Particle–Antiparticle Conjugation; Time Reversal and CPT

The construction that produced parity survives one more repetition and then fails: time reversal is antiunitary, so it never becomes a label a particle can carry.

🎯 Why this matters

Every experimental statement about time reversal is therefore indirect. What gets measured is a rate, a frequency ratio or an equality that CPT protects, and T is inferred from it rather than read off a state the way parity is.

§3.2 built parity: a discrete operator, squaring to one, whose eigenvalue labels a particle. This page runs the same construction twice more and then stops. C — swapping every particle for its antiparticle — behaves almost exactly like P and gives a second multiplicative quantum number. T looks like it should give a third and does not, for a reason worth understanding. What T gives instead is bigger than a quantum number: the CPT theorem, the one symmetry you cannot break without breaking special relativity itself.

3.3 Particle–antiparticle conjugation

The particle–antiparticle conjugation operator turns a particle into its antiparticle and touches nothing else — not position, not time, not spin.

\htmlClass{t-C}{C}\,\bigl|\,\htmlClass{t-p}{\mathbf p},\htmlClass{t-s}{\mathbf s},\htmlClass{t-Q}{\{Q\}}\,\bigr\rangle = \bigl|\,\htmlClass{t-p}{\mathbf p},\htmlClass{t-s}{\mathbf s},\htmlClass{t-Qm}{\{-Q\}}\,\bigr\rangle

(3.9)

Bettini p. 110. Everything kinematic survives; every charge flips.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — C flips every additive quantum number, not just the charge

The name “charge conjugation”, which everyone uses including this site, is slightly wrong and the book says so. C does not flip the electric charge — it flips every additive quantum number simultaneously. Electric charge, baryon number, the three lepton flavours, strangeness, charm: all of them, in one stroke.

The test for “is this state an eigenstate of C?” is therefore not “is it electrically neutral?” but “is every one of its charges zero?” Those are different questions, and the neutron is where they come apart: $Q = 0$ but $\mathcal{B} = 1$ , so C sends the neutron to the antineutron, which is a different particle. The neutron has no C eigenvalue.

⚙️ Engineer’s bridge — C is a sign flip on the whole charge word

Think of a particle’s identity as a fixed-width record of signed integer fields: {Q, 𝓑, L_e, L_μ, L_τ, S, C, B̃, T}. Every field is additive — a two-particle state’s record is the element-wise sum — and every conservation law in §3.1 is the statement that some field is invariant across a reaction.

C is then a single operation: negate the entire record. Two consequences follow immediately, and neither needs any physics:

C is an involution. Negating twice is the identity, so $C^2 = 1$ and the eigenvalues are $\pm 1$ . Same argument, same conclusion as parity.
The only fixed point of negation is zero. A state can be an eigenstate of C only if its record is all zeros — the “completely neutral” condition. Any non-zero field, of any kind, and the state is mapped somewhere else entirely.

Where the analogy breaks: the record is not the whole state. C leaves $\mathbf p$ and $\mathbf s$ alone, so for a composite state the eigenvalue is not simply the product of the constituents’ — the positions and spins get exchanged as well, and that exchange contributes the $(-1)^{l+s}$ derived below. The record model gets you the eigenvalue’s existence; it does not get you its value.

Where it breaks: and the gap between existence and value is the whole difficulty. A record with a sign flag has one bit; a C eigenvalue is only defined for a self-conjugate state, so most particles have no C at all — the π⁺ has none, only the π⁰ and genuinely neutral systems do. Bookkeeping intuition suggests every record carries every field; here the field simply does not exist for most rows, and the ones that carry it get their value from spatial exchange rather than from anything you could store.

The three operators act on the same short list of physical quantities, and the widget from §3.2 tabulates all three. Its C column is the subject of this page: notice that C acts on nothing geometric at all, which is why every entry in it is either “unchanged” or “flips because the quantity is built out of a charge”.

How P, C and T act — and where the four-way classification comes from

quantity	P	C	T	it is a…
ttime	+	+	−	scalar
rposition	−	+	+	vector
plinear momentum	−	+	−	vector
Lorbital angular momentum	+	+	−	axial vector
sspin	+	+	−	axial vector
Eenergy	+	+	+	scalar
Qelectric charge	+	−	+	scalar
jcurrent density	−	−	−	vector
𝐄electric field	−	−	+	vector
𝐁magnetic field	+	−	−	axial vector
σ·p̂helicity	−	+	+	pseudoscalar
L·sspin–orbit term	+	+	+	scalar
𝐄·𝐁the E·B invariant	−	+	−	pseudoscalar
s·𝐄electric dipole moment term	−	−	−	pseudoscalar

σ·p̂ — helicity · built from axial vector · polar vector

The most important row on the list. An axial vector dotted into a polar one is a PSEUDOscalar: rotationally invariant, but it changes sign in a mirror. A left-handed particle looks right-handed to its own reflection, which is why "the weak interaction is left-handed" and "the weak interaction violates parity" are the same sentence.

