When can a global symmetry be gauged?

Question

Take a classical field theory described by a local Lagrangian depending on a set of fields and their derivatives. Suppose that the action possesses some global symmetry. What conditions have to be satisfied so that this symmetry can be gauged? To give an example, the free Schrödinger field theory is given by the Lagrangian $$ \mathscr L=\psi^\dagger\biggl(i\partial_t+\frac{\nabla^2}{2m}\biggr)\psi. $$ Apart from the usual $U(1)$ phase transformation, the action is also invariant under independent shifts, $\psi\to\psi+\theta_1+i\theta_2$, of the real and imaginary parts of $\psi$. It seems that such shift symmetry cannot be gauged, although I cannot prove this claim (correct me if I am wrong). This seems to be related to the fact that the Lie algebra of the symmetry necessarily contains a central charge.

So my questions are: (i) how is the (im)possibility to gauge a global symmetry related to central charges in its Lie algebra?; (ii) can one formulate the criterion for "gaugability" directly in terms of the Lagrangian, without referring to the canonical structure such as Poisson brackets of the generators? (I have in mind Lagrangians with higher field derivatives.)

N.B.: By gauging the symmetry I mean adding a background, not dynamical, gauge field which makes the action gauge invariant.

Answer 1

The guiding principle is: "Anomalous symmetries cannot be gauged". The phenomenon of anomalies is not confined to quantum field theories. Anomalies exist also in classical field theories (I tried to emphasize this point in my answer on this question).

(As already mentioned in the question), in the classical level, a symmetry is anomalous when the Lie algebra of its realization in terms of the fields and their conjugate momenta (i.e., in term of the Poisson algebra of the Lagrangian field theory) develops an extension with respect to its action on the fields. This is exactly the case of the complex field shift on the Schrödinger Lagrangian.

In Galilean (classical) field theories, the existence of anomalies is accompanied by the generation of a total derivative increment to the Lagrangian, which is again manifested in the case of the field shift symmetry of the Schrödinger Lagrangian, but this is not a general requirement. (please see again my answer above referring to the generation of mass as a central extension in Galilean mechanics).

In a more modern terminology, the impossibility of gauging is termed as an obstruction to equivariant extensions of the given Lagrangians. A nontrivial family of classical Lagrangians, exhibiting nontrivial obstructions, are Lagrangians containing Wess-Zumino-Witten terms. Given these terms only anomaly free subgroups of the symmetry groups can be gauged (classically). These subgroups consist exactly of the anomaly free ones. The gauging and the obstruction to it can be obtained using the theory of equivariant cohomology, please see the following article by Compean and Paniagua and its reference list.

Answer 2

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:

1. Peter West, Introduction to Supersymmetry and Supergravity, 1990, Chap. 7.

2. Henning Samtleben, Lectures on Gauged Supergravity and Flux Compactifications, Class. Quant. Grav. 25 (2008) 214002, arXiv:0808.4076.

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$^1$ 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.

Answer 3

First of all, one can't gauge a symmetry without modifying (enriching) the field contents. Gauging a symmetry means to add a gauge field and the appropriate interactions (e.g. by covariantize all terms with derivatives, in the case of both Yang-Mills and diffeomorphism symmetries).

Global and gauge symmetries are different entities when it comes to their physical interpretations; but they're also different entities when it comes to the degree of symmetry they actually carry.

Concerning the latter difference, a symmetry is a gauge symmetry if the parameters of the transformations $\lambda$ may depend on the spacetime coordinates, $\lambda = \lambda(\vec x,t)$. If they can, they can and the theory has a gauge symmetry; if they can't, they can't and the theory has at most a global symmetry. There can't be any ambiguity here; you can't "gauge a symmetry by changing nothing at all".

Concerning the former difference, the gauge symmetries must be treated as redundancies: the physical configurations (classically) or quantum states (quantum mechanically) must be considered physically identical if they only differ by a gauge transformation. For Lie gauge symmetries, this is equivalent to saying that physical states must be annihilated by the generators of the gauge symmetries. For any local symmetry as described in the previous paragraph, one typically generates unphysical states (of negative norm etc.) and they have to be decoupled – by being classified as unphysical ones.

In the Yang-Mills case, a global symmetry may be gauged but the final spectrum should be anomaly-free because gauge anomalies are physical inconsistencies exactly because gauge symmetries are just redundancies and one isn't allowed to "break" them spontaneously because they really reduce the physical spectrum down to a consistent one. In this respect, they differ from the global symmetries that can be broken. Of course, even an anomalous global symmetry may be gauged by adding the gauge fields and other fields that are capable of canceling the gauge anomaly.

