Dirac Equation from the Spinor Lagrangian

Derivation

Dirac Equation from the Spinor Lagrangian

D-372 Home PU-402 Threads matter · fields · symmetry Depends on Euler-Lagrange Equations for Fields, Representations of the Lorentz Group

Statement

Starting from the Lorentz-scalar Lagrangian density \(\mathcal{L}=\bar\psi\left(i\gamma^\mu\partial_\mu-m\right)\psi\) for a four-component spinor field \(\psi\), with \(\bar\psi=\psi^\dagger\gamma^0\) and the Clifford algebra \(\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf{1}\), the Euler–Lagrange field equations yield the Dirac equation \((i\gamma^\mu\partial_\mu-m)\psi=0\), its adjoint, and — via the global \(U(1)\) phase symmetry — a conserved four-current \(j^\mu=\bar\psi\gamma^\mu\psi\) whose time component \(j^0=\psi^\dagger\psi\ge 0\) is positive-definite. Natural units \(\hbar=c=1\), metric \(\eta^{\mu\nu}=\mathrm{diag}(+,-,-,-)\).

Why it matters

The Dirac equation was engineered to cure the two diseases of the Klein–Gordon equation as a single-particle wave equation: a second-order time derivative that admits a probability density which is not positive-definite. By demanding a first-order equation whose square reproduces the relativistic dispersion \(E^2=\mathbf{p}^2+m^2\), Dirac was forced to introduce matrices obeying an anticommutation algebra — the Clifford algebra — and hence a multi-component field. Spin \(\tfrac12\), antiparticles, and the gyromagnetic ratio \(g\approx 2\) all fall out of this structure without being put in by hand.

Casting it as a variational principle is what makes it a field theory rather than a wave equation. The same Lagrangian delivers the equation of motion, its adjoint, the Noether current that becomes electric charge once gauged, and the canonical stress tensor — a unified accounting that is the entry point to quantum electrodynamics.

Assumptions

The field carries a finite-dimensional spinor index on which the \(\gamma^\mu\) act as constant matrices.If \(\psi\) were a scalar there is no non-trivial \(\gamma^\mu\); a first-order Lorentz-invariant equation with the correct dispersion becomes impossible and one is thrown back to Klein–Gordon. The gamma matrices satisfy \(\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf{1}\).Without this the squared operator does not collapse to \(\Box+m^2\); individual components fail to obey Klein–Gordon and the relativistic energy–momentum relation is lost. The action is real, enforced through \(\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0\) (equivalently \(\gamma^0\) Hermitian, \(\gamma^i\) anti-Hermitian).If the Dirac adjoint \(\bar\psi=\psi^\dagger\gamma^0\) is replaced by a naive \(\psi^\dagger\), the Lagrangian is not Lorentz-invariant and the conserved density is not \(\psi^\dagger\psi\); positive-definiteness of \(j^0\) is forfeited. \(\psi\) and \(\bar\psi\) are varied as independent fields in the action.Legitimate because the map \((\psi,\bar\psi)\leftrightarrow(\mathrm{Re}\,\psi,\mathrm{Im}\,\psi)\) is invertible; if one mistakenly imposes the reality constraint during variation, the equation of motion and its adjoint get entangled and the Noether analysis breaks.

Derivation

\[ S=\int d^4x\;\mathcal{L},\qquad \mathcal{L}=\bar\psi\left(i\gamma^\mu\partial_\mu-m\right)\psi,\qquad \bar\psi\equiv\psi^\dagger\gamma^0 \]

Write the action. Treat \(\psi\) and \(\bar\psi\) as independent field variables; the change of variables from \((\mathrm{Re}\,\psi,\mathrm{Im}\,\psi)\) is invertible so the stationarity conditions are equivalent. A

\[ \frac{\partial\mathcal{L}}{\partial\bar\psi}=\left(i\gamma^\mu\partial_\mu-m\right)\psi,\qquad \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar\psi)}=0 \]

Differentiate \(\mathcal{L}\) with respect to \(\bar\psi\) and its derivative. The Lagrangian contains \(\bar\psi\) undifferentiated only, so the momentum conjugate to \(\bar\psi\) vanishes. A

\[ \frac{\partial\mathcal{L}}{\partial\bar\psi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar\psi)}=0 \;\;\Longrightarrow\;\; \boxed{\left(i\gamma^\mu\partial_\mu-m\right)\psi=0} \]

