# Unveiling the Higgs mechanism to students

Notes:

• Charged particle between two charged plates.
• Energy due to electric potential V:
$U = qV\;$
• Energy due to electric field itself:
$U = {\epsilon_0 \over 2} E^2 \mathcal{V}\;$
• Combine:
$U = qV + {\epsilon_0 \over 2}E^2 \mathcal{V}\;$
• Divide by volume to get energy for a particular point
$u = {U \over \mathcal{V}} = {q \over \mathcal{V}} V + {\epsilon_0 \over 2}E^2\;$
• Comment: Energy caused by:
1. charge interacting with field potential
2. field interacting with itself
• Note symmetry
• Note this is true for ALL conservative forces
1. source of field times potential of that field
2. a field times another field, possibly itself.
• Introduce Relativity:
$u = {q \over \mathcal{V}}V + {\epsilon_0 \over 2}E^2 + {mc^2 \over \mathcal{V}}\;$
• Note that symmetry is broken!
1. Mass is source of gravity
2. But gravitational potential is nowhere to be found?!?
3. What in the world? Our beautiful universe is now no longer beautiful.
4. It looks like m is doing something by itself, while q needs the field / potential.
• We're going to "fix" this by breaking it.
• Note minimum energy: charge, field, mass all 0.
• This is important: We are going to see something where the things are not zero to get minimum energy.

3. Introducing the Higgs field

• Let's drop $mc^2$. We're going to try and get it back by introducing a new field.
$u = {q \over \mathcal{V}} V + {\epsilon_0 \over 2}E^2\;$
• Let's invent a new field, $\phi$ with its potential $\Phi$.
• It interacts with E, itself, and some new source "a".
$u = {q \over \mathcal{V}} V + {\epsilon_0 \over 2}E^2 + {a \over \mathcal{V}} \Phi + gE\phi + g'\phi^2\;$
• Symmetry is preserved!
• Explain:
• q and ${\epsilon_0 \over 2}$ are "coupling constants", describe how tightly the things interact.
• ${1 \over \mathcal{V}}$ is the density of the particle in the volume, rather, the matter.
• a, g, and g' are the coupling constants of the particle, the electric-higgs interaction, and the higgs-higgs interaction.
• Note g' is not the derivative of g!
• Let's substitute to show the symmetry:
• $\mathcal{P} = {1 \over \mathcal{V}}$
• $F_p$ for the potential of a field F
• $c_i$ the coupling constants.
$u = c_1 \mathcal{P} E_p + c_2 E E + c_3 \mathcal{P}\phi_p + c_4 E \phi + c_5 \phi \phi\;$
• Beautiful!

• Note: When all the c's are 0, the energy is at a minimum.
• Let's add a $\phi^4$ term.
• energy density of $phi$ is now:
$u &= -g'\phi^2 + \phi^4$
• Where is the minimum?
$\phi_0 = \sqrtTemplate:G' \over 2 \ne 0$
• Draw graph!
• We used to define a vacuum as "no fields, no particles".
• Let's redefine it as "minimum energy", ie, we sucked all the energy out we can.
• That means even if you remove everything, if you can make the energy even lower by adding a field or a particle, then you need to.
• Note! minimum energy is where $\phi = \sqrtTemplate:G' /over 2$
• So, in a vacuum, $\phi$ is not zero!
• Our new energy density:
• Note: g' < 0!
$u = {q \over \mathcal{V}} V + {\epsilon_0 \over 2}E^2 + {a \over \mathcal{V}} \Phi + gE\phi - |g'|\phi^2 | \phi^4\;$
• Note that adding $\phi^4$ doesn't really break the symmetry. It is still the product of 2, or rather, 2 or more fields.
• In a vacuum, there is a Higgs field!

4. Higgs Boson

• Let's rewrite $\phi = \phi_0 + \eta$
• $\phi_0$ is minimum field, the vacuum field, the field that exists when you have nothing.
• $\eta$ is now the changing bit, ranges from 0 to negative to very big as we increase. We call this the Higgs field.
• Now we need to rewrite $\Phi = \Phi_0 + \Phi_1$
• $\Phi_0$ is potential of Higgs field at its minimum
• $\Phi_1$ is the changing bit.
• Now we have:
$u = {q \over \mathcal{V}} V + {\epsilon_0 \over 2}E^2 + {a \over \mathcal{V}} (\Phi_0 + \Phi_1) + gE(\phi_0 + \eta) - |g'|(\phi_0 + \eta)^2 + (\phi_0 + \eta)^4\;$

• Now we have something for the mass to interact in a vacuum (because we've redefined what a vacuum is.)
• We want:

${mc^2 \over \mathcal{V}} = {a \over \mathcal{V}} (\Phi_0 + \Phi_1)$

• So we get:

$m = {a \over c^2} \Phi_0$

• Progress!
• But now we have this leftover term:
${a \over \mathcal{V}}\Phi_1$
• That represents the interaction of a particle with the field $\eta$.
• We can calculate a:
$a = {mc^2 \over \Phi_0}$
• Same story for the Electric Field
• If g is not 0, then we have a mass for the Electric Field. Meaning, if we have an Electric Field with no charge or masses present, then the Electric Field -Higgs Field interaction looks like a mass-Higgs interaction. We know this isn't the case, Electric Fields do not have mass, so we must set g = 0.
• The final two terms can be expanded:
\begin{align} -|g'|\phi^2 + \phi^4 &= -|g'|(\phi_0 + \eta)^2 + (\phi_0 + \eta)^4 \\ &= -|g'|(\phi_0^2 + 2\phi_0\eta + \eta^2) + (\phi_0^4 + 4\phi_0^3\eta + 6\phi_0^2\eta_2 + 4\phi_0^3\eta + \eta^4) \\ &= -|g'|\phi_0^2 -2|g'|\phi_0\eta -|g'|\eta^2 + \phi_0^4 + 4\phi_0^3\eta + 6\phi_0^2\eta_2 + 4\phi_0^3\eta + \eta^4 \\  \end{align}

• $-|g'|\phi_0^2$ is a constant, so it's irrelevant. That is, we can change the energy up or down everywhere and it doesn't mean a thing.
• $-2|g'|\phi_0\eta$ is a mass term for the Higgs field.
• $-|g'|\eta^2$ is the Higgs field interacting with itself.
• THe higher order terms are just the Higgs field interacting with itself or the vacuum. They are really combinations of the lower-order terms.
• Review:
• We said that the vacuum is not where nothing exists, but where the energy is minimized.
• We introduced a new field, the Higgs field.
• The Higgs field doesn't interact with the electric field.
• The Higgs field does interact with itself and with the vacuum.
• The Higgs field does interact with mass.
• This is not a very good description of what's going on
• It is the best you are going to get without doing Quantum Field Theory.
• It is a pretty accurate