Ten episodes concluded that only dimensionless quantities are physics. But there is a dimensionless quantity we never once touched — the phase.
The electron's Yukawa coupling is \(y_e=2.94\times10^{-6}\); the top's is \(y_t=0.99\). Two couplings of exactly the same form, one of order 1 and one of three parts in a million. Why is the electron alone so light? "Maybe the mass has an imaginary component, and that is why it looks small" — we start from that suspicion. The verdict, up front: in its most naive form, no. An imaginary part makes the mass heavier. But the story does not end there. For neutrinos, masses cancelling against each other because of a phase is something that can genuinely happen, and experiments are looking for it. And on the charged-lepton side there is one relation left over that a phase does explain.
Episode 10 closed the series with this: only dimensionless invariants are entitled to answer. If it moves it is physics; if it does not, it is bookkeeping.
And yet, across ten episodes, there is one dimensionless quantity we never used.
A phase is dimensionless from the outset. A radian is not a unit.
So by the series' own decision procedure, the phase of a mass was always a candidate for the "physics" column. And yet we only ever talked about \(|m|\).
Written in full generality, a mass term actually has two components.
$$\mathcal{L}_{\rm mass}=-\bar\psi\,(m_1+i\,m_2\gamma_5)\,\psi$$\(m_1\) is the ordinary (scalar) mass and \(m_2\) the pseudoscalar one. Package them as a complex number \(m=m_1+im_2\) and the mass has the shape of a complex number from the start. Textbooks write \(m_2=0\) because you may choose it, not because you must.
So is \(m_2\) bookkeeping, or physics? Run it through the procedure.
First, let the star of this series have a go. As we saw in episode 7, the Weyl weight of a mass is
$$\tilde m=\Omega^{-1}m$$The crucial point is that \(\Omega\) is a real, positive function. That is not a choice, it is the definition — if \(\Omega\) were complex in \(\tilde g_{\mu\nu}=\Omega^2 g_{\mu\nu}\), the metric would stop being real and the signature (one time, three space) would break. So \(\Omega>0\), and therefore
A conformal transformation can move the magnitude of a mass as far as you like. It cannot lay a finger on the phase.
That is a strong result. In episode 4, turning the Weyl knob \(s\) rescaled masses by any factor you wanted — which is exactly why \(|m|\) was bookkeeping. The phase, by contrast, has the same value in every gauge. Put it in the series' table and the phase falls into the "physics" column.
Rewrite the complex number in polar form.
$$m=m_1+im_2=|m|\,e^{i\theta},\qquad \theta=\arctan\frac{m_2}{m_1}$$Now you can see that a mass has two independently turnable knobs: one for the magnitude and one for the phase.
| Knob | Transformation | Moves | Leaves alone | Residue that cannot be erased |
|---|---|---|---|---|
| Weyl \(\Omega\) | \(g\to\Omega^2g\) | \(|m|\) | phase \(\theta\), ratios | conformal anomaly (ep. 8) |
| Chiral \(\alpha\) | \(\psi\to e^{i\alpha\gamma_5/2}\psi\) | phase \(\theta\) | \(|m|\) | chiral anomaly (strong CP) |
The two are independent, and both are gauge-like. Even the structure matches: classically you may turn either freely, but quantise, and each leaves a trace. What episode 8 did for the magnitude happens once more on the phase side.
Apply the chiral rotation
$$-\bar\psi(m_1+im_2\gamma_5)\psi\ \xrightarrow{\ \psi\to e^{i\alpha\gamma_5/2}\psi\ }\ m_1+im_2\ \longrightarrow\ (m_1+im_2)\,e^{i\alpha}$$Choose \(\alpha=-\arctan(m_2/m_1)\) and the imaginary part is gone
$$m_{\rm phys}=|m|=\sqrt{m_1^2+m_2^2}$$Check it in the dispersion relation
$$E^2=p^2c^2+(m_1^2+m_2^2)c^4$$The imaginary part always enters as a positive contribution.
Adding an imaginary part always makes the physical mass heavier, never lighter. And the phase can be rotated away anyway, so it was bookkeeping to begin with. For a single fermion, the sign is backwards.
