Higgs Mechanism

see also Symmetry Breaking

Intuitive

If you'd like, you can think in terms of a 'caramel pool', when we say that a particle 'couples' to the Higgs field, we mean to say that this particle 'sees' this 'caramel pool', and this makes it "harder" for it to move, which we measure as this particle's "mass". And, as you can imagine, there are particles that do not couple to the Higgs field, meaning to say they do not "see" it: therefore, they move much more easily.

https://physics.stackexchange.com/a/6451/37286


Concrete

Spontaneous symmetry breaking means that the vacuum configuration of the Higgs field is no longer invariant under the complete group $G$, but only under some subgroup $H \subset G$. In mathematical terms, this means the vacuum expectation value (VEV) $\langle \Phi_i \rangle$ is not invariant under the action of some elements of the group $g \in G$

\begin{equation} g(\alpha) \langle \Phi_i \rangle \neq \langle \Phi_i \rangle \, , \end{equation}

where $\alpha$ is a parameter that denotes how much we want transform, for example, the angle we want to rotate. We can use the generators $T_a$ and the exponential map to write elements of the group as

\begin{equation} g(\alpha)= \mathrm{e}^{ \mathrm{i} \alpha_a T_a } \end{equation}

and therefore the condition above can be rewritten as \begin{equation} g(\alpha) \langle \Phi_i \rangle = \mathrm{e}^{\mathrm{i} \alpha_a T_a } \langle \Phi_i \rangle = (1+ \mathrm{i}\alpha_a T_a + \ldots) \langle \Phi_i \rangle \neq \langle \Phi_i \rangle \end{equation} \begin{equation} \rightarrow \mathrm{i}\alpha_a T_a \langle \Phi_i \rangle \neq 0 \, . \end{equation}

Some elements of the group leave the vacuum configuration still invariant while others do not. Therefore we get two sets of generators:

Those which annihilate the vacuum expectation value $T_a \langle \Phi_i \rangle =0$, because then $$g(\alpha) \langle \Phi_i \rangle = \mathrm{e}^{\mathrm{i} \alpha_a T_a } \langle \Phi_i \rangle = (1+ \mathrm{i}\alpha_a T_a + \ldots) \langle \Phi_i \rangle = \langle \Phi_i \rangle .$$ These generators are called unbroken generators.

Those which do not annihilate the vacuum expectation value $T_a \langle \Phi_i \rangle \neq 0$. These generators are called broken.

The unbroken generators generate the unbroken subgroup $H \subset G$. Therefore given a VEV we can immediately check which generators annihilate it and then compute which group these unbroken generators generate. Of course, there are several subtleties one must take into account and these are discussed in subsequent sections. It follows directly from this definition of symmetry breaking that the corresponding gauge bosons become massive and this helps to understand what it means that a generator annihilates the VEV. Gauge bosons get their mass from the kinetic term in the Higgs field Lagrangian

\begin{equation} \mathscr{L} = (D_\mu \Phi)^\dagger (D_\mu \Phi), \end{equation}

where $D_\mu$ denotes the covariant derivative $D_\mu = \partial_\mu - \mathrm{i} g A_\mu$ and $A_\mu$ is the gauge field tensor $A_\mu = A_\mu^a T^a$, with $T_a$ the generators of the group as above. If some Higgs field now gets a VEV, we get terms in the Lagrangian proportional to

\begin{equation} A_\mu A^\mu \langle \Phi_i \rangle \langle \Phi_i \rangle. \end{equation}

This is a mass term for the gauge bosons. Of course, not all gauge bosons get massive and thus we need to compute which remain massless. Gauge bosons are described by objects that transform according to the adjoint representation $A$ of the gauge group $G$. The scalar fields $\Phi$ transform according to some representation $R$. A term like

\begin{equation} D_\mu \Phi = (\partial_\mu - \mathrm{i} g A_\mu) \Phi \end{equation}

therefore requires that we know how to compute the product of two different group representations. In general, one has

\begin{equation} R_1 \otimes R_2 = R_3 \oplus R_4 \oplus R_5 \oplus \ldots \, . \end{equation}

For products involving the adjoint $A$ one always has

\begin{equation} A \otimes R = R \oplus R_2 \oplus R_3 \oplus \ldots \, , \end{equation}

i.e. the representation on the left-hand side appears again on the right-hand side. We need exactly this product of the adjoint representation $A$ with our Higgs representation $R$, that yields again $R$ and not something different. This can be seen by rewriting $(\partial_\mu - \mathrm{i} g A_\mu) \Phi$ using the corresponding representation names

\begin{equation} (\partial_\mu - \mathrm{i} g A_\mu) \Phi = (\partial_\mu 1 - \mathrm{i} g A ) R = \partial_\mu \underbrace{1 R}_{=R} - \mathrm{i} g \underbrace{A R}_{\stackrel{!}{=} R}, \end{equation} where $1$ denotes the one-dimensional representation of the gauge group. Here we can see that $A R\stackrel{!}{=} R $, because otherwise we would add apples to oranges. Therefore if we want to know which gauge fields become massive we need to compute the product $A\otimes R =R$. Then we can check which generators (= elements of $A$) can be combined with the element of the scalar representation $R$ that gets a VEV and yield again an element of $R$. These generators become massive and all others remain massless. This means, we need to check which generators can act on our VEV $\langle \Phi_i \rangle$ and yield again an element of the scalar representation. If we act with some generator on $\langle \Phi_i \rangle$ and get something that is not an element of the scalar representation, then the corresponding generator annihilates the VEV. (This is analogous to the angular momentum ladder operators acting on the highest state of a representation. The raising operator annihilates the state, because going any higher would be no longer a state inside the original representation. In contrast, the lowering operator simply yields another state of the representation when one acts on the highest state and therefore does not annihilate the state.) This means the generator is unbroken.