Compose a term — the signs multiply

s·pP −C +T +pseudoscalar

P-odd. A matrix element containing this term changes sign in a mirror, so its interference with any P-even term produces an observable that must vanish if parity is conserved. Measuring that observable to be zero is how the limits in this chapter were set; measuring it to be large is how parity violation was discovered.

The C and T columns belong to §3.3 and §3.4; they are shown here so the table is met once and only once. Nothing in this widget is tabulated data — every sign is the product of the signs of the pieces the quantity is built from, which is what "P, C and T are multiplicative" means.

Which particles have a C eigenvalue at all

Only the completely neutral ones — and there are very few.

Every C eigenstate in the book so far, and the two traps

particle	C	$J^{PC}$	how it is known
$\gamma$		1^{--}	The field A is sourced by charges, so flipping every charge flips A. Eq. (3.10).
$\pi^0$	+1	0^{-+}	It decays to γγ electromagnetically, and the electromagnetic interaction conserves C. So C(π⁰) = (−1)² = +1. Eq. (3.12).
$\eta$	+1	0^{-+}	Same argument, same answer — η → γγ. Eq. (3.14).
$\eta'$	+1	0^{-+}	Also decays to two photons. The book flags the η′ here and returns to it in Chapter 4.
$\pi^\pm$	—	—
$n$	—	—	The trap. Electrically neutral, but 𝓑 = 1, so C(n) = n̄ — a different particle with a different record.

Two more C eigenstates appear later and are worth keeping in mind: the ρ⁰ and ω (both 1−−, both reachable in e+e− annihilation because the photon has those very quantum numbers) and the J/ψ. Every one of them is a q q̄ state with the quark and the antiquark of the same flavour — which is the only way a hadron can have all its charge fields zero.

The photon, and the counting rule that follows

\htmlClass{t-C}{C}\,|\gamma\rangle = -|\gamma\rangle \qquad\Longrightarrow\qquad \htmlClass{t-C}{C}\,|\htmlClass{t-n}{n}\gamma\rangle = (-1)^{\htmlClass{t-n}{n}}\,|\htmlClass{t-n}{n}\gamma\rangle

(3.10, 3.11)

Bettini p. 110. The right-hand equation is the left one plus the fact that C is multiplicative.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — C(γ) = −1 in one line, and the line is classical

The argument for $C(\gamma) = -1$ is one line and entirely classical. The macroscopic vector potential $\mathbf A$ is produced by moving charges. Replace every source by its antiparticle and every charge changes sign, so $\mathbf A$ changes sign — and the photon is the quantum of $\mathbf A$ . Hence $C = -1$ .

The consequence is a selection rule you can count on your fingers: a C-conserving decay of a state with $C = +1$ must produce an even number of photons, and a state with $C = -1$ an odd number. That is why $\pi^0$ → γγ and not γγγ; why the observation of a three-photon π⁰ decay would be a discovery rather than a footnote; and why positronium comes in two varieties with lifetimes a thousand times apart.

🔢 Worked example — positronium’s two lifetimes, from one eigenvalue

Positronium is an $e^+e^-$ atom, so it is exactly the fermion–antifermion system of Example 3.1. Its ground state comes in two flavours, and $C = (-1)^{l+s}$ tells them apart with $l = 0$ :

para-positronium, $^1S_0$ ( $s = 0$ ): $C = (-1)^0 = +1$ → an even number of photons → 2γ.
ortho-positronium, $^3S_1$ ( $s = 1$ ): $C = (-1)^1 = -1$ → an odd number → 3γ, because one photon cannot conserve momentum.

Two states of the same atom, differing only in whether the spins are parallel, forced into different final states by a sign. And the cost of the extra photon is visible in the measured lifetimes: each additional vertex costs a factor $\alpha \approx 1/137$ , plus the phase-space penalty of sharing the energy three ways.

	term	$C$	photons	measured $\tau$
para-Ps	$^1S_0$	$+1$	2	0.1244 ns
ortho-Ps	$^3S_1$	$-1$	3	142.05 ns

The ratio is 1142 — close to the naive $1/\alpha$ times a phase-space factor of order ten, which is exactly the size an extra electromagnetic vertex should cost. The book returns to positronium in Problem 3.18; §5.7 is where counting vertices becomes systematic.

