Dirac adjoint & bilinear covariants

Write down the Dirac equation and you immediately own a four-component object $\psi$ — but four components is not the same as a four-vector, and that is the trap the whole chapter turns on. Spinors rotate into themselves under a representation $S(\Lambda)$ of the Lorentz group that is not unitary: boosts are generated by anti-Hermitian matrices, so $S^\dagger S \ne 1$ . The consequence is blunt — the naïve probability density $\psi^\dagger\psi$ , which served Schrödinger theory faithfully, is no longer a Lorentz scalar. It is the time slot of something larger, and it changes when you run past the particle. The fix is one factor of $\gamma^0$ . Define the Dirac adjoint $\bar\psi \equiv \psi^\dagger\gamma^0$ , and the metric-aware $\gamma^0$ exactly undoes the non-unitarity of $S$ : where $\psi \to S\psi$ , the adjoint runs the other way, $\bar\psi \to \bar\psi\,S^{-1}$ . Sandwich any fixed matrix $\Gamma$ between $\bar\psi$ and $\psi$ and the two $S$ 's meet in the middle, leaving a clean tensor law. These sandwiches are the bilinear covariants — exactly sixteen of them, a scalar, a pseudoscalar, a four-vector, an axial vector and an antisymmetric tensor. They are the alphabet from which every Lorentz-invariant fermion coupling is spelled: the mass term $m\bar\psi\psi$ , the electromagnetic current $\bar\psi\gamma^\mu\psi$ , the weak $V\!-\!A$ current $\bar\psi\gamma^\mu(1-\gamma^5)\psi$ , the Pauli moment $\bar\psi\sigma^{\mu\nu}\psi\,F_{\mu\nu}$ . Get the adjoint right and relativistic covariance takes care of itself.

Key takeaways

The Dirac adjoint is $\bar\psi \equiv \psi^\dagger\gamma^0$ ; the extra $\gamma^0$ is what makes spinor products transform as Lorentz tensors.
The spinor representation is non-unitary ( $S^\dagger S\ne1$ for boosts), so $\psi^\dagger\psi$ is not invariant — it is the time component of the four-vector current $\bar\psi\gamma^\mu\psi$ .
The master identity $\gamma^0 S^\dagger\gamma^0 = S^{-1}$ gives $\bar\psi\to\bar\psi\,S^{-1}$ , so $\bar\psi\psi$ is a scalar and $\bar\psi\gamma^\mu\psi$ a four-vector.
There are exactly $16$ bilinear covariants: $1$ scalar + $1$ pseudoscalar + $4$ vector + $4$ axial + $6$ tensor, a complete basis of $4\times4$ Dirac matrices.
The conserved current $j^\mu=\bar\psi\gamma^\mu\psi$ satisfies $\partial_\mu j^\mu=0$ ; its five siblings build every fermion interaction in the Standard Model.
Key idea: spinors do not transform like vectors, so the naïve product $\psi^\dagger\psi$ has no clean Lorentz behaviour. Inserting the metric-aware factor $\gamma^0$ — using $\bar\psi=\psi^\dagger\gamma^0$ — exactly compensates for the spinor transformation, turning $\bar\psi\,\Gamma\,\psi$ into objects with definite tensor character.

The core insight

Spinors do not transform like vectors, so naïve $\psi^\dagger\psi$ -type products have no clean Lorentz behaviour. Inserting the metric-aware factor $\gamma^0$ — using $\bar\psi=\psi^\dagger\gamma^0$ instead of $\psi^\dagger$ — exactly compensates for the spinor transformation, turning every $\bar\psi\,\Gamma\,\psi$ into an object with a definite Lorentz-tensor transformation law.

Where this lives today

The bilinear covariants are the literal grammar of every calculation in particle physics. In any Standard Model Lagrangian each fermion interaction is one of the five families: the QED vertex is the vector current $\bar\psi\gamma^\mu\psi$ coupled to the photon, the mass and Higgs–Yukawa terms are the scalar $\bar\psi\psi$ , and the charged-current weak interaction is the parity-violating $V\!-\!A$ form $\bar\psi\gamma^\mu(1-\gamma^5)\psi$ coupled to the $W$ boson — tested to sub-percent precision at the LHC and in beta-decay experiments. The tensor $\bar\psi\sigma^{\mu\nu}\psi$ has a vivid modern life: as the Pauli term $\bar\psi\sigma^{\mu\nu}\psi\,F_{\mu\nu}$ it encodes the anomalous magnetic moment, which the Fermilab Muon $g\!-\!2$ experiment (final result 2023) measures for the muon to about $0.2$ ppm, where a tension with theory hunts for new physics. Pseudoscalar bilinears $\bar\psi\gamma^5\psi$ govern pion couplings and drive electric-dipole-moment searches — a nonzero neutron or electron EDM would signal CP-violating scalar–pseudoscalar interference. Even lattice QCD extracts hadron structure by evaluating these five currents between quark fields. The adjoint Dirac introduced to make a probability come out positive is the alphabet in which the whole matter sector is written.

