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

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1. May 2, 2015

PerilousGourd

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. May 2, 2015

Staff: Mentor

$e^{i \theta}$ 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.

3. May 2, 2015

ChrisVer

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

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

Last edited: May 2, 2015