The other reading — "complex mass = the pole of an unstable particle", \(m\to m-i\Gamma/2\) — dies instantly for the electron too, because charge conservation makes the electron exactly stable. Borexino puts a lower bound of \(\tau>6.6\times10^{28}\) years on \(e\to\nu\gamma\); as a width that is
$$\Gamma_e<3.2\times10^{-52}\ \mathrm{eV}\qquad\Longrightarrow\qquad \frac{\Gamma_e}{2m_e}<3.1\times10^{-58}$$One of the most precisely measured zeros in all of physics. In this sense there is no imaginary part.
── Stop here and this is an episode that only says no. But it does not stop, because what we just killed was only the one-particle case.
The chiral knob comes one per species. \(n\) phases and \(n\) knobs, so you can erase them all — except that when the mass is a matrix, the knobs run short. If all you can do is rotate the overall phase, the remaining \(n-1\) phase differences survive.
And this is the second copy of what the series has been doing all along.
| Bookkeeping (moves with the gauge) | Physics (does not move) | Measured value | |
|---|---|---|---|
| Magnitude | individual \(m_i\) | ratios \(m_i/m_j\) | \(m_\mu/m_e=206.768\) |
| Phase | individual \(\theta_i\) | differences \(\theta_i-\theta_j\ \ (=\arg\det M)\) | \(\bar\theta<10^{-10}\) |
That bottom-right entry, \(\bar\theta=\theta_{\rm QCD}+\arg\det M_{\rm quark}\), has a name — the strong CP problem. The neutron's electric dipole moment has been measured, and the phase sits within \(10^{-10}\) of zero. Why it is that close is still unsolved.
The electron's phase has been measured directly too. A phase makes the charge distribution front-back asymmetric, showing up as an electric dipole moment. The JILA HfF⁺ experiment (2023) gives
Raw value
$$|d_e|<4.1\times10^{-30}\ e\cdot\mathrm{cm}$$Make it dimensionless (divide by the Compton wavelength)
$$\frac{|d_e|}{e\,\lambda_C}<1.1\times10^{-19}$$The centre of the electron's charge is displaced by less than \(10^{-19}\) of its own Compton wavelength.
Translated into a phase: if the phase lives at the electroweak scale and acts at tree level, \(\sin\varphi<2.5\times10^{-8}\); conversely, if you allow an \(O(1)\) phase, new physics must sit above \(\Lambda>1.6\times10^{3}\) TeV (125 TeV even with a one-loop suppression). The imaginary part is measurable. And measured, it is essentially zero.
Now the real subject. Rewrite the original question — why is the electron light — in dimensionless form.
$$y_e=\frac{\sqrt2\,m_e}{v}=2.935\times10^{-6},\qquad y_t=0.9919,\qquad \frac{y_e}{y_t}=2.96\times10^{-6}$$That is the whole mystery. The MeV and the GeV are stage scenery; only this number is left.
So take the square root of the masses. This is a relation Yoshio Koide found in 1982.
Koide's relation
$$Q=\frac{m_e+m_\mu+m_\tau}{\left(\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau}\right)^2}$$Substitute (\(m_e=0.51099895\), \(m_\mu=105.6583755\), \(m_\tau=1776.86\) MeV)
$$Q=\frac{1883.029}{2824.570}=0.66666051\qquad\text{vs}\qquad \frac23=0.66666667$$A relative discrepancy of \(9.2\times10^{-6}\) — comfortably inside the \(\pm0.12\) MeV experimental error on \(m_\tau\).
The relation only holds for pole masses, i.e. in the infrared. Masses run, and evaluating the same \(Q\) with the \(\overline{\rm MS}\) masses at \(M_Z\) (\(0.48657,\ 102.718,\ 1746.24\) MeV) gives
\(Q=0.6679286\) — the discrepancy from \(2/3\) swells to \(+1.9\times10^{-3}\), 205 times worse than for pole masses. Running the pole masses ourselves at one-loop QED gives \(+1.8\times10^{-3}\), in agreement. A fundamental relation ought to be clean in the ultraviolet; this one is clean only at low energy. Sumino's gauge model was built precisely to cancel that QED running. Read on with this weakness in mind.
