I) The topic of gauging global symmetries is a quite large subject, which is difficult to fit in a Phys.SE answer. Let us for simplicity only consider a single (and thus necessarily Abelian) continuous infinitesimal transformation$^1$
$$ \tag{1} \delta \phi^{\alpha}(x)~=~\varepsilon(x) Y^{\alpha}(\phi(x),x), $$
where $\varepsilon$ is an infinitesimal real parameter, and $Y^{\alpha}(\phi(x),x)$ is a generator, such that the transformation (1) is a quasi-symmetry$^2$ of
the Lagrangian density
$$ \tag{2} \delta {\cal L} ~=~ \varepsilon d_{\mu} f^{\mu} + j^{\mu} d_{\mu}\varepsilon $$
whenever $\varepsilon$ an $x$-independent global parameter, such that the last term on the rhs. of eq. (2) vanishes. Here $j^{\mu}$ and
$$ \tag{3} J^{\mu}~:=~j^{\mu}-f^{\mu}$$
are the bare and the full Noether currents, respectively. The corresponding on-shell conservation law reads$^3$
$$ \tag{4} d_{\mu}J^{\mu}~\approx~ 0, $$
cf. Noether's first Theorem.
Here $f^{\mu}$ are so-called improvement terms, which are not uniquely defined from eq. (2). Under mild assumptions, it is possible to partially fix this ambiguity by assuming the following technical condition
$$ \tag{5}\sum_{\alpha}\frac{\partial f^{\mu}}{\partial(\partial_{\nu}\phi^{\alpha})}Y^{\alpha}~=~(\mu \leftrightarrow \nu), $$
which will be important for the Theorem 1 below. We may assume without loss of generality that the original Lagrangian density
$$ \tag{6} {\cal L}~=~{\cal L}(\phi(x), \partial \phi(x); A(x),F(x);x). $$
depends already (possibly trivially) on the $U(1)$ gauge field $A_{\mu}$ and its Abelian field strength
$$\tag{7} F_{\mu\nu}~:=~\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$$
The infinitesimal Abelian gauge transformation is defined to be
$$ \tag{8} \delta A_{\mu}~=~d_{\mu}\varepsilon. $$
Let us introduce the covariant derivative
$$ \tag{9} D_{\mu}\phi^{\alpha}~=~\partial_{\mu}\phi^{\alpha} - A_{\mu}Y^{\alpha}, $$
which transforms covariantly
$$ \tag{10} \delta (D_{\mu}\phi)^{\alpha}~=~\varepsilon (D_{\mu}Y)^{\alpha}$$
under gauge transformations (1) and (8). One may then prove under mild assumptions the following Theorem 1.
Theorem 1. The gauge transformations (1) and (8) are a quasi-symmetry for the following so-called gauged Lagrangian density
$$ \tag{11} \widetilde{\cal L}~:=~ \left.{\cal L}\right|_{\partial\phi\to D\phi}+\left. A_{\mu} f^{\mu}\right|_{\partial\phi\to D\phi}. $$
II) Example: Free Schrödinger field theory. The wavefunction $\phi$ is a complex (Grassmann-even) field. The Lagrangian density reads (putting $\hbar=1$):
$$ \tag{12} {\cal L}
~=~ \frac{i}{2}(\phi^*\partial_0\phi - \phi \partial_0\phi^*)
- \frac{1}{2m} \sum_{k=1}^3(\partial_k\phi)^*\partial^k\phi. $$
The corresponding Euler-Lagrange equation is the free Schrödinger equation
$$ 0~\approx~\frac{\delta S}{\delta\phi^*}
~=~ i\partial_0\phi~+\frac{1}{2m}\partial_k\partial^k\phi $$
$$ \tag{13} \qquad \Leftrightarrow \qquad
0~\approx~\frac{\delta S}{\delta\phi}
~=~ -i\partial_0\phi^*~+\frac{1}{2m}\partial_k\partial^k\phi^*. $$
The infinitesimal transformation is
$$ \tag{14} \delta \phi~=~Y\varepsilon
\qquad \Leftrightarrow \qquad
\delta \phi^*~=~Y^*\varepsilon^*, $$
where $Y\in\mathbb{C}\backslash\{0\}$ is a fixed non-zero complex number. Bear in mind that the above Theorem 1 is only applicable to a single real transformation (1). Here we are trying to apply Theorem 1 to a complex transformation, so we may not succeed, but let's see how far we get. The complexified Noether currents are