Insert into the Euler–Lagrange field equation (assumed prior result). The second term is zero, leaving the Dirac equation directly. A

\[ \frac{\partial\mathcal{L}}{\partial\psi}=-m\bar\psi,\qquad \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}=i\bar\psi\gamma^\mu \;\;\Longrightarrow\;\; i\,\partial_\mu\bar\psi\,\gamma^\mu+m\bar\psi=0 \]

Now vary \(\psi\). Here \(\psi\) appears both bare (mass term) and differentiated (kinetic term), so both Euler–Lagrange terms contribute; the result is the adjoint Dirac equation. B

\[ \left(i\gamma^\nu\partial_\nu+m\right)\left(i\gamma^\mu\partial_\mu-m\right)\psi=-\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu\psi-m^2\psi=0 \]

Act with the conjugate operator on the Dirac equation. The cross terms \(\pm im\gamma^\mu\partial_\mu\) cancel identically, isolating the second-order piece. B

\[ \gamma^\nu\gamma^\mu\partial_\nu\partial_\mu=\tfrac12\{\gamma^\nu,\gamma^\mu\}\partial_\nu\partial_\mu=\eta^{\nu\mu}\partial_\nu\partial_\mu=\Box \;\;\Longrightarrow\;\; \left(\Box+m^2\right)\psi=0 \]

Symmetrise \(\partial_\nu\partial_\mu\) so only the symmetric part of \(\gamma^\nu\gamma^\mu\) survives, then apply the Clifford algebra. Each component obeys Klein–Gordon, confirming the dispersion \(E^2=\mathbf{p}^2+m^2\). This step is exactly where \(\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\) earns its keep. B

\[ \psi\to e^{-i\alpha}\psi,\quad \bar\psi\to e^{+i\alpha}\bar\psi\;\;(\alpha\ \text{const}) \;\;\Longrightarrow\;\; j^\mu=\bar\psi\gamma^\mu\psi \]

The Lagrangian is invariant under the global \(U(1)\) phase. Noether's theorem for an internal symmetry gives the current \(j^\mu=\dfrac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi/\delta\alpha=\bar\psi\gamma^\mu\psi\) (overall constant absorbed). B

\[ \partial_\mu j^\mu=(\partial_\mu\bar\psi)\gamma^\mu\psi+\bar\psi\gamma^\mu(\partial_\mu\psi)=(im\bar\psi)\psi+\bar\psi(-im\psi)=0 \]

Use the adjoint equation \(\partial_\mu\bar\psi\,\gamma^\mu=im\bar\psi\) and the Dirac equation \(\gamma^\mu\partial_\mu\psi=-im\psi\). The two mass terms cancel, so the current is conserved on-shell. B

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

Evaluate the time component using \((\gamma^0)^2=\eta^{00}\mathbf 1=\mathbf 1\). The density is a sum of squared moduli, hence positive-definite — the property Klein–Gordon lacked. A

Result

\[ \left(i\gamma^\mu\partial_\mu-m\right)\psi=0,\qquad \partial_\mu j^\mu=0,\qquad j^\mu=\bar\psi\gamma^\mu\psi,\qquad j^0=\psi^\dagger\psi\ge0 \]

Reading. One real Lagrangian produces three linked facts: the field's equation of motion (Dirac), a locally conserved four-current, and the guarantee that its density is a genuine probability — never negative. The first-order form and the anticommuting \(\gamma^\mu\) are not decorative; they are precisely what makes \(j^0\) a sum of squares while still reproducing the relativistic energy–momentum relation of Klein–Gordon.

Units check. With \(\hbar=c=1\), the action is dimensionless so \([\mathcal L]=M^4\); a spinor field carries \([\psi]=M^{3/2}\). Then \([\bar\psi\gamma^\mu\partial_\mu\psi]=M^{3/2}\cdot M\cdot M^{3/2}=M^4\) and \([m\bar\psi\psi]=M\cdot M^3=M^4\) match. The current has \([j^\mu]=M^3=L^{-3}\), a number density, so \(\int d^3x\,j^0\) is dimensionless — a pure count of probability. Restoring constants, the equation reads \((i\hbar\gamma^\mu\partial_\mu-mc)\psi=0\), each term having the dimension of momentum.

