# What role does “spontaneously symmetry breaking” played in the “Higgs Mechanism”?

In talking about Higgs mechanism, the first part is always some introduction to the concept of spontaneously symmetry breaking (SSB), some people saying that Higgs mechanism is the results of SSB of local gauge symmetry, some people says that we can formulate Higgs mechanism in a gauge invariant way, some people also says that we need only a non-zero vaccum expectation value... I am confused about this different or maybe same point of views.

In this post: How does the Higgs mechanism work? , the most highly voted answer, I still can't feel how SSB worked in Higgs mechanism. It seem that the validity of last part, the appearance of a mass term for $A$, is guaranteed if we have a non-zero equilibrium value $\phi_0$ to expand around. I do not see that the requirement that the phase of the field $\phi$ need to be fixed at some particular value to generate mass term. Thus it seems to me it is not true that SSB is really indispensable for Higgs mechanism.

To put it simply:

The spontanously breaking of what is attributed to Higgs mechanism?

1. local gauge symmetry

2. global symmetry, since breaking of a "gauge symmetry" should not have any effect on physics. In higgs mechanism, the really broken symmetry is a global one. Mathematically, it is similar in looking as fixing a gauge, but one should not think it as a spontanously breakdown of local gauge symmetry.

3. other.

Is SSB really indispensable for Higgs mechanism?

1. yes, Higgs mechanism is relied on the SSB of some symmetry (above question), the other approches of description eventually has spontanously broke some symmetry.

2. No, the SSB is just one way to describe Higgs mechanism (or even not a complete way), what is really need is the non-zero vaccum expectation value, for example in the linked post the requirement for the mass term to occur is to have some non-zero expectation value of $\phi$ to expand around, we do not need the phase of the field to be fixed, thus the symmetry is not broken.

3. Other.

# some reference materials:

1. Is Elitzur's theorem valid only in lattice field theory? States that SSB of local gauge symmetry is impossible.

2. Gauge invariant accounts of the Higgs mechanism in the abstract states that:

gauge symmetries merely reflect a redundancy in the state description and therefore the spontaneous breaking can not be an essential ingredient. Indeed, as already shown by Higgs and Kibble, the mechanism can be explained in terms of gauge invariant variables, without invoking spontaneous symmetry breaking

1. Is electromagnetic gauge invariance spontaneously violated in superconductors? In the introduction it says:

In particular, we emphasize that global U(1) phase rotation symmetry, and not gauge symmetry, is spontaneously violated, and show that the BCS wave function is, contrary to claims in the literature, fully gauge invariant

It is frequently stated the Higgs mechanism involves spontaneous breaking of the gauge symmetry. This is, however, entirely wrong. In fact, gauge symmetries cannot be spontaneously broken.

A standard argument for this is that gauge symmetries are not actual symmetries, they are just a reflection of a redundancy in our description the system; two states related by a gauge transformation are actually the same physical state. Thus, a gauge symmetry is physically a "do-nothing transformation" and thus it does not make sense for it to be spontaneously broken.

This argument does seem like a bit of a cop-out, though -- I could just declare any symmetry to be a do-nothing transformation'' by fiat if I wanted to. A more satisfying explanation is: even if we interpret gauge symmetries as real symmetries, they can never be spontaneously broken. This result is known as Elitzur's theorem, and it's quite easy to understand why it should be true. Let's focus on classical thermal systems -- quantum systems at zero temperature map onto classical thermal systems in one higher space dimension so the argument should carry over.

First recall the hand-waving argument for why spontaneous symmetry breaking can take place in, say, the 2-D Ising model at finite temperature. The 2-D Ising model has two symmetry-breaking ground states: all $\uparrow$ and all $\downarrow$. But, if I want to get between them by local thermal fluctuations then I have to create a domain and grow it until it encompasses the whole system, which implies an extensive energy penalty due to the energy cost of the domain wall. Thus, at low temperatures transitions between the two ground states are exponentially suppressed in the system size and so the system gets stuck in either all $\uparrow$ or all $\downarrow$, so the symmetry is spontaneously broken. (The same argument shows why the 1-D Ising model cannot have spontaneous symmetry breaking at finite temperature, because there is no extensive energy penalty to get from all $\uparrow$ to all $\downarrow$.)

