# The Unapologetic Mathematician

## The Higgs Mechanism part 3: Gauge Symmetries

This is part three of a four-part discussion of the idea behind how the Higgs field does its thing. Read Part 1 and Part 2 first.

Now we’re starting to get to the really meaty stuff. We talked about the phase symmetry of the complex scalar field:

\displaystyle\begin{aligned}\phi'(x)&=e^{i\alpha}\phi(x)\\\phi'^*(x)&=e^{-i\alpha}\phi^*(x)\end{aligned}

which basically wants to express the idea that the physics of this field only really depends on the length of the complex field values $\phi(x)$ and not on their phases. But another big principle of physics is locality — what happens here doesn’t instantly affect what happens elsewhere — so why should the phase change be global?

To answer this, we “gauge” the symmetry and make it local. The origin of the term is fascinating, but takes us too far afield. The upshot is that we now have the symmetry transformation:

\displaystyle\begin{aligned}\phi'(x)&=e^{i\alpha(x)}\phi(x)\\\phi'^*(x)&=e^{-i\alpha(x)}\phi^*(x)\end{aligned}

where $\alpha$ is no longer a constant, but a function of the spacetime point $x$.

And here’s the big problem: since $\alpha$ varies from point to point, it now affects our derivative terms! Before we had

\displaystyle\begin{aligned}\partial_\mu\phi'(x)&=\partial_\mu\left(e^{i\alpha}\phi(x)\right)\\&=e^{i\alpha}\partial_\mu\phi(x)\end{aligned}

and similarly for $\phi^*$. We say that the derivatives are “covariant” under the transformation; they transform in the same way as the underlying fields. And this is what lets us say that

$\displaystyle\partial_\mu\phi'^*\partial_\mu\phi'=\partial_\mu\phi^*\partial_\mu\phi$

and makes the whole Lagrangian symmetric.

On the other hand, what do we see now?

\displaystyle\begin{aligned}\partial_\mu\phi'(x)&=\partial_\mu\left(e^{i\alpha(x)}\phi(x)\right)\\&=e^{i\alpha(x)}\partial_\mu\phi(x)+i\partial_\mu\alpha(x)e^{i\alpha(x)}\phi(x)\\&=e^{i\alpha(x)}\left[\partial_\mu\phi(x)+i\partial_\mu\alpha(x)\phi(x)\right]\end{aligned}

We pick up this extra term when we differentiate, and it ruins the symmetry.

The way out is to add another field that can “soak up” this extra term. Since the derivative is a vector, we introduce a vector field $A_\mu$ and say that it transforms as

$\displaystyle A_\mu'(x)=A_\mu(x)+\frac{1}{e}\partial_\mu\alpha(x)$

Next, we introduce a new derivative operator: $D_\mu=\partial_\mu-ieA_\mu$. That is:

$\displaystyle D_\mu\phi(x)=\partial_\mu\phi(x)-ieA_\mu(x)\phi(x)$

And we calculate

\displaystyle\begin{aligned}D_\mu\phi'(x)&=\partial_\mu\left(e^{i\alpha(x)}\phi(x)\right)-ieA_\mu'(x)e^{i\alpha(x)}\phi(x)\\&=e^{i\alpha(x)}\partial_\mu\phi(x)+i\partial_\mu\alpha(x)e^{i\alpha(x)}\phi(x)-ieA_\mu(x)e^{i\alpha(x)}\phi(x)-i\partial_\mu\alpha(x)e^{i\alpha(x)}\phi(x)\\&=e^{i\alpha(x)}\left[\partial_\mu\phi(x)-ieA_\mu(x)\phi(x)\right]\\&=e^{i\alpha(x)}D_\mu\phi(x)\end{aligned}

So the derivative $D_\mu\phi(x)$ does vary the same way as the underlying field $\phi(x)$ does! We call $D_\mu$ the “covariant derivative”. If we use it in our Lagrangian, we do recover our symmetry, though now we’ve got a new field $A_\mu$ to contend with. Just like the electromagnetic potential we use the derivative $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ to write

$\displaystyle L=-\frac{1}{4}F_{\mu\nu}F_{\mu\nu}+(D_\mu\phi)^*D_\mu\phi-m^2\phi^*\phi$

which is now symmetric under the gauged symmetry transformations.