Testing C: look for what should never happen

The book gives two limits, both on decays that C forbids outright.

\frac{\Gamma(\htmlClass{t-pi}{\pi^0} \to 3\gamma)}{\Gamma_{\text{tot}}} \le \htmlClass{t-lim}{3.1\times 10^{-8}} \qquad \frac{\Gamma(\htmlClass{t-eta}{\eta} \to 3\gamma)}{\Gamma_{\text{tot}}} \le \htmlClass{t-lim2}{4\times 10^{-5}}

(3.15)

Bettini p. 111 — McDonough et al. (1988) for the π⁰, Nefkens et al. (2005) for the η. Both are ratios of a partial width to the total, i.e. branching ratios.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — a conservation law is never proved, only bounded

A conservation law is never proved; it is bounded. Nobody can show that C is exactly conserved by the electromagnetic interaction. What an experiment does is produce an enormous number of π⁰ mesons, look for the one decay the law forbids, find none, and quote how many it looked at.

So (3.15) is not really a statement about pions. It is a statement about how hard we have looked, and the honest way to read any such limit is to invert it into a count: how many decays had to be examined for that number to be meaningful?

🔢 Worked example — turning the limits into widths, and into a count

The $\pi^0$ lives $8.43 \times 10^{-17}$ s, so its total width follows from $\Gamma\tau = \hbar$ :

\Gamma_{\text{tot}}(\pi^0) = \frac{\hbar}{\tau} = \frac{6.582\times 10^{-22}\ \text{MeV s}}{8.43\times 10^{-17}\ \text{s}} = 7.81\ \text{eV},

and the limit on the C-violating channel is $3.1 \times 10^{-8}$ of that: $\Gamma(\pi^0 \to 3\gamma) \le 2.4 \times 10^{-7}$ eV. For the $\eta$ , whose width is quoted directly as 1.31 keV, the weaker limit still gives $\Gamma(\eta \to 3\gamma) \le 0.052$ eV.

And the count. Seeing zero events out of $N$ sets a 95 % upper limit of about $3/N$ (the Poisson rule of three). To reach $3.1 \times 10^{-8}$ you therefore need to have watched roughly

N \gtrsim \frac{3}{3.1\times 10^{-8}} \approx 1 \times 10^{8}

π⁰ decays with an apparatus able to recognise a three-photon final state and reject the overwhelming two-photon one. That is the actual content of the number: a hundred million decays, and nothing.

Reproduce it

hbar = 6.582119569e-22                      # MeV s
tau0, lim0 = 8.43e-17, 3.1e-8               # pi0 lifetime (s), limit
G0 = hbar / tau0 * 1e6                      # -> eV
print(f"pi0: tau = {tau0:.3g} s -> Gamma_tot = {G0:.3f} eV")
print(f"     limit {lim0:.1e} -> Gamma(pi0 -> 3 gamma) <= {lim0*G0:.3e} eV")

Ge, lime = 1.31e3, 4e-5                     # eta width in eV, limit
print(f"eta: Gamma_tot = {Ge:.0f} eV (quoted as a width, not a lifetime)")
print(f"     limit {lime:.1e} -> Gamma(eta -> 3 gamma) <= {lime*Ge:.3e} eV")

print(f"Poisson rule of three: {lim0:.1e} needs N >= {3/lim0:.1e} decays watched")
print(f"C of the final states: C(2 gamma) = {(-1)**2:+d}, C(3 gamma) = {(-1)**3:+d}")

prints

pi0: tau = 8.43e-17 s -> Gamma_tot = 7.808 eV
   limit 3.1e-08 -> Gamma(pi0 -> 3 gamma) <= 2.420e-07 eV
eta: Gamma_tot = 1310 eV (quoted as a width, not a lifetime)
   limit 4.0e-05 -> Gamma(eta -> 3 gamma) <= 5.240e-02 eV
Poisson rule of three: 3.1e-08 needs N >= 9.7e+07 decays watched
C of the final states: C(2 gamma) = +1, C(3 gamma) = -1

⚠️ Natural units — where the ħ went

$\Gamma\tau = \hbar$ is the only place on this page where the natural-units convention hides a constant. In $\hbar = c = 1$ units the relation reads $\Gamma = 1/\tau$ and a width is an inverse time; to get a number in eV you must put $\hbar = 6.582 \times 10^{-16}$ eV s back in.

