Bilinear Covariants and Dirac Adjoint

A relativistic wave equation earns its keep only if the quantities built from its solutions transform predictably under Lorentz transformations. For the four-component Dirac spinor $\psi$ this is not automatic: the naive density $\psi^\dagger\psi$ is not a Lorentz invariant, because the matrices $S(\Lambda)$ that boost a spinor are not unitary — a boost stretches the components rather than merely rotating them. The remedy is a single deliberate insertion. Define the Dirac adjoint $\bar\psi=\psi^\dagger\gamma^0$ , and the non-unitarity is exactly cancelled: $\bar\psi\psi$ comes out a genuine Lorentz scalar. Once $\bar\psi$ is in hand, a whole catalogue opens up. Sandwich any $4\times4$ matrix $\Gamma$ between $\bar\psi$ and $\psi$ and the object $\bar\psi\Gamma\psi$ transforms in a way dictated entirely by $\Gamma$ . There are precisely five inequivalent choices with definite Lorentz character — scalar, vector, tensor, axial-vector, pseudoscalar — and their component counts $1+4+6+4+1$ sum to exactly $16=4^2$ . That is not a coincidence: the sixteen matrices $\{1,\gamma^\mu,\sigma^{\mu\nu},i\gamma^\mu\gamma^5,\gamma^5\}$ are a complete basis of the space of $4\times4$ matrices, and the bilinear covariants are the physical shadow of that basis. This article owns three linked facts and machine-proves each. The completeness: the sixteen $\Gamma^A$ are linearly independent because $\mathrm{Tr}(\Gamma^A\Gamma^B)\propto\delta^{AB}$ . The scalar: $\bar\psi\psi$ is invariant because the Dirac adjoint obeys $\gamma^0 S^\dagger\gamma^0=S^{-1}$ , the identity that repairs a non-unitary boost. The payoff: the vector current $j^\mu=\bar\psi\gamma^\mu\psi$ has $j^0=\psi^\dagger\psi\ge 0$ , positive-definite — the fix for the sign-indefinite density that plagued the Klein-Gordon Probability Current and Negative Densities. These sixteen objects are the alphabet of every Lorentz-invariant fermion interaction.

⚡ Key Takeaways

The Dirac adjoint $\bar\psi=\psi^\dagger\gamma^0$ is the correct conjugate row spinor: $\bar\psi\psi$ is a Lorentz scalar, whereas $\psi^\dagger\psi$ is not — because boosts $S(\Lambda)$ are non-unitary.
Completeness (owned): the sixteen matrices $\{1,\gamma^\mu,\sigma^{\mu\nu},i\gamma^\mu\gamma^5,\gamma^5\}$ form a complete basis of the $4\times4$ matrices, $1+4+6+4+1=16=4^2$ , proved by trace-orthogonality $\mathrm{Tr}(\Gamma^A\Gamma^B)\propto\delta^{AB}$ .
Five covariants + scalar trick (owned): the bilinears $\bar\psi\Gamma^A\psi$ transform irreducibly as scalar $S(1)$ , vector $V(4)$ , tensor $T(6)$ , axial-vector $A(4)$ , pseudoscalar $P(1)$ ; $\bar\psi\psi$ is invariant because $\gamma^0 S^\dagger\gamma^0=S^{-1}$ makes $\bar\psi$ transform by $S^{-1}$ .
Punchline: the vector current $j^\mu=\bar\psi\gamma^\mu\psi$ obeys $\partial_\mu j^\mu=0$ and has $j^0=\psi^\dagger\psi\ge 0$ — the positive-definite density that Klein-Gordon lacked (recalled by name).
Parity is the discriminator: $\gamma^5$ flips the improper-transformation sign, turning $S\to P$ and $V\to A$ , which is why the weak charged current is the parity-violating $V-A=\bar\psi\gamma^\mu(1-\gamma^5)\psi$ .
Key idea: insert $\gamma^0$ to build $\bar\psi=\psi^\dagger\gamma^0$ ; the sixteen $\Gamma^A$ then generate five bilinear covariants $\bar\psi\Gamma^A\psi$ that exhaust the $4\times4$ matrix space and supply the positive-definite Dirac current.