The physical picture

The whole subject sits inside one thought experiment: take an electron at rest, then run past it. The density $\psi^\dagger\psi$ is a charge density, and density is length-contracted — sprint past the particle and you sweep up more per unit volume, so $\psi^\dagger\psi$ grows. That growth is the signature that $\psi^\dagger\psi$ was never a scalar: it is the time component of the current $j^\mu=\bar\psi\gamma^\mu\psi$ , and like any four-vector its pieces reshuffle under a boost while one combination — its Minkowski length — stays nailed down. That fixed length is what the scalar $\bar\psi\psi$ measures. The first figure follows a real electron through a boost.

Figure

ψ̄ψ is the same number in every frame while ψ†ψ = 2E is the time component of the current four-vector jᵘ = ū γᵘ u = 2pᵘ; boosting an electron from rest E = m = 0.511 MeV to E = 2m = 1.022 MeV grows u†u from 1.022 to 2.044 MeV but leaves ū u = 2m = 1.022 MeV fixed

Why sixteen and not more? Because $\bar\psi\,\Gamma\,\psi$ is built from one $4\times4$ matrix $\Gamma$ , and the space of $4\times4$ matrices is sixteen-dimensional. A chosen basis of those matrices — the identity, $\gamma^5$ , the four $\gamma^\mu$ , the four $\gamma^\mu\gamma^5$ , the six $\sigma^{\mu\nu}$ — sorts the sixteen sandwiches into five Lorentz families that never mix under proper transformations. The second figure maps that alphabet, worth memorising because every fermion coupling you will write is one row of it.

Figure

The complete set of 16 bilinear covariants ψ̄Γψ splits into five Lorentz types — scalar 1, pseudoscalar 1, vector 4, axial-vector 4, antisymmetric tensor 6 — which sum to 16 and exhaust the space of 4×4 Dirac matrices; the last column shows the sign each picks up under a parity flip

The core equations

Everything begins with a definition. The Dirac adjoint is the row spinor you get by taking the Hermitian conjugate and reinserting one factor of $\gamma^0$ ; that factor is the entire trick, because $\gamma^0$ is the object that knows the Minkowski metric.

\bar\psi \equiv \psi^\dagger\gamma^0 \qquad (1)

where: $\psi$ — the four-component Dirac spinor (a column); $\psi^\dagger$ — its Hermitian conjugate (a row of complex conjugates); $\gamma^0$ — the timelike Dirac matrix, Hermitian with $(\gamma^0)^2 = 1$ ; $\bar\psi$ — the Dirac adjoint, a row spinor. Read it as "conjugate-transpose, then reflect through the time axis." The reason $\gamma^0$ is the right insertion is a two-line matrix identity and how it propagates to the finite transformation $S(\Lambda)$ . The hermiticity conditions $\gamma^{0\dagger}=\gamma^0$ , $\gamma^{k\dagger}=-\gamma^k$ package into one covariant statement that forces the adjoint to transform inversely to $\psi$ .

(\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0, \qquad \gamma^0 S^\dagger\gamma^0 = S^{-1} \;\;\Longrightarrow\;\; \bar\psi \to \bar\psi\,S^{-1} \qquad (2)

where: $S=S(\Lambda)$ — the $4\times4$ spinor representation of a Lorentz transformation, satisfying $S^{-1}\gamma^\mu S = \Lambda^\mu{}_\nu\gamma^\nu$ , with $S^\dagger S\ne1$ for boosts. The middle identity is the workhorse — $\gamma^0$ turns a dagger into an inverse: since $\psi\to S\psi$ , $\bar\psi=\psi^\dagger\gamma^0\to\psi^\dagger S^\dagger\gamma^0=\psi^\dagger\gamma^0 S^{-1}=\bar\psi\,S^{-1}$ . With the adjoint transforming as $\bar\psi\,S^{-1}$ , any fixed matrix between $\bar\psi$ and $\psi$ yields a definite tensor. A complete basis of $4\times4$ matrices sorts the sixteen sandwiches into five families.