On the other hand, since a gauge symmetry is a local symmetry, this argument breaks down. Any two symmetry-breaking ground states are related by a sequence of local gauge transformations, which (since they commute with the Hamiltonian) have exactly zero energy penalty. Thus, there is no energy barrier between different ground states, and the system will explore the entire space of ground states -- so no symmetry-breaking. We expressed everything here in terms of classical thermal systems, but it will be important for later that the quantum version of no symmetry breaking is that the Hamiltonian must have a unique ground state (at least with appropriate boundary conditions), because degenerate ground states can always couple to each other through quantum fluctuations to create a superposition state with lower energy.

So now that we have established that the Higgs mechanism does not, and cannot, correspond to spontaneous symmetry breaking, let's take a look at what's really happening. For simplicity we will look at the simplest case, namely (quantum, $T = 0$) $\mathbb{Z}_2$ lattice gauge theory. This comprises two-dimensional quantum systems on all the vertices and links of a square lattice. The ones on the vertices comprise the "matter field" and the ones on the links comprise the "gauge field". We denote the Pauli matrices on the links by $\sigma_{ab}^x$, etc. and on the vertices by $\tau_{a}^x$, etc. The Hamiltonian is $$H = -g \sum_{\langle a, b\rangle} \sigma^x_{ab} - \frac{1}{g} \sum_{\square} \sigma^z \sigma^z \sigma^z \sigma^z - \lambda \sum_{a} \tau^x_a - \frac{1}{\lambda} \sum_{\langle a, b \rangle} \tau_a^z \sigma^z_{ab} \tau_b^z$$ [the second-term is a sum of four-body $\sigma^z$ interactions on squares of the lattice ("plaquettes"), and $\langle a, b \rangle$ means a sum over nearest neighbor pairs of vertices.] This Hamiltonian has a gauge symmetry $\tau^x_a \prod_{\langle a, b \rangle} \sigma^x_b$ for each vertex $a$.

One can map out the phase diagram of this Hamiltonian in detail, but here we will just want to focus on the "Higgs" phase, which occurs when $g$ and $\lambda$ are small so that the second and fourth terms dominate. We will take the limit $g \to 0$, claiming without proof that the $g$ small but not zero case is qualitatively similar. In this limit the ground state must be a $+1$ eigenstate of the product of $\sigma^z$ around every plaquette ("no-flux" condition). If the model is defined on a space with no non-contractible loops, this implies that we can write, for every no-flux'' configuration, $\sigma^z_{ab} = \widetilde{\sigma}^z_a \widetilde{\sigma}^z_b$ for some choice of $\{ \widetilde{\sigma}^z_a \} = \pm 1$. Hence, all "no-flux" configurations can be made to satisfy $\sigma^z_{ab} = 1$ by an appropriate gauge transformation. Thus, under this gauge-fixing condition, the Hamiltonian reduces to the transverse-field quantum Ising model on the matter fields: $$H_{gf} = -\lambda \sum_{a} \tau^x_a - \frac{1}{\lambda} \sum_{\langle a,b \rangle} \tau_a^z \tau_b^z$$ which we know will have a symmetry-breaking phase (i.e. a two-fold degenerate ground state) for small $\lambda$. This is the Higgs phase.

Q: But hang on, now, doesn't Elitzur's theorem say that gauge symmetries can't be spontaneously broken?

A: Well, actually in fixing the gauge we used up the local part of the gauge symmetry, and the above Hamiltonian $H_{gf}$ only has a $\mathbb{Z}_2$ global symmetry. Thus, it does not violate Elitzur's theorem for it to have spontaneous symmetry breaking.

Q: But what about the original Hamiltonian, $H$? It had a gauge symmetry, and it's equivalent to the new Hamiltonian $H_{gf}$, which has spontaneous symmetry-breaking, so the original Hamiltonian must have spontaneous symmetry-breaking too?

A: You have to be very careful about the sense in which $H$ and $H_{gf}$ are equivalent, because the "gauge-fixing" transformation which relates them isn't unitary (since it's many-to-one). Still, if one thinks hard enough and uses the fact that $H$ is invariant under the gauge symmetry, it is not hard to show that there is a correspondence between eigenstates of $H$ and of $H_{gf}$. However, because the two degenerate ground states of $H_{gf}$ are related by a gauge transformation, they actually correspond only to a single unique ground state of $H$, in accordance with Elitzur's theorem. This unique ground state $|\Psi\rangle_H$ of $H$ can be found in terms of the ground states $|\Psi\rangle_{H_{gf}}$ of $H_{gf}$ by symmetrizing them to make them gauge-invariant, i.e. $$|\Psi\rangle_H = \sum_{\mathcal{G}} \mathcal{G} |\Psi\rangle_{H_{gf}},$$ where the sum is over all possible gauge transformations $\mathcal{G}$ (since the two degenerate ground states are related by a gauge transformation, this gives the same $|\Psi\rangle_H$ regardless of which one you choose to be $|\Psi\rangle_{H_{gf}}$.)

