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  When can a global symmetry be gauged?

+ 9 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Tomáš Brauner
asked Apr 16, 2013 in Theoretical Physics by Tomáš Brauner (115 points) [ no revision ]
Most voted comments show all comments
@DavidBarMoshe: Thanks for clarifying this point! Yes, I chose the Schrödinger Lagrangian precisely because the commutator of the two shift symmetries has a central charge, unlike e.g. the Klein-Gordon Lagrangian for a massless complex scalar field.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Tomáš Brauner
Thanks and sorry, but I do not understand that. 1)The non-relativistic limit of a real Klein-Gordon field is a (complex) Schrödinger field. The former obeys a 2nd order EOM, while the latter a 1st EOM. So one physical degree of freedom in both cases. 2) The transformation is uniparametric so that there is only one generator ($\int \, \theta \, \psi^{\dagger} + h.c.$). Sorry about the brevity, I'm in a rush. @DavidBarMoshe

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user drake
@drake: The nonrelativistic limit has a two-parametric shift symmetry: $i\int(\psi-\psi^\dagger)$ generates real shifts of $\psi$ while $\int(\psi+\psi^\dagger)$ imaginary shifts. The commutator of these two generators is a constant, i.e., a central charge, which represents the obstruction to gauging this symmetry. This comes from the fact that $\psi$ and $\psi^\dagger$ are canonically conjugate. In the case of a complex Klein-Gordon field, $\psi$ and $\psi^\dagger$ are independent dynamical variables; classically, the symmetry is the same, but now there is no central charge in the commutator.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Tomáš Brauner
Let me point out that the reason why my argument does not work is because one would need a real, massless Klein-Gordon field. Real and Klein-Gordon are no problem. The problem is "massless" since then one cannot take the non-relativistic limit.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user drake
Related preprint by OP: arxiv.org/abs/1001.5212 page 6.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Qmechanic
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I think that one can gauge this symmetry, which is the responsible of the renormalizability of Proca theory physics.stackexchange.com/q/16931 Have you read the opposite? Ref or link?

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user drake
@drake: Sorry, but I don't see the connection of my example to Proca's theory. I am talking about a classical nonrelativistic scalar field. If you think the global symmetry in this example can be gauged, can you write down the gauge-invariant Lagrangian? Anyway, thanks for the comment!

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Tomáš Brauner

4 Answers

+ 7 like - 0 dislike

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 Schrodinger Lagrangian.

In Galilean (classical) field theoties, 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 Schrodinger 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.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user David Bar Moshe
answered Apr 17, 2013 by David Bar Moshe (4,355 points) [ no revision ]
I see, so my guess that the obstruction to gauging is related to the central charge (in the commutator of the generators of real and imaginary shifts of $\psi$) was in the right direction. Thanks for a clear and general statement! I will study your previous answer carefully. Apparently, there is a lot they didn't tell me at school :)

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
I read your previous answer and got the point with the classical anomaly, thanks! Before I get into the mathematical details, is there an intuitive way to understand why an obstruction like a central charge in the Poisson brackets of the symmetry generators prevents gauging the symmetry on the classical level? Also, is there an elementary way to decide whether the symmetry can be gauged which does not use the canonical structure? (I think of low-energy effective Lagrangians which can contain an arbitrary number of derivatives.)

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
@Tomáš Brauner: Yes, the obstruction to gauging can be understood intuitively by means of the Dirac's constraint theory. If the theory can be gauged, the currents couple to the gauge field. Since the time component of the gauge field is non-dynamical, the charge densities of the symmetry currents will become constraints, i.e., must become zero on the gauge surface which can be thought as a reformulation of the theory with gauge invariant fields. But, then how can the bracket of two vanishing quantities give a nonzero constant.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user David Bar Moshe
I think I have understood your point, thanks once again!