Higgs mechanism in Superconductivity

A physical example in which the Higgs mechanism actually takes place is superconductivity. The Lagrangian of the theory is invariant under local phase changes of the electron field, but the ground state is not, owing to a condensation of Cooper pairs made up of two electrons. As a consequence, the photon becomes massive inside a superconducting body. In particular, an externally applied magnetic field can penetrate the body only to a finite depth equal to the inverse mass (Meissner effect). A phenomenological way to describe the condensation phenomenon is to introduce an "order parameter" $\phi(x)$ to describe the condensate, as done in the Landau-Ginzburg theory.

page 55 in Quarks, Leptons & Gauge Fields BY K. Huang


Reading Recommendations

Another nice, non-technical introduction is the series by Flip Tanedo:

Then, to get a deeper understanding it is really helpful to see how symmetry breaking happens in a simple model like the Ising model. For a nice introduction see the chapter "Global Symmetry, Local Symmetry, and the Lattice" in "An Introduction to the Confinement Problem" by Jeff Greensite


“We conclude, then, that the Goldstone zero-mass difficulty is not a serious one, because we can probably cancel it off against an equal Yang-Mills zero-mass problem” (1963, p. 442). Anderson

The Higgs mechanism described in §8.4 is sometimes expressed in an alternative way: Goldstone bosons can be made to disappear in the presence of long-range forces (Anderson, 1963; Guralnik, Hagen, and Kibble, 1968). The connection between the two descriptions comes through the Yukawa-Wick interpretation that long-range forces like the Coulomb interaction are mediated by massless exchange particles (gauge fields are massless). In this picture, the long-range force is shielded and becomes short-ranged, which is equivalent to the generation of an effective mass for the gauge boson. One nonrelativistic example of the Higgs mechanism for which this shielding description is illuminating is the Meissner effect in superconductivity. Aitchison and Hey (1982, Ch. 9) give a nice heuristic discussion of how the condensate of electron pairs in the ground state of a superconductor plays the role of a Higgs field, and how this gives photons an effective mass inside the superconductor. As a direct consequence, a magnetic field can penetrate only exponentially into the superconductor, with a range proportional to the in- verse of the effective photon mass. Thus a superconductor expels a magnetic field from its interior (the Meissner effect-see Fig. 12.3), except for a thin layer at the surface (the London penetration depth) over which the field decreases exponentially. The microscopic origin of the Higgs phenomenon in this case lies in resistanceless screening currents that are produced in the superconductor to compensate the external field. The finite range of the magnetic field that results is just what would be expected if the photon had acquired a mass inside the superconductor. Furthermore, it can be shown that the superconductor with a “massive” photon is gauge invariant, despite our remarks in connection with eq. (2.149). That is because the photon mass has come by the Higgs mechanism-not by an explicit mass term in the Lagrangian. This is our first example of a real physical system in which gauge bosons acquire an effective mass without breaking gauge invariance.page 261 in Gauge Field Theories: An Introduction with Applications by Mike Guidry

The Higgs mechanism is a type of superconductivity which occurs in the vacuum. It occurs when all of space is filled with a sea of particles which are charged, or, in field language, when a charged field has a nonzero vacuum expectation value. Interaction with the quantum fluid filling the space prevents certain forces from propagating over long distances (as it does in a superconducting medium; e.g., in the Ginzburg–Landau theory). http://vixra.org/pdf/1607.0448v1.pdf

Abstract

Why is it interesting?

The Higgs mechanism is a crucial ingredient of the standard model of particle physics. Without the Higgs mechanism, all particles are not allowed to have a mass, because such terms would violate the gauge symmetry. (Breaking of gauge symmetry is a bad thing, because the renormalizability, i.e. the removing of the infinities that pop-up in most quantum field theory calculations, depends on the existence of gauge symmetry.)

However, we know from experiments that some elementary particles, like the electron or also the W-bosons that mediate weak interactions, are massive. These masses can be explained thanks to the Higgs mechanism. The masses then arise as a result of the coupling of the massive particles to the Higgs field and this is possible without breaking gauge symmetry.

The Higgs mechanism is no longer just a theory but was confirmed by the discovery of the Higgs particle through the LHC experiment.

Research

One sure priority is the research program in Higgs physics. The discovery of the Higgs boson [7] opened the door to the understanding of the mechanism of electroweak breaking. This mechanism plays a crucial role in nature, giving rise to a fundamental scale that rules not only the microscopic world but many physical properties of our universe, from the size of atoms to the timescale of the processes that make the sun shine. Missing the opportunity to study in depth the mechanism of electroweak breaking would be like giving up the exploration of a new continent in the planet of knowledge. Although a fundamental milestone, the discovery of the Higgs cannot be regarded as the final resolution to the enigma of electroweak breaking. If anything, its discovery has rendered unavoidable the need to address some of the open problems in particle physics. Indeed, every single Higgs interaction introduces its own puzzle. The Yukawa couplings express the flavour problem, i.e. our inability to compute quark and lepton masses and mixing angles. The Higgs quartic coupling raises issues with the stability of the SM electroweak vacuum. The Higgs quadratic term incarnates the naturalness problem. The constant term in the Higgs potential is an expression of the cosmological constant problem. The root of these problems is that the Higgs boson introduces 14 new forces (without taking into account other forces associated with neutrinos) besides the 3 known fundamental forces of the Standard Model. Unlike strong, weak, and electromagnetic forces, these new 14 forces are not gauge-like. Not surprisingly, they do not share the properties of elegance, simplicity, robustness, and predictivity that characterise the gauge forces of the Standard Model. Accustomed with the conceptual perfection of the gauge sector of the Standard Model, it is difficult for a theorist to believe that a structure so arbitrary and provisional as the Higgs sector could be the final word on electroweak breaking. There must be more still to be discovered. The way to get to the bottom of the question is a program of precision measurements on the Higgs boson properties. This is already underway, but future high-energy colliders will be able to bring these precision studies to unprecedented levels of precision. Much can be learned from these studies about the Higgs boson, the phenomenon of electroweak breaking, the fundamental laws that govern nature, the early stages of the Universe and its ultimate fate.

