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Lorentz transformation of a scalar field

  1. Aug 5, 2015 #1
    Hello,

    I'm reading Tong's lecture notes on QFT and I'm stuck on the following problem, found on p.11-12.

    A scalar field [itex] \phi [/itex], under a Lorentz transformation, [itex] x \to
    \Lambda x [/itex], transforms as

    [itex] \phi(x) \to \phi'(x) = \phi(\Lambda^{-1} x) [/itex]

    and the derivative of the scalar field transforms as a vector, meaning

    [itex] (\partial_\mu \phi)(x) \to (\Lambda^{-1})^\nu{}_\mu (\partial_\nu \phi) (y). [/itex]

    where [itex] y = \Lambda^{-1}x [/itex]

    Could someone please explain the steps above? If the transformation is an active one, meaning that the field itself is rotated, then how does [itex] x \to \Lambda x [/itex] make sense. I don't get how he got the transformation property of the derivative either.
     
  2. jcsd
  3. Aug 5, 2015 #2

    Avodyne

    User Avatar
    Science Advisor

    A (classical) scalar field assigns a number ##\Phi(P)## to each spacetime point ##P##. (I'll generalize to quantum fields at the end; it's easier to think about numbers for now.) Suppose that Alice uses coordinates ##x## to label a particular spacetime point ##P##, and that Bob uses coordinates ##x'## to label the same spacetime point, and that their coordinates are related by a Lorentz transformation: ##x'{}^\mu=\Lambda^\mu{}_\nu x^\nu##. Alice then uses a function ##\phi## of her coordinates for the field, with the property that ##\phi(x)=\Phi(P)## when ##x## is Alice's label for ##P##. Bob uses a different function ##\phi'## of his coordinates for the field, with the property that ##\phi'(x')=\Phi(P)## when ##x'## is Bob's label for ##P##. From this we have ##\phi'(x')=\phi(x)##. Since ##x'=\Lambda x##, we also have ##x=\Lambda^{-1}x'##. Substituting for ##x## in ##\phi'(x')=\phi(x)##, we get ##\phi'(x')=\phi(\Lambda^{-1}x')##. We now change the dummy label ##x'## to ##x##, and we get ##\phi'(x)=\phi(\Lambda^{-1}x)##, which is Tong's first formula.

    Tong's second formula then follows from the chain rule for derivatives. Let ##\partial_\mu## denote a derivative with respect to ##x^\mu##, and ##\bar\partial_\nu## denote a derivative with respect to ##y^\nu=(\Lambda^{-1})^\nu{}_\rho x^\rho##. We want to compute ##\partial_\mu\phi'(x)##. Using ##\phi'(x)=\phi(y)## and the chain rule, we have ##\partial_\mu\phi'(x)=\partial_\mu\phi(y)=\bar\partial_\nu\phi(y)\partial_\mu y^\nu##. Then we have ##\partial_\mu y^\nu = \partial_\mu[(\Lambda^{-1})^\nu{}_\rho x^\rho] = (\Lambda^{-1})^\nu{}_\rho \partial_\mu x^\rho = (\Lambda^{-1})^\nu{}_\rho \delta_\mu{}^\rho = (\Lambda^{-1})^\nu{}_\mu##. Hence we get ##\partial_\mu\phi'(x)=(\Lambda^{-1})^\nu{}_\mu\bar\partial_\nu\phi(y)##, which is Tong's second formula.

    Note that it is a property of the Lorentz transformation matrices that ##(\Lambda^{-1})^\nu{}_\mu=\Lambda_\mu{}^\nu##, so we can also write the second formula as ##\partial_\mu\phi'(x)=\Lambda_\mu{}^\nu\bar\partial_\nu\phi(y)##, which looks nicer to me.

    In QFT, ##\Phi(P)## is an operator instead of a number, but the key formula ##\phi'(x')=\phi(x)=\Phi(P)## still holds, so all of the above goes through.
     
    Last edited: Aug 5, 2015
  4. Aug 6, 2015 #3
    Ah, I see it now. Thanks a lot for the help.
     
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