The geometric restatement is even sharper. Compute the angle between the vector \((\sqrt{m_e},\sqrt{m_\mu},\sqrt{m_\tau})\) and the diagonal \((1,1,1)\):
$$\theta=44.99974^\circ\qquad(0.00026^\circ\text{ away from }45^\circ)$$And the condition "\(45^\circ\)" is exactly equivalent to the following form.
\(\sqrt{m}\) is the real part of a complex number. The three generations are three projections spaced \(120^\circ\) apart.
Why does \(Q=2/3\) come out automatically? Only two facts are needed: \(\sum_k\cos(\delta+2\pi k/3)=0\) and \(\sum_k\cos^2(\cdots)=3/2\). The numerator is \(M^2\sum(1+\sqrt2\cos)^2=M^2(3+0+3)=6M^2\), the denominator \(\left(M\sum(1+\sqrt2\cos)\right)^2=9M^2\). Hence \(Q=6/9=2/3\) — always, independently of \(\delta\) and of \(M\).
Solve for \(M\) and \(\delta\) from the measured masses.
| \(\sqrt{m}/M\) | Angle \(\theta_k\) | Where it sits | |
|---|---|---|---|
| \(\tau\) | 2.379423 | 12.7352° | near the crest of the cosine |
| \(\mu\) | 0.580226 | 107.2670° (\(\tau-120.0022^\circ\)) | partway down the slope |
| \(e\) | 0.040351 | 132.7323° (\(\tau+119.9972^\circ\)) | just short of the zero |
\(M=17.7156\,\sqrt{\rm MeV}\), \(\delta=12.7352^\circ\). The \(120^\circ\) spacing holds to three digits.
And here the reason the electron is light becomes a single line. \(1+\sqrt2\cos\theta\) vanishes at \(\theta=135^\circ\). The electron sits at \(132.73^\circ\).
The electron is \(2.27^\circ\) away from being exactly massless.
The sensitivity is \(d\ln m_e/d\delta=0.90\) per degree — shift \(\delta\) by \(1^\circ\) and the electron mass changes by a factor of 2.5. The tau stands on the crest (\(12.7^\circ\)); the electron stands on the lip of the valley. Same \(M\), same formula, differing only by \(120^\circ\).
"The electron is light because of a phase" — that is correct. Only, the phase in question is not the chiral \(\gamma_5\) phase but a rotation angle in \(\sqrt{m}\) space. The address was wrong; the instinct was right.
For charged leptons the phase was, in the end, a number inside an empirical formula. Move to neutrinos and the phase becomes impossible to remove, as a matter of theory. The reason is sharp.
First the order of magnitude. Oscillation data put the heaviest neutrino at \(m_3\gtrsim\sqrt{\Delta m^2_{31}}=0.0501\) eV; as a Yukawa that is
$$y_\nu=2.9\times10^{-13},\qquad \frac{y_\nu}{y_e}=9.8\times10^{-8}$$The electron already looked absurdly light, and the neutrino is another hundred-millionth below it. Writing a coupling that small by hand takes some nerve. Use the structure of a ratio instead.
Mass matrix (light component on the left, heavy right-handed one on the right)
$$M=\begin{pmatrix}0 & m_D\\ m_D & M_R\end{pmatrix},\qquad \det M=-m_D^2$$Diagonalise for \(M_R\gg m_D\)
$$\lambda_+\simeq M_R,\qquad \lambda_-\simeq-\frac{m_D^2}{M_R}$$The determinant is fixed, so making one side heavy necessarily makes the other light. A seesaw.
Here is the beautiful part in dimensionless terms. Rewrite \(m_\nu=m_D^2/M_R\):
The Dirac mass is the geometric mean of the neutrino mass and the heavy scale.
