# 4. Why the Higgs Field is Necessary

© Matt Strassler [November 27, 2012]

This is article 4 in the sequence entitled How the Higgs Field Works: with Math. Here is the previous article.

Up to this point in this series of articles on How the Higgs Field Works, I’ve explained to you the basic idea behind how the Higgs field works, and I’ve described how the Higgs field becomes non-zero and how the Higgs particle arises, at least for the simplest type of Higgs field and particle (that of the “Standard Model.” But I haven’t explained to you why there’s no alternative to introducing something like a Higgs field — why there’s a fundamental impediment to introducing masses for the known particles in the absence of this field. That’s the goal of this article.

I’ve explained that all the elementary “particles” (i.e. quanta) of nature are quanta of waves in fields. And naively, all of these fields satisfy Class 1 equation of the form

• d/dt (d Z(x,t)/dt) – c2 d/dx (d Z(x,t)/dx) = – (2π c2/h) 2 m2 Z(x,t)

where Z(x,t) is a field, m is the corresponding particle’s mass, c is the speed of light, and h is Planck’s quantum mechanics constant. If the particle is massless, then the corresponding field satisfies the same equation with m=0, which I called a Class 0 equation.

Cases with m=0 include photons and gluons and gravitons, which are the quanta of the electric, chromoelectric (or “gluon”) and gravitational fields; they are all massless quanta (“particles”) traveling at the universal speed limit c. For electrons, muons, taus, all the quarks, all the neutrinos, and the W, Z and Higgs bosons, each one with its own mass, the corresponding field satisfies a Class 1 equation with the corresponding mass inserted.

Unfortunately, this isn’t the full story. You see, for all of the known elementary fields of nature that correspond to massive quanta, the violet equation that I’ve written above is illegal, at least as I’ve written it so far. Why? The problem is that we haven’t put in the weak nuclear force into our equations. And when we do, as we’ll see now, these simple equations can’t be used. More clever equations that lead to the same physical result are needed instead.

Why?

The problem is this; the equation we have just written is necessary but not sufficient. We need it to be true, but it’s not the only thing we need to be true. We’re leaving something out: the weak nuclear force. And the weak nuclear force and the violet equation above are not going to get along.

Now this could get rather technical if I went into it in detail, so I won’t. I’m going to explain this using equations that are similar to the ones that are actually used, but without giving the full story.

The Electron’s More Elaborate Equations

To see the problem, let’s consider it in the context of a particular field — we will take the electron field as an example. The problem is that the electron field doesn’t quite satisfy the equation I wrote above. An electron is a spin-1/2 particle, which means it not only moves but also incessantly rotates, in a way that is impossible to visualize — and it turns out the previous equation is only enough to describe how its position is changing, but not enough to describe what can happen to its spin. In the end it turns out the electron is actually formed from two fields ψ(x,t) and χ(x,t) satisfying two equations, of the form

• dψ/dt – c dψ/dx = μ χ
• dχ/dt + c dχ/dx = – μ ψ

where I’ve introduced the constant μ = 2π mc²/h to keep the equations short. Again, I’m lying to you slightly, because this is the equation for motion only along one direction of space, the x direction; the full form of the equations is more complicated. But the point is right; we’ll check in a moment that these two equations imply the violet equation at the beginning of this article. [Note: ψ and χ are often called the “left-handed electron” and “right-handed electron” fields, but without more math, this nomenclature is more confusing than illuminating, so I’m avoiding it for now. See this link.]

Now the sense in which these two fields jointly make up the electron field is that in an electron wave, the amplitude of χ and ψ have to be proportional to one another. In fact you can check that if you make a wave in both of them

• ψ = ψ0 cos (2π [ν t + x/λ])
• χ = χ0 sin (2π [ν t + x/λ])

where ψ0 and χ0 are the amplitudes of the waves, and ν and λ are their frequency and wavelength (which I’ve assumed are equal), we get the equations

• (2π) (ν -c/λ) ψ0 sin(2π [ν t + x/λ]) = μχ0 sin(2π [ν t + x/λ]) 0
• -(2π) (ν +c/λ) χ0 cos(2π [ν t + x/λ]) = -μψ0 cos(2π [ν t + x/λ])

which means

• (ν + c/λ) ψ0 = (μ/2π) χ0
• (ν – c/λ) χ0 = (μ/2π) ψ0

These equations show ψ0 and χ0 are proportional; generally, if one is non-zero, so the other must be too, and if you make one larger, the other has to become larger too.

But look carefully: there are two equations, giving two relations that could easily contradict each other. The only way the two equations can be consistent is if there is an additional relation between ν, -c/λ and μ. What is that relation? Multiply the two equations together and divide by ψ0χ0 (which we’re allowed to do as long as both ψ0 and χ0 are non-zero — let’s assume that here) and we find

• ν2 – (c/λ)2 = (μ/2π)2

What is the implication of this equation? Suppose that we have a single quantum of a wave in the ψ and χ fields — a wave of minimal amplitude — in other words, an electron! Then the energy E = hν and momentum p = h/λ of that quantum can be obtained by multiplying this equation by h² and substituting μ = 2π mc²/h, giving

• E2 – (pc)2 = (mc2)2

which is Einstein’s relation between an object’s energy, momentum and mass, which of course an electron of mass m should satisfy.

This is no accident, because Einstein’s relation is true for a quantum of a wave that satisfies a Class 1 equation, and the two green equations for ψ and χ secretly imply that both ψ and χ satisfy a Class 1 equation! To see this, multiply the first equation by -μ and substitute the second equation

• -μ(dψ/dt – c dψ/dx) = (d/dt- c d/dx)(dχ/dt + c dχ/dx) = -μ² χ

which gives [using d/dx(dχ/dt) = d/dt(dχ/dx)] a Class 1 equation for χ (and a similar trick gives a Class 1 equation for ψ):

• d/dt(dχ/dt) – c² d/dx (dχ/dx) = – μ² χ

Having two equations instead of one is a clever way (invented by Dirac) of having spin-1/2 particles that satisfy Einstein’s relation for energy and momentum and mass. An electron is a quantum of a wave in the ψ and χ fields, which jointly make up the electron field, and that quantum acts as a particle with mass m and spin 1/2. The same is true for the muon, the tau, and the six quarks.