\bar\psi\psi,\quad \bar\psi\gamma^5\psi,\quad \bar\psi\gamma^\mu\psi,\quad \bar\psi\gamma^\mu\gamma^5\psi,\quad \bar\psi\sigma^{\mu\nu}\psi \qquad (3)

where: $\bar\psi\psi$ — scalar ( $1$ component); $\bar\psi\gamma^5\psi$ — pseudoscalar ( $1$ ), with $\gamma^5\equiv i\gamma^0\gamma^1\gamma^2\gamma^3$ , $(\gamma^5)^2=1$ , $\{\gamma^5,\gamma^\mu\}=0$ ; $\bar\psi\gamma^\mu\psi$ — vector ( $4$ ); $\bar\psi\gamma^\mu\gamma^5\psi$ — axial vector ( $4$ ); $\bar\psi\sigma^{\mu\nu}\psi$ — antisymmetric tensor ( $6$ ), with $\sigma^{\mu\nu}\equiv\tfrac{i}{2}[\gamma^\mu,\gamma^\nu]$ . The component counts $1+1+4+4+6=16$ exhaust the sixteen dimensions of the Dirac-matrix space. The most important covariant is the vector one: the conserved probability-and-charge current, whose conservation is now manifestly Lorentz-invariant. The scalar covariant supplies the mass term of the Dirac Lagrangian.

j^\mu=\bar\psi\gamma^\mu\psi,\qquad \partial_\mu j^\mu=0,\qquad \mathcal{L}=\bar\psi\,(i\gamma^\mu\partial_\mu-m)\,\psi \qquad (4)

where: $j^\mu$ — the conserved four-current, with density $j^0=\psi^\dagger\psi\ge0$ and flux $\mathbf{j}=\psi^\dagger\boldsymbol\alpha\psi$ ; $\partial_\mu j^\mu=0$ — the continuity equation, a scalar and hence frame-independent; $\mathcal{L}$ — the Dirac Lagrangian, whose mass term $-m\bar\psi\psi$ is a scalar. Coupling $j^\mu$ to the photon, $-e\,\bar\psi\gamma^\mu\psi\,A_\mu$ , is the QED interaction.

Derivation AWhy the adjoint carries a factor of gamma-zero

The goal is to show, with no hand-waving, that $\psi^\dagger\psi$ fails to be a Lorentz scalar while $\bar\psi\psi=\psi^\dagger\gamma^0\psi$ succeeds, and that the difference is forced by one algebraic fact — the spinor representation is non-unitary. We start from the spinor transformation law, establish the master identity $\gamma^0 S^\dagger\gamma^0=S^{-1}$ from the hermiticity of the gammas, and feed it through the two simplest sandwiches to read off a scalar and a four-vector.

Step 1. Under $x^\mu\to\Lambda^\mu{}_\nu x^\nu$ the spinor transforms as $\psi\to S(\Lambda)\psi$ , with $S$ fixed (up to sign) by demanding the Dirac equation keep its form, which requires $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ . For a rotation the generator is Hermitian and $S$ unitary, but for a boost it is anti-Hermitian, so $S^\dagger=S$ and $S^\dagger S=S^2\ne1$ . Why: this non-unitarity is the whole reason $\psi^\dagger\psi$ misbehaves — a unitary $S$ would leave $\psi^\dagger\psi$ invariant, but boosts are not unitary in this finite-dimensional representation.

Step 2. Establish the master identity. The conditions $\gamma^{0\dagger}=\gamma^0$ , $\gamma^{k\dagger}=-\gamma^k$ combine into $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$ . Since $S$ is built by exponentiating $\sigma^{\mu\nu}$ and $(\sigma^{\mu\nu})^\dagger=\gamma^0\sigma^{\mu\nu}\gamma^0$ follows from the same relation, exponentiation gives $\gamma^0 S^\dagger\gamma^0=S^{-1}$ , i.e. $S^\dagger\gamma^0=\gamma^0 S^{-1}$ . Why: this is the exact statement that sandwiching with $\gamma^0$ converts a Hermitian conjugate into an inverse — precisely what cancels the offending $S^\dagger$ .

Step 3. Transform the adjoint. Using Step 2, $\bar\psi=\psi^\dagger\gamma^0\to\psi^\dagger S^\dagger\gamma^0=\psi^\dagger\gamma^0 S^{-1}=\bar\psi\,S^{-1}$ , whereas the bare conjugate $\psi^\dagger\to\psi^\dagger S^\dagger$ keeps the uncancelled non-unitary factor. Why: the adjoint runs inversely to $\psi$ , so the two transformations annihilate each other — the single fact that makes every bilinear covariant work.

