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Gauge invariance

  1. Dec 15, 2008 #1
    In classical e&m, for gauge invariance you can choose div[A]=0 or div[A]=dV/dt, where A is vector potential and V is the scalar potential; however, in qft you multiply your wavefunction by a phase factor that is dependent on space time. My question is that is there any parallel that can be drawn between these two different processes, if there is then I am not seeing it. I mean I can understand there may be the difference between the fact that one is for a classical picture of a field and the other in a quantum sense. Also I am having a hard time really understand how attaching a phase factor dependent on spacetime can really lead to the feynman rules for interaction terms or in the sense what it really does (like how are you able to understand that doing this gives a mathematical description of how particles couple with the gauge bosons), (I do understand that the equations should be invariant under this transformation). In the event that there is no connection between these two different ways of expressing gauge invariance, why do they differ (but I still think that they are connected somehow). Thanks in advance to whoever answers this post.
     
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  3. Dec 15, 2008 #2

    malawi_glenn

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    The phase of a (quantum) field is not an observable. Recall from QM that only the modulus square of the wave function is an observable, not the wave function itself. Therefore, assignation of an arbitrary phase to your field/wave function does not change anything.

    Now comes the concept of LOCAL gauge invariance, which is crucial. EACH observer may choose his OWN phase, i.e the phase is a function of space-time coordinates!

    Now applying this to your lagrangian, that the fields transforms under local gauge as:
    [tex] \phi (t,\vec{x}) \rightarrow \phi (t,\vec{x}) e^{i\theta (t,\vec{x})} [/tex]

    But now you are in trouble, your (free) lagrangian will NOT be invariant under this transformation (since derivative operation dont commute), UNLESS, the derivatives ALSO changes under guage transformation.

    [tex] \partial _{\mu} \rightarrow D_{\mu} [/tex]

    So now the problem is to find out what [tex] D_{\mu} [/tex] should be, and you'll find that it will be something like (for details, look in any souce of QED) [tex] D_{\mu} = \partial _{\mu} + \vec{A} (t,\vec{x}} )[/tex]

    Where A is a field, a potential - interactions! i.e the Lagrangian you started out with, which was free, has now interactions

    Now this was QED, the gauge transformation symmetry is U(1). Let's take QCD, there the gauge transformation symmetry is SU(3), and the fields transforms as:

    [tex] \phi (t,\vec{x}) \rightarrow \phi (t,\vec{x}) e^{i T^a\theta _a(t,\vec{x})} [/tex]

    Where T^a is a generator of the SU(3) lie group, and now the problem is again to find out how the derivatives should transform to leave the Lagrangian invariant :-)

    Summary:
    - Fields and Wave functions has a phase which is non observable
    - EACH observer can choose his/her own phase convention, physics should not depend on choice of phase!
    - Doing the local phase transformation of the fields, the lagrangian is not invariant unless you find out how the derivatives should transform under same symmetry transformation of the fields
    - By finding how the derivatives should transfrom, a field - potential- interatcions!! is "generated"
     
  4. Dec 15, 2008 #3

    siddharth

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    To add to what malawi said, you can have a classical field theory where a local gauge transformation gives rise to field-potential interactions too. For example, consider a scalar field [itex]\Phi=(\phi_1(x),\phi_2(x),\phi_3(x),...,\phi_n(x))^T[/itex] in the fundamental representation of SU(n).

    For a local gauge transformation, the field transforms as,

    [tex]\Phi^\prime=g.\Phi[/tex]

    where, g is an element of the lie group. ie,

    [tex]g(x)=e^{i\theta^a(x)T_a}[/tex]

    If you consider a normal Lagrangian, such as,

    [tex]\mathcal{L}=\partial^{\mu}\Phi^\dagger \partial_{\mu}\Phi + U(\Phi^\dagger \Phi)[/tex]

    the lagrangian is not invariant under the local gauge transformation, because

    [tex]\partial_{\mu}\Phi^\prime=g(\partial_{\mu}\Phi})+(\partial_{\mu}g})\Phi[/tex]

    However, the lagrangian will be invariant if the normal derivatives [itex]\partial_{\mu}[/itex] are replaced by covariant derivatives [itex]D_{\mu}[/itex] defined by,

    [tex]D_{\mu}\Phi=(\partial_{\mu}-iA_{\mu})\Phi[/tex]

    and where the transformation of the derivative is chosen as [tex]D_{\mu}^\prime \Phi^\prime=g.D_{\mu}\Phi[/tex], which fixes the transformation of [itex]A_{\mu}^\prime[/itex]

    Now, the Lagrangian is invariant under the local gauge transformation if we replace all the normal derivatives with the covariant derivative which gives rise to field potential interactions. I think this is what malawi was stating when he pointed out that the derivatives are replaced by covariant derivatives, which transform such that the lagrangian is invariant
     
    Last edited: Dec 15, 2008
  5. Dec 15, 2008 #4

    malawi_glenn

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    Exactly, thanx for the complementary post siddharth.
     
  6. Dec 17, 2008 #5
    I think that you have gotten very clear and good answer for your problem of the classical gauge theory.
    However, from your original question, I guess you haven't been familiar with some content of classical field theory, so I want to say something maybe helpful for you hopefully.

    First of all, as they have already pointed out, the gauge theory you considered actually is just "classical" field theory. It has not been quantized yet.

    Second, the gauge freedom you learned in classical electrodynamics is totally the same as the U(1) gauge symmetry (the physics is not changed by a local choice of phase). I guess you haven't been familiar with the tensor formulation of electromagnetics, i.e. stuff like [tex]F_{\mu\nu}[/tex] and so on. Once you understood that, it's manifest that the gauge transformation by a phase factor is totally the same as the gauge transformation in four Maxwell equations. [tex]A_\mu \rightarrow A_\mu + \partial_\mu\Lambda(x)[/tex]

    Moreover, the examples you gave, e.g. the Coulomb gauge, is a choice of gauge, i.e. gauge fixing! In quantum field theory, once we have gauge theory at hand, due to the gauge freedom, the system is constrained, the old way to quantize the system is to choose a gauge, so we also perform gauge-fixing in quantum field theory.
     
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