Naive Electron Mass and Weak Nuclear Force are Inconsistent

But unfortunately, this beautiful set of equations, set up in the 1930s, turns out to be inconsistent with experimental data. What we learned in the 1950s and 1960s is that the weak nuclear force affects only χ and not ψ! So that means the equation

• dχ/dt – dχ/dx = -μ ψ

makes no sense; the change in time of a field χ that is affected by the weak nuclear force cannot be proportional to a field ψ that is not affected by the weak nuclear force. A different way to say this is that the W field can convert the field χ(x,t) into the neutrino field ν(x,t), but it can’t convert ψ(x,t) into anything, so the version of this equation that arises when a W field is combined with it is undefined and meaningless:

• dχ/dt – dχ/dx = -μ ψ

W field

• dν/dt – dν/dx = ???

So this failing of the equations when combined with the weak nuclear force tells you (as it told physicists of the 1960s) that a different set of equations is needed… and the solution to this problem is going to require a new idea. That idea is the Higgs field.

Enter the Higgs Field: The Right Equations for the Electron’s Mass

At this point, the equations are going to become a little more tricky (which is why I didn’t explain this stuff right at the beginning.) You may want to read my non-technical article on what the world would be like if the Higgs field were zero. The structure described there is going to appear within the equations I’m about to write down.

We need to have equations for electrons and for neutrinos that allow for the possibility that a W particle turns an electron into a neutrino or vice versa… but only by interacting with χ (the so-called “left-handed electron field”) and not with ψ.

To do this we need to recall a subtle point: that before the Higgs field is non-zero, there are actually four Higgs fields, not one — three of them disappear in the end. What’s a bit confusing is that there are several ways to name them, each naming convention being useful in different contexts. In my post about the world with a Higgs field that’s zero, I called the four fields, each of which is a real number at each point in space and time, by the names H°, A°, H+, and H; the Higgs field H(x,t) that I’ve referred to throughout previous articles in this series is H0(x,t). Here I’m going to name them as two complex fields — i.e., functions that have a real and an imaginary value at each point in space and time. I’ll call these two complex fields H+ and H0; the Higgs field H(x,t) that I’ve referred to throughout previous articles in this series is the real part of H0(x,t). After the Higgs field becomes non-zero, H+ gets absorbed into what we call the W+ field, and the imaginary part of H0 gets absorbed into what we call the Z field. [The complex conjugate of H+ is called H; and since W+ absorbs H+, its complex conjugate W absorbs H.]

Now here’s a fact about the weak nuclear force: the particles in nature, and the equations they satisfy, have to be symmetric under the exchange of some of the fields with each other. The full symmetry is a bit complicated, but the part of the symmetry we need is the following:

• ψ is unchanged
• χ ⇆ ν
• H+H0
• HH0* (just the complex conjugate of the previous line)
• W+⇆ W

The fact that χ ⇆ ν reflects the fact that these fields are affected by the weak nuclear force, while the fact that ψ is unchanged reflects the fact that it is not affected by this force. [Without this symmetry (and without the larger one of which it is a part) the quantum versions of the equations for the weak nuclear force simply don’t make sense: they lead to predictions that the probabilities for certain events to occur are greater than one, or negative.]

The required equations turn out to be (here y is the Yukawa coupling for the electron, and g is a constant which determines how strong the weak nuclear force is)

• dψ/dt – dψ/dx = (2π c2/h) y (H0* χ + H ν)
• dχ/dt + dχ/dx + g W ν = – (2π c2/h) y H0 ψ
• dν/dt + dν/dx + g W+ χ = – (2π c2/h) y H+ ψ

Notice that these equations satisfy the symmetry listed above. [Experts will note I have slightly over-simplified in multiple ways; but I hope they will agree that the essence of the issues is captured by these equations.] Note again that t and x are time and space (though I’m simplifying, since I’m only keeping track of one of our three space dimensions); c, h, y, and g are constants that don’t depend on space or time; and ψ, χ, W, H etc. are fields — functions of space and time.

Now, what happens when the Higgs field becomes non-zero? The H field and the imaginary part of H0 disappear (in a way that I won’t explain here) from the equations, absorbed into other fields. The real part of H0 becomes non-zero, with an average value v ; as described in my overview of how the Higgs field works, we write

• Real[H0(x,t)] = H(x,t) = v + h(x,t),

where h(x,t) is the field whose quanta are the physical Higgs particles we observe in nature. And the equations then become

• dψ/dt – dψ/dx = (2π c2/h) y (v + h) χ
• dχ/dt + dχ/dx + g W ν = – (2π c2/h) y (v + h) ψ
• dν/dt + dν/dx + g W+ χ = 0

These are the equations that, after the Higgs field takes a non-zero value v, describe the interactions among

• the electron field, whose quanta are electrons with a mass me = y v;
• one of the three neutrino fields, whose quanta are neutrinos (which are massless in these equations — putting in their masses requires small modifications that I won’t describe here);
• W fields, whose quanta are W particles, and whose presence implies the involvement of the weak nuclear force
• Higgs fields h(x,t), whose quanta are Higgs particles

Notice the equations no longer appear to satisfy the symmetry shown above in the red equations. This symmetry is “hidden”, or “broken”; its presence is no longer obvious once the Higgs field is non-zero. And yet, everything works the way it must to match what is observed in experiments:

• if the fields h and W and ν are zero in some region of space and time, the equations become the original green equations for the electron field, built as a combination of ψ and χ;
• if the W field is zero in some region, the terms involving h show that the interaction between electrons and Higgs particles are proportional to y, and therefore proportional to the electron’s mass
• if the h field is zero in some region, the terms involving W and W+ indicate the weak nuclear force can convert electrons to neutrinos and vice versa, specifically by converting χ to ν while leaving ψ unaffected.