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Luboš Motl
answered Apr 16, 2013 by Luboš Motl (10,278 points) [ no revision ]
Hi Luboš, thanks for the comprehensive answer! Concerning the first six paragraphs: I do know these things :) As to the rest: I'm afraid I had something else in mind. I want to add an auxiliary gauge field, i.e. no new physical degrees of freedom. The point is that I want to construct a generating functional for the conserved currents of the theory. This is just a mathematical trick; the physical symmetry of the theory remains global. E.g. for a massless Lorentz scalar, the shift symmetry can be gauged as $\mathscr L=\frac12(\partial_\mu\phi-A_\mu)^2$, but this doesn't work in my case.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
Thanks, Tomáš! Unfortunately, I can't help you with this because I don't understand your theory that is both gauge-symmetric as well as physically globally symmetric only. It's like the Princess Koloběžka (Scooter?) the First, right? ;-) youtube.com/watch?v=mBC9vr3nuiI What does it mean for a field to be called a "gauge field" if the normally associated with it gauge symmetry doesn't exist at all?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Luboš Motl
Suppose you have an action $S[\phi]$ with a global symmetry. I'm looking for an action $S'[\phi,A_\mu]$ such that $S'[\phi,0]=S[\phi]$. If I manage to make $S'[\phi,A_\mu]$ gauge-invariant, then $e^{-W[A_\mu]}=\int[d\phi]e^{-S'[\phi,A_\mu]}$ will be a generating functional of Green's functions of conserved currents of the original theory. That's why I say that $A_\mu$ is not a dynamical field and the physical symmetry, obtained by setting it to zero, is still global. You see the scooter there? :)

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
Dear Tomáši, first, it's easy to add fields so that the new theory will have a gauge symmetry but in a typical case, one doesn't get an interesting theory because the new gauge symmetry just removes some degrees of freedom and it's more useful to erase them immediately with the symmetry, anyway. Second, I don't understand why you would consider extra fields to find out whether the original theory has conserved currents. If $S$ admits conserved currents, it does, otherwise it doesn't.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Luboš Motl
Third, you may always add $J\cdot A$ fields to an action so that you obtain a generating functional for correlation functions of $J$ in the original theory: the theory with the extra term doesn't have to have a gauge symmetry. The field $A$ is auxiliary. Fourth, treating it as auxiliary is less constraning because if it is dynamical, you have to impose the eqn of motion from varying $A$. Fifth, the equation of motion is $J=0$ unless you manage to write new "kinetic" terms for $A$ as well which is what makes the gauge symmetry physical and interesting but it's not guaranteed to exist.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Luboš Motl
Thanks for the comments, Luboš! Here is an answer to some of your points. (2) One can derive Noether currents by variation w.r.t. the gauge field, although that's not my main reason to gauge the symmetry, see next. (3) You can do this, but it's much less restrictive than gauge invariance and tells you nothing about how the external fields such as $A$ appear in a low-energy EFT, see e.g. the paper hep-ph/9311264 by Leutwyler. One can also add other external fields; the reason why I couple "gauge" fields to conserved currents is that I'm interested in low-energy EFTs for Goldstone bosons.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
+ 6 like - 0 dislike

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.


$^1$ 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
answered Apr 16, 2013 by Qmechanic (3,120 points) [ no revision ]
Thanks a lot, this is very helpful! I still have to check how (and whether) this works in the non-Abelian case, but I have already learned quite something from you :)

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
I updated the answer.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Qmechanic
This is again very instructive, thanks! So to summarize this example, the free Schrödinger field theory has a global ISO(2) symmetry (phase rotations plus two translations in the plane of complex $\psi$). Any of the three one-parametric Abelian subgroups can be gauged, but this breaks the remaining global symmetry. Two whole ISO(2) group, or even the normal subgroup of translations, cannot be gauged. It's good to see this so explicitly.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner
+ 4 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user10001
answered Apr 17, 2013 by user10001 (635 points) [ no revision ]
Nice construction! In effect, it boils down to introducing a extra field $\chi$ which transforms as $\chi\to\chi-\theta_1-i\theta_2$, and replacing $\psi^\dagger i\partial_t\psi$ with $\frac i2[(\psi^\dagger+\chi^\dagger)\partial_t\psi-\partial_t\psi^\dagger(\psi+\chi)]$ and subsequently ordinary derivatives with covariant ones. I'm not sure if this is what I want though, since even after setting $A=0$, this theory has a different global symmetry than the theory I started with. Thanks anyway, you made me formulate more precisely what I'm looking for!

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Tomáš Brauner

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