Tag Info

New answers tagged

3

Note that the finite transformation of: $$ W^a_\mu \to W^a_\mu + \frac{1}{g} \partial_\mu \theta^a + \epsilon^{abc} \theta^b W^c_\mu $$ is: $$ W^a_\mu t^a \to g W_\mu^a t^a g^{-1} + \frac{i}{g} \partial_\mu g \tag{1} $$ where: $$ g = \exp(-i \theta^a t^a) \;\;\; \text{and} \;\;\; [t^a,t^b] = i \epsilon^{abc} t^c $$ Thus, the first term on the right-hand ...


5

Yes, you would have to introduce another gauge field. For example in the Standard Model there is gauge invariance under $SU(3)\times SU(2) \times U(1)$, and so there are three gauge fields: the gluons, the $W^\pm, Z$ weak gauge bosons and the photon. In general terms, it is simpler to argue like this: if you have gauge invariance under a Lie group $G$, the ...


2

We interpret OP's question (v4) as: How do we recover the phase ambiguity from the generator of translation method in Ref. 1? Recall that an eigenvector for an operator can be rescaled with a non-zero multiplicative factor. The main point is that the position eigenket $| x \rangle$, which satisfies $$\tag{A} \hat{x}| x \rangle~=~ x| x \rangle, $$ ...


0

I think because the wavefunctions are required to be normalized so that $\psi^{*}\psi$ represents the probability or probability density of finding the particle, so their amplitude are not allowed to scale arbitrarily. That's why the gauge field can only be real.


1

First of all there is no proof of this statement. It is just a general expectation that the more symmetries you have the more reason to expect better quantum properties. This works with SUSY, the more SUSY you have the better the theory is at the quantum level, say $N=4$ SYM, or $N=8$ SUGRA that some people still have hope to be well-defined. If you involve ...


2

OP is pondering if extremization of the action commutes with the minimal coupling (MC) recipe, i.e., if the following diagram would commute: $$ \begin{array}{ccc} \text{Lagrangian density} && \text{Lagrangian density} \cr {\cal L}(\phi(x),\partial\phi(x),A(x),F(x),x) &\stackrel{\text{MC}}{\longrightarrow} & {\cal ...


5

I am not sure of what you mean by "make a difference". The covariant derivative is introduced in the Lagrangian density to add Local Gauge Symmetry to the Action. A given field theory is described by its Action. Hence if the Action does not have a particular symmetry, you cannot introduce the symmetry later. By "making a difference", if you mean a ...


1

There are a lot of rather different aspects being touched on in the question. I'll try to give some indications. But I notice that the relation of this question to actual physics is not very strong, instead the question seems to be more generally after getting a feeling for identity types in homotopy type theory (HoTT). I imagine there are other discussion ...


2

In quantum field theory, when manipulating the path integral, we naively assume the measures (or strictly speaking the product of the measures and integrand) are invariant under the gauge transformations. In a fundamental paper, Fujikawa demonstrated the flaw in this assumption (in certain cases), and how to rigorously compute the analogue of a Jacobian ...



Top 50 recent answers are included