Summing Up

Let me bring this to a close through a quick summary. For spin-1/2 particles, the simple Class 1 equations

• d/dt (d Z(x,t)/dt) – c2 d/dx (d Z(x,t)/dx) = – (2π c2/h) 2 m2 Z(x,t)

that we studied up to now have to be made more elaborate, as Dirac realized; describing the electron and its mass requires multiple equations that imply the Class 1 equation but have more going on. Unfortunately Dirac’s simple equations aren’t enough, because their structure is inconsistent with the behavior of the weak nuclear force. The solution is to make the equations more complicated and introduce a Higgs field, which, once it is non-zero on average, can give the electron its mass without messing up the workings of the weak nuclear force.

We’ve seen how this works for the mass of the electron, as far as the equations for the electron field. Similar equations work for the electron’s cousins, the muon and the tau, and for all of the quark fields; a slight modification works for the neutrino fields. The masses of the W and Z particles arise through different equations, but some of the same concerns — the need to maintain certain symmetries in order that the weak nuclear force can make sense — play a role there too.

In any case, the behavior of the weak nuclear force, as we observe it in experiments, and the masses of the known apparently-elementary particles, as we observe them in experiments, would be completely inconsistent with each other if it weren’t for something like the Higgs field. Recent experiments at the Large Hadron Collider have provided what appears to be good evidence that the equations that I have described to you here, and the concepts that go with them, are more or less correct. We await further experimental study of the newly found Higgs-like particle, to see if there are more Higgs fields, and/or whether the Higgs field is more complicated, than I’ve described here.

### 72 responses to “4. Why the Higgs Field is Necessary”

1. Acai

Very interesting article. Where can one learn more about weak interactions? Are there any good books you could recommend?

Also I wonder why if electrons have to be described by this set of equations are they considered elementary? Wouldn’t it be more natural to treat them as composite products of interaction of those component fields?

• Matt Strassler

Depends on your math and physics background, but a famous book (out of print??) is Commins and Bucksbaum, http://www.abebooks.com/9780521273701/Weak-Interactions-Leptons-Quarks-Commins-0521273706/plp, or Georgi, http://www.people.fas.harvard.edu/~hgeorgi/weak.pdf . A little lower in difficulty would be classic particle physics books like http://www.amazon.com/Introduction-Energy-Physics-Donald-Perkins/dp/0521621968, although one problem that you have to be aware of is that the less technical books also tend to have some conceptual errors floating around (poor treatment of concepts like virtual particles, symmetry “hiding”, confinement of quarks, and other concepts; various archaic notions will still be in there; etc.) In general the books tend to run about 10-20 years behind knowledge in the field, and that’s a problem… one that my website is trying to address.

Also — electrons are *not* *composites* of the two fields; they are *mixtures*. I’ve emphasized the difference between these two things a few times. A hydrogen atom is a composite made from an electron and a proton; as a consequence it has a size, and with some energy the electron and proton can be moved further apart, exciting the atom. But electrons are not atom-like things made from ψ and χ particles, rather they are sometimes χ and sometimes ψ. The same notion of *mixing* comes into neutrino oscillations and into many other quantum mechanical phenomena.

• Acai

Thanks a lot for the links, I will definitely check them out.

Yes, I meant mixtures, as in white light, not as in hydrogen.

• Matt Strassler

“Mixture” in quantum mechanics is not “as in white light”. It’s a technical term that means something else. (Thank you for illuminating this ambiguity in the language; I’m going to have to think about how to improve my terminology.)

White light can be pried apart into its parts by passing it through a prism; that tells you that it can be decomposed into constituents. In this sense it is a bit more like hydrogen.

This is not true of a quantum mechanical state that is a quantum mechanical mixture of two others. There’s just one electron; it doesn’t have constituents that you can pry apart — except by turning off the value of the Higgs field, making it zero.

• TimG

But is not (classical) white light a superposition of different frequencies? It seems to me that classical wave theories and quantum mechanics both have a concept of superposition, with the difference being that in QM a measurement destroys the superposition and collapses it to a single eigenstate. I also find your use of “mixture” a bit confusing; I think you’re using it as a layman’s term for “superposition”, but in QM there is also such a thing as a “mixed state”, which (of course you know) is quite different. In general I think it’d be nice when you’re using your own terms if you included parenthetically the real word for it, for those who may know (or be learning) this terminology.

• Matt Strassler

You’re right that I’m not entirely consistent on my terminology (it is a big challenge when writing multiple articles over many months, during which one learns about one’s pedagogical flaws.) And I should from the beginning have done what you say (include parenthetically the technical term where appropriate). But an additional problem is that particle physicists do use terms that are often inconsistent (“mixing” and “mixed state” mean very different things; don’t get me started on “multiple interactions”) and of course we know what we’re talking about from context, but non-experts can’t possibly figure that out. So this also is part of the pedagogical calculus…

Anyway, this will be something I have to think over very carefully when I rewrite/revise this as a more coherent package. Gotta love the word “coherent”.

2. martenvandijk

Prof. Strassler,

The electron’s mass-to-charge ratio being a constant, how is this reconcilable with the above?

• Kudzu

It’s only constant when the Higgs field is constant, and is a simple result of the fact that the electron’s mass and charge aren’t changing; the same applies to quarks say or W+ particles or ions. If the Higgs field fluctuated, so would the the mass-to-charge-ratio.

• Matt Strassler

It’s not a constant in principle; if you changed the Higgs field’s value somehow, then the electron’s mass would change but not its charge. Or if you changed the electron’s Yukawa coupling, that would change the mass and not the charge. And indeed, in the very early universe, one of these might have been the case. Not sure where you got the impression that mass-to-charge ratio is fixed. You can see from the equations that it’s only fixed in nature because right now, in our phase of the universe, the Higgs field’s value is constant and because the Yukawa coupling is constant and because the strength of electromagnetism is fixed. But that’s not a matter of principle, it’s just a matter of current fact.