Step 4. Read off two covariants. The scalar sandwich gives $\bar\psi\psi\to\bar\psi\,S^{-1}S\,\psi=\bar\psi\psi$ — invariant. The vector sandwich gives $\bar\psi\gamma^\mu\psi\to\bar\psi\,S^{-1}\gamma^\mu S\,\psi=\Lambda^\mu{}_\nu\,\bar\psi\gamma^\nu\psi$ — a genuine four-vector, using $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ from Step 1. Meanwhile $\psi^\dagger\psi\to\psi^\dagger S^\dagger S\psi\ne\psi^\dagger\psi$ . Why: the same $S^{-1}\gamma^\mu S$ that makes the Dirac equation covariant promotes the current to a four-vector — covariance of the equation and tensor character of the bilinears are one fact seen twice.

Go deeper · Derivation B · 1 minThe conserved probability current

Derivation BThe conserved probability current

The aim is to prove that $j^\mu=\bar\psi\gamma^\mu\psi$ is conserved, $\partial_\mu j^\mu=0$ , working directly from the Dirac equation and its adjoint, and to see that conservation now holds as a Lorentz-scalar statement rather than the frame-dependent one of Schrödinger theory. We derive the equation obeyed by $\bar\psi$ , differentiate the current, and watch the mass terms cancel — which is what lets $j^0=\psi^\dagger\psi$ serve as a positive-definite density with a consistent flux in every frame.

Step 1. Start from $(i\gamma^\mu\partial_\mu-m)\psi=0$ and take the Hermitian conjugate: $-i\,\partial_\mu\psi^\dagger(\gamma^\mu)^\dagger-m\psi^\dagger=0$ . Why: to build $\partial_\mu(\bar\psi\gamma^\mu\psi)$ we need an equation of motion for $\bar\psi$ , obtained by conjugating the equation for $\psi$ .

Step 2. Right-multiply by $\gamma^0$ and use $(\gamma^\mu)^\dagger\gamma^0=\gamma^0\gamma^\mu$ (a rearrangement of the master identity), turning $\psi^\dagger(\gamma^\mu)^\dagger\gamma^0$ into $\bar\psi\gamma^\mu$ . This gives the adjoint Dirac equation $i\,\partial_\mu\bar\psi\,\gamma^\mu+m\bar\psi=0$ , i.e. $\partial_\mu\bar\psi\,\gamma^\mu=im\bar\psi$ . Why: the factor of $\gamma^0$ converts the awkward $(\gamma^\mu)^\dagger$ into a clean $\gamma^\mu$ acting from the right — again the adjoint is engineered to make the algebra close.

Step 3. Differentiate the current and substitute both equations of motion: $\partial_\mu j^\mu=(\partial_\mu\bar\psi)\gamma^\mu\psi+\bar\psi\gamma^\mu(\partial_\mu\psi)=(im\bar\psi)\psi+\bar\psi(-im\psi)=0$ . Why: the two mass terms are equal and opposite by construction, so the current is conserved for any solution — and because $j^\mu$ is a four-vector, $\partial_\mu j^\mu$ is a scalar, so conservation in one frame holds in all frames.

Go deeper · Derivation C · 1 minSixteen matrices, sixteen covariants

Derivation CSixteen matrices, sixteen covariants

The goal here is to explain the number sixteen — why there are exactly that many independent bilinear covariants, why they organise into the five families scalar, pseudoscalar, vector, axial and tensor, and how a parity flip is the quickest way to tell the "true" tensors from the "pseudo" ones. We count the dimension of the matrix space, exhibit a basis, use trace-orthogonality to prove independence, and act with parity to split each pair — the backbone guaranteeing the alphabet is complete and non-redundant.

Step 1. A bilinear $\bar\psi\,\Gamma\,\psi$ is linear in a single $4\times4$ matrix $\Gamma$ , and that complex vector space has dimension $16$ , so there can be at most sixteen independent covariants — a basis of the matrix space yields a basis of covariants. Why: the counting is pure linear algebra; the physics only enters in choosing a basis whose members transform simply.

