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I've been watching Leonard Susskind's particle physics lectures and in one lecture, he discusses a very simple gauge theory. We have a complex scalar field $\phi(x)$ with Lagrangian

$$\mathscr{L} = \partial_\mu \phi^* \partial^\mu \phi - V(\phi^* \phi)$$

and we want to do the gauge transformation $\phi(x) \to e^{i\theta(x)} \phi(x)$. But the derivative term in the Lagrangian is not invariant if $\theta$ varies from place to place, so we add a new vector field $A_\mu$, and define it to transform like $$A_\mu \to A_\mu + \partial_\mu \theta$$ under the gauge transformation. Then we change the Lagrangian to use covariant derivatives $D_\mu = \partial_\mu + iA_\mu$ instead of ordinary ones; now when the gauge transform is done all the derivatives of $\theta$ cancel out, and the result is invariant.

My question is: was there any freedom in choosing to add a vector field, with that specific gauge transformation law? Are there any other ways we could get gauge invariance here—by using a different transformation law, or with a different type of field (say, a scalar, tensor, or spinor) with some other transformation law?

More generally, how can we tell what type of field is needed and what its gauge transformation law should be, to get some particular Lagrangian to be invariant under some particular gauge transformation?

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We have no choice.

Let $G$ be our gauge group and $\Sigma$ our spacetime. Then, for the theory to actually be gauge invariant, every field must have a defined action of the gauge group upon it, i.e. every field must transform in a representation of this group:

$$\phi : \Sigma \to V_\rho \text{ where there is a group morphism } \rho : G \to \mathrm{GL}(V_\rho)$$

We want a derivative $\mathrm{d}_A : \Omega^k(\Sigma) \to \Omega^{k+1}(\Sigma)$ acting upon the fields (more generally, upon $k$-forms producing $k+1$-forms) such that for every gauge transformation $g : \Sigma \to G$ we have $\mathrm{d}_A(\rho(g)\phi) = \rho(g)\mathrm{d}_A\phi$, i.e. the derivative must also transform in the representation.

Now, the forms only come with two natural operations producing a $k+1$-form out of a $k$-form: The exterior derivative $\mathrm{d}$, which fails miserably on its own, and the wedge product of the $k$-form with some $1$-form. Therefore, the only natural way of searching for the exterior derivative is

$$\mathrm{d}_A \omega := \mathrm{d}\omega + A \wedge \omega$$

for some $1$-form (i.e. dual vector field) $A$. It must be stressed that, though $A$ is, as a $1$-form, indeed a (dual) vector field with regards to the Lorentz group, it is not transforming in a proper linear representation of $G$, since its transformation law is (in order to make the group action and the derivative commute)

$$ A \mapsto gAg^{-1} + g\mathrm{d}g^{-1}$$

It is called a connection form and corresponds to a particular choice of an orthogonal subspace of a tangent space of the underlying principal bundle. From this, it is also seen that $A$ must take values in the Lie algebra $\mathfrak{g}$ (also since else the transformation law above would make litte sense). For all gauge theories, this choice of subspace (called an Ehresmann connection) is in bijection with choosing a $1$-form $A$ on the bundle, which projects down to a (local!) $1$-form on $\Sigma$, giving again the vector field we heuristically found by searching for the covariant derivative.


However, it must be remarked that if we relax our notions of what a gauge theory is, then there are "connection forms" that are no vector fields. The best example (and the only one I know) is that of the Christoffel symbols in GR, being sections of the tangent bundle of the frame bundle, the jet bundle (in contrast to the gauge fields being sections of the tangent bundle of the principal bundle), which can be seen as being the connection forms determining the Levi-Civita connection, analogous to the gauge fields determining the Ehresmann connection.

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  • $\begingroup$ I'm still digesting this (you did it in a bit more generality than I needed :)), but let me see if I've got the general idea. The problem is that because $g$ varies in spacetime, derivative terms in the Lagrangian pick up unwanted derivatives of $\rho(g)$. To cancel them, we must use a field that transforms under the Lorentz group the same way that derivatives of $\rho(g)$ do—that is, as a dual 4-vector whose components are members of (the $\rho$ representation of) the Lie algebra of $G$. Is that about right? $\endgroup$ – Nathan Reed Aug 3 '14 at 5:51
  • $\begingroup$ @Nathan Reed: Yup. :) $\endgroup$ – ACuriousMind Aug 3 '14 at 13:22

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