Spontaneous Symmetry Breaking : a crash course

Here I give a brief overview of spontaneous symmetry breaking (SSB) in the way I would present it to experimental particle physicists, assuming only little QFT background and barely even mentioning group theory. We will start with one-dimensional discrete symmetry breaking, extend it to continuous SB, mention Goldstone’s theorem and introduce the Higgs mechanism, before jumping to electroweak theory and obtaining the main result of interest : the generation of masses for the $W^\pm$ and $Z^0$ gauge bosons in the Standard Model.

Note : this is all standard material, but for more lectures on the Standard Model you can have a look under my teaching page.

Discrete symmetry breaking

We’ll start off by considering the simplest case in QFT : the scalar Lagrangian in $\phi^4$ theory, written

We note straight away that $\mathcal{L}$ is invariant under $\phi\to -\phi$ (sometimes referred to as “flip-symmetry”). The physical case is of course that where $m^2>0$ (that is, the real scalar field $\phi$ has positive definite mass $m$), for which the potential admits a unique, global minimum at $\phi=0$ (and we can freely perturbate around this vacuum) :

Unbroken symmetry

But what about the case where $\mathbf{m^2<0}$? Of course we lose, or at least we seem to lose, the physicality of our scalar field – how do we interpret imaginary mass? Looking at the modified potential should give some clues on how to make sense of the situation :

Broken symmetry

We readily observe the transition to a double vacuum system, with local minima located at the vacuum expectation values (VEV) $\pm v$, where we’ve naturally defined $v^2\equiv -\frac{m^2}{\lambda}$ by completing the square in the potential :

(Recall that shifting the potential by a constant has a no effect on the equations of motions). We should already have a feeling that perturbations around either of those vacua are no longer going to be symmetric : on one side the potential goes up to infinity, on the other we could classically (quantum mechanically) jump (tunnel) to the second vacuum. We are now going to make this intuition more precise, by considering small perturbations $h(x)$ around the minimum :

(Again, we have ignored the purely constant terms). It should now be clear that the perturbed Lagrangian is that of a scalar field $h(x)$ representing massive excitations with $m_h^2=2\lambda v^2>0$ around the vacuum – we have effectively restored a physical positive mass. And because of the third order term in $h$, the Lagrangian is no longer invariant under flip symmetry $h\to -h$; the symmetry was spontaneously broken by the VEV!

Continuous symmetry breaking

Instead of a single real scalar field, let’s now consider $N$ of them : $\phi=\left(\phi_1,\phi_2,\ldots,\phi_N\right)^\text{T}$. We still only look at $\phi^4$ theory and write the Lagrangian as

where dot products are understood between the $\phi_i$’s. In fact, all of our previous definitions (especially that of the VEV $v$) still apply, except that now we have a continuum of vacua $\phi^2=v^2$, which live on a circle in the diagram below :

Broken symmetry and the Mexican hat potential

(This particular potential, by the way, is often referred to as the “Mexican hat potential”, for obvious reasons…). For simplicity, and without loss of generality, we’ll pick a particular vacuum configuration $\phi_0=\left(0,0,\ldots,v\right)^\text{T}$, and pertubate around it, making a difference between radial (i.e. along the potential gradient) and angular excitations (along the circle of vacuua) :

where $m_\sigma$ is defined as in the discrete case by $m_\sigma^2=2\lambda v^2>0$. We have here another important result : only the radial mode is massive, and the $N-1$ angular $\pi$ modes are left massless. Indeed, only the $\sigma$ field loses energy to the potential when excited.

A tiny bit of group theory, and the shortest statement of Goldstone’s theorem ever…

(OK, I lied, but I think this is an important enough point to deserve a paragraph of its own). The physical statement that we have reduced (or broken) the $N$-dimensional rotational symmetry of the Lagrangian to only $N-1$ directions (the vacuum circle) is equivalent to saying the group of orthogonal matrices (which describe rotations in space) $O(N)$ has been broken down to $O(N-1)$. I’ll probably come back some day to write a proper article on the group theoretic treatment of SSB here it is, but the only fact we need right now is that $O(N)$ has $\frac{N(N-1)}{2}$ generators (a sort of basis of matrices, much like unit vectors). A quick calculation will convince you that if we start with $O(N)$ and end up with $O(N-1)$, we’ve lost $N-1$ generators along the way. This is where Goldstone’s theorem come into play :

GOLDSTONE : broken generators $\sim$ massless modes

This is perhaps the crudest, least formal statement of this theorem I’ve ever come across, and it completely by-passes many of its subtleties (which, as it turns out, are the reason we need the Higgs mechanism in the first place), but at least it gets the point across : what we lose in symmetry (generators), we gain in massless excitations; and every other component of the original field (in our case, $N-(N-1)=1$) has to be massive.


