The Unapologetic Mathematician

Mathematics for the interested outsider

The Higgs Mechanism part 4: Symmetry Breaking

This is part four of a four-part discussion of the idea behind how the Higgs field does its thing. Read Part 1, Part 2, and Part 3 first.

At last we’re ready to explain the Higgs mechanism. We start where we left off last time: a complex scalar field \phi with a gauged phase symmetry that brings in a (massless) gauge field A_\mu. The difference is that now we add a new self-interaction term to the Lagrangian:

\displaystyle L=-\frac{1}{4}F_{\mu\nu}F_{\mu\nu}+(D_\mu\phi)^*D_\mu\phi-\left[-m^2\phi^*\phi+\lambda(\phi^*\phi)^2\right]

where \lambda is a constant that determines the strength of the self-interaction. We recall the gauged symmetry transformations:

\displaystyle\begin{aligned}\phi'(x)&=e^{i\alpha(x)}\phi(x)\\A_\mu'(x)&=A_\mu(x)+\frac{1}{e}\partial_\mu\alpha(x)\end{aligned}

If we write down an expression for the energy of a field configuration we get a bunch of derivative terms — basically like kinetic energy — that all occur with positive signs and then the potential energy term that comes in the brackets above:

\displaystyle V(\phi^*\phi)=-m^2\phi^*\phi+\lambda(\phi^*\phi)^2

Now, the “ground state” of the system should be one that minimizes the total energy, but the usual choice of setting all the fields equal to zero doesn’t do that here. The potential has a “bump” in the center, like the punt in the bottom of a wine bottle, or like a sombrero.

So instead of using that as our ground state, we’ll choose one. It doesn’t matter which, but it will be convenient to pick:

\displaystyle\begin{aligned}A_\mu^{(v)}&=0\\\phi^{(v)}=\frac{1}{\sqrt{2}}\phi_0\end{aligned}

where \phi_0=\frac{m}{\sqrt{\lambda}} is chosen to minimize the potential. We can still use the same field A_\mu as before, but now we will write

\displaystyle\phi(x)=\frac{1}{\sqrt{2}}\left(\phi_0+\chi(x)+i\theta(x)\right)

Since the ground state \phi_0 is a point along the real axis in the complex plane, vibrations in the field \chi measure movement that changes the length of \phi, while vibrations in \theta measure movement that changes the phase.

We want to consider the case where these vibrations are small — the field \phi basically sticks near its ground state — because when they get big enough we have enough energy flying around in the system that we may as well just work in the more symmetric case anyway. So we are justified in only working out our new Lagrangian in terms up to quadratic order in the fields. This will also make our calculations a lot simpler. Indeed, to quadratic order (and ignoring an irrelevant additive constant) we have

\displaystyle V(\phi^*\phi)=m^2\chi^2

so vibrations of the \theta field don’t show up at all in quadratic interactions.

We should also write out our covariant derivative up to linear terms:

\displaystyle D_\mu\phi=\frac{1}{\sqrt{2}}\left(\partial_\mu\chi+i\partial_\mu\theta-ie\phi_uA_\mu\right)

so that the quadratic Lagrangian is

\displaystyle\begin{aligned}L^{(2)}&=-\frac{1}{4}F_{\mu\nu}F_{\mu\nu}+\frac{1}{2}\lvert\partial_\mu\chi+i\partial_\mu\theta-ie\phi_uA_\mu\rvert^2-m^2\chi^2\\&=-\frac{1}{4}F_{\mu\nu}F_{\mu\nu}+\left[\frac{1}{2}\partial_\mu\chi\partial_\mu\chi-m^2\chi^2\right]+\frac{e^2\phi_0^2}{2}\left(A_\mu-\frac{1}{e\phi_0}\partial_\mu\theta\right)^2\end{aligned}

Now, the term in parentheses on the right looks like the mass term of a vector field B_\mu with mass e\phi_0. But what is the kinetic term of this field?

