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##\bar{\psi}=e^{i\theta}\psi## global gauge transformation

  1. May 2, 2015 #1
    Why is ##\bar{\psi}=e^{i\theta}\psi##, where ##\theta## is a real number, used as the global gauge transformation? Why ##e^{i \theta}##; what's the physical significance or benefit?

    Why is ##\bar{\psi} = e^{i \theta(x)} \psi## the local gauge transformation? What does ##\theta## being a function of ##x## instead of a real number change such that it no longer applies globally?

    Thank you in advance for any explanations. I'm sorry if my questions are trivial!
  2. jcsd
  3. May 2, 2015 #2


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    ##e^{i \theta}## with a real ##\theta## is a general way to express a complex number z with |z|=1. You don't want to change the magnitude of the numbers (that is not covered by the gauge), just their phase.

    The other direction: This local transformation is the more general case. The special case of the same θ everywhere is called global gauge transformation because the same phase is applied everywhere. A global transformation is easier to describe than a local transformation, but considering local transformations is important in QFT.
  4. May 2, 2015 #3


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    The [itex]e^{i \theta}[/itex] is used for a global transformation. Global because at any point you just apply a [itex]\theta[/itex] transformation. Transformation, well , that depends... for a U(1) where the generators are just numbers you can use the [itex]e^{i \theta}[/itex], whereas for some other symmetry the transformation should change and you use instead the generators [itex]T^a[/itex] on the exponential [itex]e^{i \theta_a T^a}[/itex].
    For a simple case, consider that you have a complex number [itex]\psi = a + ib[/itex], transforming the [itex]\psi[/itex] itself will lead you to a [itex]\psi'= a' + i b'[/itex]. How are the two ([itex]\psi,\psi'[/itex]) connected? Well you can use that the multiplication of two complex numbers is a complex number so you can write:
    [itex]\psi' = z \psi[/itex]
    With [itex]z[/itex] an appropriate number that can apply for your given transformation. A complex number can be written as [itex]z= r e^{i \theta}[/itex] , and so an appropriate choice of ##z## would mean an appropriate choice for ##r## and ##\theta##.
    Now if you are looking for a unitary transformation the [itex]r=1[/itex], otherwise you would change the magnitudes. And so you have:
    [itex]\psi' = e^{i \theta} \psi[/itex].
    So in fact what you are doing is getting [itex]\psi[/itex] as a "vector" on the complex plane, and you are making a rotation to it.

    Now what would be the case if [itex]\theta[/itex] was in fact [itex]\theta(x)[/itex]?
    Obviously this rotation doesn't happen anywhere on space [itex]x^\mu[/itex] in the same way. At a spacetime point [itex]x_1[/itex] you apply a rotation [itex]\theta_1= \theta(x_1)[/itex] whereas in a point [itex]x_2[/itex] you can apply a rotation [itex]\theta_2= \theta(x_2)[/itex].
    It's naturally leading you to insert a "connection" (term from GR) into the game when you try to make invariant objects. The connection [itex]\Gamma_{\mu \nu}^\rho[/itex] in GR for example appears when you go from a global metric [itex]n_{\mu \nu}[/itex] to a local metric [itex]g_{\mu \nu}(x)[/itex]. Here the "ranks" are smaller and you end up with an "electromagnetic field" (better a U(1) gauge field) [itex]A_\mu[/itex].
    Last edited: May 2, 2015
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