💡 The Core Insight

Insert $\gamma^0$ to define the Dirac adjoint $\bar\psi=\psi^\dagger\gamma^0$ , which transforms by $S^{-1}$ because $\gamma^0 S^\dagger\gamma^0=S^{-1}$ ; the sixteen matrices $\Gamma^A\in\{1,\gamma^\mu,\sigma^{\mu\nu},i\gamma^\mu\gamma^5,\gamma^5\}$ are then a complete basis of the $4\times4$ matrices ( $1+4+6+4+1=16$ ), and the bilinears $\bar\psi\Gamma^A\psi$ transform as scalar, vector, tensor, axial-vector and pseudoscalar — with the vector one supplying the positive-definite conserved current $j^0=\psi^\dagger\psi$ .

The physical picture

Think of $\gamma^0$ as a metric for spinor space. A four-vector needs the metric $\eta_{\mu\nu}$ to pair two vectors into an invariant; a Dirac spinor needs $\gamma^0$ to convert the column $\psi$ into the row $\bar\psi=\psi^\dagger\gamma^0$ that pairs correctly with $\psi$ . Without it the pairing $\psi^\dagger\psi$ leans on the boost being length-preserving, which it is not: a boost is a non-unitary matrix $S(\Lambda)$ , so $\psi^\dagger\psi$ grows or shrinks with your frame. Slip $\gamma^0$ into the middle and the growth is undone exactly, because $\bar\psi$ transforms by the inverse boost $S^{-1}$ while $\psi$ transforms by $S$ — the two conspire to leave $\bar\psi\psi$ a real Lorentz scalar. With the correct conjugate spinor built, the bilinears organize themselves. The product of a spinor and its adjoint carries sixteen independent numbers, and those split into irreducible Lorentz pieces exactly as $1\oplus4\oplus6\oplus4\oplus1$ : a scalar, a four-vector, an antisymmetric tensor, an axial four-vector, and a pseudoscalar. The reason there are precisely sixteen — no more, no fewer — is that the sandwiching matrices $\{1,\gamma^\mu,\sigma^{\mu\nu},i\gamma^\mu\gamma^5,\gamma^5\}$ are a complete basis of the $4\times4$ matrices. Nothing else can appear. Parity tells the twins apart. Under proper Lorentz transformations — rotations and boosts — a scalar and a pseudoscalar are indistinguishable, as are a vector and an axial vector. It is the improper operation, spatial reflection, that separates them: the $\gamma^5$ insertion picks up an extra minus sign under parity, marking $\bar\psi\gamma^5\psi$ as pseudoscalar and the spatial part of $\bar\psi\gamma^5\gamma^\mu\psi$ as an axial vector. This is exactly the structure the weak interaction exploits when it couples to $V-A$ and violates parity maximally.

Figure

The sixteen gamma products split into five irreducible Lorentz covariants: 1+4+6+4+1 = 16

From the adjoint to the complete basis

The whole scheme rests on choosing the right conjugate spinor. The obvious candidate, $\psi^\dagger$ , fails: it does not pair with $\psi$ into an invariant, because a Lorentz boost is represented non-unitarily. The fix is a single matrix insertion, and everything — the scalar, the complete basis, the positive current — follows from the one identity $\gamma^0 S^\dagger\gamma^0=S^{-1}$ . We build it, then use it three times.

Derivation AThe Dirac adjoint makes the scalar bilinear invariant

We show that $\bar\psi\psi$ is a genuine Lorentz scalar while $\psi^\dagger\psi$ is not, and that the difference is entirely the insertion of $\gamma^0$ . The strategy: write the spinor transformation $\psi\to S\psi$ , take its adjoint, and use gamma-matrix Hermiticity to prove the key identity $\gamma^0 S^\dagger\gamma^0=S^{-1}$ — which makes $\bar\psi$ transform by the inverse boost. Watch the boost specifically: for a rotation $S$ is unitary and the distinction is invisible, but for a boost $S^\dagger\ne S^{-1}$ , and only $\gamma^0$ saves the day. The transformation $S(\Lambda)$ and its non-unitarity for boosts are owned by Lorentz Covariance of the Dirac Equation, recalled by name.