Put numbers in. With \(m_\nu=0.0501\) eV and \(M_R=5.95\times10^{14}\) GeV the geometric mean is
$$\sqrt{m_\nu M_R}=\sqrt{5.01\times10^{-11}\,\mathrm{GeV}\times5.95\times10^{14}\,\mathrm{GeV}}=172.7\ \mathrm{GeV}$$Exactly the top quark mass. Turn it round: if \(m_D\) is the electroweak scale, then \(M_R\sim10^{15}\) GeV — the grand unification tick mark. The lightest thing there is tells you about the heaviest scale there is. (Set \(M_R\) to the Planck mass instead and you get \(m_\nu=v^2/M_{\rm Pl}=5.0\times10^{-6}\) eV, far too small. Not Planck — GUT.)
Look at the seesaw again. The light eigenvalue is
$$\lambda_-\simeq-\frac{m_D^2}{M_R}\qquad\text{(negative)}$$A negative mass is \(|m|e^{i\pi}\) — a phase of \(\pi\). For a Dirac particle a chiral rotation absorbs the sign and that is the end of it. But if the neutrino is a Majorana particle (its own antiparticle), the mass term has the form \(\nu^T C\nu\), so rotating the field's phase moves the mass term's phase twice as fast, and there is not enough rotation freedom left. The phase can no longer be erased.
| Three-generation mixing matrix | Angles | Phases | Meaning |
|---|---|---|---|
| Dirac | 3 | 1 | the CP phase \(\delta_{\rm CP}\) only |
| Majorana | 3 | 3 | \(\delta_{\rm CP}\) plus two Majorana phases |
In general, for \(n\) generations Dirac gives \((n-1)(n-2)/2\) phases and Majorana \(n(n-1)/2\). The difference, \(n-1\), is the count of phases that survive.
The Dirac phase is being measured, and the dimensionless invariant used for it is the Jarlskog invariant — individual phases move under rephasing (bookkeeping), but this one does not.
$$J_{\rm PMNS}=0.0330\,\sin\delta_{\rm CP}\approx-0.026\qquad\text{vs}\qquad J_{\rm CKM}=3.08\times10^{-5}$$The leptonic CP phase is about 845 times more effective than the quark one. (Though \(\delta_{\rm CP}\) is still moving experimentally, wobbling somewhere around \(200^\circ\)–\(250^\circ\).)
This is the crux. How does a Majorana phase show up in an observation? The rate of neutrinoless double beta decay (\(0\nu\beta\beta\)) is set by
Three masses added in the complex plane. If the phases do not line up, they cancel.
Which is precisely the shape of the original suspicion — mass looks small because of a phase. And this is not a thought experiment; it can actually happen. For normal ordering (\(m_1<m_2<m_3\)) we scan \(m_1\) and record the range of \(m_{\beta\beta}\) as the phases vary.
| \(m_1\) [meV] | min \(m_{\beta\beta}\) [meV] | max [meV] | Effect of misaligned phases |
|---|---|---|---|
| 1.0 | 0.78 | 4.36 | suppression by more than 5× |
| 2.0 | 0.15 | 5.09 | more than 30× |
| 3.0 | 0.01 | 5.85 | essentially complete cancellation |
| 6.0 | 0.02 | 8.31 | essentially complete cancellation |
| 10.0 | 1.78 | 11.85 | 7× |
Around \(m_1\approx2\)–\(6\) meV, all three neutrinos have mass and yet \(m_{\beta\beta}\) goes to zero. This valley is called the normal-ordering funnel, and for experimentalists it is a nightmare region — the neutrino can be Majorana and the decay still invisible.
The best current bound is KamLAND-Zen's \(m_{\beta\beta}<28\)–\(122\) meV (the spread is nuclear-matrix-element uncertainty). Still far above the funnel.
Finally, extrapolate the relation of section 06 to neutrinos. From here on this is an empirical rule pushed well past its evidence, so read it in that spirit. But what comes out is interesting.
Oscillation experiments measure squared differences, not masses (\(\Delta m^2_{21}=7.41\times10^{-5}\), \(\Delta m^2_{31}=2.51\times10^{-3}\ \mathrm{eV}^2\)). So there is exactly one unknown — the lightest mass. Impose \(Q=2/3\) and solve.
