20, 2008 #4
    Yes, sorry. The Lagrangian is [tex]\bar\psi^\dagger \gamma^0 \partial_0 \psi + ...[/tex], so the canonical momentum to [tex]\psi[/tex] is simply [tex]\psi^\dagger[/tex]. My bad.

    In other words:

    [tex]\left\{ \Psi^*_i(x), \Psi_k(y) \right\} = \delta_{ik} \delta(x-y) [/tex]
    Last edited: May 20, 2008
  6. May 20, 2008 #5
    No problem, just wanted to make sure I had things straight. I'll probably have more questions soon :)
  7. May 26, 2008 #6
    Ok, new question, this time from Steve Martin's phenomenology notes (http://zippy.physics.niu.edu/phys586_spring2002.html [Broken]). On page 160, eq 9.78, the claim is that the first term transforms as an adjoint representation and the second as a singlet. I'm trying to verify this by using eq 9.29, the definition of the tensor product, and then using the definition of the complex conjugate representation and the fact that the matrices are hermitian. I seem to indeed get zero when acting on the second term, but since the operators are linear doesn't this mean that when acting on the first term (that is, the thing enclosed in the first parentheses) I should get zero for the part that is subtracted, and so just end up with the same transformation I already had? (If this is unclear to anyone let me know and I'll try to tex up my work.)

    Last edited by a moderator: May 3, 2017
  8. May 26, 2008 #7
    One index transforms as [tex]N[/tex], and the other one as [tex]\overline{N}[/tex]. The whole thing therefore transforms "like" an adjoint object. The point is to break it up into irreducible representations. The trace doesn't mix with the traceless part, and so you can break the two apart. The trace transforms as a singlet because the trace of a matrix is invariant under similarity transformations, and so is the Kronecker delta.

    I have some notes here: http://www.mathematics.thetangentbundle.net/wiki/Lie_group/reducing_direct_products_of_irreducible_representations [Broken] Unfortunately it is for SO(N), but similar ideas hold.
    Last edited by a moderator: May 3, 2017
  9. May 26, 2008 #8
    I guess I'm not entirely sure what it means to say that it transforms as the adjoint representation. Doesn't this mean that if I restrict the map [tex]T_{N\times\bar{N}}^a[/tex] to the subspace defined by the first parenthesis, it should act as the adjoint, which is defined by [tex](T^a)^b_c=-if^{abc}[/tex] (where f denotes the structure constants)? I know it can't be so simple, because the subspace defined by the first parenthesis is still described by two indices instead of one. Is there some obvious isomorphism from this subspace to the vector space that the adjoint is defined over?
  10. May 27, 2008 #9
    Ok generally the adjoint representation of a group acts on a vector in the Lie algebra as [tex]g X g^{-1}[/tex], so the adjoint is defined as having an [tex]N[/tex] component, and it's inverse. For unitary groups, that inverse is also it's complex conjugate, so we get [tex]\overline{N}[/tex] for the second index. (Here are some more useless notes: http://www.mathematics.thetangentbundle.net/wiki/Lie_algebra/adjoint_representation [Broken]). The point is that this is only a necessary, but not sufficient condition to give a representation of the adjoint representation. The adjoint representation is also irreducible, so you don't want bits of the trivial representation hanging around. That's why you subtract off the trace.

    It's more a group theory thing than anything else. I wish I had a better explanation.
    Last edited by a moderator: May 3, 2017
  11. May 29, 2008 #10
    Dear All,

    I have a question regarding adjoint representation. Adj representation is a real representation. Does it mean that the gauge field under this transformation have to be necessarily real? if I take some other field like Higgs triplet or fermionic triplet ( let say I have three fermions which transform in adjoint representation of SU(2)_L gauge group of standard model ), then what would be the situation? Fermionic field is a spinor field and complex.
  12. May 29, 2008 #11
    Adjoint representation

    Dear All,

    I'm a bit confused. Is Adjoint representation a real representation? The definition of a real representation is the generator T^a and Conjugate of T^aare related by similarity transformation for every a. The definition of Adj representation is gXg^-1 . But does it ensure that it is a real representation?
  13. May 29, 2008 #12
    So, my question is does Adjoint representation has something to do with the real fields? or can we take fields transforming in Adjoint representation as complex also.What about gauge fiels? are they real or complex?
  14. May 29, 2008 #13


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    Gauge fields must be real in order for the gauge-covariant derivative to be hermitian as a differential operator. I'm not sure off-hand what would go wrong if this is not the case. EDIT: for one thing, the gauge-invariant fermion kinetic operator would not be hermitian. I'm sure other stuff goes wrong as well.
    Spinor and complex scalar fields can be in real representations, but do not have to be. A real scalar field must be in a real representation.
    Last edited: May 29, 2008
  15. May 29, 2008 #14


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    A more pedestrian definition of the adjoint rep is that it is the rep whose generator matrices are [itex](T^a)^{bc}=-if^{abc}[/itex], where [itex]f^{abc}[/itex] are the structure coefficients that appear in the commutation relations [itex][T^a,T^b]=if^{abc}T^c[/itex]. Then it is obviously true that the adjoint-rep generator matrices satisfy the reality condition [itex](T^a)^*=-T^a[/itex].
  16. May 30, 2008 #15
    Thanks Avodyne. It's clear.
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