https://arxiv.org/pdf/1710.07663.pdf#page23

But is the gauge symmetry actually broken spontaneously? In the above exposition of the Higgs mechanism, there were two instances when a symmetry was broken. First, when we selected one minimum out of the infinite number of equivalent minima, a spontaneous breaking indeed took place, but only of a global symmetry. This minimum represents a vacuum, and in order to perturbatively describe the quantum field theory, we need to quantize the fields. Quantization of gauge field theories requires the introduction of a gauge-fixing procedure, and during this procedure, we break the gauge symmetry by hand, explicitly, not spontaneously. Thus, the two notions, EWSB and SSB, are in certain sense correct, but they do not refer to the same symmetry. […]

As Englert says in his Nobel lecture [54]: “… The vacuum is no more degenerate and strictly speaking there is no spontaneous symmetry breaking of a local symmetry. The reason why the phase with nonvanishing scalar expectation value is often labeled SSB is that one uses perturbation theory to select at zero coupling with the gauge fields a scalar field configuration from global SSB; but this preferred choice is only a convenient one.”.

https://arxiv.org/pdf/1702.03776.pdf

It is well known that gauge symmetries do not actually get broken, but rather become concealed in what is called—in an abuse of language—the “broken phase” of a theory [28–31].

https://arxiv.org/pdf/1703.02964.pdf

Symmetry breaking in QFT results from a mismatch between variational symmetries of the Lagrangian and symmetries that can be implemented as unitary transformations on the Hilbert space of states. (The inapt adjective ‘spontaneous’ differentiates symmetry breaking that arises due to the noninvariance of the vacuum state from that due to explicitly adding asymmetric terms to the Lagrangian.) […] Fabri and Picasso (1966) showed that if the vacuum state $|0\rangle$ is translationally invariant, then the vacuum is either invariant under the internal symmetry, $Q |0\rangle = 0$ , or there is no state corresponding to $Q |0\rangle$ in the Hilbert space. The second case corresponds to SSB. The symmetry is hidden in that there is no unitary operator to map a physical state to its symmetric counterparts; instead, the symmetry is (roughly speaking) a map from one Hilbert space of states to an entirely distinct space. This is usually described as ‘vacuum degeneracy’, although each distinct Hilbert space has a unique vacuum state. […] Parenti, Strocchi, and Velo (1977) study the features of SSB in classical, nonlinear field theories; in these theories, solutions to the equations of motion fall into distinct “sectors,” corresponding to global field configurations that cannot be transformed into each other via local perturbations. The variational symmetries of the Lagrangian then fall into the unbroken symmetries, for which $Q_V$ converges in the limit, and broken symmetries, for which $Q_V$ fails to converge. The broken symmetries map between the “physically disjoint worlds” represented by the distinct global field configurations. Similarly, in QFT, the degenerate vacua correspond to distinct global field configurations with minimum energy, with Hilbert spaces built up from a particular vacuum state.

http://publish.uwo.ca/~csmeenk2/files/HiggsMechanism.pdf


See also:

FAQ

Can the Higgs interactions be considered a fifth fundamental force?
Yes! See https://physics.stackexchange.com/questions/1080/why-isnt-higgs-coupling-considered-a-fifth-fundamental-force and also THE FIFTH FORCE by James D. Bjorken
Is there an "inverse" Higgs mechanism?
Yes, this is known as symmetry restoration.

Associated with spontaneous symmetry breaking is the phenomenon of symmetry restoration. If one heats a system that possesses a broken symmetry it tends to be restored at high temperature. Thus a ferromagnetic material can be magnetized at low temperature (or even at room temperature) with all the little atomic magnets aligned in the same direction. This is a state of broken rotational symmetry. As the temperature increases the atoms vibrate more and more. Finally when the temperature is greater than a certain critical value the fluctuations win out over the forces that tend to align the atomic magnets and the average magnetization vanishes. Above the critical temperature the system exhibits rotational symmetry. Such a transition from a state of broken symmetry to one where the symmetry is restored is a phase transition. We believe that the same phenomenon occurs in the case of the symmetries of the fundamental forces of nature. Many of these are broken at low temperatures. Very early in the history of the universe, when the temperature was very high, all of these symmetries of nature were presumably restored. The resulting phase transitions, as the universe expanded and cooled, from symmetric states to those of broken symmetry have important cosmological implications.

The role of symmetry in fundamental physics by David J. Gross

See also: http://physics.stackexchange.com/a/241615/37286

Do all particles become massless at energies above the symmetry breaking scale?
No! High-energy is not the same as high temperature. See:

http://physics.stackexchange.com/questions/119921/are-there-massless-bosons-at-scales-above-electroweak-scale

Does the Higgs mechanism explain the origin of mass?

“In summary, the Higgs mechanism accounts for about 1 per cent of the mass of ordinary matter, and for only 0.2 per cent of the mass of the universe. This is not nearly enough to justify the claim of explaining the origin of mass”.

A Zeptospace Odyssey by Gian Francesco Giudice

Is the Higgs Theory of the Standard Model Trivial?
Yes!

