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Derivative inside, outside time ordering

  1. Jul 24, 2008 #1
    Dear guys,

    In GSW's string theory book, p.43, he derived and demonstrated the Ward identity. Start from the M-point function,
    [tex]A_{\mu_1m_2\cdots\mu_M}(k_1,k_2,\cdots,k_M) = \int d^4x_1\cdots d^4x_M\,e^{ik_i\cdot x_i}\,\left\langle T\big[J_{\mu_1}(x_1)J_{\mu_2}(x_2)\cdots J_{\mu_M}(x_M)\big]\right\rangle[/tex]
    where [tex]\langle\cdots\rangle[/tex] is the vacuum expectation value. If we have a gauge symmetry, such that the current is conserved, we have
    [tex]\left\langle T\big[\partial_{\mu_1}J^{\mu_1}(x_1)J_{\mu_2}(x_2)\cdots J_{\mu_M}(x_M)\big]\right\rangle = 0[/tex]
    The key point(which I don't quite understand) is, if the currents commute, we can move the derivative [tex]\partial_{\mu_1}[/tex] outside the time ordering, hence
    [tex]0 = \partial_{\mu_1}\left\langle T\big[J_{\mu_1}(x_1)J_{\mu_2}(x_2)\cdots J_{\mu_M}(x_M)\big]\right\rangle\quad\cdots(*)[/tex]
    so that, with (*), we can easily derive,
    [tex]k^{\mu_1}A_{\mu_1\mu_2\cdots\mu_M}(k_1,k_2\cdots,k_M) = 0[/tex], i.e. the Ward identity.

    And subsequently he said, "In the nonabelian case the structure of Ward identities is more complicated. Roughly speaking, the electromagnetic current is replaced by Yang-Mills currents [tex]J^a_\mu[/tex], which, while still conserved, do not commute with each other. Hence, in trying to remove the derivative from inside the T-product, one picks up extra terms involving involving equal-time commutators."

    What I don't understand is,
    (1) [tex]J^a_\mu[/tex] are just composed of the matter fields, say [tex]\psi[/tex], [tex]J^a_\mu[/tex] have nothing to do with the generators, why are they related to the nonabelian nature of the gauge group?
    (2) I guess it's due to that [tex]J^a_\mu[/tex] are composed of the matter field and the canonical momentum, so that to change the order of [tex]J^a_\mu[/tex], we have "equal-time" commutators. But this is the feature of both abelian and nonabelian gauge theory. why we only have this problem is abelian gauge theory?
    (3) It not clear to me why we have to change the order of currents after moving derivative to the outside? Can anyone give me an example of how to do this?

    Thanks so much for any instructions!
  2. jcsd
  3. Jul 24, 2008 #2
    In QED, the EM current is conserved [itex]\partial_\mu j^\mu=0[/itex] But in QCD, the current [itex]j^{a\mu}=\bar{q}\gamma^\mu t^a q[/itex] requires a covariant derivative
    [tex]{\cal D}_\mu j^{\alpha\mu}=(\partial_\mu\delta^{ac}+gf^{abc}A^b_\mu)j^{c\mu}=0[/tex]
    Just as a local conserved energy momentum-tensor does not exist in general curved space in GR, this does not describe any local conservation law. Therefore, there is no reason for [itex]k_\mu A^{\mu\nu}(k_1,k_2,\cdots)[/itex] to be zero, and in fact it is not. The Ward-Takahaski identities are replaced in QCD with the Slavnov-Taylor identities, and can not be expressed in a simple compact way on one-particle irreducible functions. You need to introduce ghost fields and make a Becchi Rouet Stora transformation.