Take every \(\sqrt{m}\) positive (normal ordering)
$$Q\ \text{only ranges over}\ 0.582\to0.333\qquad\Longrightarrow\qquad \boxed{\text{no solution}}$$Flip the sign of \(\sqrt{m_1}\) alone
$$m=(0.360,\ 8.616,\ 50.101)\ \mathrm{meV},\qquad \sum m_\nu=59.08\ \mathrm{meV}$$Inverted ordering behaves the same way — all-positive only ranges over \(0.498\to0.333\) and never reaches \(2/3\), while flipping one sign does give a solution (\(\sum m_\nu=103.9\) meV).
And here is the most satisfying point of the whole episode.
That minus sign is a Majorana phase of \(\pi\).
Charged leptons are not allowed it — they are Dirac particles, so the phase is rotated away (section 04). Only neutrinos cannot erase it. And only for neutrinos does the relation close with it.
So section 07's "a Majorana phase cannot be erased" and section 09's "no solution unless you flip a sign" are pointing at the same spot. For particles that forbid a phase, Koide's formula holds as it stands; for particles that permit one, it holds only with the phase included.
Measure the phase of that solution and you get \(\delta_\nu=27.38^\circ\). Its difference from the charged-lepton value \(\delta_\ell=12.74^\circ\) is \(0.2558\) rad — only 2.3% away from \(\pi/12=0.2618\) rad. Brannen (2006) imposed \(\delta_\nu=\delta_\ell+\pi/12\) exactly and predicted the masses. Checking that version:
Fix the scale using \(\Delta m^2_{31}\) alone
$$m=(0.379,\ 8.808,\ 50.101)\ \mathrm{meV},\qquad \sum m_\nu=59.29\ \mathrm{meV}$$Then \(\Delta m^2_{21}\) becomes a prediction. Compare with the measurement
$$\Delta m^2_{21}\big|_{\text{pred}}=7.74\times10^{-5}\ \mathrm{eV}^2\qquad\text{vs}\qquad (7.41\pm0.21)\times10^{-5}$$Off by +4.5%, or \(1.6\sigma\). ── Neither a confirmation nor an exclusion.
Either version, and sweeping the oscillation parameters from one end of their errors to the other, lands in the same place.
The cosmological bound is \(\sum m_\nu<120\) meV (Planck + BAO). Inside it — but the room is gone. Recent galaxy surveys keep pushing that ceiling down, which puts this prediction firmly on the side of things that will be settled soon.
While we are here, we can also ask what this solution says about \(0\nu\beta\beta\). The CP parity of \(m_1\) is negative (phase \(\pi\)), so
$$m_{\beta\beta}=1.20\ \mathrm{meV}\qquad(\text{3.90 meV if the phases were aligned})$$The phase suppresses \(0\nu\beta\beta\) by a factor of 3.2. And 1.2 meV is out of reach even for the next generation of experiments (target \(\sim10\) meV). ── As predictions go, this one points in a slightly depressing direction.
Following episode 10's method, here are today's items in two columns.
| Sense of "imaginary part of the mass" | Verdict | Value |
|---|---|---|
| ① Instability \(m-i\Gamma/2\) | physics, but zero for the electron | \(\Gamma/2m<3.1\times10^{-58}\) |
| ② Chiral phase (single field) | bookkeeping; rotated away — and it makes \(|m|\) larger | \(\sqrt{m_1^2+m_2^2}\ge|m_1|\) |
| ③ Phase differences \(\arg\det M\) | physics; cannot be erased; measured | \(\bar\theta<10^{-10}\), \(J\), \(d_e\) |
| ④ Phase \(\delta\) in \(\sqrt{m}\) space | empirical — but it explains the lightness and yields a prediction | \(\delta_\ell=12.74^\circ\), \(\sum m_\nu\simeq59\) meV |
And where did the conformal transformation do its work? In the one line of section 02 — \(\Omega\) is a real positive function, so a Weyl transformation does not touch the phase. Ten episodes of stripping away the bookkeeping on the magnitude side left the phase standing there as what remains. One more row for episode 10's table:
The Koide relation used in sections 06 and 09 is an empirical formula with no derivation. Why \(\sqrt{m}\)? Why \(2/3\)? Nobody can derive it. Nothing stronger can be said than "it sits inside the \(\pm0.12\) MeV error on \(m_\tau\)". The "\(45^\circ\)" and the "\(120^\circ\) spacing" are restatements of \(Q=2/3\), not independent evidence (writing three masses with two parameters \(M,\delta\) is one constraint, and that constraint is \(Q=2/3\)). And as flagged in section 06, it holds only for pole masses: run to \(M_Z\) and the discrepancy grows by a factor of 205. A fundamental relation should hold in the ultraviolet, and this one does the opposite — that is its most serious weakness.
