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Contrary to popular belief, it is not necessary to choose a gauge to quantize a gauge theory. It is just convenient, since the non-gauge-fixing approaches are often difficult to implement for all but the simplest cases. Gauge theories are, in the Hamiltonian picture, certain kinds of constrained Hamiltonian systems. Dirac's canonical quantization procedure, ...

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An experimentalist's answer, Our observations tell us that baryon and lepton number are conserved, within the accuracies of our experiments and observations. This means we have chosen as a standard model SU(3)xSU(2)xU(1) because in the group structure of the possible representations of all the quantum numbers assigned to the particles and resonances we ...

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A negative coupling leads to a Hamiltonian that is unbounded from below, and hence unphysical, since there is no lowest-lying energy state. Similarily, a phase would mean the Hamiltonian is not self-adjoint, and time evolution would not be unitary.

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When you introduce a new field to make the Lagrangian gauge invariant, then you are at liberty to choose the transformation behaviour of the new field such that the Lagrangian becomes gauge invariant. If $\sigma\mapsto\sigma+\alpha$ leads to an invariant Lagrangian, then you are free to choose $\sigma$ as a field transforming such. In the situation you ...

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Suppose you have an abelian gauge theory (forget about $\psi$ for now) in a pure gauge state, that is using a gauge trasformation $A_\mu\rightarrow A_\mu + \partial_\mu\alpha$, you can go to $A_\mu = 0$ (call this gauge 1), i.e. $A_\mu = \partial_\mu \alpha$ (call this gauge 2). Now let's assume that $\alpha$ is not differentiable and see what goes wrong. ...

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Gauge theories become constrained Hamiltonian systems when passing from the Lagrangian $L(q,\dot{q},t)$ to the Hamiltonian $H(q,p,t)$ where $p = \frac{\partial L}{\partial \dot{q}}$. Generically, you get a constrained Hamiltonian system whenever the matrix/operator with components $$\frac{\partial^2 L}{\partial \dot{q}^i\partial\dot{q}^j}$$ is singular, ...

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The algebraic approach gives the better idea of what the states and observables of a quantum theory are, and this holds in infinite dimensional systems as well. In the modern mathematical terminology, observables of quantum mechanics are the elements of a topological $*$-algebra, and states are objects of its topological dual that are positive and have norm ...

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The change of the overall phase of the wave function, $\left|\Phi \right> \to e^{i\phi}\left|\Phi \right>$ is unphysical since they give the same expectation value for any observable operator. So, they represent the same physical state: indeed the space of physical states is not the Hilbert space, but rather it is the space of equivalence classes of ...

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1) Every "bare charge" contains an infinite part. The reason why we introduce the infinite part is precisely so that the divergences in loop integrals cancel. Physically you never measure bare charges: you always measured a suitably "dressed" charge that is typically scale dependent and is subject to screening effects, etc. 2) The finite part is always ...

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It seems to be an error (due the the units mismatch). You should doublecheck the errata page and perhaps contact the publishers. http://www.umsl.edu/~chengt/gaugebooks.html The errata is for the 1997 edition. It doesn't seem to mention this page.

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Suppose I have a Lagrangian density $\mathcal{L}(\phi^\mu,\sigma)$ depending on vector fields $\phi^\mu$ and their derivatives and a scalar field $\sigma$ and its derivatives. If I make a gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ does the field $\sigma$ transform? I've seen notes claiming \$\sigma\rightarrow \sigma + ...

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