# Unitary transformation behind gauge transformation

It is very well-known that for bosonic operators a Gauge transformation can always be associated with it $$a\rightarrow e^{i\phi}a.$$ Obviously this is a Unitary transformation. Something like $$a^{\prime}=\mathcal{U}^{\dagger}a\mathcal{U}$$

I want to know what is $\mathcal{U}$?

• I don't understand your question. First, are we talking about $\mathrm{U}(1)$ gauge or arbitrary gauge theories? Second, what do you mean by "giving rise" to the phase factor? The "phase factor" is, if we are in the fundamental representation, as you seem to imply, just the element of the gauge group corresponding to the gauge trafo. Aug 5, 2014 at 16:15
• I am considering U(1) gauge. But I want to know what is the form of this U(1)? Aug 5, 2014 at 16:29
• $\mathrm{U}(1)$ is the circle group. I'm still uncertain what your question really is. You might be interested in What is the basis of gauge theory? Aug 5, 2014 at 16:38

Mathematically speaking, U(n) is what is known as a unitary group. In particular, when n = 1, as Acuriousmind stated, we get what is known as the "circle group". This group in particular is composed of all numbers on the complex plane {with asb(1) under the multiplication}. However, for all values of n, the unitary groups contain copies of n = 1 (or U(1)). Actually, during the developement of quantum mechanics, a decision was made by Weyl and Fritz London to modify gauge theory by changing the scalar factor into a complex quantity, which in effect,turned the scale transformation into a change of phase (which is U(1) gauge symmetry).

Ok, eventually I figured out the answer.

$$\mathcal{U}=e^{-i\phi a^{\dagger}a}$$