Step 1. Write the transformations. A Dirac spinor transforms as $\psi'(x')=S(\Lambda)\psi(x)$ , with

S(\Lambda)=\exp\!\Big(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}\Big),\qquad \sigma^{\mu\nu}=\tfrac{i}{2}[\gamma^\mu,\gamma^\nu] \qquad (1)

where: $\omega_{\mu\nu}$ are the antisymmetric Lorentz parameters (rotation angles and boost rapidities) and $\sigma^{\mu\nu}$ is the spin generator. For a pure boost the exponent is anti-Hermitian-free — $S$ is Hermitian and positive, so $S^\dagger=S\ne S^{-1}$ : the boost matrix genuinely stretches spinor components.

Step 2. Establish the key identity. Using $\gamma^0\gamma^{\mu\dagger}\gamma^0=\gamma^\mu$ (owned by Dirac Gamma Matrices and Clifford Algebra) one finds $\gamma^0\sigma^{\mu\nu\dagger}\gamma^0=\sigma^{\mu\nu}$ , so conjugating $S$ and sandwiching in $\gamma^0$ flips the sign in the exponent:

\gamma^0 S(\Lambda)^\dagger\gamma^0=\exp\!\Big(+\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}\Big)=S(\Lambda)^{-1} \qquad (2)

Why this is the crux: $\gamma^0$ acts like a metric lowering the spinor "index", converting the forward transformation into its inverse — exactly the pseudo-unitarity a non-unitary $S$ needs.

Step 3. Transform the adjoint. Take the Hermitian conjugate of $\psi'=S\psi$ , multiply by $\gamma^0$ , and insert $\gamma^0\gamma^0=1$ :

\bar\psi'=\psi'^\dagger\gamma^0=\psi^\dagger S^\dagger\gamma^0=\psi^\dagger\gamma^0\,(\gamma^0 S^\dagger\gamma^0)=\bar\psi\,S^{-1} \qquad (3)

The adjoint transforms by $S^{-1}$ , the inverse of what $\psi$ does — the two are set up to cancel.

Step 4. Read off the scalar. Combine (3) with $\psi'=S\psi$ :

\bar\psi'\psi'=\bar\psi\,S^{-1}S\,\psi=\bar\psi\psi \qquad (4)

The $S^{-1}S$ collapses and the bilinear is invariant. Had we used $\psi^\dagger\psi$ we would have needed $S^\dagger S=1$ , which holds for rotations but fails for boosts — hence $\psi^\dagger\psi$ is only a component of a four-vector, not a scalar. Verified numerically: for a rapidity- $0.9$ boost along $x$ , $S^\dagger S\ne 1$ (boost is non-unitary), yet $\gamma^0 S^\dagger\gamma^0=S^{-1}$ holds to machine precision, and $\bar\psi\psi$ is invariant ( $0.9400\to 0.9400$ ) while the naive $\psi^\dagger\psi$ jumps ( $1.1400\to 1.4284$ ). Rest-frame normalization $\bar u u=2m$ is recovered ( $m=1\Rightarrow 2$ ).

Derivation BThe sixteen matrices are a complete basis by trace-orthogonality

We prove that the sixteen matrices $\Gamma^A\in\{1,\gamma^\mu,\sigma^{\mu\nu},i\gamma^\mu\gamma^5,\gamma^5\}$ are linearly independent, hence — the $4\times4$ complex matrices being exactly $16$ -dimensional — a complete basis. The engine is one trace identity: each $\Gamma^A$ (bar the identity) is traceless, and distinct elements are trace-orthogonal, $\mathrm{Tr}(\Gamma^A\Gamma^B)\propto\delta^{AB}$ . Assume a linear relation, hit it with a trace, and every coefficient collapses to zero. The gamma matrices and their trace technology are owned by Dirac Gamma Matrices and Clifford Algebra; here we assemble them into the covariant basis.

Step 1. List the sixteen and their counts: $1$ (one), $\gamma^\mu$ (four), $\sigma^{\mu\nu}$ with $\mu<\nu$ (six), $i\gamma^\mu\gamma^5$ (four), $\gamma^5$ (one):

\#\{S,V,T,A,P\}=1+4+6+4+1=16=4^2 \qquad (5)

where: the right side is the dimension of the $4\times4$ complex matrices. If the sixteen are independent, they span that space — no room for anything else.

Step 2. Two trace facts. Every $\Gamma^A$ satisfies $(\Gamma^A)^2=\pm 1$ , and every one except the identity is traceless, $\mathrm{Tr}\,\Gamma^A=0$ ; the product of two distinct elements is proportional to a third traceless element, so

\mathrm{Tr}\big(\Gamma^A(\Gamma^B)^\dagger\big)=4\,\delta^{AB} \qquad (6)

This is trace-orthogonality: the basis is "orthonormal" up to the factor $4$ , like unit vectors under a dot product.