\displaystyle\begin{aligned}B_{\mu\nu}&=\partial_\mu B_\nu-\partial_\nu B_\mu\\&=\partial_\mu\left(A_\nu-\frac{1}{e\phi_0}\partial_\nu\theta\right)-\partial_\nu\left(A_\mu-\frac{1}{e\phi_0}\partial_\mu\theta\right)\\&=\partial_\mu A_\nu-\partial_\nu A_\mu-\frac{1}{e\phi_0}\left(\partial_\mu\partial_\nu\theta-\partial_\nu\partial_\mu\theta\right)\\&=F_{\mu\nu}-0=F_{\mu\nu}\end{aligned}

And so we can write down the final form of our quadratic Lagrangian:

\displaystyle L^{(2)}=\left[-\frac{1}{4}B_{\mu\nu}B_{\mu\nu}+\frac{e^2\phi_0^2}{2}B_\mu B_\mu\right]+\left[\frac{1}{2}\partial_\mu\chi\partial_\mu\chi-m^2\chi^2\right]

In order to deal with the fact that our normal vacuum was not a minimum for the energy, we picked a new ground state that did minimize energy. But the new ground state doesn’t have the same symmetry the old one did — we have broken the symmetry — and when we write down the Lagrangian in terms of excitations around the new ground state, we find it convenient to change variables. The previously massless gauge field “eats” part of the scalar field and gains a mass, leaving behind the Higgs field.

This is essentially what’s going on in the Standard Model. The biggest difference is that instead of the initial symmetry being a simple phase, which just amounts to rotations around a circle, we have a (slightly) more complicated symmetry to deal with. For those that are familiar with some classical groups, we start with an action of SU(2)\times U(1) on a column vector \phi made of two complex scalar fields with a potential of the form:

\displaystyle V(\phi)=\lambda\left(\phi^\dagger\phi-\frac{v^2}{2}\right)^2

which is invariant under the obvious action of SU(2) and a phase action of U. Since the group SU(2) is three-dimensional there are three gauge fields to introduce for its symmetry and one more for the U(1) symmetry.

When we pick a ground state that breaks the symmetry it doesn’t completely break; a one-dimensional subgroup U(1)\subseteq SU(2)\times U(1) still leaves the new ground state invariant — though it’s important to notice that this is not just the U(1) factor, but rather a mixture of this factor and a U(1) subgroup of SU(2). Thus only three of these gauge fields gain mass; they become the W^\pm and Z^0 bosons that carry the weak force. The other gauge field remains massless, and becomes \gamma — the photon.

At high enough energies — when the fields bounce around enough that the bump doesn’t really affect them — then the symmetry comes back and we see that the electromagnetic and weak interactions are really two different aspects of the same, unified phenomenon, just like electricity and magnetism are really two different aspects of electromagnetism.

July 19, 2012 - Posted by | Higgs Mechanism, Mathematical Physics, Special Topics

4 Comments »

  1. […] July 17, “The Higgs Mechanism part 3: Gauge Symmetries,” July 18, y “The Higgs Mechanism part 4: Symmetry Breaking,” July 19. Recomiendo una lectura a estas […]

    Pingback by Las matemáticas del bosón de Higgs, para las abuelas cansadas de cháchara (Parte I) « Francis (th)E mule Science's News | July 19, 2012 | Reply

  2. […] The Higgs Mechanism part 4: Symmetry Breaking Share this:FacebookRedditTwitterStumbleUponEmailLike this:LikeBe the first to like this. […]

    Pingback by The Unapologetic Mathematician Tackles the Higgs Mechanism | Whiskey…Tango…Foxtrot? | July 20, 2012 | Reply

  3. I must comment you for a warts-and-all presentation. Here you can see for what it is, a jimcrack device to force the physics community to swallow the universal term mass, a move which is not licensed under reigning Medieval theology! Decades of ranting against Platonist infiltration by Wittgenstein down the drain…if they can find the missing isospin that is.

    Comment by Orwin O'Dowd | October 31, 2012 | Reply

  4. […] July 17, “The Higgs Mechanism part 3: Gauge Symmetries,” July 18, y “The Higgs Mechanism part 4: Symmetry Breaking,” July 19. Recomiendo una lectura a estas […]

    Pingback by Las matemáticas del bosón de Higgs, para las abuelas cansadas de cháchara (Parte I) | Bosón de Higgs | La Ciencia de la Mula Francis | November 1, 2017 | Reply


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