The Higgs mechanism

Let’s make things slightly more complicated by now considering a complex scalar field $\phi$ with the usual photon ($U(1)$ gauge field) $A_\mu(x)$ (this is sometimes called “scalar electrodynamics”). We assume here some QED priors and write :

We also know that under $U(1)$ gauge transformations :

As before, the “physical” case $\mu^2>0$ preserves the $U(1)$ (phase-shift) symmetry of the Lagrangian, and gives us a massive scalar $\phi$ and a massless photon, as expected. Yet, if we choose $\mu^2<0$ instead, we obtain again a circle of vacua $\lvert\phi_0\rvert^2=-\frac{\mu^2}{2\lambda}\equiv\frac{v^2}{2}>0$. This time, we’ll write

(This is just the radial-angular distinction we made in the last section). Pertubating in all directions around this particular choice of vacuum :

where the last approximation comes from the assumption that $\lvert\xi/v\rvert\ll 1$. We rewrite the Lagrangian as , where contains higher-order interaction terms :

We recognize here some usual terms (field strength tensors, kinematic terms, masses), but $A_\mu\partial^\mu\xi$ stands out. It is particularly illuminating to rewrite the last three terms above as:

Indeed, this looks very much like a $U(1)$ gauge transformation! Going to unitary gauge (just a $U(1)$ transformation, as previously defined) :

Effectively, we lost the angular mode $\xi$ and made sure the fluctuations are real. The final Lagrangian is

We started with a massive complex scalar $\phi$ and a massless gauge field $A_\mu$, broke the $U(1)$ symmetry, produced one massless Goldstone mode $\xi$, and ended up with one massive real scalar $\eta$ (the Higgs of this theory) and a now massive gauge field $A_\mu$; in colloquial terms, we say that the gauge field ate the Goldstone boson. For completeness :


Electroweak theory

We now have all the tools we need to apply the Higgs mechanism to the EW sector of the Standard Model and generate the experimentally observed masses of the otherwise massless EW gauge bosons.

Setup

The gauge group of the EW sector is , and we introduce a massive complex scalar, the Higgs, in the doublet representation of $SU(2)$ with $U(1)$ hypercharge $\Upsilon=1$ (this last fact will be relevant at the very end, the rest is largely for informative purposes). Under a combined $SU(2)\times U(1)$ transformation :

where the $SU(2)$ generators are $\tau^a=\frac{\sigma^a}{2}$ (and the $\sigma^a$ are the usual Pauli matrices). In the unbroken theory, of course, the $3+1$ gauge bosons are massless; but now we let the Higgs acquire a non-zero VEV $\phi_0=\frac{1}{\sqrt{2}}\left(0,v\right)^\text{T}$ and “mass” $\mu^2=-\lambda v^2<0$, effectively enforcing the SSB – we’ll come back to this at the end. The part of the Lagrangian we’re concerned with is :

We now move on to generating masses for the physical gauge bosons $W^\pm,Z^0$, which, we point out, are not the massless gauge fields $W^{1,2,3},B$! (but mixing thereof, defined in due course).

Mass generation

The relevant terms originate from the derivatives of the Higgs field :

And there we have it : mass terms for the physical gauge bosons! We can write

And we usually express the mixing between the $W^3$ and $B$ fields through the Weinberg angle $\theta_W$ :

such that the relation $m_W=m_Z\cos\theta_W$ holds. Experimentally, $\theta_W\sim 0.49$ rad.

Generators and broken symmetries

We conclude with final remarks on the generators of $G$ and how they relate to what might otherwise seem like an arbitrary definition of the physical gauge bosons. Generally speaking, we’ll say that a vacuum configuration is invariant under the action of a rotation (in internal space, of course) operator $\hat{\omega}$ if and only if $e^{i\alpha\hat{\omega}}\phi=\phi$. That is, for small enough $\alpha$ that we can expand around the vacuum, $\hat{\omega}\phi=0$. It is then straightforward to see how the Higgs vacuum breaks $G$ :

and so both and are broken. However, the combination $\mathcal{Q}=\frac{1}{2}(\sigma_3+\Upsilon)$ leaves the vacuum invariant! (as $\mathcal{Q}\phi=0$). This is exactly the generator of $U(1)_{EM}$, which is the left-over symmetry after SSB, and corresponds precisely to the mixing of $W^3$ and $B$, i.e. gives us a massless photon, as expected. Again, we usually say that the physical gauge bosons ate the Goldstone modes of the SSB.

Final note : counting the number of degrees of freedom (or real components) of our field content before and after SSB is a very pedagogical exercise. We started with $4$ gauge fields ($W^{1,2,3}$ and $B$), each having $2$ real components; and a complex doublet scalar field $\phi$ with $2\times 2$ real components, for a grand total of $12$. At the end, only one real, massive Higgs is left ($1$ component), we have a massless photon $A_\mu$ with two polarizations ($2$ components), and three gauge bosons $W^\pm,Z^0$ which have acquired an additional longitudinal mode (the Goldstone mode they “ate”), all of which tally up to $1+2+3\times 3=12$.

Final final note : in the same way we mentioned the number of generators of $O(N)$ above, the number of those of $SU(2)$ and $U(N)$ can be obtained by $N^2-1$ and $N-1$ respectively. That is, $SU(2)$ has 3 generators and $U(1)$ only 1 (as we already know from the above). The SSB of $G$ to $U(1)_{EM}$ guarantees $3+1-1=3$ massless modes through Goldstone’s theorem, which are effectively the other three components of the Higgs complex doublet that get “eaten” by the gauge fields.

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