The Higgs sector of the Standard Model of particle physics represents a 4d Yukawa model. A careful analysis of this sector reveals that it defines a trivial theory. Triviality refers to the behavior of the renormalized quartic coupling constant λr of the scalar field in dependence on the cutoff parameter Λ. The cutoff has to be introduced to regularize the theory. In a renormalizable theory the cutoff parameter can be sent to infinity while holding all physical observables constant, making the physical predictions arising from such a theory eventually independent of the previously introduced auxiliary parameter Λ, as desired. In a trivial theory, however, all renormalized coupling constants vanish as function of the cutoff parameter in the limit Λ → ∞, leading to a free, non-interacting theory when we try to remove the cutoff. Thus the Higgs sector of the Standard Model can only be considered as an effective theory connected with a non-removable cutoff parameter, which can be interpreted as the maximal scale up to which the underlying effective theory can be trusted. As a consequence the renormalized quartic coupling constant at a given cutoff Λ is bounded from above according to λr(Λ) ≤ λup,r(Λ). This bound translates into an upper bound mup,H (Λ) on the Higgs boson mass and this bound decreases with increasing cutoff [43]. This remarkable fact tells us that once the Higgs boson has been discovered and its physical mass mH measured, it would be possible to infer the scale up to which the Standard Model can be valid at most from the comparison of mH with its cutoff-dependent upper bound.

… More recently P. GERHOLD AND K. JANSEN employed the overlap construction for the free Dirac operator D to simulate the chirally invariant Higgs–Yukawa model. They obtained reliable upper (and lower) bounds on the Higgs boson mass as a function of the cutoff parameter [44, 45]. For example, they conclude that if the cutoff is at Λ = 1.5 TeV then the mass must be in the range between 50 GeV and 650 GeV. On the other hand, for a Higgs boson mass of about 125 GeV the Standard Model can be valid up to very large cutoff scales. Statistical Approach to Quantum Field Theoryby Andreas Wipf

See also: http://alumnus.caltech.edu/~callaway/trivpurs.pdf

Spontaneous Breaking of a Local Gauge Symmetry is Impossible... What does the Higgs Mechanism really do?
It is well known that breaking of a local gauge symmetry is impossible (see the quotes below.) However, there is symmetry breaking when the Higgs field acquires a non-zero vev. It is a global part of the gauge group that gets broken. However, there are no Goldstone bosons, because these correspond to the gauge degrees of freedom and become the longitudinal polarizations of the gauge bosons that become massive.

This is discussed nicely in "Quantum Field Theory - A Modern Perspective" by V. P. Nair at page 268:

We have already seen that the true gauge group of the theory is the set of all gauge transformations which go to the identity at spatial infinity, namely, G∗ in the notation of Chapter 10. It should be emphasized that there is no breakdown of this true gauge symmetry here. In fact gauge symmetry is crucial in removing the massless modes. What is broken is the global part of the symmetry, corresponding to G/G∗. To show how this works out in some detail, we will…

The definitions of G and G* can be found at page 188.

This breaking of a global symmetry is also mentioned at page 533 in Duncan's "Conceptual Framework of Quantum Field Theory":

Once a gauge is fixed, however, to remove the redundant degrees of freedom, the remaining (discrete!) global symmetry may undergo spontaneous symmetry-breaking exactly along the lines discussed in the previous chapter. The phrase "spontaneous breaking of local gauge symmetry" is therefore in some sense a misnomer, but a convenient one, if we think of it as a short circumlocution for "spontaneous breaking of remnant global symmetry after removal of redundant gauge degrees of freedom by appropriate gauge fixing".

This is also nicely summmarized in Is electromagnetic gauge invariance spontaneously violated in superconductors? by Martin Greiter:

In fact, a gauge symmetry cannot spontaneously break down as a matter of principle, since it is not a physical symmetry of the system to begin with, but merely an invariance of description [8]. The only way to violate a gauge symmetry is by choosing a gauge, which again has only an effect on our description, but not on the physical system it- self.

We may conclude at this point that in a superfluid or superconductor, a symmetry is spontaneously violated, but this symmetry is not gauge invariance, but global U (1) phase rotation symmetry. This is already evident from the fact that the discussion above made no reference to whether the order parameter field W^yðxÞ is charged or not, and equally well applies to neutral superfluids, where W^yðxÞ carries no charge.

There is, however, a very important difference between these two cases. If the or- der parameter field is neutral, the excitation spectrum of the system contains a gap- less (or in the language of particle physics "massless") mode, a so-called Goldstone boson [1], which physically corresponds to very slow spatial variations in the direction (as for the case of broken rotational invariance) or phase (as for the case of a superfluid) of the classical order parameter field. If the order parameter field is charged, however, it couples to the electromagnetic gauge field, and the Goldstone boson is absent due to the Higgs mechanism. The physical principle underlying this mechanism was discovered by Anderson [13] in the context of superconductivity: as the electromagnetic interaction is long-ranged, the mode corresponding to very slow spatial variations in the phase / of the superconducting order parameter, which implies currents by the equation of motion and hence also variations in the density of the superfluid by the continuity equation, acquires a gap (or "mass") given by the plasma frequency.

[…]