It is the time–bandwidth relation and nothing more: a state that lives for a time $\tau$ has an energy uncertain by $\hbar/\tau$ , exactly as a pulse of duration $\tau$ occupies a bandwidth $\sim 1/\tau$ . The particle tables exploit this by quoting whichever of the two is easier to measure — a lifetime for the π⁰, a width for the η — and the particle explorer always shows both.

A particle and its antiparticle, together

A pair is where C stops being bookkeeping. Because C swaps the two constituents, its eigenvalue picks up whatever the exchange of positions and spins costs — and that is a piece of physics, not a convention.

🪜 Why C(pair) = (−1)^(l+s)

Step 1 of 6 — C on a pair is an exchange

Why you may do this: Applying C to |particle at r₁, spin up; antiparticle at r₂, spin down⟩ turns the first into an antiparticle and the second into a particle. Relabelling them back into the standard order is exactly the exchange operator. So the eigenvalue is whatever exchanging the two constituents costs.

This is the whole idea. Everything below is bookkeeping on "what does exchange cost?"

Bettini pp. 111–112, Eqs. (3.16)–(3.18). The three cases look different and end identically; the reason is in step 5.

Aside — $l + s$ is a plain sum, not a composition

The book stops to warn about this and it is worth repeating. In $(-1)^{l+s}$ the quantity $l + s$ is arithmetic addition of two numbers, not the angular-momentum composition $\mathbf l \oplus \mathbf s$ . It is not $J$ , and it is not any of the allowed values of $J$ .

For $^3P_2$ , for instance: $l = 1$ , $s = 1$ , so $l + s = 2$ and $C = +1$ — while $J$ ranges over $0, 1, 2$ and takes the value 2 in this particular state. The two happen to coincide here and do not in $^3P_0$ , where $l + s = 2$ still but $J = 0$ . Use the exponent, not the total angular momentum.

Table 3.1, and the holes in it

Table 3.1, computed · J^PC of a particle–antiparticle pair

orbital ltotal spin s

term	J	P	C	J^PC
¹S₀	0	−	+	0⁻⁺

P = (−1)^(l+1) from Eq. (3.6) · C = (−1)^(l+s) from Eq. (3.18). Set l = 0 and l = 1 with both spins to rebuild Table 3.1 exactly.

💡 What this really says — six entries from two formulas, and the formulas are the content

The six entries the book prints — $^1S_0 = 0^{-+}$ , $^3S_1 = 1^{--}$ , $^1P_1 = 1^{+-}$ , $^3P_0 = 0^{++}$ , $^3P_1 = 1^{++}$ , $^3P_2 = 2^{++}$ — are the easy part, and the widget rebuilds them from the two formulas rather than quoting them.

The interesting content is the lattice underneath, because it shows what is not there. Set the mode to $f\bar f$ and read the hatched cells: $0^{+-}$ , $0^{--}$ , $1^{-+}$ — the three the book names — and then $2^{+-}$ and $3^{-+}$ , which it does not, because the pattern continues forever. A meson with any of those quantum numbers cannot be a quark and an antiquark, whatever its mass. That is what exotic means, and hunting for such states is a live experimental programme that §4.5 picks up.

The structure behind the holes is simple once you see it. Spin-singlet states ( $s = 0$ ) have $J = l$ and $C = -P$ , so they give $J^{-+}$ for even $J$ and $J^{+-}$ for odd $J$ — and nothing else. Spin-triplet states ( $s = 1$ ) always have $C = +P$ . So a combination with $C = -P$ and the wrong $J$ parity has no source at all.

⚙️ Engineer’s bridge — the exotic set is an unreachable state in a type system

You have met this shape of argument. A set of construction rules generates a set of reachable values; anything outside that set is not “rare”, it is not expressible. $J^{PC} = 1^{-+}$ is unreachable from $f\bar f$ in the same sense that a type signature can make an invalid state unrepresentable, or a finite state machine can have states no input sequence reaches.

The engineering payoff is the same too: observing an unreachable value is not noise, it is proof that your model of the system is incomplete. A meson at $1^{-+}$ does not mean the formulas are wrong — they follow from angular momentum and Fermi statistics. It means the object is not the two-body system you assumed: a hybrid with an excited gluon field, a four-quark state, a molecule of two mesons. The rules stay; the inventory of constituents grows.