Also, the phase of section 06 and the phase of section 03 are different objects: the former is a rotation angle in \(\sqrt{m}\) space, the latter the chiral \(\gamma_5\) phase. They must not be joined together just because both are called "phase". Models that do connect them (Sumino's gauge model, for instance) exist but are not established. Identifying the Majorana phase \(\pi\) with the sign of \(\sqrt{m}\) in section 09 is this article's reading, not a standard claim.
Section 09 is an extrapolation of an extrapolation. Whether neutrinos are Majorana is unconfirmed (\(0\nu\beta\beta\) has not been seen). \(\delta_{\rm CP}\) is not pinned down either. \(\sum m_\nu\simeq59\) meV is a prediction conditional on "if Koide's formula also holds for neutrinos", not a prediction of established theory. It also inherits the running problem above, so at which scale the relation should be imposed on neutrino masses is undetermined (here they were treated as not running).
The bound \(\Lambda>1.6\times10^3\) TeV in section 05 comes from dimensional analysis alone, \(d_e\sim e\,m_e/\Lambda^2\); the coefficient is strongly model-dependent. The table in section 08 uses \(\theta_{12}=33.41^\circ\), \(\theta_{13}=8.54^\circ\) and current \(\Delta m^2\) values, and moves within their errors.
"The electron is light because its mass has an imaginary component" — we chased that all the way. In its naive form the answer is no. A chiral rotation turns \(m_1+im_2\) into \(\sqrt{m_1^2+m_2^2}\), so the phase is bookkeeping; and the imaginary part makes the mass heavier. The instability reading is zero as well, \(\Gamma/2m<3.1\times10^{-58}\). For a single fermion, the sign runs backwards.
But the one line of section 02 — \(\Omega\) is real and positive, so a conformal transformation does not touch the phase — keeps the phase alive to the end. There were two knobs: Weyl moves the magnitude, chiral moves the phase, and each has its own quantum breakdown (the conformal anomaly and strong CP). Exactly parallel to "individual \(m_i\) are bookkeeping, ratios are physics" on the magnitude side, on the phase side we get "individual \(\theta_i\) are bookkeeping, the difference \(\arg\det M\) is physics". A second copy of the work of the first ten episodes.
And the lightness itself. Take \(\sqrt{m}\) and the charged leptons land dead on \(45^\circ\) (\(44.99974^\circ\)), the three masses writable as projections \(120^\circ\) apart. The electron stands \(2.27^\circ\) short of a zero of the cosine — which is why it is light. Move \(\delta\) by \(1^\circ\) and the mass changes by 2.5×. In the sense that a phase sets the magnitude of the mass, the suspicion was on target. The address was \(\sqrt{m}\) space, not \(\gamma_5\).
Move to neutrinos and the phase becomes theoretically unerasable (Majorana). The lightness is not a small coupling but a geometric mean, \(m_D=\sqrt{m_\nu M_R}\), and indeed \(\sqrt{0.05\,\mathrm{eV}\times5.95\times10^{14}\,\mathrm{GeV}}=172.7\) GeV — the top mass. In \(0\nu\beta\beta\) there is a real region where masses cancel to zero because of a phase: the shape of the original suspicion, turned into an experimental target. Apply Koide to neutrinos and there is no solution unless one sign is flipped — and that sign is a Majorana phase of \(\pi\). The prediction that falls out is \(\sum m_\nu\simeq59\) meV. Inside the cosmological bound of 120 meV, with no room to spare.
Print / save as PDF: ⌘+P (Ctrl+P on Windows). On screen, the slider in figure 1 turns the phase δ and figure 2 turns the Majorana phases. "Show the answer" opens each solution.