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Lorentz invariance of a field

  1. May 15, 2012 #1
    Hi all,

    I'm studying quantum field theory and I'm watching video lectures on Harward University web site (Professor Colemann's lectures). Now, in lesson number six at 1h-6 minute a student asks why after trasforming field by a Lorentz transformation he doesn't transform also integration variable (I do not catch very well student's voice...) and Professor Colemann explains that is we transform field AND observer we are not applying Lorentz transformation but a coordinates system change..could you clarify me this point?


    Many thanks,
     
  2. jcsd
  3. May 15, 2012 #2

    Bill_K

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    Wow. Despite Sidney Coleman's well-deserved reputation, the poor A-V quality of these lectures IMO is a real obstacle to learning physics from them. I'd get a book.
     
  4. May 15, 2012 #3
    Yes I know.
    Let's suppose I have an object, a four vector or something else, and we have defined a quantity as for example an integral in all space (spatial) of this object or a function of this objects..we want to investigate lorentz properties of this integral..I lorentz transform the function to be integrated but not the volume element or also the volume element?
     
  5. May 15, 2012 #4

    vanhees71

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    Standard relativistic quantum field theory is by definition constructed as a local quantum field theory, i.e., a free elementary particle is described by a certain unitary representation of the proper orthochronous Poincare group (more precisely by its covering group, where the proper orthochronous Lorentz subgroup is substituted with its covering group, SL(2,C)).

    This means that a field operator in the Heisenberg picture transforms under Poincare transformations with a unitary matrix, representing the Poincare group such as a corresponding classical field transforms under the same transformation. E.g., a scalar field operator (Klein-Gordon field operator) transforms under a proper orthochronous Lorentz transform [itex]x'=\Lambda x[/itex] as

    [tex]\hat{\Phi}'(x')=\hat{U}(\Lambda) \hat{\Phi}(x') \hat{U}^{\dagger}(\Lambda)=\hat{\Phi}(\Lambda^{-1} x').[/tex]

    A vector field, transforms like

    [tex]{\hat{A}'}^{\mu}(x')=\hat{U}(\Lambda) \hat{A}^{\mu}(x') \hat{U}^{\dagger}(\Lambda)={\Lambda^{\mu}}_{\nu} \hat{A}_{\nu}(\Lambda^{-1} x').[/tex]
     
  6. May 15, 2012 #5

    dextercioby

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    There's also the active vs. passive viewpoint when it comes to spacetime symmetries. The best source on this that I know of are the book by Fonda & Ghirardi ('Symmetry principles in quantum physics', M. Dekker, 1970) and of course the 2nd chapter of 1st volume by Weinberg.

    BTW, Hendrik has some excellent notes on formal QFT at http://theorie.physik.uni-giessen.de/~hees/
     
  7. May 16, 2012 #6

    samalkhaiat

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    This is one of the most "FAQ". I have already answered similar equations on these forums in the past. Give me some time and I will post a good answer for you. It might be a bit different from Sidney's method, but I will make sure to "change integration variable" so that it wont be that defferent from Sidney's.

    Sam
     
  8. May 16, 2012 #7

    samalkhaiat

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    Let us define the functional

    [tex]Q[\Sigma]=\int_{\Sigma}d\sigma_{a}(x) V^{a}(x).[/tex]

    where [itex]\Sigma[/itex] denotes an arbitrary space-like hypersurface in space-time [itex](M^{4},\eta^{ab})[/itex], and

    [tex]d\sigma_{a}(x) = \frac{1}{3!}\epsilon_{abcd} dx^{b}dx^{c}dx^{d},[/tex]

    is a 4-vector differential at x. The functional derivative at some point [itex]x[/itex] is defined by

    [tex]\frac{\delta Q[\Sigma]}{\delta \sigma (x)} = \lim_{\omega (x)\rightarrow0}\frac{Q[\bar{\Sigma}]-Q[\Sigma]}{\omega (x)},[/tex]

    where [itex]\omega (x)[/itex] is the volume enclosed between [itex]\bar{\Sigma}[/itex] and [itex]\Sigma[/itex]. Therefore, according to Gauss' theorem, we have

    [tex]\frac{\delta Q[\Sigma]}{\delta\sigma (x)}= \partial_{a}V^{a}.[/tex]

