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Phase and Gauge Transformations

  1. Nov 2, 2012 #1
    I'm currently working through Griffith's book Introduction to Elementary Particles, and studying the chapters on gauge theories. From classical E&M, I understand what we mean by a gauge transformation and why the Lagrangian must be invariant under such a transformation, but what I don't understand is what a phase transformation of the form [itex]\psi\to e^{i\theta(x)}\psi[/itex] represents and how the gauge field [itex]A_{\mu}[/itex] necessarily arises from such a transformation. It seems like this is a potentially deep connection between rotating the fields [itex]\psi[/itex] and introducing a vector potential, but I don't understand what is actually going on when we apply a phase transformation. If anyone could elaborate or provide some intuition for what a phase transformation really is I would greatly appreciate it.
  2. jcsd
  3. Nov 2, 2012 #2


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    hi unchained1978! :smile:
    i don't think it does mean anything physical :redface:

    i think it's just a mathematical expression of a symmetry that any field theory has to satisfy
  4. Nov 2, 2012 #3
    The gauge field is a way to introduce a local phase shift symmetry into a theory which already has a global phase shift symmetry, by a process called the "minimal coupling prescription."

    Say you have a Lagrangian for a free electron and a free photon: [itex]\mathcal{L} = \bar{\psi}(i\not{\partial} - m)\psi + \frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/itex] (where [itex]F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu[/itex]). This Lagrangian has two symmetries: the local gauge transformation of the photon field [itex]A^\mu(x) \rightarrow A^\mu(x) + \partial^\mu f(x)[/itex], and a global phase symmetry [itex]\psi(x) \rightarrow e^{i\theta}\psi(x)[/itex].

    Now say we would like the phase symmetry to be local instead--that is, we want to turn the constant [itex]\theta[/itex] into a field [itex]\theta(x)[/itex]. If you try to make this transformation, you will see that the Lagrangian is not invariant under it--the problem is that the derivative causes you to pick up the extra term [itex]-\bar{\psi}(\not{\partial}\theta)\psi[/itex].

    We can fix this problem by coupling the two fields together: [itex]\mathcal{L} = \bar{\psi}(i\not{\partial} - m)\psi + \frac{1}{4}F_{\mu\nu}F^{\mu\nu} + e\bar{\psi}\not{A}\psi[/itex]. If you substitute the local phase shift into this Lagrangian, you will see that we can "eat" the troublesome extra term by also making a gauge transformation on the photon field: [itex]A^\mu(x) \rightarrow A^\mu(x) + \frac{1}{e}\partial^\mu \theta(x)[/itex]. This causes the interaction term to shift by exactly the same amount as the derivative term, canceling it out. Additionally, because the [itex]\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/itex] term has a gauge symmetry, it remains invariant under this transformation. Therefore, we have now created a theory which is invariant under a local transformation of both fields together.

    Now, you can run this argument the other way--say you have a theory with a free electron only, and you want to make its phase symmetry local. You can do so by introducing a new field, and coupling it to the electron field by the above procedure. Thus, you can also view the minimal coupling prescription as a way to create a photon field out of the air, so that you can make an electron with a local phase symmetry.

    Why would you want the electron to have a local phase symmetry? The answer is that if you have a local phase symmetry on the electron, then Noether's Theorem says that there is a conserved current associated with the electron field--that is, the electron has a charge. Since this is an observed property of the electron in the real world, we need to construct a Lagrangian which has that symmetry, and this procedure gives us a way to do it.
    Last edited: Nov 2, 2012
  5. Nov 2, 2012 #4
    I understand the machinery that goes into constructing a lagrangian that is invariant under local phase transformations, but I don't understand what a local phase transformation is. What does it mean to rotate the electron field by some phase angle? Could you please explain?
    Last edited: Nov 2, 2012
  6. Nov 3, 2012 #5
    It might be good to look at gauge symmetry in nonrelativistic quantum mechanics first, in which gauge transformations involve a change in the vector potential and simultaneously a rotation of the familiar wave function. Here's the first google hit, not necessarily the best reference: http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html
  7. Nov 3, 2012 #6
    It just means that you multiply the field by a complex number--it's a purely mathematical operation. The key point is that there's nothing requiring the Lagrangian to be invariant under this operation--for instance, you might have [itex]\mathcal{L} = Re\: \psi[/itex] or something silly like that. In cases where the Lagrangian is invariant under that operation, though, then you get a conserved current.
  8. Nov 3, 2012 #7


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    i stand by my previous answer :biggrin:

    forget about it for a couple of years :wink:

    maybe in two years time, with all your extra knowledge of the maths, you'll look back and think "oh yes, that was really beautiful, now i see what it is!" :smile:
  9. Nov 3, 2012 #8
    I don't mean to sound unappreciative, but it's not the math that's confusing me at all. It seems to me that applying a local phase transformation to the electron field is more than just a mathematical trick, because without the requirement of invariance under local transformations we would have no photons! The vector potential is added so that it counters the offending terms in [itex]\mathcal{L}[/itex] and thus the lagrangian remains invariant. I guess a better question would be why would you apply any sort of phase transformation in the first place? Thanks for your replies.
  10. Nov 3, 2012 #9


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    Because, as Chopin stated,
  11. Nov 3, 2012 #10
    You're perfectly free to have photons without a locally phase-invariant electron, they just wouldn't couple to the electron field.

    Quantum field theory is a general mathematical framework that allows you to construct all kinds of theories, including ones that aren't physically realistic. Indeed, a beginning QFT student will spend a lot of time studying theories of spinless particles interacting with each other, spinless particles interacting with themselves, electrons interacting with massive photons, etc. None of these exist in real life, but they're all mathematically describable by QFT.

    So nothing is stopping you from constructing a theory with seven different massless vector fields of which only one couples to the electron field, except that this isn't what we observe to be true in real life. In real life we observe an electron field and a photon field which interact, which we recognize by the mathematical gymnastics above to mean that it can be described by a Lagrangian with a local U(1) symmetry.

    So I guess the answer to your question of why you would want a local U(1) symmetry is "because it gives us results consistent with the physical world."
    Last edited: Nov 3, 2012
  12. Nov 4, 2012 #11
    Thanks for your replies. They've helped considerably.
  13. Nov 5, 2012 #12


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    Here's a simple overview:

    * The potential formulation of classical electrodynamics - Lorenz 1867.
    * Resolution of field equations by "choosing a gauge" - Lorenz 1867.
    * Weyl defines <local gauge transformations> - 1918 - gravity and electromagnetism as gauge theories.
    * Dirac creates a general theory of gauge transformations as dynamical systems with 1st class constraints - 1950 (Dirac conjecture).
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