Where the analogy breaks: a type system rejects at compile time, whereas nature just does not populate the state. Nothing forbids you from writing down the label $1^{-+}$ — the book does it — and searching for something that carries it.

Where it breaks: “unreachable state” suggests the label is meaningless, and it is not. $1^{-+}$ cannot be built from a $q\bar q$ pair, but it can be built from other things — a hybrid with an excited gluon field, or a four-quark state — so the exotic set is unreachable by one construction, not forbidden. Candidates ( $\pi_1$ (1600), and more recently $\eta_1$ (1855)) are exactly why anyone searches there. An unreachable region in a type system means no valid program; here it means no ordinary meson, which is a much weaker and much more interesting statement.

🔢 Worked example — Example 3.2, the C eigenvalues of positronium

Problem. Find the eigenvalues of C for a spin-½ particle and its antiparticle in an $S$ wave and in a $P$ wave.

Setup. $C = (-1)^{l+s}$ with $l = 0$ ( $S$ ) or $l = 1$ ( $P$ ), and $s = 0$ (singlet) or $s = 1$ (triplet). No other input is needed — $J$ does not appear in the formula at all, which is why every state within one $^{2s+1}L$ multiplet shares a C.

Reproduce it

sign = lambda x: '+' if x > 0 else '-'
L, rows, reach = 'SPDF', [], set()
for l in (0, 1):
    for s in (0, 1):
        Pp, Cc = (-1)**(l + 1), (-1)**(l + s)          # Eqs. (3.6) and (3.18)
        for J in range(abs(l - s), l + s + 1):
            note = ' s=1/2+1/2 -> s=0' if (l, s) == (0, 0) else ' s=%d             ' % s
            rows.append(f"{2*s+1}{L[l]}{J}   l={l}{note} J={J}   "
                        f"P={sign(Pp)}  C={sign(Cc)}   J^PC = {J}{sign(Pp)}{sign(Cc)}")
print('\n'.join(rows))

for l in range(5):                                      # now sweep for the holes
    for s in (0, 1):
        Pp, Cc = (-1)**(l + 1), (-1)**(l + s)
        reach |= {(J, Pp, Cc) for J in range(abs(l - s), l + s + 1)}
holes = [(J, p, c) for J in range(4) for p in (1, -1) for c in (1, -1) if (J, p, c) not in reach]
print("unreachable with J <= 3: ", ", ".join(f"{J}{sign(p)}{sign(c)}" for J, p, c in holes))

prints

1S0   l=0 s=1/2+1/2 -> s=0 J=0   P=-  C=+   J^PC = 0-+
3S1   l=0 s=1              J=1   P=-  C=-   J^PC = 1--
1P1   l=1 s=0              J=1   P=+  C=-   J^PC = 1+-
3P0   l=1 s=1              J=0   P=+  C=+   J^PC = 0++
3P1   l=1 s=1              J=1   P=+  C=+   J^PC = 1++
3P2   l=1 s=1              J=2   P=+  C=+   J^PC = 2++
unreachable with J <= 3:  0+-, 0--, 1-+, 2+-, 3-+

Answer. The singlets ( $s = 0$ ) have $C = +1$ in the $S$ wave and $C = -1$ in the $P$ wave; the triplets ( $s = 1$ ) have $C = -1$ in the $S$ wave and $C = +1$ in the $P$ wave. Attaching these to the parities of Example 3.1 gives Table 3.1 exactly.

Sanity check. $C$ depends on $l + s$ , $P$ on $l$ alone. So the two quantities agree for triplets and disagree for singlets — which is precisely why all four $P$ -wave triplet states are $J^{++}$ while $^1P_1$ stands apart at $1^{+-}$ .

Question 3.1 — the same table for Majorana fermions

A Majorana fermion is its own antiparticle (§2.9), and the book asks for Table 3.1 rebuilt for a pair of them. Switch the widget above to the third mode and two things change at once.

C is always +1. For a Dirac pair, C had to swap the two constituents, and the exchange cost is where $(-1)^{l+s}$ came from. For two identical Majorana fermions, C maps each particle to itself — nothing is exchanged — so the eigenvalue is just the product of the two intrinsic ones, $(\pm 1)^2 = +1$ , independently of $l$ and $s$ . This is the book’s statement, and it is the same argument that gives $C(n\gamma) = (-1)^n$ for photons, which are also their own antiparticles.