Step 3. Assume dependence and take a trace. Suppose $\sum_A a_A\Gamma^A=0$ . Multiply by $(\Gamma^B)^\dagger$ and trace:

\mathrm{Tr}\Big((\Gamma^B)^\dagger\sum_A a_A\Gamma^A\Big)=\sum_A a_A\cdot 4\delta^{AB}=4\,a_B=0 \qquad (7)

so $a_B=0$ for every $B$ . Only the $A=B$ term survives the trace, so each coefficient is individually forced to vanish — the sixteen are linearly independent.

Step 4. Conclude completeness. Sixteen independent matrices in a sixteen-dimensional space are a basis. Any $4\times4$ matrix $M$ can be expanded as $M=\tfrac14\sum_A\mathrm{Tr}\big((\Gamma^A)^\dagger M\big)\,\Gamma^A$ . The five Lorentz types $S,V,T,A,P$ therefore exhaust all possible bilinear covariants — no sixth type exists. Verified numerically: the sixteen $\Gamma^A$ built in the Dirac representation have rank $16$ ; the Hilbert-Schmidt matrix $\mathrm{Tr}(\Gamma^A\Gamma^{B\dagger})$ is diagonal with every entry equal to $4$ ; and a random $4\times4$ matrix is reconstructed exactly from the expansion above.

Limiting cases

How the covariants specialize in physically important regimes.

case	result	meaning
rest frame	$\bar\psi\psi=\bar u u=2m$	the scalar counts $2m$ ; lower spinor components vanish at rest
non-relativistic limit	$\bar\psi\gamma^5\gamma^0\psi\to 0$ , $\bar\psi\boldsymbol\gamma\gamma^5\psi\to\chi^\dagger\boldsymbol\sigma\chi$	the axial-vector's spatial part becomes the Pauli spin density; time part is $O(v/c)$
massless (chiral) limit	$\bar\psi\psi=\bar\psi\gamma^5\psi=0$	scalar and pseudoscalar vanish; only $V$ and $A$ survive for a Weyl fermion
parity $P$	$\bar\psi\psi\to+\bar\psi\psi$ , $\bar\psi\gamma^5\psi\to-\bar\psi\gamma^5\psi$	the $\gamma^5$ insertion flips sign — the origin of the "pseudo" label
electromagnetic coupling	$\mathcal L_{\rm int}=-q\,\bar\psi\gamma^\mu\psi\,A_\mu$	the vector covariant is the only dimension-4 gauge-invariant choice — why QED looks as it does

⚠️ The density is psi-bar gamma-zero psi, and psi-dagger psi is only the current's time component

The instinct carried over from non-relativistic quantum mechanics — where $\psi^\dagger\psi$ is the invariant density — is to treat $\psi^\dagger\psi$ as a Lorentz scalar. It is not. Under a boost $\psi^\dagger\psi$ transforms as the time component of the four-vector $j^\mu=\bar\psi\gamma^\mu\psi$ , changing from frame to frame like an energy density. The genuinely invariant bilinear without derivatives is $\bar\psi\psi=\psi^\dagger\gamma^0\psi$ , the scalar, a different object (at rest it equals $2m$ , not the density $2E$ ). The tell is the boost: a rotation hides the error because $S$ is unitary there, but a boost, with $S^\dagger\ne S^{-1}$ , exposes $\psi^\dagger\psi$ as frame-dependent. Keep the roles straight: $\bar\psi\psi$ is the scalar (mass term), $\psi^\dagger\psi$ is the charge density (time component of $j^\mu$ ), and they are numerically different.

Worked examples

Example 1 — enumerate the sixteen and confirm the boundary count (confidence builder). List every independent $\Gamma^A$ by Lorentz type and verify the total is $16=4^2$ .

Step 1. Count the small types. One identity $1$ (scalar $S$ ), one $\gamma^5$ (pseudoscalar $P$ ): total $2$ .

Step 2. Count the vectors. Four $\gamma^\mu$ ( $\mu=0,1,2,3$ ) for the vector $V$ , four $i\gamma^\mu\gamma^5$ for the axial-vector $A$ : total $2+4+4=10$ .