It may not be apparent, but this Lagrangian does contain interaction terms. We can expand out the second term to find:

\displaystyle\begin{aligned}(D_\mu\phi)^*D_\mu\phi&=\left(\partial_\mu\phi^*-ieA_\mu\phi^*\right)\left(\partial_\mu\phi-ieA_\mu\phi\right)\\&=\partial_\mu\phi^*\partial_\mu\phi-ieA_\mu\partial_\mu\phi^*\phi-ieA_\mu\phi^*\partial_\mu\phi-e^2A_\mu A_\mu\phi^*\phi\end{aligned}

Our rules of thumb tell us that if we vary the Lagrangian with respect to $A_\mu$ we get the field equation

$\displaystyle\partial_\mu F_{\mu\nu}=ej_\mu$

which — if we expand out $F_{\mu\nu}$ as if it’s the Faraday field into “electric” and “magnetic” fields — give us Gauss’ and Ampère’s law in the presence of a charge-current density $j_\mu$.

The charge-current, in particular, we can write as

$\displaystyle j_\mu=-i\left(\phi^*\partial_\mu\phi-\partial_\mu\phi^*\phi\right)-2eA_\mu\phi^*\phi$

or, in a gauge-invariant manner, as

$\displaystyle j_\mu=-i\left[\phi^*D_\mu\phi-(D_\mu\phi)^*\phi\right]$

which is just the conserved current from last time with the regular derivatives replaced by covariant ones. Similarly, varying with respect to the field $\phi$ we find the “covariant” Klein-Gordon equation:

$\displaystyle D_\mu D_\mu\phi+m^2\phi=0$

and, when this holds, we can show that $\partial_\mu j_\mu=0$.

So we’ve found that if we take the global symmetry of the complex scalar field and “gauge” it, something like electromagnetism naturally pops out, and the particle of the complex scalar field interacts with it like charged particles interact with the real electromagnetic field.

July 18, 2012 -

1. That derivative operator is reminiscent of canonical momentum. Are they related?

Comment by Bluescreenofdebt | July 18, 2012 | Reply

2. When quantizing, the derivative with respect to a coordinate is indeed related to the “momentum operator” conjugate to that coordinate. I’m sticking to classical field theory here, though.

Comment by John Armstrong | July 18, 2012 | Reply

3. What does this “covariant derivative” have to do with the covariant derivative as usually defined on a bundle with a connection over a manifold? As far as I can tell, you’re still working in Lorentzian space-time.

Comment by Greg Friedman | July 18, 2012 | Reply

4. That is an excellent question, Greg. I’ll start answering it by saying we’re actually working over Lorentzian spacetime, in that our fields are the sections of some fiber bundle over this space, and in this case the gauge group is $U(1)$. The $A_\mu$ fields are actually the components — analogous to the Christoffel symbols — of a connection on the associated principal fiber bundle.

Comment by John Armstrong | July 18, 2012 | Reply

5. Incidentally, that’s why there’s the $i$ in the covariant derivative formula: $iA_\mu$ is really a $\mathfrak{u}(1)$-valued field.

Comment by John Armstrong | July 18, 2012 | Reply

6. […] discussion of the idea behind how the Higgs field does its thing. Read Part 1, Part 2, and Part 3 […]

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7. […] “The Higgs Mechanism part 2: Examples of Lagrangian Field Equations,” July 17, “The Higgs Mechanism part 3: Gauge Symmetries,” July 18, y “The Higgs Mechanism part 4: Symmetry Breaking,” July 19. […]

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