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Gauge invariance is not normal invariance?

  1. May 8, 2015 #1
    I recently learned that with (local) gauge invariance, functional quantization needs to factor out volume factor(Faddeev-Popov procedure).

    Why does this has to be done?Just to remove infinity? As far as I am concerned, ##\phi^4## theory contains invariance(for example ##\phi\to\phi\cdot e^{i \alpha}##) but do not need such procedure.

    What is the difference between the invariance in ##\phi^4## theory and that of Yang-mills theory? I learned that guage invariance is redundant freedom but what's the exact meaning of redundant freedom?
     
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  3. May 9, 2015 #2

    vanhees71

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    It's indeed to cancel an infinity. The most simple analogue I can come up with is an integral like

    \begin{equation}
    \label{1}
    I=\int_{\mathbb{R}^2} \mathrm{d}^2 \vec{x} \exp[-(x_1-x_2)^2].
    \end{equation}

    Obviously that's divergent, because you can just do the ##x_2## integration and get an expression that is independent of ##x_1##. Thus the ##x_1## integral is divergent.

    Now, as in the case of gauge invariance, there's first of all a symmetry here: The integrand is obviously independent under translations, i.e., under ##x_1 \rightarrow x_1+a##, ##x_2 \rightarrow x_2+a##, which is the reason for the divergence from the point of view of symmetries. Second it might be that the situation you describe does not care about the individual values of ##x_1## and ##x_2## but only their difference, and what you really want to do is to integrate over this difference. Of course, here it's simple to just do so, but for the gauge invariance it's not that simple, and you have to do this in an indirect way. Indeed, for a gauge theory, it's not the gauge-boson vector fields ##A^{\mu}## but only the whole equivalence class of these fields which are defined as being connected by a gauge transformation. So only these "gauge orbits" are the distinct physical states, and you want to count each gauge orbit only once in the path integral. So instead of the divergent quantity (\ref{1}) we want
    \begin{equation}
    \tilde{I}=\int_{\mathbb{R}} \mathrm{d} x_1 \exp(-x_1^2), \label{2}
    \end{equation}
    which is well-defined. So here we choose the gauge fixing ##x_2=0## and integrated only over all distinct situation, given by the values of ##x_1 \in \mathbb{R}##.

    So in our analogue situation, the Faddeev-Popov procedure is achieved as follows: First we choose a "gauge condition". Here, we have not much choices, because of the very simple symmetry. We can just fix ##x_2=0## as the gauge condition, and any configuration, equivalent to this one representative of the corresponding gauge orbit is given by the translation,
    $$\vec{x}^{(a)}=\begin{pmatrix} x_1 \\ 0 \end{pmatrix}+\begin{pmatrix} a \\ a \end{pmatrix}.$$
    The gauge fixing function is
    ##g(\vec{x})=x_2##.
    Now we define
    $$\Delta^{-1}(\vec{x})=\int_\mathbb{R} \mathrm{d} a \delta[g(x^a)]=\int_\mathbb{R} \mathrm{d} a \delta(x_2+a)=1.$$
    Then we can write
    $$I=\int_{\mathbb{R}^2} \mathrm{d}^2 \vec{x} \Delta(\vec{x}) \int_{\mathbb{R}} \mathrm{d} a \delta[g(\vec{x}^{(a)}] \exp[-(x_1-x_2)^2].$$
    Now the integration measure ##\mathrm{d}^2 \vec{x}##, the original integrand (the exponential of the "action"), and ##\Delta(\vec{x})## are "gauge invariant". So by making the substitution ##x=y^{(-a)}##, we get
    $$I=\int_{\mathbb{R}} \mathrm{d} a \int_{\mathbb{R}^2} \mathrm{d}^2 \vec{y} \Delta(\vec{y}) \delta[g(\vec{y})] \exp[-(y_1-y_2)^2].$$
    Now it's easy to see that you get simply the desired value ##\tilde{I}## defined in (\ref{2}) multiplied by a diverging trivial factor, and that's precisely what's also achieved for the functional integrals used in QFT of gauge models, and that's the only thing you need for all practical purposes.
     
    Last edited: May 9, 2015
  4. May 9, 2015 #3

    atyy

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    Gauge invariance means that there is more than one way of describing a physical situation. A simple example is the electric potential in classical electrostatics - one can add any constant to the potential without changing the physical situation that is described. So the "addition of a constant" is analogous to gauge covariance, and only the potential difference is physical or invariant. In gauge theories, the physical quantity or gauge invariant quantity is (at least heuristically) the Wilson loop.
     
  5. May 10, 2015 #4
    Thanks vanhees71 and atyy for illuminating illustration.

    However, what I still don't understand is what is the exact meaning of physical situation. Electrodynamics is easy, but when you look at an unfamiliar Lagrangian, how can you deduce whether it has redundant degree of freedom?
     
  6. May 10, 2015 #5

    atyy

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    In general, one has to be told whether Lagrangian or the variables being used are gauged or not.

    For example, if in wave mechanics we have ψ(x), we have to be told whether x is gauged by 2π. If it is not gauged, then it is a particle on a line, if it is gauged, it is a particle on a circle.

    In the standard model, gauge covariance is related to renormalizability. However, it is difficult to argue that renormalizability is a fundamental requirement, since gravity is not renormalizable. http://isites.harvard.edu/fs/docs/icb.topic473482.files/21-renormalizability.pdf

    The most common argument as to why we use gauge redundant variables (A(x)) rather than gauge invariant variables (Wilson loops) is that the gauge redundant variables allow us to write a theory in manfestly Lorentz covariant form.
     
    Last edited: May 10, 2015
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