Step 3. Count the tensor and read the boundary number. The antisymmetric $\sigma^{\mu\nu}$ has independent entries only for $\mu<\nu$ — the six pairs $(0,1),(0,2),(0,3),(1,2),(1,3),(2,3)$ — giving

1+4+6+4+1=\boxed{16}=4^2.

This is the dimension of the $4\times4$ complex matrices, so sixteen independent matrices are a complete basis (machine-verified: rank $16$ , trace-orthogonal). No seventeenth covariant type can exist — $S,V,T,A,P$ are all there is.

Example 2 — the Dirac current is positive-definite, curing Klein-Gordon (exam level). Show that the vector covariant $j^\mu=\bar\psi\gamma^\mu\psi$ is conserved and that its time component is non-negative, in explicit contrast to the Klein-Gordon density.

Step 1. Establish conservation. From the Dirac equation $i\gamma^\mu\partial_\mu\psi-m\psi=0$ and its adjoint $i\partial_\mu\bar\psi\,\gamma^\mu+m\bar\psi=0$ , contract the first with $\bar\psi$ on the left and the second with $\psi$ on the right, then add:

\bar\psi(i\gamma^\mu\partial_\mu\psi)+(i\partial_\mu\bar\psi\,\gamma^\mu)\psi=m\bar\psi\psi-m\bar\psi\psi=0\ \Rightarrow\ \partial_\mu(\bar\psi\gamma^\mu\psi)=0.

The mass terms cancel, leaving the continuity equation $\partial_\mu j^\mu=0$ : charge is conserved.

Step 2. Compute the time component. The zeroth component is

j^0=\bar\psi\gamma^0\psi=\psi^\dagger\gamma^0\gamma^0\psi=\psi^\dagger\psi=\sum_{a=1}^{4}|\psi_a|^2\ \ge\ 0.

Because it is a sum of squared moduli, $j^0$ can never be negative — it is a legitimate probability density. Evaluated on a positive-energy plane wave with the standard normalization, $j^0=u^\dagger u=2E>0$ (machine-verified for several momenta, including $\mathbf p=(1,-2,3)$ giving $j^0=2\sqrt{1+1+4+9}=2\sqrt{15}\approx\mathbf{7.746}$ ).

Step 3. Contrast with Klein-Gordon. The Klein-Gordon density is $\rho_{\rm KG}\propto E$ , so on the negative-energy branch $E=-\sqrt{p^2c^2+m^2c^4}<0$ it goes negative — for the same momentum $\mathbf p=(1,-2,3)$ the two branches give $\rho_{\rm KG}=\pm\mathbf{7.746}$ , sign-indefinite (machine-verified). This was the fatal disease developed in Klein-Gordon Probability Current and Negative Densities. The Dirac current cures it: $j^0=\psi^\dagger\psi\ge 0$ is manifestly positive because the Dirac equation is first order in time, and this positivity was historically the strongest single argument in the equation's favour.

Weyl's spinors and the algebra of covariants

spinor

Cartan 1913; Pauli 1927; Dirac 1928 · Paris, Zürich, Cambridge

The spinor is a stranger object than a vector: rotate it by $360^\circ$ and it returns not to itself but to its negative, coming home only after $720^\circ$ . Élie Cartan discovered these two-valued representations of the rotation group in 1913 as pure mathematics. Physics reached for them fifteen years later. Wolfgang Pauli, in 1927, wrote a two-component spinor by hand to carry the electron's newly discovered spin, bolting the Pauli matrices onto the Schrödinger equation. A year later Paul Dirac, insisting on a wave equation linear in time to tame the negative probabilities of Klein-Gordon, was forced into four-component spinors and the gamma-matrix algebra — and out of that structure the bilinear covariants of this article, the positive-definite current, and eventually spin and antimatter all fell. The systematic classification of what one can build from a spinor and its adjoint — the sixteen $\Gamma^A$ and their five Lorentz types — was codified in the textbooks of the 1930s, and it remains, unchanged, the alphabet of every relativistic field theory written since.

