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2

The canonical momentum is always (in position "$q$" basis) given by $-i\hbar\partial_q$ so the mapping of the commutator to the Poisson bracket $$[q,p]=i\hbar \leftrightarrow \{q,p\}=1$$ stays true. The nice covariant object however involves the covaraint derivative $\nabla_q$ as $$-i\hbar\nabla_q =-i\hbar\left(\partial_q-\frac{i}{\hbar}qA\right)... 0 So, to complete the half solution in the answer one need only notice that multiplying the \frac{\partial S}{\partial\phi^i} the expression for F^{ij} in terms of F^{ia} and F^{ijk}, one is left with$$\frac{\partial S}{\partial\phi^i}F^{ij}=-\frac{\partial S}{\partial\phi^i}F^{ia}R^j_a.$$However, recall that this trivial gauge transformation was ... 0 \newcommand{\rto}{\overset{\scriptscriptstyle r\to\infty}{\longrightarrow}} \newcommand{\v}{\boldsymbol{#1}} \newcommand{\t}{\tau} \newcommand{\pd}{\partial} \newcommand{\demeqq}{\overset{!}{=}} One should make a distinction between time-dependent and time-independent gauge transformations. I will denote x=(\t,\v x). What is described in the book is ... 1 Briefly, Ref. 1 considers generalizations of Yang-Mills-type gauge theories based on a Leibniz algebra structure. Concretely, Ref. 1 defines a covariant variation as$$ \Delta {\cal A}~:=~e^{-{\cal A}}\delta e^{\cal A}. \tag{3.33}$$In physics jargon, {\cal A} is a gauge field; \Lambda=\delta {\cal A} plays the role of an infinitesimal gauge ... 0 The issue is that the scalar has to be transformed as well when you perform a gauge transformation. Namely if you want to perform the change$$A_\mu + \partial_\mu \theta(x)$$it has to come along with a gauge transformation of the scalar field by the same parameter, i.e.$$\phi \rightarrow e^{-i\theta(x)}\phi$$I would recommend writing the complete ... 1 Indeed you're doing something wrong, or at least confusing, when you say "...by gauge invariance A_\mu and A_\mu+\partial_\mu \theta are exactly the same field..." and then discard the \partial_\mu \theta: You're implictly imposing a gauge transformation with parameter \theta, which absorbs the phase into the gaueg field. Also, your ... 1 You have a single electromagnetic field described by single potentials \vec{A}(\vec{r},t) and V(\vec{r},t). These potentials extend over the whole space (all \vec{r}). And you have N particles represented by the wave function \Psi(\vec{r}_1,\dots,\vec{r}_N,t). Then the Hamiltonian is$$H=\sum_{n=1}^N\left (\frac{1}{2m}(\vec{p}_n-q\vec{A}(\vec{r}_n,...

3

In electromagnetism there is one vector potential, $\vec{A}$. So the correct way to write the Hamiltonian is \begin{equation} H = \sum_n \frac{(p_n-q_n \vec{A})^2}{2m_n} + V(\vec{r}_1, \vec{r}_2,...\vec{r}_N) \end{equation} This Hamiltonian is invariant under gauge transformations of the form (apologies if I get a sign wrong) \begin{eqnarray} \Psi_n &\...

1

Consider first a time-independent gauge transformation. Then the vanishing of the gradient implies only that $$\theta'-\theta+\frac{q}{\hbar}\chi= {\rm constant}.$$ But we also know that $\theta-\theta'=0$ if $\chi=0$. Therefore the constant is zero. For a time dependent gauge transformation, one needs to include the gauge covariance of the Josephson ...

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