• martenvandijk

The electron’s mass-to-charge ratio respectively charge-to-mass ratio are mentioned in lists of constants in physics with their values and references on the internet, but not with your proviso that these ratio’s being constant is only a current fact and not a principle. Thank you for telling me.

• Matt Strassler

All masses for all particles in the world are contingent, directly or indirectly, on the Higgs field; they shift continuously as the Higgs field changes its value. Masses are delicate and easily changed; this is presumably why the pattern of masses of the known particles is so bizarre. Charges are much harder to change because charge ratios *are* fixed, because of deep mathematical requirements on quantum field theory; this is presumably part or all of why charges are quantized in units of the down-quark’s charge.

There’s a caveat to this, and I have neglected to warn you that the Higgs field *does* change the way charges are computed once it becomes non-zero. See my answer to Giovanni below. But it doesn’t change the fact that, for instance, the muon and electron have exactly the same charges.

• Parlyne

To add to Matt’s answer, while the charge to mass ratios are not any fixed in any fundamental way, it’s worth understanding that the reason they are so often given in lists of constants is that they are actually quite a bit easier to measure than either the charge or the mass alone. This means that it is quite possible for a particle’s charge to mass ratio to be known significantly more precisely than either its charge or its mass. And, in fact, the electron’s charge to mass ratio was known before either the charge or mass were measured independently.

• Matt Strassler

thank you for this wise and helpful amendment.

3. Tim Preece

Previously you described in figure 5 https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-known-apparently-elementary-particles/the-known-particles-if-the-higgs-field-were-zero/ how the top quark flips between right and left versions.

Is this related to these equations being out of phase?

ψ = ψ0 cos (2π [ν t + x/λ])
χ = χ0 sin (2π [ν t + x/λ])

Is it easy to intuitively understand why inertial mass is related to the flip flopping? ( Sorry if you have explained this already ).

• Matt Strassler

Yes, the flip-flopping or “mixing” is directly related to the out-of-phase waves.

However, the inertial mass of the electron is not directly related to the flip-flopping (“mixing”). The details of the two equations are important. The fact that the two equations for ψ and χ can be combined to form the Class 1 equation is the point — because, as we saw in the Particles and Fields series, the inertial or “rest” mass of a particle is simply the energy associated to a quantum of a (standing) wave in a field that satisfies a Class 1 equation, divided by c2 .

4. Frank

According to Sidney Coleman in his book Aspects of Symmetry and his lectures, the Higgs is needed because continuous symmetry gauge groups lead to Goldstone bosons, whereas discrete symmetry groups do not. Wouldn’t it all be ironic if the underlying lepton and quark particle states actually represent discrete symmetry groups, such as discrete subgroups of the Standard Model gauge group, and no Higgs is required for electroweak symmetry breaking?
Particle mass values would originate from another source, perhaps as mathematical constants related to these discrete subgroups or some larger group.

Then the tremendous success of the SM would be because it acts sort of as a “covering group” for these discrete symmetries, i.e., being an excellent approximation at lower energies. Of course, if the 125 GeV is truly the Higgs, which needs to be verified by its parity sign, etc., then the discrete symmetry particle groups as mentioned above would not enter the game.

• Matt Strassler

It would be ironic, but it would also be wrong, because the symmetry groups involved here are necessary to get the weak nuclear force in the first place. Your suggestion would eliminate the weak nuclear force from nature.

• Frank

It seems that one should get the same 4 vector fields as for the Standard Model if one has a discrete subgroup of the local gauge symmetry SU(2)L x U(1)Y, a discrete subgroup that also has 4 generators. The electroweak symmetry breaking is then from the continuous SU(2) x U(1) to the discrete symmetry. That is, the ground state with discrete symmetry we observe as leptons and quarks is like the ferromagnet with T < Tc; the SU(2) x U(1) symmetry is hidden.

Or do I not understand your reasons for saying there would be no weak nuclear interaction?

• Matt Strassler

The vector fields arise from the local symmetry. Make it a discrete symmetry and there’s no W or Z field. This would be the same for electromagnetism; make the local U(1) symmetry into a discrete symmetry, and there’s no photon.

5. thetasteofscience

Thank you, thank you, thank you!

6. Kudzu

So, are spin 1/2 particles due to said particles being composed of two fields, or is it possible to have an ‘elementary’ spin 1/2 particle not created via such mixing? And is spin rotational in nature or is it just a property named as it is for historical reasons?

7. Martin

If spin 1/2 particle must be made from 2 fields, does it mean that each of the 2 fields does not have spin 1/2? So the left-handed electron does not have spin 1/2? What is the spin of the left-handed electron? Is it still a fermion?

• Matt Strassler

“If spin 1/2 particle must be made from 2 fields, does it mean that each of the 2 fields does not have spin 1/2?”

No. A massless spin 1/2 particle (Weyl fermion) may be made from 1 field. A neutral massive spin 1/2 particle (Majorana) may be made from 1 field; neutrinos may be examples. But an electrically charged [or otherwise charged] spin 1/2 particle (Dirac fermion) must be made from two. This is because there has to be two helicity states for a massive fermion, and because the field must create both the fermion and the (distinct, because of the electric charge) anti-fermion.

• Martin

Thank you for the clarification.

• Giovanni

Nice article! I have a few questions:
In a zero-Higgs-field universe, how would be the left- and right-electrons be related? Are they two distinct fields in which has nothing to do with one another except their charge and spin or they are two ‘twin’ fields?

And in the “the known particles if the higgs field were zero” article you said:

“[…] you have to recalculate everything carefully — and you would find, after recalculating, that the strong nuclear force would be weaker than it is in our world, so the effect would be much smaller in such a world than it is in this one.”

Could you please explain why (how) a non-zero higgs field “strenghtens” the strong nuclear force? I find this a bit confusing…

Thanks.