In the literature, one sometimes finds the statement that "the gauge field acquires a mass" due to the Higgs mechanism. This is not exactly to the point, as it suggests that the massive vector field $A_\mu'$ is still a gauge field, while we have just seen that it is gauge invariant. In the case of a superconductor, we even know how to interpret the individual components of $A_\mu'$ physically. According to (\ref{e:newfields}), (\ref{e:mu}), and (\ref{e:vs}), \begin{equation} \label{eq:a'} -\frac{e^*}{c}\left(A_\mu'\right) =\bigl(-\frac{\mu}{c},m^* {v}_{\text s}\bigr). \end{equation} The Higgs mechanism hence does not imply that "the electromagnetic gauge field acquires a mass", but only that we can describe the superconductor in terms of gauge invariant fields, that is, in terms of the chemical potential $\mu(x)$ and the superfluid velocity ${v}_{\text s}(x)$. If we do this, we also have to express the Maxwell Lagrange density (\ref{e:lmax}) in terms of $\mu$ and ${v}_{\text s}$. With \begin{equation} -\frac{1}{2}F_{\mu\nu}F^{\mu\nu}={E}^2-{B}^2 \end{equation} we obtain for the total Lagrange density \begin{eqnarray} \mathcal{L}%_{\text{tot}} \!&\!=\!&\! \frac{1}{8\pi {e^*}^2} \Bigl\{(\nabla\mu+m^*\partial_t{v}_{\text s})^2- c^2{m^*}^2 (\nabla\times{v}_{\text s})^2\Bigr\} \nonumber \\ \nonumber &&\hspace{-7mm} -n_{\text{s}}\biggl\{ \mu+\Bigl(-\mu+\frac{1}{2}m^* {v}_{\text s}^2\Bigr) +\frac{1}{2m^*}\frac{1}{v^2} \Bigl(-\mu+\frac{1}{2}m^* {v}_{\text s}^2\Bigr)^2\biggr\}.\\ \label{e:l2} \end{eqnarray} The Euler–Lagrange equations we obtain from (\ref{e:l2}) for $\mu$ and ${v}_{\text s}$ are equivalent to (\ref{e:max1})–(\ref{e:j4}), and yield exactly the same solution as above. % Writing the Lagrangian in terms of $\mu$ and ${v}_{\text s}$ does not yield any practical advantage, but clearly illustrates that gauge invariance has become irrelevant—it is not broken, but has simply left the stage. Since all the fields are gauge invariant, (\ref{e:l2}) does not even provide a framework to think of a spontaneous violation of a gauge invariance.

These considerations apply to every field theory which displays the Higgs mechanism. In any such theory, the Lagrange density is invariant under a global physical symmetry for a matter field, and invariant under a local gauge symmetry, which affects both the matter field and the gauge field. The global symmetry is "physical" as we can classify the states of matter according to their transformation properties, while the gauge symmetry is "unphysical" as gauge transformations have no effect on the states of matter, but only on our description of these states. In our example of a superfluid, charged or neutral, the global symmetry transformation is \begin{equation} \label{e:global} \phi(x)\rightarrow\phi(x)+\lambda, \end{equation} where $\lambda$ is independent of spacetime. This symmetry is spontaneously violated, which means that there are many degenerate ground states which map into each other under (\ref{e:global}). For a neutral superfluid, we obtain a massless mode according to Goldstone's theorem. The situation is more subtle for a superconductor, as the matter field is coupled to a gauge field and the Lagrange density is also invariant under the gauge transformation (\ref{e:gauge}). This "unphysical" symmetry, however, seems to contain the physical symmetry as the special case \begin{equation} \label{e:globalgauge} \Lambda(x)=-\frac{\hbar c}{e^*}\lambda. \end{equation} {The formal equivalence of the transformation (\ref{e:global}) and (\ref{e:gauge}) with (\ref{e:globalgauge}) is at the root of the widely established but incorrect interpretation of (\ref{e:global}) as a gauge transformation, and in particular of the spontaneous violation of (\ref{e:global}) as a spontaneous violation of a gauge symmetry.} (This is presumably the reason why particle physicists like Steven Weinberg speak of "spontaneously broken gauge symmetries" interchangeably with "the Higgs mechanism".) The problem here is that the equivalence is only formal. The gauge transformation (\ref{e:globalgauge}) represents a transformation of our description, similar to a rotation of a coordinate system we use to describe a physical state, while the transformation (\ref{e:global}) corresponds to a transformation of our physical state, like a rotation of a physical system. Clearly, a (counterclockwise) rotation of the coordinate system has the same effect on our equations as a (clockwise) rotation of the physical system we describe with these equations, but the transformations are all but equivalent. It is hence incorrect to refer to the spontaneous violation of (\ref{e:global}) as a spontaneous violation of gauge symmetry. A gauge symmetry cannot be spontaneously violated as a matter of principle.

The difference between the "physical" symmetry (\ref{e:global}) and the gauge symmetry (\ref{e:gauge}) can also be appreciated at the level of conservation laws. The former yields particle number (or charge) as a conserved quantity, according to (\ref{e:cont}), while there is no conservation law associated with the latter. In the literature, (\ref{e:global}) is often referred to as a global gauge transformation, and the conservation of charge attributed to gauge invariance. This view, however, is not consistent. If one speaks of a global gauge symmetry, this symmetry has to be a proper subgroup of the local gauge symmetry group. The alleged global gauge symmetry hence cannot be a "physical" symmetry while the local gauge symmetry is an invariance of description, or be spontaneously violated while the local symmetry is fully intact. The difference between the global phase rotation (\ref{e:global}) and a global gauge rotation (\ref{e:globalgauge}) is even more at evident at the level of quantum states. The BCS ground state (\ref{e:bcs}) is, for example, not invariant under (\ref{e:global}), while it is fully gauge invariant, as we have seen in Section \ref{sec:gau}

The conclusions regarding the physical significance (or maybe better insignificance) of gauge transformations we reached here for superconductors hold for any field theory which displays the Higgs mechanism.

Is electromagnetic gauge invariance spontaneously violated in superconductors? by Martin Greiter

The fact that spontaneous breaking of a local symmetry is impossible is the message of Elitzur's Theorem. For a gauge symmetry, we have a copy of the symmetry group $G$ at each spacetime point. Thus symmetry breaking would need to happen at each spacetime point individual, i.e. in each zero-dimensional subsystem. A spacetime point is zero-dimensional and there is no symmetry breaking in systems of dimension lower than 2. This is known as the Mermin-Wagner theorem.

However, strictly speaking a local gauge symmetry can never be broken because we would want and observable to serve as an order parameter, but all such are necessarily gauge-invariant; this is the content of Elitzur's Theorem.