So in summary, the Higgs mechanism appears to resemble spontaneous symmetry breaking in a particular choice of gauge, but this is an illusion. The true ground state is unique and gauge-invariant.

• Perfect, thanks! This finally made things click for me. If I can add a comment: usually we say the quantum Ising chain has SSB since the symmetric state $|\uparrow \uparrow \cdots\rangle +|\downarrow \downarrow \cdots \rangle$ has long-range entanglement which would instantaneously decohere upon the smallest interaction of one our spins to a single fixed external spin. However in your case, such a coupling is excluded by gauge invariance and so the long-range entanglement of the cat state is in fact undetectable and so there is no SSB in your Ising chain! – Ruben Verresen Dec 30 '16 at 22:39
• Equivalently, for the same reason, one can in fact pretend that the Ising chain breaks the symmetry, the main point being that there is no physical (i.e. gauge-invariant observable) which could tell the difference between the 'SSB' state and the symmetric cat state. Theoretically it is of course more pleasing to work with the cat state, since that is the only consistent option with (unbreakable) gauge symmetry. – Ruben Verresen Dec 30 '16 at 22:42
• @SRS It is true for the 2D Ising model. I don’t know why you would think otherwise. – Dominic Else Jun 8 at 12:10
• I got it... Sorry – SRS Jun 8 at 12:17

In short:The spontaneously breaking of global U(1) symmetry, rather than local 'gauge symmetry', give rise to the non-zero vacuum expectation value of Higgs field. This non-zero VEV is the essential part of Higgs mechanism, which describe how Higgs field give mass to other particles, and its value is proportional to the generated mass.

To study Higgs mechanism, we can use a Lagrangian of the form: \begin{align} \mathcal{L}=(D_{\mu}\phi)^2-\frac{1}{4} F_{\mu\nu}F^{\mu\nu}-V(|\phi|) \end{align}

Where: \begin{align} V(|\phi|)&=-2v^2|\phi|^2+|\phi|^4 \\ &=(|\phi|^2-v^2)^2-v^4 \\ D_\mu \phi&=\partial_{\mu}\phi+\mathrm{i}e A_{\mu} \phi \\ F_{\mu\nu}&=\partial_{\mu} A_{\nu}-\partial_{\nu}A_{\mu} \end{align}

This Lagrangian is gauge invariant and have global $U(1)$ symmetry. Note that I am not speaking they have local $U(1)$ gauge symmetry nor local U(1) symmetry because:

We say the Lagrangian is gauge invariant in the sense that: \begin{align} \phi(x) \to \phi(x) e^{i\alpha(x)} \quad A_{\mu} \to A_{\mu}(x) -\frac{1}{e}\partial_{\mu} \alpha(x) \end{align} keep the Lagrangian unchanged.

The first step towards to the Higgs mechanism is a spontaneously symmetry breaking(SSB) of global U(1) symmetry. i.e. We are going to choose a particular value of $\phi$ in the "minimum circle" of potential $V(|\phi|)$. By doing this, the ground state lost the global $U(1)$ symmetry the Lagrangian has. With loss of generality, we assume it spontaneously broken into $\phi_0 = v$, a real value.

The next step, we expand our field around $\phi_0=v$, we assume $\phi=(v+h)e^{i\xi}$. Substitute it into the Lagrangian, we get:

\begin{align} \mathcal{L}=(\partial_{\mu}h)^2+e^2(v+h)^2(A_{\mu}+\frac{1}{e}\partial_{\mu}\xi)^2 -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} -4v^2h^2-4vh^3-h^4+v^4 \end{align}

We can see this Lagrangian is still gauge invariant, recall our definition of gauge transformation above, it means: $$\xi \to \xi +\alpha \quad A_{\mu}\to A_{\mu} -\frac{1}{e}\partial_{\mu}\alpha$$

So under gauge transformation, we have: \begin{align} A_{\mu}+\frac{1}{e}\partial_{\mu}\xi \to A_{\mu}+\frac{1}{e}\partial_{\mu}\xi \end{align}

Which means the Lagrangian is still gauge invariant.