And half the states cease to exist. This part the book does not spell out here, though it uses the identical argument in §3.5: two identical fermions must have a totally antisymmetric wave function. Space contributes $(-1)^l$ and spin $(-1)^{s+1}$ , so their product is $-1$ only when $l + s$ is even. $^3S_1$ and $^1P_1$ are therefore not “states with $C = +1$ ” — they are not states at all. What survives of Table 3.1 is

^1S_0 = 0^{-+}, \qquad ^3P_0 = 0^{++}, \qquad ^3P_1 = 1^{++}, \qquad ^3P_2 = 2^{++} .

Four states instead of six, every one of them $C = +1$ . The same counting rule reappears immediately in §3.5, where it leaves $\pi^-d \to nn$ with exactly one possible final state, and again in §10.7, where whether the neutrino is a Majorana particle is still an open experimental question.

3.4 Time reversal and CPT

The time reversal operator inverts $t$ and leaves the space coordinates alone. The book spends half a page on it, and the brevity is the message: T gives no quantum number. Understanding why is more useful than any table.

📐 Physics you need first — unitary, antiunitary, and why only one of them labels a particle

Every symmetry in quantum mechanics is represented by an operator that preserves transition probabilities, $|\langle \psi | \phi \rangle|^2$ . Wigner’s theorem says there are exactly two ways to do that:

unitary — $\langle U\psi | U\phi\rangle = \langle \psi|\phi\rangle$ ;
antiunitary — $\langle A\psi | A\phi\rangle = \langle \psi|\phi\rangle^*$ , i.e. the inner product is preserved up to complex conjugation.

P and C are unitary. T must be antiunitary, and the reason is one line of the Schrödinger equation. Under $t \to -t$ ,

i\hbar\frac{\partial \psi}{\partial t} = H\psi \qquad\longrightarrow\qquad -i\hbar\frac{\partial \psi}{\partial t} = H\psi ,

which is a different equation — unless you also conjugate the $i$ . So T must carry a complex conjugation inside it, and that makes it antiunitary.

And an antiunitary operator has no useful eigenvalues. Multiply a state by a phase, $|\psi\rangle \to e^{i\alpha}|\psi\rangle$ , and an eigenvalue of a unitary operator is unchanged, while an antiunitary one picks up $e^{-2i\alpha}$ — you can tune it to anything you like. An “eigenvalue” you can set by choosing a phase convention is not a property of the particle, so there is no $T$ quantum number to conserve, no $T$ -parity to look up in a table. That is what the book means by ” $T$ does not transform as an observable under unitary transformations”.

⚙️ Engineer’s bridge — time reversal is complex conjugation, and you already knew that

The complex conjugation in T is not a technicality; it is the same fact you use whenever you handle a real-valued signal.

For a real impulse response $h(t)$ , reversing time in the time domain is conjugation in the frequency domain: $h(-t) \leftrightarrow H^*(\omega)$ . A minimum-phase filter and its time-reverse have the same magnitude response and opposite phase. Running a system backwards means conjugating its transfer function — which is exactly the operation that turns $i\hbar\,\partial_t\psi = H\psi$ back into itself.

Two consequences carry straight over:

Magnitudes survive, phases do not. $|H|$ is invariant under time reversal and $\arg H$ flips. In quantum mechanics that is why T-violation shows up in interference between two amplitudes and never in a single rate — the same reason a phase error is invisible in a power spectrum.
A real system is time-reversal symmetric until something breaks it. In circuits the breaker is dissipation; in particle physics it is a complex phase in the mixing matrix, which is the entire subject of Chapter 8.

Where it breaks: conjugating $H(\omega)$ is a passive relabelling, whereas T also reverses momenta and spins. Position stays, motion reverses — a film run backwards, not a mirror.

Where it breaks: complex conjugation is an antiunitary operation, and that is not a detail. Every other symmetry in this chapter is a unitary operator with eigenvalues you can measure; T is not, so there is no conserved quantum number associated with it and no “T-parity” to tabulate. That is why T violation has to be looked for in rate asymmetries and in the phase of ε rather than in a forbidden decay, and why CPT — which is a theorem rather than an observation — does the work T alone cannot.

The CPT theorem

What T does provide, in combination, is the strongest statement in this chapter.

💡 What this really says — Lüders’ theorem

If a theory of interacting fields is invariant under the proper Lorentz group, it is invariant under C, P and T applied in succession, in any order.