Step 2. Take the sixteen matrices $\{\,I,\ \gamma^5,\ \gamma^\mu,\ \gamma^\mu\gamma^5,\ \sigma^{\mu\nu}\,\}$ ( $\mu<\nu$ in the tensor), counts $1+1+4+4+6=16$ . They are linearly independent because trace-orthogonal: $\mathrm{Tr}(\Gamma^A\Gamma_B)=4\,\delta^A{}_B$ , so no nontrivial combination vanishes. Why: trace-orthogonality is the fermionic analogue of an orthonormal basis — any $4\times4$ matrix expands as $M=\tfrac14\sum_A \mathrm{Tr}(\Gamma^A M)\,\Gamma_A$ , confirming completeness.

Step 3. Assign Lorentz character via $S^{-1}\Gamma S$ : the identity gives a scalar, $\gamma^\mu$ a vector, $\sigma^{\mu\nu}$ an antisymmetric rank-2 tensor; the two "pseudo" families differ from their partners by one $\gamma^5$ . Why: $\gamma^5$ commutes with $S$ (a scalar under proper transformations), so multiplying by it does not change boost/rotation behaviour — the distinction only appears under parity.

Step 4. Act with parity $P:\psi(t,\mathbf{x})\to\gamma^0\psi(t,-\mathbf{x})$ . Then $\bar\psi\psi\to+\bar\psi\psi$ (scalar) but $\bar\psi\gamma^5\psi\to-\bar\psi\gamma^5\psi$ (pseudoscalar), since $\gamma^0\gamma^5\gamma^0=-\gamma^5$ . The vector current has $j^0\to+j^0,\ j^i\to-j^i$ (polar) while the axial current flips oppositely, $A^0\to-A^0,\ A^i\to+A^i$ . Why: parity separates the five families into their observable roles — scalar mass term, pseudoscalar pion coupling, polar electromagnetic current, and the axial current whose interference with the vector current is the fingerprint of weak parity violation.

Limiting cases

Case	Result	Meaning
Rest frame ( $\mathbf p=0$ )	$\psi^\dagger\psi=\bar\psi\psi=2m$	density and scalar coincide; the current has no spatial part
General boost	$\psi^\dagger\psi=2E$ , $\bar\psi\psi=2m$	density grows as $E$ ; the scalar is pinned at $2m$ — the Minkowski length of $j^\mu=2p^\mu$
Massless / ultra-relativistic	$\bar\psi\psi,\ \bar\psi\gamma^5\psi\to0$	scalar and pseudoscalar vanish; only $V$ , $A$ , $T$ survive and chirality becomes exact
Non-relativistic limit	$j^\mu\to(\rho,\ \rho\mathbf v)$	upper two components dominate; the current reduces to the Schrödinger probability current
Parity flip $P$	$S\,+$ , $P\,-$ , $V$ polar, $A$ axial	the five families sort by their sign under $\mathbf x\to-\mathbf x$
Charge conjugation $C$	$\bar\psi\gamma^\mu\psi\to-\bar\psi\gamma^\mu\psi$	the vector current flips sign — particle and antiparticle carry opposite charge

Common trapThe gamma matrices are not a four-vector

The single most common error is to read the index on $\gamma^\mu$ as a Lorentz index and conclude the gammas are a four-vector. They are not. The four $\gamma^\mu$ are fixed, Lorentz-invariant constant matrices — the same in every frame — as guaranteed by $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ : the matrices do not change; the transformation is absorbed by how the spinors on either side rotate. The genuine four-vector is the bilinear $j^\mu=\bar\psi\gamma^\mu\psi$ , whose components mix because $\psi$ and $\bar\psi$ transform. The mirror-image mistake is equally common: believing $\psi^\dagger\psi$ is a scalar because it "looks like a modulus." It is not — it is $j^0$ , the time component of the current, and it grows under a boost as a charge density should. Only the $\gamma^0$ -corrected $\bar\psi\psi=\psi^\dagger\gamma^0\psi$ is the true scalar.

Worked examples

1 · The two invariants of a plane-wave electron (fully numerical)

Take an electron of mass $m=0.511\ \mathrm{MeV}$ prepared in a momentum eigenstate with $E=2m$ . Using the Dirac-representation spinor $u^s(p)=\sqrt{E+m}\,\big(\xi^s,\ \tfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\xi^s\big)^{\!\top}$ with $\xi^\dagger\xi=1$ and the normalisation $u^\dagger u=2E$ , evaluate $u^\dagger u$ , $\bar u u$ , and the current $\bar u\gamma^\mu u$ . Confirm which one is frame-independent.

Step 1. Kinematics. With $E=2m$ , $|\mathbf p|=\sqrt{E^2-m^2}=\sqrt{3}\,m=1.732\times0.511=\mathbf{0.885\ MeV}$ . Point $\mathbf p$ along $z$ , so $(\boldsymbol\sigma\cdot\mathbf p)^2=|\mathbf p|^2=E^2-m^2$ .