Classification of topological defects and their relevance to cosmology and elsewhere by T.W.B. Kibble

See also https://youtu.be/XM4rsPnlZyg?t=18m38s

But there are several reasons not to accept this view. First of all terminology. When we say gauge symmetry, this is really a misnomer. It's a misnomer because in physics gauge symmetry is not a symmetry. It is not a symmetry of anything. Symmetry is a set of transformations that act on physical observables. They act on the Hilbert space. The Hilbert space is always gauge invariant. So the gauge symmetry doesn't even act on the Hilbert space. So it's not a symmetry of anything. […] Second, gauge symmetry can be made to look trivial. So, I'll give one trivial example and then I'll make it more elaborate… [explains the Stückelberg mechanism, where one introduces a Stückelberg field to make a non U(1) gauge invariant Lagrangian, gauge invariant] This is almost like a fake… This gauge symmetry is what we would call emergent, except that in this case it is completely trivial. The second thing which is wrong about gauge symmetry, which suggests that it's not fundamental is that, it started in condensed matter physics, people talked about spontaneous symmetry breaking. That was crucial in the context of superconductivity and superfluidity and so forth. And the recent Nobel price in physics was also associated with spontaneous gauge symmetry breaking. That of Higgs, and Englert. This is all very nice and physicists love to talk about spontaneous symmetry breaking, but this is a bit too naive. First of all I've already emphasized that a gauge symmetry is not a symmetry. And since it is not a symmetry, how could it possibly be broken. You can break a symmetry that exists, but you cannot break a symmetry that does not exist. Second, the phenomenon of spontaneous symmetry breaking is often associated with the fact that the system goes to infinity. Concretely in quantum mechanics, you never have symmetry breaking. It is only in quantum field theory or statistical mechanics, where we have volume going to infinity we have an infinite number of degrees of freedom and there we have this phenomenon of spontaneous symmetry breaking. That's not true for gauge theories. For gauge theories, we have a lot of symmetry. At every point of space we have a separate symmetry. But the number of degrees of freedom that transform under a given symmetry transformation is always finite. Nothing goes off to infinity. So the gauge symmetry cannot be spontaneously broken. The ground state is always unique. Or if you wish, all these would-be separate ground states are all related to each other by a gauge transformation. […] I said that gauge symmetry cannot be ultimate symmetry because it's so big, there is a separate transformation at every point in space. So the breaking of a gauge theory cannot happen, I can use a phrase from the financial crisis in 2008 that a gauge symmetry is so big, it's too big to fail.

Duality and emergent gauge symmetry - Nathan Seiberg

Happily, the Higgs mechanism has nothing to do with spontaneous symmetry breaking, and therefore there is no real problem here. (See the answer to the next question below).

Fröhlich, Morchio and Strocchi, and, independently, Banks and Rabinovici and 't Hooft proposed that all the physics of these models be described in terms of gauge independent fields. Fröhlich, Morchio and Strocchi proved that a complete set of such gauge invariant fields exists and outlined how to express interesting quantities such as the mass of the vector meson in terms of them. Thus, it is not necessary to use gauge dependent local order parameters in order to describe the physical predictions for a state which is a Higgs phase.

Classification of topological defects and their relevance to cosmology and elsewhere by Fundamental Problems of Gauge Field Theory herausgegeben by A.S. Wightman

In the various models presented in previous chapters, finitely many local order parameters discriminate among the possible phases of a statistical system. For instance, the existence of a nonvanishing magnetization signals a spontaneous breaking of global symmetry in a spin model. One might imagine a similar mechanism in gauge theories, and study for instance the mean value $\langle U_{ij} \rangle$ of the field. This simple idea does not work. Elitzur has proved very generally that any local quantity (as $\langle U_{ij} \rangle$) has a vanishing mean value at any temperature, and therefore cannot be used as an order parameter. In other words, local observables cannot exhibit spontaneous breaking of local gauge symmetry.

section 6.1.3. in Statistical Field Theory: Volume 1 by Claude Itzykson,Jean-Michel Drouffe

We can understand the Brout–Englert–Higgs mechanism in the light of our general discussion of gauge symmetry in Chap. 4 (see Sect. 4.7). Remember that there we had considered the set G of all gauge transformations g(x) ∈ G approaching the identity at infinity and the subset G 0 ⊂ G formed by those contractible to the identity. These latter are the ones generated by Gauss’ law, [(D · E) A − ρ A ]|phys = 0 where ρ a represents the matter contribution. The set of all gauge transformations also contains elements g(x) approaching any other element of G as |x| → ∞. This differs from G by a copy of the gauge group G at infinity. This is identified as the group of global transformations associated with the existence of conserved charges via Noether’s theorem. When the gauge symmetry is spontaneously broken, the invariance of the theory under G is nevertheless preserved, while the invariance under global transformations (i.e. the copy of G at infinity) is broken. Notice that this in no way poses a threat to the consistency of the theory since properties like the decoupling of unphysical states are guaranteed by the fact that Gauss’ law is satisfied quantum mechanically. This follows from the invariance under G 0 .

page 143 in An Invitation to Quantum Field Theory by Alvarez-Gaume and Vazguez-Mozo

In gauge theories it has long been known that their phases cannot characterized by a local order parameter, since local symmetries cannot be spontaneously broken. The phases of gauge theories are understood instead in terms of the behavior of generally non-local operators such as Wilson loops and disorder operators [30,31], a concept borrowed from the theory of the two-dimensional Ising magnet [32]. http://www.fysik.su.se/~ardonne/files/papers/AnnPhys_310_493_2004.pdf

Does the Higgs mechanism break a gauge symmetry spontaneously?
No! The gauge symmetry only gets hidden by the Higgs mechanism and is still present. This is shown nicely and explicitly in "An introduction to gauge theories and modern particle physics" by Leader and Predazzi, in section 3.2 at page 45. Other authors that emphasize this point that the gauge symmetry does not get broken, but only hidden are Chris Quigg in his textbook "Gauge Theories Of Strong, Weak, And Electromagnetic Interactions" and Coleman in his lecture notes.