• Matt Strassler

They are, indeed, two distinct field which have nothing to do with each other — and in fact, they don’t even have the same charge, because in a zero-Higgs-field universe there is no distinct electromagnetic force and weak nuclear force; instead there are weak-isospin and hypercharge forces (see my article https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-known-apparently-elementary-particles/the-known-particles-if-the-higgs-field-were-zero/ ) and the left-electron and the electron-neutrino are affected by the weak-isospin force and have hypercharge 1/2, while the right-electron is unaffected by the weak isospin force and has hypercharge -1. The Higgs field being non-zero causes a significant rearrangement of the particles, marrying some of them together that naively would have been unrelated (and thereby creating a tricky puzzle for particle physicists that took decades to unlock!)

The last thing you ask about — how the strong nuclear force is affected by a change in the Higgs field — occurs through a subtle quantum effect, whereby quantum fluctuations (“virtual particles”, but they aren’t really particles at all) in fields affected by the strong nuclear force react back on that force to make it stronger. The degree of the effect depends on the masses of the fields’ particles, and therefore on the Higgs field. I have not yet explained this on the website, but I expect to do so sometime in the next six months or so.

• Giovanni

Is there a particular reason why the left-electron mixes with the right-electron and not, lets say, the right-muon or the top-left? I ask the same for the case of a collision left-electron + H -> right-electron for a average zero Higgs field.

And if all the leptons were massless, what would be the difference between a left-electron and a left-muon or left-tau? They are different fields, yes, but they all would look pretty much the same, dont they?
The same for the corresponding neutrinos?
Thanks.

• Matt Strassler

Excellent question!

The left-electron can’t mix with top-left because they have completely different conserved charges. The top-left feels the strong nuclear force and the left-electron does not. The left-electron and right-electron can mix because AFTER the Higgs field becomes non-zero, and weak nuclear charge is no longer explicitly conserved, they have the same conserved charges; electric charge -1 (in units of “e”) and no strong nuclear force or any other force.

The left-electron and the right-electron start off as completely different particles, and at least as far as their interactions with the non-Higgs forces — with everything except the Higgs field — there is no difference between the three fields that we call the left-electron, left-muon and left-tau. Only the Higgs field, even before it is non-zero, through the fact that it has different Yukawa couplings to the three fields, explicitly ruins the symmetry that relates the three fields. Each field has its associated neutrino, and again these are almost symmetric, except for those Yukawa couplings.

As I emphasized in the question that follows this one (which you should read), left-electron and right-electron (or electron-left and electron-right — it doesn’t matter) are names that we assign to these distinct fields χ and ψ only after they get married by the Higgs field. We call ψ “electron-right” only because (1) we’ve chosen, arbitrarily, to call χ “electron-left”, and (2) the Higgs field married χ to ψ, and not to some other field. Had the Higgs field married χ to some other field similar to ψ but by the name η, we would have called η “electron-right”. These are married-names, not names of birth.

• Tim Preece

Would it be right to consider that χ and ψ are related by a space-time transformation ( reflection ) and hence the use of the naming convention left-right ( but they are not transformed into each other by any Lorentz boost ) ?

• hpi

“two helicity states for a “massive” fermion”?

Is it not the case that only a “massless” particle can have two helicity states(=spin directions?) + and – along its traveling direction such as a photon or gluon? But, any massive particle also has two helicity states although its spin axis can point to any direction? If so, why “massless particle’s spin axis” is always found along its traveling direction?

• Matt Strassler

No. Particles with different spin have different counting rules, and massless particles have different counting rules from massive ones. It is not the case that all massless particles have the same number of helicity states. The counting you are giving is for massless spin-one particles. Massless spin-zero particles have 1 helicity state, for instance.

As for why (or if) the axis is along or opposite the traveling direction; this is quantum mechanics, and spin is very much a quantum mechanical effect, and very non-intuitive. You cannot make all the measurements you could make in non-quantum physics, say, on a baseball, which can spin however it likes relative to its travel direction, and whose spin and travel direction you can measure with complete freedom. You asked, `why “massless particle’s spin axis” is always found along its traveling direction?` But that wasn’t the question, or the assumption. The question was “if we measure the massless particle’s spin via its component along the traveling direction, what do we find?” The answer is the helicity. No assumption was made about what the massless particle’s spin axis was before the measurement. Quantum mechanics tells us that the answer to this measurement will only have one of a finite number of values; the number of possible values is the set of possible helicities for a particle of a particular spin, massless or massive.

• hpi

Matt. Thanks for your explanation. yes, for massless spin 0 particle seems to have 1 helicity state. what I meant was, is it not the case that only a massless particle with non-zero spin can have only two helicity states (+ or – of spin direction along its traveling axis) but what about massive ones? Does a massive particle with non-zero spin also always have two helicity states? or does it always (or except specific cases) have more than two?

You said “No assumption was made about what the massless particle’s spin axis was before the measurement.” but what do you mean by this? The question was “if we measure the massless particle’s non-zero spin, is it always found along its traveling direction”? (or does it make sense theoretically that it always points to its traveling direction even if we do not measure it?) If it is always found along its traveling direction, why? because massless particles cannot have any observable components (vibration/amplitude of wave etc.) along its traveling direction?? A massless particle (=wave) cannot vibrate along its traveling direction because it(=the wave) “cannot have (observable?) thickness” along its traveling direction due to length (Lorentz) contraction? What is the explanation/reason behind this?

8. John

Nice article, thanks Matt. Just a tiny note: “were non-zero. ” should read “were zero”. Cheers.

9. Tim Preece

“the weak nuclear force affects only χ and not ψ”

Is this a result of spontaneous symmetry breaking?

I had a feeling it was the strong force breaks the left/right symmetry but it would seem more natural if the Higgs field was responsible.

• Matt Strassler

No. There never was a left-right symmetry, so it didn’t have to be broken, and nothing had to break it. That symmetry is called “parity”, and it is *not* a symmetry of nature, though it might have been. Our world simply looks different in a mirror than it does in reality.

This was one of the greatest discoveries of the last century, in the 1950s.

Actually the effect of the Higgs field being non-zero is to make this lack of parity less obvious, because at first glance ordinary matter is parity-symmetric. Once the Higgs field is non-zero, only the weak interactions (which, being weak, took centuries to discover) violate this symmetry. If the Higgs field had been zero, the lack of parity would have been much easier to observe (except for the fact that had the Higgs field been zero, there’d be no atoms and we wouldn’t be here to observe it.)

