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

[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!

 

[mfb]

$e^{i \theta}$ 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.

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?

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.

 

[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.