Step 2. The density $u^\dagger u$ . Squaring, $u^\dagger u=(E+m)\big[1+\tfrac{E^2-m^2}{(E+m)^2}\big]=(E+m)+(E-m)=2E$ . Numerically $u^\dagger u=2E=\mathbf{2.044\ MeV}$ . This is $j^0$ — a density, depending on the frame through $E$ .

Step 3. The scalar $\bar u u$ . Insert $\gamma^0=\mathrm{diag}(I,-I)$ , flipping the sign of the lower components: $\bar u u=u^\dagger\gamma^0 u=(E+m)\big[1-\tfrac{E^2-m^2}{(E+m)^2}\big]=(E+m)-(E-m)=2m$ . Numerically $\bar u u=2m=\mathbf{1.022\ MeV}$ — and crucially it carries no $E$ , so it is the same in every frame. The scalar $\bar u u=2m$ is frame-independent; the density $u^\dagger u=2E$ is not.

Step 4. The current $\bar u\gamma^\mu u$ . A short calculation (the Gordon identity for equal spins) gives $\bar u\gamma^\mu u=2p^\mu$ . The time slot is $\bar u\gamma^0 u=u^\dagger u=2E=2p^0$ ; the spatial slot is $\bar u\gamma^3 u=u^\dagger\alpha^3 u=2|\mathbf p|=2\times0.885=\mathbf{1.771\ MeV}=2p^3$ . So $j^\mu=(2.044,\,0,\,0,\,1.771)\ \mathrm{MeV}$ , with Minkowski length $\sqrt{j^\mu j_\mu}=\sqrt{2.044^2-1.771^2}=\sqrt{1.041}=1.020\approx2m$ . The vector's components shuffle with the frame, but its invariant length equals $2m$ — the very number the scalar $\bar u u$ reports.

2 · Scalar versus pseudoscalar under a mirror (sign-tracking)

Show that $\bar\psi\psi$ and $\bar\psi\gamma^5\psi$ behave oppositely under a parity flip, and explain why this forces the pion and a hypothetical scalar meson to couple to nucleons through different bilinears.

Step 1. Parity acts as $P:\psi(t,\mathbf x)\to\gamma^0\psi(t,-\mathbf x)$ , so the adjoint becomes $\bar\psi\,'=(\gamma^0\psi)^\dagger\gamma^0=\psi^\dagger\gamma^{0\dagger}\gamma^0=\bar\psi\,\gamma^0$ (at $-\mathbf x$ ). Keep the $\gamma^0$ 's explicit through the sandwiches.

Step 2. The scalar. $\bar\psi\psi\to\bar\psi\gamma^0\cdot\gamma^0\psi=\bar\psi(\gamma^0)^2\psi=\bar\psi\psi$ (at $-\mathbf x$ ). Even under parity — a true scalar.

Step 3. The pseudoscalar. $\bar\psi\gamma^5\psi\to\bar\psi\gamma^0\gamma^5\gamma^0\psi$ . Because $\{\gamma^5,\gamma^0\}=0$ , we have $\gamma^0\gamma^5\gamma^0=-\gamma^5(\gamma^0)^2=-\gamma^5$ , so $\bar\psi\gamma^5\psi\to-\bar\psi\gamma^5\psi$ . Odd under parity — a pseudoscalar.

Step 4. Physics. A pion has parity $J^P=0^-$ , so a parity-conserving pion–nucleon coupling must pair the pion with a pseudoscalar bilinear, $g\,\bar\psi\gamma^5\psi\,\phi_\pi$ ; a genuine scalar meson ( $0^+$ ) couples through $g\,\bar\psi\psi\,\phi$ . The one factor of $\gamma^5$ is the difference between a $0^+$ and a $0^-$ coupling — the bilinears carry the quantum numbers. The same $V$ -versus- $A$ bookkeeping applied to the currents makes the weak $\bar\psi\gamma^\mu(1-\gamma^5)\psi$ violate parity maximally.