Read the logical direction carefully, because it is what makes CPT different from everything else on this page. P, C and T individually are hypotheses: you assume one, look for a violation, and the weak interaction duly supplies one for each. CPT is a theorem — it follows from Lorentz invariance and locality, the assumptions underneath every calculation in the book.

So a CPT test is not a test of CPT. It is a test of special relativity and locality, using particles as the apparatus. If CPT ever failed, the response would not be to add a small CPT-violating term; it would be to rebuild quantum field theory from the foundations.

Its two concrete predictions are unusually blunt: a particle and its antiparticle have exactly the same mass and the same lifetime, and exactly opposite charges. No free parameter, no small correction, no “to a good approximation”.

⚙️ Engineer’s bridge — a checked invariant versus an asserted one

The distinction between CPT and the individual symmetries is one you enforce in code every day.

P, C and T are runtime assertions: statements you hope hold, that you instrument and watch, and that can fail — and in the weak interaction they fail loudly, P and C maximally so (§7.2).

CPT is a guarantee of the construction: it holds because of how the theory is built, in the same way that a value’s type is guaranteed by the compiler rather than checked by an if. You do not test it because you doubt the antiproton’s mass; you test it because a failure would prove the construction is wrong — that locality, or Lorentz invariance, is not what we think.

That is also why the tests are so aggressive. When an invariant is supposed to be exact, the only interesting measurement is the most precise one you can build, and every additional digit is a stronger statement about the foundations.

Where it breaks: “the only interesting measurement is the most precise one” holds while you are testing a foundation nobody expects to fail, and it is bad advice everywhere else. Precision buys nothing if the systematic floor is reached first, and it buys nothing if the prediction it tests is uncertain — the muon anomaly of §5.9b is limited by a lattice number, not by the experiment. CPT is worth pushing because the prediction is exactly zero with no theory error; most quantities are not in that position.

🔬 Experiment card — Ulmer et al. 2015, the antiproton in a Penning trap

Apparatus

A Penning trap: a strong uniform magnetic field confines a charged particle radially, while a set of ring and endcap electrodes holds it axially. A single antiproton is stored, alone, for months. The comparison particle is not a proton but a negatively charged hydrogen ion H⁻ — same sign of charge, so it sits in exactly the same trap with exactly the same voltages, with a small and precisely calculable correction for its two electrons.

What is measured

One frequency. In a field

B

a particle of charge

q

and mass

m

circulates at the cyclotron frequency

\nu_c = qB/2\pi m

, so measuring

\nu_c

measures

q/m

— and comparing the two species in the same field cancels

B

, which is the quantity nobody can know to eleven digits.

The result

The charge-to-mass ratios agree to

(0.1 \pm 6.9) \times 10^{-11}

, Eq. (3.19).

What it proved

CPT holds for the proton–antiproton system at the

10^{-10}

level — which, since CPT is a consequence of Lorentz invariance and locality, is a test of those. On a 938 MeV particle, a fractional limit of

6.9\times 10^{-11}

is a statement about matter–antimatter asymmetry at the level of 0.065 eV.

🛠️ A Penning trap, and why the comparison is a frequency comparison

Click a numbered marker for what that piece does.

Schematic, redrawn from the description on p. 113. The measurement is a comparison of one frequency against another in the same magnetic field — which is why it reaches eleven digits.

\frac{|q_{\htmlClass{t-pb}{\bar p}}|}{m_{\htmlClass{t-pb}{\bar p}}} \Big/ \frac{|q_{\htmlClass{t-p}{p}}|}{m_{\htmlClass{t-p}{p}}} - 1 = \htmlClass{t-res}{(0.1 \pm 6.9)\times 10^{-11}}

(3.19)

Bettini p. 113, S. Ulmer et al. (2015). A ratio of ratios, constructed so that everything hard to measure cancels.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — the equation is built so that the hard quantities cancel

The equation is built to make the hard quantities disappear. Each species gives a cyclotron frequency $\nu_c = qB/2\pi m$ ; divide one by the other in the same trap and $B$ cancels identically. What is left is a pure ratio of $q/m$ values, which is why the measurement can reach a precision no absolute determination of $q$ , $m$ or $B$ could approach.

You have seen this design twice already on this site: the detailed-balance spin measurement of §2.3 cancelled an incalculable matrix element by taking a ratio, and the antiproton spectrometer of §2.6–2.7 used two independent velocity handles rather than one better one. When the quantity you cannot control appears identically in two measurements, divide.