Limiting cases

Massless limit \(m\to0\): the equation decouples into two Weyl equations \(i\bar\sigma^\mu\partial_\mu\psi_L=0\), \(i\sigma^\mu\partial_\mu\psi_R=0\); chirality becomes conserved and left/right handed fields evolve independently.
Non-relativistic limit \(|\mathbf p|\ll m\): expanding the positive-energy solution to leading order in \(v/c\) reproduces the Pauli equation for a two-component spinor, and \(j^0=\psi^\dagger\psi\) reduces to the Schrödinger probability density \(|\psi|^2\).
Plane-wave / free field: \(\psi=u(p)e^{-ip\cdot x}\) gives \((\gamma^\mu p_\mu-m)u=0\), forcing \(p^2=m^2\); solutions organise into two positive- and two negative-energy branches (particle/antiparticle).
Static/zero-momentum \(\mathbf p=0\): the equation becomes \(i\gamma^0\partial_t\psi=m\psi\), whose four solutions are the rest-frame spin-up/down particle and antiparticle states with \(E=\pm m\).

Breaks when

Strong or rapidly varying external fields — the single-particle interpretation collapses. When field energies approach \(2m\) (the Schwinger regime \(E\sim m^2/e\)), pair creation makes particle number ambiguous; \(j^0\) is still conserved but can no longer be read as a one-particle probability, and only the quantised field theory is consistent. The Klein paradox at a step of height \(>2m\) is the sharp signal.
Curved spacetime with generic geometry — the flat-space \(\gamma^\mu\) fail. \(\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\) must be promoted to position-dependent \(\{\gamma^a,\gamma^b\}=2\eta^{ab}\) via a vierbein \(e^\mu{}_a\), and \(\partial_\mu\) to a spin-covariant derivative \(D_\mu=\partial_\mu+\tfrac14\omega_{\mu ab}\gamma^a\gamma^b\); the naive equation is neither generally covariant nor unique.
Interactions turned on without gauging — the current is no longer conserved. Adding a term that breaks the global \(U(1)\) (e.g. a Majorana mass \(\bar\psi^c\psi\)) destroys \(\partial_\mu j^\mu=0\); conservation returns only for symmetries the full Lagrangian actually possesses.

Failure modes

Using \(\psi^\dagger\) instead of \(\bar\psi=\psi^\dagger\gamma^0\). The bilinear \(\psi^\dagger\psi\) is not a Lorentz scalar; the Lagrangian would not be invariant and the mass term would be wrong. \(\bar\psi\psi\) is the scalar, \(\psi^\dagger\psi=j^0\) is the density (time component of a vector).
Dropping the second Euler–Lagrange term when varying \(\psi\). Because \(\psi\) is differentiated in the kinetic term, forgetting \(-\partial_\mu[\partial\mathcal L/\partial(\partial_\mu\psi)]\) loses the derivative in the adjoint equation and gives the algebraically wrong \(m\bar\psi=0\).
Treating \(\gamma^\mu\) as commuting. Writing \(\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu=\gamma^\mu\gamma^\mu\partial_\mu\partial_\mu\) and cancelling by hand ignores the anticommutator; the symmetrisation step and the appearance of \(\eta^{\mu\nu}\) are the whole point.
Sign errors from the metric convention. With \((-,+,+,+)\) the Lagrangian and Clifford algebra pick up signs; mixing conventions mid-derivation flips the mass term or ruins \((\gamma^0)^2=+\mathbf 1\), which then breaks \(j^0\ge0\).
Confusing spinor index with spacetime index. \(\gamma^\mu\) carries one spacetime index \(\mu\) and two suppressed spinor indices; summing \(\mu\) over spinor components, or vice versa, is dimensionally and structurally meaningless.

Discussion

The deepest lesson of this derivation is that positivity of probability and the correct relativistic dispersion are not independent inputs — they are reconciled by a single algebraic demand. Requiring a first-order operator whose square is \(\Box+m^2\) forces \(\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\), and once the \(\gamma^\mu\) are anticommuting matrices the natural inner product \(\psi^\dagger\psi\) is automatically a sum of squares. Klein–Gordon fails precisely because its first integral of motion, being built from a second-order equation, mixes \(\psi\) and \(\partial_t\psi\) and so cannot be sign-definite.

The four components are not four particles. Under the Lorentz group \(\psi\) transforms in the reducible \((\tfrac12,0)\oplus(0,\tfrac12)\) representation (the prior "Lorentz Lie-algebra representations" result), packaging a left- and a right-handed Weyl spinor. The mass term \(m\bar\psi\psi\) is exactly the coupling that ties these two chiralities together; setting \(m=0\) unlinks them, which is why chirality is a good quantum number only for massless fermions and why the Standard Model's chiral gauge structure is so tightly constrained.