    Now, if [itex]V^{a}(x)[/itex] is a conserved vector field, then [itex]\delta Q / \delta\sigma = 0[/itex] and therefore [itex]Q[\Sigma][/itex] is independent of [itex]\sigma (x)[/itex]. This means that we are free to pick a particular hypersurface to evaluate [itex]Q[/itex]. So, we choose the hyperplane [itex]\Sigma : x^{0}= t =\mbox{const.}[/itex] to evaluate [itex]Q[/itex];

    [tex]Q(t) = \int_{t = \mbox{const.}} d^{3}x\, V^{0}(t,\mathbf{x}).[/tex]

    Clearly, this integral is time-independent iff the conserved vector field satisfies the boundary condition

    [tex]|\mathbf{x}|^{2}V^{i}(x) \rightarrow 0 \ \ \mbox{as} \ |\mathbf{x}| \rightarrow \infty . \ \ (1)[/tex]

    Indeed; [itex] dQ/dt = \int d^{3}x\ \partial_{0}V^{0} = -\int d^{3}x \ \partial_{i}V^{i}(x) = 0[/itex].

    So let us summarize what we have done by the following: If [itex]V^{a}(x)[/itex] is a conserved vector field (i.e., [itex]\partial_{a}V^{a}=0[/itex]) satisfying the boundary condition in eq(1), then [itex]Q[/itex] is independent of the hypersurface on which it is evaluated, i.e., time independent;
    [tex]
    Q = \int_{\Sigma}d\sigma_{a}(y) \ V^{a}(y) = \int_{t = \mbox{const.}}d\sigma_{a}(y) \ V^{a}(y), \ \ (2)
    [/tex]

    This result will be used below to prove that the charge [itex]Q[/itex] is a Lorentz invariant quantity.
    The Lorentz transform of [itex]Q[/itex] is obtained by conjugating it with [itex]U(\Lambda) \in SO(1,3)[/itex];

    [tex]
    \bar{Q} = U^{-1}QU = \int_{t = \mbox{const.}} d\sigma_{a}(x) U^{-1}(\Lambda)V^{a}(x)U(\Lambda).
    [/tex]
    Since (vector representation of SO(1,3))
    [tex]
    U^{-1}(\Lambda)V^{a}(x)U(\Lambda) = \Lambda^{a}{}_{c}V^{c}(\Lambda^{-1}x).
    [/tex]
    Thus
    [tex]
    \bar{Q} = \int_{t = \mbox{const.}}d\sigma_{a}(x) \ \Lambda^{a}{}_{c}V^{c}(\Lambda^{-1}x). \ \ (3)
    [/tex]
    Now, we change integration variables according to
    [tex]x = \Lambda y \ \ (4)[/tex]
    To find the Jacobian, we note that
    [tex]
    d\sigma_{a}(x) = \frac{1}{3!} \epsilon_{abcd} \Lambda^{b}{}_{p} \Lambda^{c}{}_{q} \Lambda^{d}{}_{r}dy^{p}dy^{q}dy^{r}.
    [/tex]
    Now, we use the identity
    [tex]
    \epsilon_{abcd}\Lambda^{b}{}_{p}\Lambda^{c}{}_{q} \Lambda^{d}{}_{r} = (\Lambda^{-1})^{s}{}_{a}\epsilon_{spqr}\det \Lambda
    [/tex]
    Since [itex]\det \Lambda = 1[/itex], we find
    [tex]
    d\sigma_{a}(x) = \frac{1}{3!}(\Lambda^{-1})^{s}{}_{a} \epsilon_{spqr} \ dy^{p}dy^{q}dy^{r} = (\Lambda^{-1})^{s}{}_{a}d\sigma_{s}(y). \ \ (5)
    [/tex]

    Inserting eq(4) and eq(5) in eq(3) (with the new integration domain [itex]\Sigma[/itex]) we find

    [tex]
    \bar{Q} = \int_{\Sigma}d\sigma_{s}(y) (\Lambda^{-1})^{s}{}_{a} \Lambda^{a}{}_{c}V^{c}(y) = \int_{\Sigma}d\sigma_{a}(y)V^{a}(y).
    [/tex]

    Thus, using eq(2), we arrive at
    [tex]\bar{Q} = \int_{t = \mbox{const.}} d\sigma_{a}(y)V^{a}(y) = Q[/tex]
    This proves that [itex]Q[/itex] is invariant under SO(1,3). qed

    Sam
     
    Last edited: May 16, 2012
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