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Invariance of a physical system (its action) under a continuous group of local transformations underlain by a global symmetry whose group parameters fixed in space-time have now been extended to vary in space-time instead. Use for buildup of the invariance, fixing the gauge, and accounting for the corresponding changes in the functional measure of the system.

2 votes
0 answers
115 views

Unitarity Gauge : how to undo the gauge transformation

I will simplify the argument. Let's consider a Gauge Boson (like the gauged one of U(1), $A_\mu$). Then, consider the Higgs boson with exponential representation, then $$H = e^{i\pi(x)/v}\left(\begin …
apt45's user avatar
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5 votes
1 answer
3k views

Conserved current in scalar QED

Consider a theory of a free massless complex scalar $\phi$ which undergoes global $U(1)$ transformations. The conserved current associated to this symmetry is the usual scalar current $$ J^\mu = i\le …
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6 votes
1 answer
725 views

Non-local field redefinition and effects on path-integral measure

Consider the partition function $$ Z[0] = \int \left[\mathcal{D}A_\mu\right]\left[\mathcal{D}\pi\right] e^{-i \int d^4x \left(-\frac{1}{2}(\partial\pi)^2-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+ \frac{a}{M^2 …
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4 votes

Why do the right-handed up quark and down quark not form an $SU(2)$ doublet?

An important feature of the models you need to specify is the symmetry group and the representations under which the fields transform. The absence of $SU(2)_R$ doublets in the Standard Model can be se …
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7 votes
1 answer
3k views

Why do we impose de Donder gauge?

In the field language, a massless particle corresponds to irreducible representations of the Lorentz group. In particular, given a spin-2 massless particle, we can embed the creation and annihilation …
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7 votes

Why do we impose de Donder gauge?

Actually this is not possible. To impose the gauge $h=0$, I have to solve $$ 2\partial\cdot \epsilon = -h\,,\qquad\qquad (1) $$ which always admits solutions. This partial gauge fixing leaves a redund …
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