$$ \tag{15} j^0~=~ \frac{i}{2}Y\phi^*, \qquad
j^k~=~-\frac{1}{2m}Y\partial^k\phi^*, \qquad k~\in~\{1,2,3\},$$
$$ \tag{16} f^0~=~ -\frac{i}{2}Y\phi^*, \qquad f^k~=~0, $$
$$ \tag{17} J^0~=~ iY\phi^*, \qquad
J^k~=~-\frac{1}{2m}Y\partial^k\phi^*, $$
and the corresponding complex conjugate relations of eqs. (15)-(17). The infinitesimal complex gauge transformation is defined to be
$$ \tag{18} \delta A_{\mu}~=~d_{\mu}\varepsilon
\qquad \Leftrightarrow \qquad
\delta A_{\mu}^*~=~d_{\mu}\varepsilon^*. $$
The Lagrangian density (11) reads
$$ \widetilde{\cal L}
~=~\frac{i}{2}(\phi^*D_0\phi - \phi D_0\phi^*)
- \frac{1}{2m} \sum_{k=1}^3(D_k\phi)^*D^k\phi
+\frac{i}{2}(\phi Y^* A_0^* - \phi^*Y A_0) $$
$$ \tag{19} ~=~\frac{i}{2}\left(\phi^*(\partial_0\phi-2Y A_0)
- \phi (\partial_0\phi-2YA_0)^*\right)
- \frac{1}{2m} \sum_{k=1}^3(D_k\phi)^*D^k\phi . $$
We emphasize that the Lagrangian density $\widetilde{\cal L}$ is not just the minimally coupled original Lagrangian density $\left.{\cal L}\right|_{\partial\phi\to D\phi}$. The last term on the rhs. of eq. (11) is important too. An infinitesimal gauge transformation of the Lagrangian density is
$$ \tag{20} \delta\widetilde{\cal L}~=~\frac{i}{2}d_0(\varepsilon^*Y^*\phi -\varepsilon Y\phi^*) + i|Y|^2(\varepsilon A_0^* - \varepsilon^* A_0) $$
for arbitrary infinitesimal $x$-dependent local gauge parameter $\varepsilon=\varepsilon(x)$. Note that the local complex transformations (14) and (18) is not a (quasi) gauge symmetry of the Lagrangian density (19). The obstruction is the second term on the rhs. of eq. (20). Only the first term on the rhs. of eq. (20) is a total time derivative. However, let us restrict the gauge parameter $\varepsilon$ and the gauge field $A_{\mu}$ to belong to a fixed complex direction in the complex plane,
$$ \tag{21}\varepsilon,A_{\mu}~\in ~ e^{i\theta}\mathbb{R}.$$
Here $e^{i\theta}$ is some fixed phase factor, i.e. we leave only a single real gauge d.o.f. Then the second term on the rhs. of eq. (20) vanishes, so the gauged Lagrangian density (19) has a real (quasi) gauge symmetry in accordance with Theorem 1. Note that the field $\phi$ is still a fully complex variable even with the restriction (21). Also note that the Lagrangian density (19) can handle both the real and the imaginary local shift transformations (14) as (quasi) gauge symmetries via the restriction construction (21), although not simultaneously.
III) An incomplete list for further studies:
Peter West, Introduction to Supersymmetry and Supergravity, 1990, Chap. 7.
Henning Samtleben, Lectures on Gauged Supergravity and Flux Compactifications, Class. Quant. Grav. 25 (2008) 214002, arXiv:0808.4076.
--
$^1$ The transformation (1) is for simplicity assumed to be a so-called vertical transformation. In general, one could also allow horizontal contributions from variation of $x$.
$^2$ For the notion of quasi-symmetry, see e.g. this Phys.SE answer.
$^3$ Here the $\approx$ symbol means equality modulo equation of motion (e.o.m). The words on-shell and off-shell refer to whether e.o.m. is satisfied or not.
This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Qmechanic
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