Now we define $A'_{\mu}=A_{\mu}+\frac{1}{e}\partial_{\mu}\xi$. We emphasis that $A'_{\mu}$ should not be called a gauge field because itself is gauge invariant. In certain physical context, this vector field can be expressed in the gauge invariant physical quantities. We also have $F'_{\mu\nu}=F_{\mu\nu}$, where: \begin{align} F'_{\mu\nu}=\partial_{\mu} A'_{\nu}-\partial_{\nu}A'_{\mu} \end{align}

Now the Lagrangian becomes: \begin{align} \mathcal{L}=(\partial_{\mu}h)^2-4v^2h^2+e^2v^2(A'_{\mu})^2 -\frac{1}{4} F'_{\mu\nu}F'^{\mu\nu} +\cdots \end{align}

Where we've omit the constant terms and interaction terms, and we get a mass term for vector field $A'$ and the original massless goldstone boson $\xi$ has just disappeared. This is the saying that "the vector field has eaten goldstone bosons and grow heavy."

At this point, we have seen that we can explain Higgs mechanism in a gauge invariant way, this is no such thing as spontaneously local gauge symmetry breaking in the above analysis, as we have just shown it is gauge invariant.

On the other hand, we can also manually fix the gauge in the former Lagrangian. The usual choice is to let $\xi = 0$, so that we have $A'=A$. For we have fixed the gauge, it is no more valid to talk about gauge invariant. This may be the reason that people talking about "spontaneously local gauge symmetry breaking". However, the manually gauge fixing process should not be considered as any spontaneous process.

• Great post. I do not see why you call the global U(1) a symmetry though. It is as much gauge as the local part. – Ruben Verresen Dec 30 '16 at 22:14
• @RubenVerresen Because it is easy to check the symmetry operation $\phi\to\phi e^{i c}$ doesn't change the Lagrangian, where $c$ is independent of $x$. – 喵喵是我的猫猫 Mar 12 '17 at 8:34
• @buzhidao Can you please explain the expression above the statement "Which means the Lagrangian is still gauge invariant." Is it some typo? – SRS Aug 7 '17 at 16:21
• @SRS You can ignore that part, I just want to confirm the Lagrangian is gauge invariant. – 喵喵是我的猫猫 Aug 11 '17 at 13:43
• @buzhidao Since we are reinterpreting the original $U(1)$ gauge transformation as a shift in the goldstone (in order to argue that gauge symmetry is intact), and since the new Lagrangian is still invariant under a global shift, does that not mean that the global $U(1)$ phase rotation symmetry is as much intact as the local gauge symmetry? In other words, why do we reinterpret the $U(1)$gauge transformation as a shift of the goldstone, but do not reinterpret the global $U(1)$ symmetry as a global shift invariance? – Nanashi No Gombe Jan 10 at 14:35

The essence of the Higgs mechanism is that it allows the breaking of the (gauge) symmetry to grow a mass for the gauge (vector) bosons, which are necessarily massless in the unbroken symmetry. The Higgs scalar and the two degrees of freedom of the massless vector boson combine to form the three degrees of freedom of a massive vector boson. Goldstone's theorem (https://en.wikipedia.org/wiki/Goldstone_boson) states that if a continuous symmetry of a system is spontaneously broken, then the ground state of the system is degenerate. In a gauge theory, the degenerate ground states are actually reachable from one another by a gauge transformation, so they are simply gauge copies on one another, and correspond to a single physical state.

• not quite right, the degenerate ground state is related by physical symmetry, not the gauge. – 喵喵是我的猫猫 Aug 13 '15 at 14:39
• ""if a continuous symmetry of a system is spontaneously broken, then the ground state of the system is degenerate"" this is statement is even worse... – 喵喵是我的猫猫 Aug 13 '15 at 14:55
• I strongly suggest looking at old papers by Strocchi where the Higgs mechanism is described in a fully gauge invariant way, as it should, in terms of a phase transition where one looks at the gauge boson correlation length. In my opinion, there are lots of gauge and perturbative artifacts in the usual description possibly related to some automatism in the way experimental people try to trivialize QFT in classical terms. A good entry point is amazon.com/… . – Matteo Beccaria May 19 '17 at 20:42