Unitary transformation behind gauge transformation

Question

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. — ACuriousMind, Aug 05 '14 at 16:15
I am considering U(1) gauge. But I want to know what is the form of this U(1)? — Pratyay Ghosh, Aug 05 '14 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? — ACuriousMind, Aug 05 '14 at 16:38

score 0 · Answer 1 · answered Aug 05 '14 at 22:07

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).

score 0 · Accepted Answer · answered Aug 06 '14 at 07:51

0

Ok, eventually I figured out the answer.

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

answered Aug 06 '14 at 07:51

Pratyay Ghosh

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Unitary transformation behind gauge transformation

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