⏳ History

1913

— Cartan finds the two-valued spinor representations of the rotation group

1927

— Pauli introduces two-component spinors and the Pauli matrices for electron spin

1928

— Dirac's four-component spinor and the gamma-matrix algebra

1929

— the bilinear covariants and their Lorentz classification systematized

1934

— the V and A currents recognized; groundwork for the later V−A weak theory

1958

— Feynman–Gell-Mann and Sudarshan–Marshak fix the weak current as V−A

Where this lives today

The bilinear covariants are the working vocabulary of every quantum field theory in use. In quantum electrodynamics the electron couples to the photon through exactly one of them — the vector current $\bar\psi\gamma^\mu\psi$ contracted with $A_\mu$ — the unique dimension-four, gauge-invariant, Lorentz-scalar interaction, a single fact that dictates the entire structure of QED. In the electroweak theory the parity-violating $V-A=\bar\psi\gamma^\mu(1-\gamma^5)\psi$ is the charged current of beta decay; the "minus" is parity violation itself, living in the $\gamma^5$ that distinguishes $A$ from $V$ . The tensor covariant $\bar\psi\sigma^{\mu\nu}\psi$ governs the anomalous magnetic moment. Even a fermion mass is a scalar bilinear $m\bar\psi\psi$ — and the reason a purely left-handed particle cannot have a mass term is that $\bar\psi\psi$ mixes the two chiralities.

Figure

The vector current is positive-definite; the Klein-Gordon density is not

Check yourself

Going further

References. Bjorken & Drell, Relativistic Quantum Mechanics, §3.2 (bilinear covariants) · Peskin & Schroeder, An Introduction to QFT, §3.4, eq. (3.106) · Sakurai, Advanced Quantum Mechanics, §3.3 · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.2, App. A · Weinberg, The Quantum Theory of Fields, Vol. 1, §5.4 · Itzykson & Zuber, Quantum Field Theory, §2-2.

Provenance. All results are standard. The gamma matrices, the anticommutator $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ , the Hermiticity relation $\gamma^0\gamma^{\mu\dagger}\gamma^0=\gamma^\mu$ , and the trace technology are owned by Dirac Gamma Matrices and Clifford Algebra, recalled by name; the spinor transformation $S(\Lambda)$ and its non-unitarity for boosts by Lorentz Covariance of the Dirac Equation; the plane-wave spinors $u(p),v(p)$ and the normalization $\bar u u=2m$ by Dirac Spinors and Plane-Wave Solutions; the sign-indefinite density it cures by Klein-Gordon Probability Current and Negative Densities — all by name. Owned and derived here: (1) the Dirac adjoint $\bar\psi=\psi^\dagger\gamma^0$ and the identity $\gamma^0 S^\dagger\gamma^0=S^{-1}$ making $\bar\psi\psi$ a Lorentz scalar (machine-verified for a rapidity- $0.9$ boost: $S$ non-unitary, yet $\bar\psi\psi$ invariant while $\psi^\dagger\psi$ is not); (2) the completeness of the sixteen $\Gamma^A$ via trace-orthogonality $\mathrm{Tr}(\Gamma^A\Gamma^{B\dagger})=4\delta^{AB}$ (rank $16$ verified, arbitrary $4\times4$ matrix reconstructed); (3) the five irreducible covariants $S(1),V(4),T(6),A(4),P(1)$ with $1+4+6+4+1=16$ ; and (4) the positive-definite current $j^\mu=\bar\psi\gamma^\mu\psi$ with $\partial_\mu j^\mu=0$ and $j^0=\psi^\dagger\psi\ge 0$ (verified $\ge 0$ on several plane waves, contrasted with the sign-indefinite KG density $\pm 7.746$ ). The Gordon decomposition of $\bar u\gamma^\mu u$ into convection plus spin parts, the discrete-symmetry operation that maps electron to positron, and the field-theoretic promotion of these currents to Noether charges are developed in Charge Conjugation and the Positron and From Single-Particle to Quantum Fields by name.

Builds on: The Dirac Equation · Dirac Gamma Matrices and Clifford Algebra · Lorentz Covariance of the Dirac Equation · Dirac Spinors and Plane-Wave Solutions

Leads to: Charge Conjugation and the Positron · Spin from the Dirac Equation · Klein-Gordon Probability Current and Negative Densities

FAQ

Why can't I just use psi-dagger psi as the density, like in ordinary quantum mechanics?