This is one of the main reasons I hate to call χ and ψ by the names “electron-left” and “electron-right” when teaching these things to students. The names makes it sound as though these two fields are fundamentally related. But this is the same mistake as assuming that a married couple by the names Barack Obama and Michelle Obama must be brother and sister, since they are of opposite gender and share the same last name. In fact, they came from entirely different places, and were only given the same name once they got married; moreover, the decision to give them the same name is a convenient but unnecessary choice. Same with χ and ψ; they were completely separate fields until they were married by the non-zero Higgs field. Once married to form a single unit — the fields whose quanta are massive electrically-charged electrons — it is convenient to give them new names, electron-left and electron-right. But they were not born this way.

10. Martin

I’m still not sure about one thing: are the ψ and χ particles massive after the Higgs field becomes non-zero? Or they stay massless? Their 2 equations together seem to make them massive and also you have indicated that: “…the two green equations for ψ and χ secretly imply that both ψ and χ satisfy a Class 1 equation.”

But you also indicated that a fermion in single field must be either massles or neutral. I understand that both ψ and χ are charged, so they need to be massless Weyl fermions.

I must have overlooked or misunderstood something…

• Matt Strassler

This IS confusing the first time through, and maybe I can improve the wording of the article.

“are the ψ and χ particles massive after the Higgs field becomes non-zero?”

Wrong question. There aren’t separate ψ and χ particles after the Higgs field becomes non-zero. You can see that in my discussion of the green equations that tie them together. Notice they have independent equations of motion if m=0, so they have independent waves, and therefore independent quanta = particles. We would call those massless ψ and χ particles, and they are Weyl fermions. But once m is not zero, their equations are tied together, and you cannot have a wave in ψ without having a wave in χ also. The quanta of this joint wave are the particles we call “electrons”, and their mass is m. An electron is a Dirac fermion. There simply aren’t separate ψ and χ particles any longer.

11. Jon Lennox

What happens when a Higgs particle (or several) collides with an electron? If the particles correspond to a wave in the Higgs field that makes its value locally zero, will the ψ and χ locally separate?

12. zeynel

Prof. Strassler wrote: “All the elementary “particles” (i.e.quanta of waves in fields) are quanta of waves in fields.”

This statement is nothing more than a meaningless tautology. [Abridged by editor]

• Matt Strassler

No, I didn’t write that. So I have removed your rant about it.

• zeynel

What am I missing here: I copied and pasted the following from your post above:

“…all the elementary “particles” (i.e. quanta) of nature are quanta of waves in fields.”

Do you deny writing this?

• Matt Strassler

You are a liar. Look at your first note. It is NOT a cut and paste; you inserted words. You said that I wrote:

“…all the elementary “particles” (i.e. quanta of waves in fields) of nature are quanta of waves in fields.”

and that obviously is a tautology; and then you went into an unpleasant, nasty rant about it.

What I wrote is a pedagogically informative statement that X = Y; many people need a reminder that the word “particle” in particle physics has the same meaning as the word “quantum”, and many people also need a reminder that that a quantum has to do with a wave in a field that has the smallest possible amplitude.

You’ve been nasty, and you’ve wasted five minutes of my time; you won’t waste a minute more until you learn to make comments that are useful to the readers of this site. You’re banned for now.

13. Giovanni

Thanks very much for the previous answers, Matt!
If Im not abusing your time:

Since we differentiate the electron from the muon by their couplings to the Higgs field, is there any chance that there are actually two different types of electrons with different couplings only to some yet-to-discover field?
– When I say ‘electron’, I mean any particle from the Standard Model.

• Giovanni

Both the right electrons and muons carry the same charge, dont they? So I still didnt get why the left-electron doesnt mix with the right-muon. Is it because the fields to be mixed must have the same coupling to the Higgs?

• Matt Strassler

What I’m saying is this: we don’t even have names for right-electron and right-muon and right-tau, initially. They are just three fields, ψ1, ψ2, ψ3, with the numbers assigned arbitrarily. After the Higgs field makes these particles into massive Dirac fermions, we NAME the lightest one the electron, and AT THAT POINT we name the ψ that forms part of the electron the “right-electron”. In other words, the answer to your question is that there’s no question here: the name right-electron and right-muon are BY DEFINITION. The χ-like field that interacts with W particles and participates in the second-heaviest of these particles, which we call the muon, is, by definition, called the left-muon (or muon-left, or left-handed-muon); and the ψ-like field that does not interact with W particles and participates in the heaviest of these (the tau) is, by definition, called the right-tau (or tau-right, or right-handed-tau.)

So there’s nothing to “get”. These names are applied after-the-fact. Have I said this clearly enough? It’s a tricky point. Maybe I need another analogy here…

• Matt Strassler

If there were two types of electrons of identical mass (that is, if the Dirac fermion we call the “electron” actually came in two types) then that would mean there are two sets of electron fields, and the Pauli exclusion principle(see this article on fermions and bosons, https://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/7-particles-are-quanta/7a-how-bosons-and-fermions-differ/) would not always apply to a pair of electrons. Atomic physics would then be completely different from what we observe in nature.

• Giovanni

14. Jon Lennox

I think what Giovanni was asking is — pre-Higgs, we have ψ1, ψ2, and ψ3, and also χ1, χ2, and χ3. Once we have the Higgs, ψ1 mixes with χ1 to make the electron, ψ2 with χ2 to make the muon, and ψ3 with χ3 to make the tau.

The question is, why do they only pair up one-to-one like this? What stops ψ1 from mixing with χ2, etc?

• Matt Strassler

Ok, since two of you are asking, let’s try to get it precisely right.

As you say, we start with ψ1, ψ2, and ψ3, and also χ1, χ2, and χ3. THE NUMBERING IS ARBITRARY. There is no relation between ψ1 and χ1; I could renumber the ψ’s and leave the χ’s the same. In fact, just to make this completely clear, let’s number the χs as χI, χII, and χIII.