Moreover, the Higgs mechanism can be formulated in a completely gauge invariant way (i.e. all quantities that appear in the theory are not influenced by a gauge transformation). This means immediately that the "symmetry breaking" part is not an essential ingredient in the mechanism.

This is shown nicely in this article. See also, the references mentioned there.

We started from a Lagrangian which was gauge invariant, but which contained a parameter that looked like a negative squared mass. We went to a new form of $L$ which was still gauge invariant, but under a more complicated type of transformation law and which had an unphysical degree of freedom. Finally, we chose a particular gauge (often called the "U gauge") where the unphysical field has been "gauged away" and we then had a form for $L$ which was no longer invariant under any gauge transformation (we had fixed the gauge) but which had a sensible looking physical spectrum of massive particles, one vector, one scalar.

chapter 3 of An Introduction to Gauge Theories and Modern Particle Physics Vol 1 by Elliot Leader,Enrico Predazzi

(This reference also shows explicitly the more complicated form of the gauge transformations after SSB)

The symmetry may be hidden. (This case, first investigated by Nambu and Goldstone, is usually called 'spontaneously broken symmetry', but this terminology is deceptive; the symmetry is not really broken - the currents associated with the symmetry are still conserved and the Ward identities are still valid - it is just that the consequences of the symmetry are less obvious than in the other case.) The ground state of the theory (the vacuum) is not invariant and particles do not arrange themselves into multiplets.

Aspects of Symmetry by Sidney Coleman

What about the analogy between Superconductivity and the Higgs Mechanism?
See the nice discussion in http://philsci-archive.pitt.edu/12449/1/Fraser&Koberinski_HiggsAnalogiesSHPMP.pdf and Is electromagnetic gauge invariance spontaneously violated in superconductors? by Martin Greiter
Criticism of the Higgs Mechanism?

The $W$ and $Z$ are massive, yet local gauge invariance prohibits a conventional vector meson mass term in the Langrangian - as mentioned earlier, $M^2 A_\mu^aA^{\mu a}$ is not gauge invariant. In order to circumvent this problem, model builders [17] have invoked the mechanism of spontaneous gauge symmetry breaking. One would like to see this breaking occur for dynamical reasons in a theory involving just gauge potentials and massless spinor fields, so that only one parameter- the gauge coupling constant - characterized the theory. Unfortunately, no realistic and convincing model has been constructed in which this attractive speculation can be plausible supported [18].

Failing at this, one arranges for the spontaneous breaking by introducing scalar fields whose gauge-invariant self-couplings are so chose that the lowest energy configuration is indeed non-symmetric; i.e., one uses the Goldstone-Higgs mechanism [17]. While there are many unattractive aspects of this procedure - the first being its ad hoc nature- it is phenomenologically successful [19], and this where the subject stands today.

Topological Investigations of Quantized Gauge Theories, by R. Jackiw (1983)

Is it completely general to use the unitary gauge?

