1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Srednicki QFT Problem 2.9

  1. Dec 20, 2016 #1

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    This is from Srednicki's QFT book, problem 2.9a:
    Let ##\Lambda = 1+\delta\omega## in the equation:
    $$ U(\Lambda)^{-1} \partial^{\mu}\varphi(x) U(\Lambda) = \Lambda^{\mu}{}_{\rho} \overline{\partial^{\rho}}\varphi(\Lambda^{-1}x) $$
    where ##\overline{\partial^{\rho}}## denotes differentiation with respect to ##\overline{x^{\rho}} = \Lambda^{\sigma}{}_{\rho} x##. Show that
    $$[\partial^{\rho}\varphi(x),M^{\mu \nu}] = \mathcal{L}^{\mu \nu} \partial^{\rho}\varphi(x) + (S^{\mu \nu}_V)^{\rho}{}_{\tau}\partial^{\tau}\varphi(x)$$

    2. Relevant equations
    $$U(1+\delta\omega) = I + \frac{i}{2\hbar}\delta\omega_{\mu \nu}M^{\mu \nu}$$
    $$\Lambda^{\rho}{}_{\tau} = \delta^{\rho}{}_{\tau} + \frac{i}{2\hbar}\delta\omega_{\mu \nu}(S^{\mu \nu}_V)^{\rho}{}_{\tau}$$
    $$\overline{\partial^{\rho}} = \Lambda^{\rho}{}_{\sigma} \partial^{\sigma}$$
    $$\mathcal{L}^{\mu \nu} = \frac{\hbar}{i}(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu})$$
    $$(S^{\mu \nu}_V)^{\rho}{}_{\tau} = \frac{\hbar}{i}(g^{\mu \rho}\delta^{\nu}{}_{\tau} - g^{\nu \rho}\delta^{\mu}{}_{\tau})$$
    3. The attempt at a solution
    The lhs of the first equation is:
    $$U(1-\delta\omega) \partial^{\alpha}\varphi(x) U(1+\delta\omega) = (I - \frac{i}{2\hbar}\delta\omega_{\mu \nu}M^{\mu \nu})\partial^{\alpha}\varphi(x)(I + \frac{i}{2\hbar}\delta\omega_{\mu \nu}M^{\mu \nu})$$
    To first order in ##\delta\omega##, this is:
    $$\partial^{\alpha}\varphi(x) + \frac{i}{2\hbar}\delta\omega_{\mu \nu}[\partial^{\alpha}\varphi(x),M^{\mu \nu}]$$
    The rhs of the first equation is:
    $$\Lambda^{\alpha}{}_{\rho} \Lambda^{\rho}{}_{\sigma}\partial^{\sigma}\varphi(x-x\delta\omega) =(\delta^{\alpha}{}_{\rho} + \frac{i}{2\hbar}\delta\omega_{\tau \nu}(S^{\tau \nu}_V)^{\alpha}{}_{\rho})(\delta^{\rho}{}_{\sigma} + \frac{i}{2\hbar}\delta\omega_{\tau \nu}(S^{\tau \nu}_V)^{\rho}{}_{\sigma})\partial^{\sigma}\varphi(x-x\delta\omega) $$
    To first order in ##\delta\omega_{\tau \nu}##, we have:
    $$\partial^{\alpha}\varphi(x-x\delta\omega) +\frac{i}{\hbar}\delta\omega_{\tau \nu}(S^{\tau \nu}_V)^{\alpha}{}_{\sigma}\partial^{\sigma}\varphi(x-x\delta\omega) $$
    Expanding ##\varphi(x-x\delta\omega)## to first order gives ##\varphi(x) + \delta\omega_{\mu \nu}x^{\mu}\partial^{\nu}\varphi(x) = \varphi(x) + \frac{1}{2}\delta\omega_{\mu \nu}(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})\varphi(x)=\varphi(x) + \frac{i}{2\hbar}\delta\omega_{\mu \nu}\mathcal{L}^{\mu\nu}\varphi(x)##, by the antisymmetry of ##\delta\omega_{\mu \nu}##. Substituting this into the above equation and retaining terms to first order in ##\delta\omega## gives:
    $$\frac{i}{2\hbar}\delta\omega_{\mu \nu}[\partial^{\alpha}\varphi(x),M^{\mu \nu}] = \frac{i}{2\hbar}\delta\omega_{\mu \nu}\partial^{\alpha}(\mathcal{L}^{\mu\nu}\varphi(x)) + \frac{i}{\hbar}\delta\omega_{\mu \nu}(S^{\mu \nu}_V)^{\alpha}{}_{\sigma}\partial^{\sigma}\varphi(x)$$
    Equating terms from each of the indices of ##\delta\omega_{\mu\nu}## gives us:
    $$[\partial^{\alpha}\varphi(x),M^{\mu \nu}] =\mathcal{L}^{\mu\nu}\partial^{\alpha}\varphi(x) + 2(S^{\mu \nu}_V)^{\alpha}{}_{\sigma}\partial^{\sigma}\varphi(x)$$
    This is almost right, but why am I getting an extra factor of two in front of the ##(S^{\mu \nu}_V)^{\alpha}{}_{\sigma}## term?

    EDIT: I'm an idiot. ##\partial^{\alpha}(\mathcal{L}^{\mu\nu}\varphi(x)) \ne \mathcal{L}^{\mu\nu}\partial^{\alpha}\varphi(x)##. I'll work on it a little more.
     