🔢 Worked example — what $6.9\times 10^{-11}$ means in hertz and in eV

Two translations make the number concrete.

As a mass. A fractional limit on $q/m$ is, with the charges known to be equal and opposite, a fractional limit on the mass. On the proton’s 938.272 MeV,

\delta m \le 6.9\times 10^{-11} \times 938.272\ \text{MeV} = 0.065\ \text{eV}.

Sixty-five millielectronvolts — about the thermal energy of a room-temperature molecule — on a particle a billion times heavier. Any difference between matter and antimatter in this system is smaller than that.

As a frequency. Take an illustrative $B = 1.9$ T, a typical superconducting trap field. Then $\nu_c = qB/2\pi m = 29.0$ MHz for a proton, and the limit corresponds to resolving

\delta\nu = 6.9\times 10^{-11} \times 29.0\ \text{MHz} = 2.0\ \text{mHz} .

Two millihertz out of twenty-nine megahertz. That is why the experiment is a single trapped particle observed for months rather than a beam: a frequency resolution of a few parts in $10^{11}$ needs an observation time of order $1/\delta\nu \sim$ minutes per measurement and an environment stable across all of them.

Reproduce it

import numpy as np
q, mp_kg, mp = 1.602176634e-19, 1.67262192369e-27, 938.27208816   # C, kg, MeV
d, B = 6.9e-11, 1.9                                                # limit, tesla

print(f"CPT limit on |q/m|:  {d:.1e}")
print(f"  as a mass on m_p = {mp:.3f} MeV      ->  delta m <= {d*mp*1e6:.4f} eV")
nu = q * B / (2 * np.pi * mp_kg)
print(f"  cyclotron frequency at B = {B} T    ->  nu_c = {nu/1e6:.2f} MHz")
print(f"  the same fraction of that frequency ->  delta nu = {d*nu*1e3:.2f} mHz")
print(f"  observation time needed, ~1/delta nu ->  {1/(d*nu):.0f} s per resolved bin")

prints

CPT limit on |q/m|:  6.9e-11
as a mass on m_p = 938.272 MeV      ->  delta m <= 0.0647 eV
cyclotron frequency at B = 1.9 T    ->  nu_c = 28.97 MHz
the same fraction of that frequency ->  delta nu = 2.00 mHz
observation time needed, ~1/delta nu ->  500 s per resolved bin

🔑 If you remember only three things

The work is in the conversion, not the measurement. A branching-ratio bound becomes a width, a width becomes a count, and only after those steps can two experiments be compared at all.
CPT cannot be tested on its own terms. The theorem assumes Lorentz invariance and local field theory, so an experiment that finds it broken has falsified those assumptions rather than a symmetry.
Only a particle that is its own antiparticle can carry this label. Everything else is moved by C to a different state, and a state that moves has no eigenvalue to quote.

Where this goes next

§3.5 The parity of the pions is the chapter’s experimental heart, and it uses both halves of this page: the identical-fermion counting rule that cut Table 3.1 in half, and P and C together on the π⁰.
§3.10 G-parity composes C with an isospin rotation to make a quantum number that charged pions can have — the natural repair of Eq. (3.13).
§4.5 Pseudoscalar and vector mesons is Table 3.1 applied: every meson multiplet in the particle tables is a row of it, and the exotic search is the hunt for something that is not.
§7.2–7.3 demolishes C and P separately in the weak interaction; §8.5 does the same to their product, CP — measured always against the CPT fixed point established here.
§9.15 determines the spin and parity of the Higgs boson by the method of §3.5, and the two-photon final state means C matters there too.

✅ Check yourself — C, T and CPT

0/5 answered · 0 correct

1.Which of these particles is an eigenstate of $C$ ?
Hint: The condition is not "electrically neutral". Look at the whole record of additive charges.
2.The measured limit is $\Gamma(\pi^0 \to 3\gamma)/\Gamma_{\text{tot}} \le 3.1 \times 10^{-8}$ . What makes this decay interesting enough to search for?
3.Set the widget above to $f\bar f$ and read the $J = 1$ column of the lattice. Which combination is hatched as unreachable?
Hint: Spin singlets ( $s=0$ ) always have $C = -P$ and $J = l$ ; spin triplets ( $s=1$ ) always have $C = +P$ .
4.Parity and $C$ each give a multiplicative quantum number. Why does time reversal not?
5.Eq. (3.19) compares the charge-to-mass ratios of $\bar p$ and $p$ to seven parts in $10^{11}$ . What is really being tested?