Physics of the portrait

The adjoint spinor and its bilinears are so woven into field theory that it is easy to forget they were one person's inventions, arrived at in a rush between 1928 and the early 1930s while the meaning of a relativistic wave equation was still molten.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · relativistic quantum mechanics and the equation of the electron

Paul Adrien Maurice Dirac, trained as an electrical engineer in Bristol before turning to physics, wrote down in 1928 the relativistic wave equation that bears his name — and was forced to invent nearly everything in this article at once. Demanding a first-order equation consistent with special relativity produced four-component wavefunctions and the four anticommuting matrices $\gamma^\mu$ ; insisting the theory carry a positive probability density forced the adjoint $\bar\psi=\psi^\dagger\gamma^0$ and the current $\bar\psi\gamma^\mu\psi$ that reduces to Schrödinger's in the slow limit. Dirac's temperament — spare, guided by an almost aesthetic conviction that fundamental equations must be beautiful — is stamped on the covariant form $(i\gamma^\mu\partial_\mu-m)\psi=0$ , which hides a mountain of structure behind one line. The "extra" negative-energy solutions troubled him until, in 1930–31, he reinterpreted them as antiparticles and predicted the positron, found by Anderson in 1932; he shared the 1933 Nobel Prize with Schrödinger. The classification of the sixteen bilinear covariants grew directly out of his formalism, and when parity violation surfaced in the 1950s it was the vector–axial distinction among his bilinears that let Feynman, Gell-Mann, Sudarshan and Marshak write down the weak interaction. Every fermion line in every Feynman diagram is, at bottom, a statement about Dirac adjoints.

History

1928

— Dirac writes the relativistic electron equation; the four-current ψ̄γᵘψ appears in the same paper

1929

— the adjoint spinor ψ̄ = ψ†γ⁰ and the covariant bilinears are formalized by Dirac and Darwin

1937

— Fierz derives the identities relating the five families of bilinear covariants

1957

— Wu's cobalt-60 experiment reveals parity violation, physically splitting the vector from the axial current

1958

— Feynman–Gell-Mann and Sudarshan–Marshak fix the weak current as the V minus A form ψ̄γᵘ(1−γ⁵)ψ

Going further

Texts. Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.2–3.4 (spinor bilinears, discrete symmetries) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.3 (bilinear covariants) · Halzen & Martin, Quarks and Leptons, §5.4–5.6 (the sixteen $\Gamma$ -matrices and their transformations) · Thomson, Modern Particle Physics, §4.5–4.7 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 10–11 (spinors and Lorentz representations); Bjorken & Drell, Relativistic Quantum Mechanics, §3–4.

Provenance. The covariant Dirac equation and $\gamma^0\equiv\beta$ , $\gamma^k\equiv\beta\alpha^k$ are recalled from aq-pp-dirac-equation and aq-pp-gamma-matrices-clifford-algebra; the density–current pair $\rho=\psi^\dagger\psi$ , $\mathbf j=\psi^\dagger\boldsymbol\alpha\psi$ from aq-pp-dirac-probability-current; the index handling of $\Lambda^\mu{}_\nu$ from aq-pp-four-vector-notation. A verifier should confirm: (1) $\gamma^0 S^\dagger\gamma^0=S^{-1}$ from $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$ ; (2) $\partial_\mu(\bar\psi\gamma^\mu\psi)=0$ ; (3) $1+1+4+4+6=16$ and $\mathrm{Tr}(\Gamma^A\Gamma_B)=4\delta^A{}_B$ ; (4) the numerics $u^\dagger u=2E=2.044$ , $\bar u u=2m=1.022$ , $\bar u\gamma^\mu u=2p^\mu$ for $E=2m$ , $m=0.511\ \mathrm{MeV}$ .

Prerequisites. The Dirac Equation · Gamma Matrices and the Clifford Algebra · The Dirac Probability Current · Four-Vector Notation. Leads to. The Covariant Dirac Equation · Feynman Rules for QED · Trace Techniques · Chirality and $\gamma^5$ .

Ask the tutor: Ask me to boost a spinor and watch $\psi^\dagger\psi$ change while $\bar\psi\psi$ stays fixed, or to build any Standard-Model coupling from the five bilinear covariants.

Why not just use ψ† instead of ψ̄ = ψ†γ⁰?

Because $\psi^\dagger$ transforms with the wrong factor. Under a boost $\psi\to S\psi$ with $S$ non-unitary, so $\psi^\dagger\to\psi^\dagger S^\dagger$ and the product $\psi^\dagger S^\dagger S\psi$ carries an uncancelled $S^\dagger S\ne1$ . Inserting $\gamma^0$ exploits $S^\dagger\gamma^0=\gamma^0 S^{-1}$ , so the adjoint runs as $\bar\psi\to\bar\psi\,S^{-1}$ and the $S^{-1}$ cancels the $S$ from $\psi$ . In short, $\psi^\dagger\psi$ is the time component of a four-vector (it grows under a boost), while $\bar\psi\psi$ is a true scalar. The $\gamma^0$ is not decoration; it is the metric doing its job in spinor space.