Even worse — our set of ψ’s form a “vector space”. I’m assuming those of you who are asking have taken linear algebra once upon a time; if not, let me know. That means that choosing the three fields to call ψ1, ψ2, and ψ3 is like choosing three axes in space to call the x-direction, the y-direction and the z-direction — it’s completely arbitrary. The χ fields form a SEPARATE vector space; we can choose χI, χII, and χIII however we like. Different choices are called different “bases”; each choice is a “basis”.

Now, one more bit of notation: let’s write ψi to represent the ψs, where i can be 1, 2 or 3. And we’ll write χa to represent the χs, where a can be I, II or III.

Finally: the Higgs interacts with all six of these fields, through Yukawa couplings. Any ψ can interact with any χ through the Higgs! So there are nine Yukawa couplings in all, labelled yia, where again i can be 1, 2 or 3 and a can be I, II or III. In short, these nine numbers form a matrix, a 3×3 matrix. The equations contain terms like y1,II H ψ1 χII and y3,I H ψ3 χI — all nine possible terms appear in the equations.

But the precise numbers yia depend on the basis you choose! If we choose different bases for the χ and ψ fields, we will get a different-looking Yukawa coupling matrix. The physics is the same; we’ve just labelled things differently.

Fact of linear algebra: There is a (nearly) unique basis for ψ and one for χ in which

* the Yukawa matrix is diagonal: y1,I, y2,II, and y3,III are all non-zero, all others Yukawa couplings are zero
* 0 < y1,I < y2,II < y3,III

WITH THIS CHOICE OF BASIS ONLY, we may identify the electron mass as y1,I v, the muon mass as y2,II v, and the tau mass as y3,III v. And what we have chosen, by selecting this basis, to call ψ1 is the part of the electron we call the electron-right, and χ1 is similarly the electron-left, etc.

So Jon, when you ask “what stops ψ1 from mixing with χ2?”, the answer is “nothing”. But if they mix, then you are choosing to label things using a basis where ψ1 isn’t the electron-right and/or χ2 isn’t the muon-left. In the basis where we identify ψ1 as the electron-right and χ2 as the muon-left, they don’t mix — by definition! Because in that very basis, the Yukawa couplings form a *diagonal* matrix, and there is no y1,II entry in the matrix to cause mixing.

• Giovanni

The matrix explanation was very helpful. Thanks a lot.
And something similar happens to the quark fields, I suppose?

• Matt Strassler

Yes. It’s the same for the up-type quarks, and separately for the down-type quarks.

15. Martin

Thank you for the last reply, professor Strassler. Now I understand the cause of my confusion: from the article “The Known Particles — If The Higgs Field Were Zero” I remember this explanation about the left- and right-handed fermions: “You never have both [top-left and top-right] at the same time, which is why the top quark remains elementary, not composite.”

However you mentioned in your last response to my question that “you cannot have a wave in ψ without having a wave in χ also.” These 2 explanations sound contradicting each other (I must have misunderstood at least one of them), so I looked at the equations. They seem to allow the 2 waves (in ψ and χ) to co-exist simultaneously and continuously, cross-tied to form one complicated field. I can see in the math how the existence of ψ automatically triggers the χ and vice versa and how it’s impossible to have only one of them without the other.

So I’m not sure how to understand the concept of flip-flopping between ψ and χ without having both of them ever simultaneously (as mentioned in the article “particles without Higgs field”). Or does the math say something else? Like – ψ will immediately decay into χ, which immediatelly decays back to ψ, and so forth. That would mean there are still distinguishable ψ and χ, which is not the case after Higgs field gets its vev.

• Matt Strassler

This relates to Tim Preece’s question from above.

The “flip-flopping” is inherent in the fact that if ψ is proportional to a cosine function, χ is proportional to a sine function. See the waves that were written after the green equations for ψ and χ. As time evolves, one of those functions is reaching maximum magnitude when the other is reaching zero. That tells you that the probability of the electron being ψ and the probability of it being χ, which are proportional to the squares of the magnitudes of the waves, oscillates back and forth.

The picture I drew in the article “The Known Particles — If The Higgs Field Were Zero” shows the flip-flopping as discrete — as though the top quark (or electron) is either ψ with probability 1 or it is χ with probability 1. But it’s a failing of the picture.

As is often the case in quantum mechanics,
*the probability of the electron being ψ generally is not zero or one, but something in between — yet
*if you ask whether a particular electron is ψ or χ at a particular time, you will always find it is either ψ or χ, not something between.
Quantum mechanics is weird. But true.

• Kudzu

It may help to produce one of your ever-illuminating animated images for this. You could show the probability of the two possibilities changing through time and thus how they were related and how one peaks when the other is zero. Once I was able to relate your explanation to a graph of sine and cosine it became quite intuitive to me. (I had a similar experience many, many years ago trying to grasp how electric and magnetic fields worked.)

• Matt Strassler

Good idea… I’ll think about how best to incorporate that.

• Martin

Thank you so much! I didn’t realize that the flip-flopping is related to the phase shift between sine and cosine. It is obvious and I should have seen it myself.

• Martin

If I may bother you with one more question: are the fields ψ and χ also giving probabilities (when their magnitude is squared) of finding the particle at a location? I was told that there is a big difference between a field (function of space and time) and probability amplitude (function of all configurations of a system). The wave function (squared magnitude) will tell me the probability of finding a field in a given “shape”. And the field itseld can tell me probability of finding the particle at a location? Sorry if I’m asking too much.

• Matt Strassler

Fields and wave functions are indeed very different things, though easy to confuse at first. A field is a function of space and time. A wave function is a function of the full set of configurations of a physical system, and time. For instance, a wave function of two particles is a function of six spatial variables and time. A wave function of five particles is a function of fifteen spatial variables, and time. A wave function of a field is a function of a function of space (!) [called a “functional”] and of time.

Moreover, the square of a wave function (integrated over a set of configurations) is always less than or equal to one. It can therefore be interpreted as a probability. The square of a field may be arbitrarily large and cannot in general be interpreted as a probability.