Earman (2002) critiques the role that the choice of the unitary gauge plays in this standard presentation of the Higgs mechanism. The problem is that a gauge symmetry (according to philosophical accounts) is supposed to represent “descriptive fluff,” a mere redundancy in the mathematical formalism of QFT. All physically relevant parameters are thus required to be gauge invariant. In particular, whether the dynamical equations describe a massive Higgs boson or a massless Goldstone boson should not depend on the choice of gauge. Earman thus rejects the “just-so stories” that particles gain mass “by eating the Higgs field” (2002, p. 1239). In order to determine the correct physical interpretation of the Higgs mechanism, he therefore demands a gauge invariant presentation. Morrison (2003), Smeenk (2006), Struyve (2011), and Lyre (2008) (among others) offer various approaches to this problem. The issues taken up in this paper are for the most part orthogonal to these issues pertaining to the choice of gauge. One reason for this is historical circumstances. Struyve (2011) points out that the early treatments of the Abelian Higgs model in Higgs (1966) and of the non-Abelian Higgs model in Kibble (1967) were in fact gauge invariant. Struvye reviews Higgs’ 1966 gauge invariant presentation of the Abelian Higgs model. According to Struyve, the presentation of the Abelian Higgs model rehearsed above with the appeal to the unitary gauge was first employed by Weinberg as a simpler way of obtaining Kibble’s results, and was then picked up in the textbook literature (p. 226). The historical period in which analogies to the BCS model influenced the formulation of the EW component of the Standard Model was c.1960-6, when the gauge invariant presentations of the Higgs mechanism were current. http://philsci-archive.pitt.edu/12449/1/Fraser&Koberinski_HiggsAnalogiesSHPMP.pdf</blockquote> ←- –> Is there also spontaneous symmetry breaking in quantum mechanics?# No, only in QFT: <blockquote> This mechanism, urged upon us by Heisenberg, Anderson, Nambu and Goldstone, is readily illustrated by the potential energy profile possessing left-right symmetry and depicted in the Figure. The left-right symmetric value at the origin is a point of unstable equilibrium; stable equilibrium is attained at the reflection unsymmetric points ±a. Once the system settles in one location or the other, left-right parity is absent. One says that the symmetry of the equations of motion is 'spontaneously' broken by the stable solution. But here we come to thefirstinstance where infinities play a crucial role. The above discussion of the asymmetric solution is appropriate to a classical physics description, where a physical state minimizes energy and is uniquely realized by one configuration or the other at ±a. However, quantum mechanically a physical state can comprise a superposition of classical states, where the necessity of superposing arises from quantum mechanical tunneling, which allows mixing between classical configurations. Therefore if the profile in the Figure describes the potential energy of a single quantum particle as a function of particle position, the barrier between the two minima carries finite energy. The particle can then tunnel between the two configurations of ±a, and the lowest quantum state is a superposition, which in the end respects the left-right symmetry. Spontaneous symmetry breaking does not occur in quantum particle mechanics. However, in a field theory, the graph in the Figure describes spatial energy density as a function of the field, and the total energy barrier is the finite amount seen in the Figure, multiplied by the infinite spatial volume in which the field theory is defined. Therefore the total energy barrier is infinite, and tunneling is impossible. Thus spontaneous symmetry breaking can occur in quantum field theory, and Weinberg as well as Salam employed this mechanism for breaking unwanted symmetries in the standard model. But we see that this crucial ingredient of our present-day theory for fundamental processes is available to us precisely because of the infinite volume of space, which is also responsible for infrared divergences! […] [I]nfrared infinity […] is a consequence of various idealizations for the physical situation: taking the region of space-time which one is studying to be infinite, and supposing that massless particles can be detected with infinitely precise energy-momentum resolution, are physically unattainable goals and lead in consequent calculations to the aforementioned infrared divergences. In quantum electrodynamics one can show that physically realizable experimental situations are described within the theory by infrared-finite quantities. The Unreasonable Effectiveness of Quantum Field Theory by R. Jackiw </blockquote> <blockquote> Relating this situation to quantum mechanics might provoke suspicion: through tunneling the real ground state could surely be a superposition of the individual ground states, which would not be degenerate. But although such a situation could occur in ordinary quantum mechanics, it does not apply to quantum field theory. The fields live in an infinite volume and thus have infinitely many degrees of freedom. Tunneling cannot happen in this case. All vacuum states are orthogonal. This is extensively discussed by, for example, Weinberg [54, pp. 163-167]. http://philsci-archive.pitt.edu/9295/1/Spontaneous_symmetry_breaking_in_the_Higgs_mechanism.pdf </blockquote> <blockquote> However, in a system with an infinite number of degrees of freedom, as in QFT, the tunneling amplitude is zero because each degree of freedom should tunnel, and therefore no SSB is possible. page 264 in A Modern Introduction to Quantum Field Theory by Michele Maggiore </blockquote> <blockquote> 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 ↑ and all ↓. 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 ↑ or all ↓, 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 ↑ to all ↓.) […] [T]he 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. https://physics.stackexchange.com/a/203739/37286 </blockquote> <blockquote> What the vacuum degeneracy represents physically is perhaps easiest to see in nonrelativistic quantum statistical mechanics, where these ideas were initially developed.5 Consider, for example, constructing the Hilbert space for an infinite chain of spin -1/2 systems interacting via a specified Hamiltonian. Operators on the Hilbert space of states are linear combinations of the spin operators $\sigma_i$ for each site, and a state specifies the spin for each system. For any particular global state of the system, such as all $\sigma_z$-spin $+$, there are other states, such as all $\sigma_z$-spin $-$, that cannot be reached by the finite application of spin operators, corresponding to “flipping” the spin of finitely many individual sites. In the case of QFT, the degeneracy corresponds to field configurations that differ globally in a similar sense; particles defined as excitations over distinct vacua cannot be transformed into each other via local operations analogous to flipping the spins at individual sites. http://www.jstor.org/stable/10.1086/518324 </blockquote> ←- –> The Higgs as "generalized Gupta-Bleuler mechanism?# See O'Raifeartaigh §8.5. Group Structure of Gauge Theories, as quoted here https://books.google.de/books?id=5jzuk_A3SP8C&lpg=PA258&ots=fhnlPUgLbd&dq=%22the%20higgs%20mechanism%20may%20be%20viewed%20as%20a%20kind%20of%20generalized%20Gupta-bleuler%22&hl=de&pg=PA258#v=onepage&q&f=false ←- —> Is the Higgs field responsible for a force?# Yes. "In the gaugeless limit g→0 the electron becomes instable! "It is surprising to me how many people are startled by this result. Most have to be told what the final state is. But upon the slightest reflection, it should not be startling at all." http://inspirehep.net/record/256768/files/Pages_from_C87-01-24_1-18.pdf ←- –>How many Higgs particles are there?# There is only one Higgs boson since we can always switch to a basis where only one linear combination of scalar fields gets a nonzero VEV. All other linear combinations are normal scalar fields and not Higgs fields since they have a zero VEV. Source: http://www.people.fas.harvard.edu/~hgeorgi/review.pdf ←- –>Do charged Higgs exist? No! The notion "charged Higgs" is actually an oxymoron. A Higgs field is a scalar field with a nonzero VEV. A charged Higgs would, therefore, be a scalar field that carries an electric charge and has a nonzero VEV. However, the electromagnetic $U(1)$ symmetry that would be broken by such a field is unbroken and hence there is no such field. Source: http://www.people.fas.harvard.edu/~hgeorgi/review.pdf ←- * Many questions and answers can be found here: https://profmattstrassler.com/articles-and-posts/the-higgs-particle/the-higgs-faq-2-0/

History

For a nice discussion of the History of the Higgs mechanism, see chapter 9 in "The Infinity Puzzle" by Close

Anatoly Larkin, posed a challenge to two outstanding undergraduate teenage theorists, Sacha Polyakov and Sasha Migdal: “In field theory the vacuum is like a substance; what happens there?”Chapter 9 in The Infinity Puzzle, by F. Close

A vacuum is a hell of a lot better than some of the stuff that nature replaces it with. Tennessee Williams

"The mass generation through an interaction with a nonempty vacuum can be traced back to the σ-model proposed by Schwinger where the σ and ϕi (i = 1, . . . , 3) lead to the appearance of three massive and one massless vector bosons." http://www.galpac.net/members/steinergroup/paper/08/tp08-8.pdf

See also: http://www.math.columbia.edu/~woit/wordpress/?p=3282 and https://cds.cern.ch/record/346612/files/9802142.pdf