Are the gamma matrices a four-vector because they carry an index μ?

No — this is the classic trap. The four $\gamma^\mu$ are fixed, frame-independent constant matrices, the same numbers in every inertial frame. The relation $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ looks like a transformation law but states the opposite: it is a consistency condition among constant matrices. The genuine four-vector is the number-valued bilinear $j^\mu=\bar\psi\gamma^\mu\psi$ ; its components mix under $\Lambda$ because $\psi$ and $\bar\psi$ transform, not because the gammas do. Think of $\gamma^\mu$ as fixed "machinery" and $\bar\psi(\cdots)\psi$ as the tensor it manufactures.

Where does the number sixteen come from?

From pure linear algebra. A bilinear $\bar\psi\,\Gamma\,\psi$ is linear in one $4\times4$ matrix $\Gamma$ , and the space of $4\times4$ complex matrices has exactly $16$ dimensions. Any basis of that space gives a basis of bilinears; the useful basis $\{I,\gamma^5,\gamma^\mu,\gamma^\mu\gamma^5,\sigma^{\mu\nu}\}$ has $1+1+4+4+6=16$ members that sort into five clean Lorentz families. Independence follows from trace-orthogonality, $\mathrm{Tr}(\Gamma^A\Gamma_B)=4\delta^A{}_B$ . So "sixteen" is not deep physics — it is the dimension of the matrix algebra — but the sorting into scalar, pseudoscalar, vector, axial and tensor is.

What is the difference between a scalar and a pseudoscalar, physically?

Both are single numbers invariant under rotations and boosts, so under proper Lorentz transformations they are indistinguishable. The difference appears only under a mirror reflection (parity): a scalar $\bar\psi\psi$ is unchanged, a pseudoscalar $\bar\psi\gamma^5\psi$ flips sign. Physically this labels the intrinsic parity of the particle a field couples to: a $0^+$ scalar meson couples through $\bar\psi\psi$ , a $0^-$ pion through $\bar\psi\gamma^5\psi$ . In a parity-conserving theory you must match the parity of the coupling to that of the particle — mixing them would let a mirror experiment tell left from right for free.

What does the axial vector current have to do with the weak force?

Everything. The charged weak current is the combination $\bar\psi\gamma^\mu(1-\gamma^5)\psi$ — a vector minus an axial vector, the famous $V\!-\!A$ structure. Because the vector part flips one way under parity and the axial the opposite way, their sum has no definite parity, so the weak interaction violates parity — maximally, since the two enter with equal weight. This is why only left-handed fermions (and right-handed antifermions) feel the charged weak force. Once Wu's 1957 experiment showed parity is violated, the question became which mix of Dirac's bilinears nature uses; the answer, $V\!-\!A$ , was pinned down within a year.

Is ψ̄ψ the same thing as the probability density?

No, and conflating them is a common slip. The probability (and charge) density is $j^0=\psi^\dagger\psi$ , the time component of $j^\mu=\bar\psi\gamma^\mu\psi$ ; it is positive-definite and frame-dependent, growing under a boost like any density. The scalar $\bar\psi\psi=\psi^\dagger\gamma^0\psi$ is a different object — Lorentz-invariant, equal to $2m$ for a plane wave in any frame, and not sign-definite (negative for antiparticle solutions). They coincide numerically only in the rest frame. In the Lagrangian $\psi^\dagger\psi$ never appears alone precisely because it is not invariant, whereas $\bar\psi\psi$ is the mass term.

How does the tensor bilinear σᵘᵛ connect to spin and magnetism?

The tensor covariant $\bar\psi\sigma^{\mu\nu}\psi$ , with $\sigma^{\mu\nu}=\tfrac{i}{2}[\gamma^\mu,\gamma^\nu]$ , is the relativistic spin operator in disguise: its spatial components reduce to the Pauli spin matrices non-relativistically. Coupled to the field strength it forms the Pauli moment term $\bar\psi\sigma^{\mu\nu}\psi\,F_{\mu\nu}$ , whose coefficient is the magnetic moment. The Dirac equation alone predicts $g=2$ ; radiative corrections add a small $\sigma^{\mu\nu}$ coupling, the anomalous moment $a=(g-2)/2$ , which the Fermilab Muon $g\!-\!2$ experiment measures to sub-ppm precision. The sixth family of bilinears is where spin, magnetism, and precision QED all meet.