So no, you cannot in general connect squares of fields to probabilities of particle positions. It only (roughly) works when you have a (roughly) single particle state of the field; then there’s a limited relationship. The case we were discussing is one of the rare cases where there is a limited relationship… and moreover, I was simplifying the discussion so as not to go into a full set of technical details.

• Martin

Thank you for the clarification.

16. Matt Strassler

Just a general comment for everyone like Tim Preece, Giovanni and Jon Lennox; you folks are asking all the right questions, exactly the questions that any good but slightly confused physics graduate student would ask. But really you’re going beyond what I can cover effectively at this site, because the answer I would give the graduate student would be more technical than is typical here. Folks like you should really take the next step and try to make your way through a more technical book, or more likely, some technical introductory lectures for beginning graduate students on this subject. Now there’s a question as to whether such a book exists that can be readable by people who haven’t had a complete physics undergrad education… maybe such a book is waiting to be written. But when I cover this subject, it takes, oh, maybe ten lectures — there’s a lot of material in the Standard Model. So I’m sorry not to do a better job for you — there just isn’t time for me to put these lectures into print right now — but really you should probably move on and learn this stuff properly.

• Giovanni

Thanks for all the work you have done and is going to do, Matt. It is very interestanting and clarifying, I am really glad for this website.
I am trying to read Cohen-Tanoudji’s QM book, but it is hard… and takes time… specially for someone who is now finishing what you may know as “High School”.

• Tim Preece

Just wondering if the other readers would like to form a study group to understand more about particle physics.
Perhaps we could hang out on the physics stackexchange ( or some other website that allow’s typing of math formulae ) and work through some SM problems.

A quick search found this problem set http://www.uni-leipzig.de/~nfp/For_Students/Teilchenphysik/ , but maybe there are better examples around.

In my case I’ve been watching lots of material lately, The Susskind video lectures ( I think aimed at curious pensioners ), and some of the introductory talks on TASI and on fermilab. Unfortunately I keep hearing the same stuff, and perhaps like a some mantra, I’m fooled into thinking that I’m beginning to understand but in reality it is just repetition
and I don’t really understand at all.

Matt,
Just to echo Giovanni and say thanks again. Reading your site I’ve finally understood some things which have confused me for years. You have re-kindled my desire to understand fundamental physics better.
I was beginning to think the Higgs field was a bit featureless and boring but now I start to see how mysterious and magical it really is.

And if you ever get round to publishing those 10 lectures ( or maybe the videos ) I would certainly buy them.

• Martin

I’m trying to study more technical books, but they often assume that the reader already understands the subject pretty well, so a lot of things is not explained in them. I already have some basic understanding of operators, eigenstates and superpositions of states, Hilbert space, Fock space and similar stuff, but at the same time I’m struggling to understand certain elementary concepts that are rarely addressed in textbooks.

This webpage helped me understand a lot more. So thanks a million for writing this for us, professor Strassler!

17. Bill Brown

I just realized I’m confused at a level that is far below what a slightly confused graduate student would encounter: By the word “after”, as in “After the Higgs field becomes non-zero”, to you refer to time or logical sequence. That is, was there a time in cosmological history when the Higgs was zero, afterwards it became non-zero; or are you simply talking about the logical/mathematical consequences of the Higgs being non-zero?

• Matt Strassler

Excellent question. I always struggle with how to say this clearly. I should probably clarify this explicitly in the article.

I do refer to logical sequence.

However, note that it is also true that the time sequence resembled the logical sequence; there was (probably! we’re not yet certain the universe was hot enough, though we suspect so) a time when the universe was so hot that the Higgs field was on average zero. (It’s similar to the fact that an ordinary magnet has a non-zero magnetization at low temperatures, but if you heat it up enough the magnetization will drop to zero.) But in that hot dense environment, not everything I said about the world with a Higgs field equal to zero would have been strictly true.

And actually a beginning graduate student might well be confused by this.

18. Kai

Hi sir, from your article above, can I deduce that weak nuclear force (interaction) is originated from weak isospin but not hypercharge?

19. Amos Dettonville

Just an editorial comment: The constant “c” seems to have been omitted from the left side of some of the later equations, such as

dψ/dt – dψ/dx = (2π c^2/h) y (v + h) χ

I know there’s a convention to choose units such that c=1, so it can be omitted to simplify equations, but the “c” appears on the right side of the equation, so it’s a bit confusing. Earlier in the article the “c” appeared on the left side too.

I also have a dumb question. This article is entitled “Why the Higgs Field is Necessary”, but it seems more like an explanation of why the neutrino is necessary. The two equations in green contain the mass factor m on the right side, and the article explains that these equations can’t be right, because they imply that the two electron fields are strictly proportional, whereas we know experimentally that the W field only interacts with one and not the other. Fair enough. The resolution is to introduce the neutrino field, and have the W field interacting with the neutrino and one of the two electron fields. Along the way, the mass factor m is replaced with y(v+h), but it isn’t clear (to me) how this replacement contributes to the resolution of the problem raised by the W interaction. I think that problem is solved just by introducing the neutrino field and the W terms. So my question is, what would be wrong with the equations

dψ/dt – c dψ/dx = (2π c2/h) m χ
dχ/dt + c dχ/dx + g W- ν = – (2π c2/h) m ψ
dν/dt + c dν/dx + g W+ χ = 0

These are the final set of equations in the article, except I’ve replaced the quantity y(v + h) with the symbol “m”. I think h is not constant, but it’s very close to zero (right?), so we essentially have just the product of the constants y and v. Why is it necessary to represent the mass m as the product of a coupling constant y and a constant average Higgs field v? (I think I understand why it’s possible, but not why it’s necessary.)

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I think that we should support a local enterprise by
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22. moving home

Why don’t we support local businesses by hiring a nearby moving company.

23. Yu-Shuan Bill Cai

Dear Dr.Strassler,
Why the particles in nature have to be symmetric under the exchange of some of the fields with each other such that H+ ⇆ H0 and H- ⇆ H0*?