  2. jcsd
  3. Dec 20, 2016 #2

    ChrisVer

    User Avatar
    Gold Member

    why 2 Lorentz transformations for the derivative?
     
  4. Dec 20, 2016 #3

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Because ##\overline{\partial^{\rho} }= \Lambda^{\rho}{}_{\sigma}\partial^{\sigma}## is the Lorentz transformed derivative with respect to ##\overline{x^{\rho}}=\Lambda^{\rho}{}_{\sigma}x^{\sigma}##, so ##\Lambda^{\rho}{}_{\sigma} \overline{\partial^{\sigma}} = \Lambda^{\rho}{}_{\sigma} \Lambda^{\sigma}{}_{\tau} \partial^{\tau}##.
     
  5. Dec 20, 2016 #4

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Ok so I got a little further, but now I have a new question. First off, notice that I messed up in my expansion of ##\varphi(x-\delta\omega x)##. It should have a minus sign in front of the ##\delta\omega_{\mu\nu}## term. This just means that the ##\partial^{\alpha}(\mathcal{L}^{\mu\nu}\varphi(x))## term in the 2nd to last equation has a negative sign in front of it. Writing this term out explicitly, we get
    $$ \frac{\hbar}{i} \partial^{\alpha}(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})\varphi(x) = \mathcal{L}^{\mu\nu}\partial^{\alpha}\varphi(x)+\frac{\hbar}{i} (g^{\mu \alpha}\partial^{\nu}\varphi(x)-g^{\nu\alpha}\partial^{\mu}\varphi(x))$$
    The second term is just ##(S^{\mu\nu}_V)^{\alpha}{}_{\tau}\partial^{\tau}\varphi(x)##, so the factor of 2 is taken care of. However, the problem now is that the ##\mathcal{L}## term is negative (due to the corrected negative sign that I mentioned at the beginning of this post). I finally went in and looked at Srednicki's notes, and the big difference is that he expands ##\varphi(x-x\delta\omega)## as
    $$\varphi(x-x\delta\omega) = \varphi(x)-\delta\omega_{\mu \nu}x^{\nu}\partial^{\mu}\varphi(x)$$
    which allows him to eliminate the superfluous negative in the offending term. What's the justification for this step? Why isn't the expansion
    $$\varphi(x-x\delta\omega) = \varphi(x)-\delta\omega_{\mu \nu}x^{\mu}\partial^{\nu}\varphi(x)$$
     
  6. Dec 21, 2016 #5

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    So I can rationalize Srednicki's expansion (##\delta\omega_{\mu\nu}x^{\nu}\partial^{\mu}## instead of ##\delta\omega_{\mu\nu}x^{\mu}\partial^{\nu}##) in the following way: If ##\delta\omega_{\mu\nu}## is a matrix and ##x^{\nu}## is a vector, then standard matrix multiplication would look like:
    $$\delta\omega_{\mu\nu}x^{\nu} = \pmatrix{\delta\omega_{00} & \delta\omega_{01} &\delta\omega_{02} &\delta\omega_{03} \\ \delta\omega_{10} &\delta\omega_{11} &\delta\omega_{12} &\delta\omega_{13} \\\delta\omega_{20} &\delta\omega_{21} &\delta\omega_{22} &\delta\omega_{23} \\\delta\omega_{30} &\delta\omega_{31} &\delta\omega_{32} &\delta\omega_{33} } \pmatrix{x^{0} \\ x^{1} \\ x^{2} \\ x^{3} }$$
    which gives a covariant vector indexed by ##\mu##. Then we take the dot product of ##\delta\omega_{\mu\nu}x^{\nu}## and ##\partial^{\mu}## to get the operator that acts on ##\varphi(x)##. The reason the expansion is ##\delta\omega_{\mu\nu}x^{\nu}\partial^{\mu}## instead of ##\delta\omega_{\mu\nu}x^{\mu}\partial^{\nu}## is because the second form isn't compatible with standard matrix multiplication. Is this the right way to think about this problem? Little errors like this are really tripping me up in trying to understand tensor index manipulations in general.

    Thanks as always for your help.

    Edit: I'm pretty sure the ordering in ##\varphi(\Lambda^{-1}x)=\varphi((1-\delta\omega)x)## is important as well (since matrix multiplication doesn't commute). However, I doubt I would have been able to come up with any of this on my own.
     
    Last edited: Dec 21, 2016
  7. Dec 21, 2016 #6

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I understand your frustration. Unfortunately, there are a lot of little details that authors leave unsaid/unexplained in advanced physics books and this makes it difficult sometimes to reproduce calculations. It seems to me that here, the notation is ambiguous, because the argument of the field, ##x- x \delta \omega## could have meant either ##x_\mu - x^\mu \delta \omega_{\mu \nu}## or ##x_\mu - x^\mu \delta \omega_{\nu \mu}##. Srednicki assumed one and worked with that (and then his equations and results are of course consistent), but without being told explicitly what his choice was, one cannot tell without more information (like for example working out the problem you worked on and imposing that the answer agrees with his).
     
  8. Dec 21, 2016 #7

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Thanks for your reply, @nrqed. I'm becoming a little more comfortable with Srednicki's convention now that I've had some time to digest it. Interestingly enough, I made the same sign error in an earlier problem, but it was specifically canceled out by the antisymmetry of ##\delta\omega##, so it's been flying under my radar for a few days now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Srednicki QFT Problem 2.9
  1. Srednicki Problem 3.4 (Replies: 3)

  2. Srednicki problem 3.3 (Replies: 28)

Loading...