    Do you have access to Faddeev's and Slavnov's book, or Itzykson's and Zuber's ?
  4. Jul 24, 2008 #3
    It seems that I have to make up lots of knowledge to fully understand this. :P
    Thanks! Originally I thought that [tex]J^\mu_a[/tex] has nothing to do with the group generator. Actually, the Noether current [tex]J^\mu_a\sim \frac{\partial L}{\partial\psi_{,\mu}}\delta\psi_a[/tex] can be calculated to be [tex]J^\mu_a\sim \bar{\psi}T_a\gamma^\mu\psi[/tex] from the Lagrangian [tex]L = \bar{\psi}(iD_\mu\gamma^\mu-m)\psi - \frac{1}{4}<F_{\mu\nu}F^{\mu\nu}>[/tex]. So, the currents in nonabelian gauge theory do actually not commute.
    Hmm...I know that in general relativity, we need covariant derivative because the spacetime curves. In the case of nonabelian gauge quantum field theory, we have similar situation? But the Noether theorem guarantees that the current is conserved in the way of [tex]\partial_\mu J^\mu_a = 0[/tex], not [tex]D_\mu J^\mu_a = 0[/tex], isn't it?
    Or...due to the quantum effect, we have to replace the partial derivative with the covariant derivative? I got confused, sorry.
    I just learned a little bit about Ward identity and BRST symmetry these days, I'm not very clear about the relation between them. I will go to study them deeper.
    I searched the library in my school just now, there is a book titled "gauge fields:quantum theory" by Faddeev and Slavnov, I will borrow it soon.
    Thank you so much!
  5. Jul 24, 2008 #4
    The analogy between QCD and GR stems mainly from the parallel transport on a fiber bundle over spacetime with SU(3) section, and the parallel transport on a curved Riemann surface. The gauge potential plays the role of the Christoffel symbols, and the curvature (of the fiber bundle in color space) is just given by the field strength [itex]F_{\mu\nu}=i[{\cal D}_{\mu},{\cal D}_{\nu}][/itex]
    The analogy is well-known for a long time. Since you are interested in string theory, let me mention that this is not the fashionable gauge/gravity duality (about which you will not find anything in GSW's book)
    The classical equation of motion would be [itex]\partial^{\mu}F^{a}_{\mu\nu}+gf^{abc}A^{b\mu}F^{c}_{\mu\nu}={\cal D}_{\mu}F^{a}_{\mu\nu}=-j^{a}_{\nu}[/itex] with the current from matter fields alone :
    [tex]j^{a}_{\nu}=-i\frac{\partial{\cal L}}{{\cal D}_\nu q}t^{a}q[/tex]
    or if you prefer
    [tex]j^{a}_{\nu}=-i\frac{\partial{\cal L}}{{\cal D}_\nu \psi}t^{a}\psi[/tex]
    The covariant derivative is not a quantum effect. It is classical gauge theory.
  6. Aug 4, 2008 #5
    Re: derivative inside, outside time ordering(Got it!)

    I think I understand the difference of derivatives inside and outside T-product. Consider the simplest case, two currents.
    [tex]\partial_{\mu_1}\text{T}[J^{\mu_1}_a(x)J^{\mu_2}_b(y)] = \partial_{\mu_1}\Big(J^{\mu_1}_a(x)J^{\mu_2}_b(y)\theta(x_0-y_0) + J^{\mu_2}_a(y)J^{\mu_1}_b(x)\theta(y_0-x_0)\Big)[/tex]
    [tex]= \text{T}\Big[\partial_{\mu_1}J^{\mu_1}_a(x)J^{\mu_2}_b(y)\Big] + (\nabla_1 J^{\mu_1}_a)(x)J^{\mu_2}_b(y)\delta(x_0-y_0) - J^{\mu_2}_b(y)(\nabla_1J^{\mu_1}_a(x))\delta(x_0-y_0)
    =\text{T}\Big[\partial_{\mu_1}J^{\mu_1}_a(x)J^{\mu_2}_b(y)\Big] + [J^{\mu_1}_a)(x),J^{\mu_2}_b(y)]\delta(x_0-y_0)[/tex]
    To sum up, we use step function to write out the T-product explicitly, and then perform the derivative. The derivative acts on the step functions arising the "equal-time" commutator, where equal-time is due to the delta function.

    Finally, the currents can be computed according to my second reply as [tex]J^a_\mu = \bar{\psi}T^a\gamma_\mu\psi[/tex], hence the equal-time commutation relations do not commute while the field theory is non-abelian.
    Therefore, there is nothing to do with the canonical commutation relation.
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