© Matt Strassler [April 30, 2022]
Please feel free to point out errors or suggest improvements; this is a dynamic document.
This page is the second of several (first one is here) that together will explain the “Triplet Model”, a very simple method for shifting the W boson mass upward relative to the predictions of the Standard Model.
A particle is simply a tiny ripple in a field. For example, suppose we have a field T with potential
- V = ½ mT2 T2
A particle will correspond to a ripple in space that moves with time, as in
- T(x,y,z,t) = f(x,t) = a cos([E t – p x] / ħ)
where the cosine function gives us a ripple moving in the positive x direction, and a, E, p and ħ are constants:
- E is the particle’s energy (all of its energy, not just its motion-energy);
- p is the particle’s momentum (and the speed of the particle is pc2/E);
- a is the amplitude of the ripple (fixed by details of quantum physics but unnecessary here);
- ħ, Planck’s constant, is a fundamental constant of nature.
[A sine function would work just as well as a cosine.] In this formula, E/ħ, which multiplies t in the cosine, determines how the ripple changes with time; the larger is E, the quicker the time dependence. Similarly, p/ħ, which multiplies x, determines the space dependence of the ripple; the larger is p, the quicker is the x dependence.
When we substitute this into the equation of motion (which I’ve now written more carefully, so as to get the units right)
- [(time variation of F)2 – c2(space variation of F)2] = – [c4 / ħ2] the variation of V with F
where c is the cosmic speed limit (a.k.a. the speed of light), we obtain
- [ (E2 – p2c2)/ ħ2 ] f(x,t) = [c4 / ħ2] mT2 f(x,t)
which is true as long as
- E2 – p2c2 = (mT c2)2
This is the formula that relates the energy E and momentum p of a particle of the field T to its rest mass mT. Notice that for p=0, we get E = mT c2, Einstein’s formula relating energy to rest mass for a stationary T particle.
We learn from this that when we see a term ½ mT2 T2 in the potential, that’s an indication that the T particle, a ripple in the T field, has a mass mT.
We might consider a more complicated potential, such as
- V = + ½ mH2 H2 + ¼ λ H4
The vacuum of this theory has <H>=0, which I called the “unbroken phase” in the previous discussion of this potential. If we put a ripple like the one we used earlier into the equation of motion, it won’t quite work, because the equation of motion will have a term proportional to H3 , and thus the ripple will appear cubed in the last term:
- (E2 – p2c2)/ ħ2 f(x,t) = mH2 [c4 / ħ2] f(x,t) + λ [c4 / ħ2] f(x,t)3
The last term would seem to mess everything up, and in fact it does; our ripple does not solve this equation.
But it almost solves it as long as f(x,t) is extremely small… and that’s the case here. To have just one H particle, we need to take the amplitude a really small. As long as λ is small too, then λ [c4 / ħ2] f(x,t)3 will be tiny compared to the other terms in the equation. (This is because if a number x << 1, then x3 << x.) The consequence of being an almost-solution to the equation, rather than a true solution, is that our ripple may or may not live forever, but it will still live for a long enough time to be meaningfully described as a particle.
However, something quite different happens in the broken phase, for which
- V = – ½ μH2 H2 + ¼ λ H4
where μH2 > 0. If we follow the logic above, a ripple in the H field of the form H = a cos([E t – p x] / ħ) would seem to have mass equal to the square root of (- μH2 ), a negative number. Would that mean the H particle’s mass is imaginary?
A particle with an imaginary mass is called a tachyon. The equation of motion for a tachyon implies that E2 – p2c2 = (mT c2)2 < 0. With energy less than its momentum (times c), such a particle naively travels faster than the cosmic speed limit. That doesn’t sound very healthy…
… and in fact, it is not. We have made a mistake. In one of the vacua of the broken phase, < H > = μH /λ1/2 . In that vacuum, we should not be looking for a particle to be a ripple of the form H = a cos([E t – p x] / ħ), as we did earlier. That would be a ripple around H=0, a field that is constant, uniform and zero.. Instead we must look for a ripple of the form
- H = <H> + a cos([E t – p x] / ħ) = μH /λ1/2 + a cos([E t – p x] / ħ)
so that it is a ripple around the actual vacuum, where the field is constant, uniform and non-zero. When we do this, we find no tachyon.
The standard way to see this is to shift H, by writing
- H(x,y,z,t) = < H > + h(x,y,z,t)
where < H > is a constant, h is the shifted H field, and <h> = 0; the ripple we want will be a cosine in h. Traditionally, since “< H >” takes up a lot of room, physicists often replace < H > with “v” to stand for “vev”. So let’s say
- H = v + h
where v is a constant, H is the original field, and h is the shifted field. Then
- V = – ½ μH2 (v+h)2 + ¼ λ (v+h)4 = – ¼ μH2 v2 + μH2 h2 + λvh3 + ¼ λh4
Remember this is the same potential as before, simply rewritten in terms of h instead of H!
In this new form, the first term is a constant and plays no role in determining the vacuum. The term proportional to h vanishes exactly, so <h>=0 is indeed the vacuum. Since the h2 term in this potential has a positive coefficient +μH2 , a ripple in the H field around the true vacuum H = v has a positive mass (√2 μH) as it should, not an imaginary one. There is no tachyon in the real vacuum.
In fact we will never find, in a real vacuum of a quantum field theory, a tachyon; a tachyon is simply a sign that we have chosen a state in which all variation of V with the fields is zero, but which is not a [local] minimum, and therefore not a vacuum.
By the way, the broken phase has another possible vacuum in which < H > = -μH /λ1/2. The logic in that vacuum is exactly the same as in the one I just considered, so all the conclusions are the same too.
The Anderson/Higgs/Brout-Englert Mass-Generation Mechanism
Now what would happen if we had fields H and W and a potential
- V = + ½ mH2 H2 + ¼ λ H4 + ½ mW2 W2 + ⅛ g2 H2 W2
- V = – ½ μH2 H2 + ¼ λ H4 + ½ mW2 W2 + ⅛ g2 H2 W2
The first case is the unbroken phase; it would have <H> = 0. Also, <W> = 0, and the coefficient of W2 is mW2, meaning the W particle has mass mW.
- Mass of h particle = mH
- Mass of W particle = mW
But in the second case, we are in the broken phase, < H > = v = μH /λ1/2. Again <W> = 0. Shifting H and writing H = v + h, we find
- V = constant + μH2 h2 + λvh3 + ¼ λh4 + ½ mW2 W2 + ⅛ g2 v2 W2 + ¼ g2 v h W2
We can collect the purely W2 terms
- V = constant + μH2 h2 + λvh3 + ¼ λh4 + ½ (mW2 + ¼ g2 v2) W2 + ¼ g2 v h W2
Now the W field’s mass – the terms multiplying W2 – has shifted
- Mass of h particle = √2 μH
- Mass of W particle = (mW2 + ¼ g2 v2)1/2
In short, a non-zero vev for H (i.e. <H>, or in other words, v) shifts the mass of the W.
And in fact, if mW were zero, then in the unbroken phase the W would have zero rest mass, while in the broken phase its mass would be ½ g v. In short, this is an example in which a massless particle has developed a mass. This is a crude cartoon, but it captures much of the essence of the Anderson/Higgs/Brout-Englert mechanism for mass generation.
To understand why certain mass parameters, such as mW here, must be zero for some fields, and why the corresponding particles therefore require this mechanism in order for them to have any mass at all, requires a longer discussion. We will get back to that later.