Finally, the shift invariance of the massless Dirac $\psi$ in your example physically corresponds to the possibility to add a zero-momentum, zero-energy fermion into the system. It's just a way to find a "new solution" of this theory which is possible because $\psi$ is only coupled via derivative terms. The symmetry wouldn't be a symmetry if there were a mass term.

You may easily gauge this symmetry by replacing $\psi$ with $\psi+\theta$ everywhere in the action and promoting $\theta$ to a new field – which plays a similar role as the new gauge field $A_\mu$ if you're gauging a Yang-Mills-like global symmetry. By doing so, you will have twice as many off-shell fermionic degrees of freedom but the action won't depend on one of them, $\psi+\theta$ (the opposite sign), at all. So this field will create ghostly particles that don't interact with anything else – in fact, they don't even have kinetic terms. Clearly, these dynamically undetermined quanta shouldn't be counted in a physical theory (although, in some sense, they "just" increase the degeneracy of each state of the physical fields by an infinite extra factor) so the right way to treat them, as always in gauge theories, is to require that physical states can't contain any such quanta.

This requirement effectively returns you to the original theory, just with $\psi$ renamed as $\psi+\theta$. You won't get a new interesting theory in this way and there's no reason why gauging a symmetry should always produce such an interesting new theory. The case of Yang-Mills theories or generally covariant theories is different because they are interesting: with a Lorentz-covariant field content, one may create theories with no ghosts (negative-norm states) despite the fact that they predict the existence of spin-one or spin-two particles (from the gauge field – which is the metric tensor in the spin-two case). But this is only possible because these theories are special and the action of the symmetry transformations is less trivial than in your case. "Shift" symmetries may only be gauged in a way that renames or erases whole fields so they just can't lead to interesting new possibilities.

Answer 4

Assuming that the following manipulations are correct the translational symmetry of your Lagrangian can be gauged by including a scalar gauge field $\phi$ and a one form gauge field $A_{\mu}$.

First of all, assuming that the boundary terms do not contribute we can write the Lagrangian density as $$ \mathscr L=\psi^\dagger i\partial_t\psi-\frac{1}{2m}(\partial^{\mu}\psi)^{\dagger}\partial_{\mu}\psi. $$

Now writing $\psi$ as $\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$ the translation of $\psi$ can be written as $\left[ \begin{array}{cc} 1 & \theta_1+i\theta_2\\ 0 & 1\\ \end{array}\right]\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$

Lie algebra of the group of matrices of the form $\left[ \begin{array}{cc} 1 & \theta_1+i\theta_2\\ 0 & 1\\ \end{array}\right]$ is the set of matrices $\left[ \begin{array}{cc} 0 & a+ib\\ 0 & 0\\ \end{array}\right]$

Now to gauge this symmetry introduce a Lie algebra valued one form $A=A_{\mu}dx^{\mu}$ which under a gauge transformation

$\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]\rightarrow \left[ \begin{array}{cc} 1 & \theta_1(x)+i\theta_2(x)\\ 0 & 1\\ \end{array}\right]\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$

transform as

$A_{\mu}\rightarrow g(x)A_{\mu}g(x)^{-1}+(\partial_{\mu}g(x))g(x)^{-1} $

Where $g(x)=\left[ \begin{array}{cc} 1 & \theta_1(x)+i\theta_2(x)\\ 0 & 1\\ \end{array}\right]$

However we note that the Lagrangian $$ \mathscr L=\psi^\dagger i(\partial_t-A_t)\psi-\frac{1}{2m}((\partial^{\mu}-A^{\mu})\psi)^{\dagger}(\partial_{\mu}-A_{\mu})\psi. $$

is not gauge invariant and neither real.

Obstruction to gauge invariance is the fact that $\psi^{\dagger}$ doesn't transform by right multiplication by $g(x)^{-1}$ but rather by right multiplication by $g(x)^{\dagger}$

To repair gauge invariance one may introduce a matrix valued scalar gauge field $\phi$ whose exponential under change of gauge changes as

$exp(\phi(x))\rightarrow (g(x)^{\dagger})^{-1}exp(\phi(x))g(x)^{-1}$

(how does $\phi$ change? I am not sure)

Then we see that the Lagrangian

$$ \mathscr L=\psi^\dagger iexp(\phi(x))(\partial_t-A_t)\psi-\frac{1}{2m}((\partial^{\mu}-A^{\mu})\psi)^{\dagger}exp(\phi(x))(\partial_{\mu}-A_{\mu})\psi. $$

is gauge invariant. However still the Lagrangian is not real. To repair that we can include complex conjugate of each term in it.