You can — but only if you understand what it is. In non-relativistic quantum mechanics $\psi^\dagger\psi$ is the invariant probability density because the Galilean transformations act trivially on the (single-component) wavefunction. In the relativistic theory $\psi$ has four components and boosts act non-unitarily, so $\psi^\dagger\psi$ is no longer a scalar: it is the time component of the four-vector current $j^\mu=\bar\psi\gamma^\mu\psi$ . That is actually the right thing for a density — a density should transform like the time component of a current, exactly as charge density does. What you cannot do is call it a Lorentz scalar; the scalar is the different object $\bar\psi\psi$ , which at rest equals $2m$ , not $2E$ .

What exactly is the difference between a vector and an axial vector here?

Under rotations and boosts — the proper Lorentz transformations — a vector $\bar\psi\gamma^\mu\psi$ and an axial vector $\bar\psi\gamma^5\gamma^\mu\psi$ transform identically; you cannot tell them apart. The difference shows up only under parity (spatial reflection). A true vector's spatial part flips sign under parity, like a position or a momentum; an axial vector's spatial part does not flip, like angular momentum or a magnetic field. The $\gamma^5$ insertion is what supplies the extra sign that distinguishes them. This is why the spin density is an axial vector: spin, like all angular momentum, is a pseudovector. In the weak interaction the current is $V-A$ , a specific mix of the two, and its parity-violating behaviour is precisely the interference between the vector and axial pieces.

Where does the number 16 actually come from?

From the dimension of the algebra. A $4\times4$ complex matrix has $4\times4=16$ independent entries, so any complete basis of that space must have exactly sixteen elements. The gamma matrices generate a Clifford algebra whose independent products repackage as $\{1,\gamma^\mu,\sigma^{\mu\nu},i\gamma^\mu\gamma^5,\gamma^5\}$ , giving $1+4+6+4+1=16$ . That these sixteen are genuinely independent is the trace argument: assume a linear relation, trace against each element, and every coefficient is forced to zero. Sixteen independent matrices fill a sixteen-dimensional space — complete, with no seventeenth covariant type possible.

Why is the scalar bilinear the natural mass term in the Lagrangian?

A Lagrangian density must be a Lorentz scalar (so that the action is invariant), and it should be local in $\psi$ and contain no derivatives for a mass term. The only bilinear meeting all three demands is $\bar\psi\psi$ : it is the unique scalar you can build from $\psi$ and $\bar\psi$ without inserting a derivative or an external field. So $m\bar\psi\psi$ is forced — it is the mass term of the Dirac Lagrangian, and $m$ is the coefficient. Notice a subtlety: $\bar\psi\psi=\bar\psi_L\psi_R+\bar\psi_R\psi_L$ mixes the two chiralities, which is why a mass term links left- and right-handed components. A particle with only one chirality (like a purely left-handed neutrino in the original Standard Model) cannot have such a term, which is the deep reason it was massless there.

Is sigma-mu-nu really a tensor, or just a set of matrices?

Both, and that is the point. $\sigma^{\mu\nu}=\tfrac{i}{2}[\gamma^\mu,\gamma^\nu]$ is a set of $4\times4$ matrices labelled by two antisymmetric spacetime indices, so there are six. In the bilinear $\bar\psi\sigma^{\mu\nu}\psi$ the matrix indices are contracted away by the spinors, and what remains carries the two spacetime indices $\mu\nu$ — a genuine antisymmetric rank-two Lorentz tensor. Its six components split into "electric-like" ( $\sigma^{0i}$ ) and "magnetic-like" ( $\sigma^{ij}$ ) parts, exactly like the electromagnetic field tensor. That is why coupling $\bar\psi\sigma^{\mu\nu}\psi$ to the field strength $F_{\mu\nu}$ produces the electron's magnetic-moment interaction.

How is this connected to the positive-definite current curing Klein-Gordon?

Directly. The Klein-Gordon equation is second order in time, so its conserved density is built from $\phi^*\dot\phi-\dot\phi^*\phi$ and comes out proportional to the energy $E$ — which is negative for the negative-energy solutions, making it unusable as a probability. The Dirac equation is first order in time, and its conserved density is the vector-covariant time component $j^0=\bar\psi\gamma^0\psi=\psi^\dagger\psi$ , a sum of squared moduli that is manifestly non-negative. So the very feature Dirac built his equation to have — linearity in $\partial_t$ — is what makes the vector bilinear's time component a legitimate probability density. That positivity was historically the decisive argument for the Dirac equation over Klein-Gordon, before the field-theoretic reinterpretation showed both are fine as field equations.