The conserved current \(j^\mu=\bar\psi\gamma^\mu\psi\) is the seed of gauge interaction. Promoting the global \(U(1)\) to a local one, \(\alpha\to\alpha(x)\), forces the replacement \(\partial_\mu\to\partial_\mu+ieA_\mu\) and generates the coupling \(-e\,j^\mu A_\mu\): electromagnetism is the minimal price of making the phase symmetry local. The same current, integrated, is conserved electric charge.

At the quantum level the "probability current" is reinterpreted. The Dirac field is quantised with anticommutators, \(\{\psi_a(\mathbf x),\psi_b^\dagger(\mathbf y)\}=\delta_{ab}\delta^3(\mathbf x-\mathbf y)\), which is what the spin-statistics theorem demands for the energy to be bounded below. Then \(Q=\int d^3x\,j^0\) becomes the charge operator counting particles minus antiparticles, not a probability, and the once-troubling negative-energy solutions are reinterpreted as positive-energy antiparticles via the Feynman–Stückelberg prescription. The single-particle equation is thus a low-energy shadow of a many-body field theory.

Common misconceptions. "\(\bar\psi\psi\) and \(\psi^\dagger\psi\) are the same thing" — no: the first is a Lorentz scalar (appears in the mass term), the second is the time component of the vector current (the density). "The Dirac equation predicts spin \(\tfrac12\) as an assumption" — spin emerges from the transformation properties forced by the \(\gamma^\mu\), it is not inserted. "Negative-energy solutions are unphysical and should be discarded" — discarding them makes the solution set incomplete; they are essential and become antiparticles.

Worked examples

\[ \text{Plane wave } \psi=u(p)\,e^{-ip\cdot x},\quad p\cdot x=p_\mu x^\mu=Et-\mathbf p\cdot\mathbf x \]

Find the energy of a free electron with momentum \(|\mathbf p|=1.00\ \text{MeV}/c\). Substitute the plane wave into the Dirac equation. A

\[ \partial_\mu\psi=-ip_\mu\psi\;\Rightarrow\;(\gamma^\mu p_\mu-m)\,u(p)=0 \]

Each derivative brings down \(-ip_\mu\); the common exponential and a factor \(i\) cancel, leaving an algebraic (momentum-space) equation for the spinor \(u\). A

\[ (\gamma^\nu p_\nu+m)(\gamma^\mu p_\mu-m)u=(p^2-m^2)u=0\;\Rightarrow\;p^2=m^2 \]

Multiply by the conjugate operator and use the Clifford algebra exactly as in Steps 5–6; a non-trivial \(u\) exists only on the mass shell. B

\[ E=\sqrt{|\mathbf p|^2c^2+m^2c^4}=\sqrt{(1.00)^2+(0.511)^2}\ \text{MeV}=\sqrt{1.261}\ \text{MeV} \]

Insert numbers: electron rest energy \(mc^2=0.511\ \text{MeV}\), \(|\mathbf p|c=1.00\ \text{MeV}\). A

\[ E=1.12\ \text{MeV} \]

Reading. The Lagrangian's kinematic content is nothing more than \(E^2=(pc)^2+(mc^2)^2\); at \(p\sim 2mc\) the electron is already relativistic, \(E\approx 2.2\,mc^2\).

\[ j^\mu=\bar\psi\gamma^\mu\psi,\qquad \bar u\,\gamma^\mu u=2p^\mu \ \ (\text{normalisation } u^\dagger u=2E) \]

For the same electron, compute the probability current and hence the transport velocity \(\mathbf v=\mathbf j/j^0\). Use the standard positive-energy spinor bilinear. B

\[ j^\mu=2p^\mu\;\Rightarrow\; j^0=2E,\quad \mathbf j=2\mathbf p \;\Rightarrow\; \mathbf v=\frac{\mathbf j}{j^0}=\frac{\mathbf p}{E} \]

The overall factor \(2\) cancels in the ratio; the group velocity of the wave equals \(\mathbf p/E\), matching the relativistic particle velocity. B

\[ \frac{v}{c}=\frac{|\mathbf p|c}{E}=\frac{1.00\ \text{MeV}}{1.12\ \text{MeV}}=0.890 \]

Insert \(E=1.12\ \text{MeV}\) from Worked Example 1 and \(|\mathbf p|c=1.00\ \text{MeV}\). A

\[ v=0.890\,c=2.67\times10^{8}\ \text{m/s} \]

Reading. The conserved current is not an abstraction: its ratio of spatial to time component is the physical velocity of the packet. Positive-definiteness of \(j^0=2E>0\) also guarantees \(|\mathbf v|<c\) for any on-shell state.

Problems

From \(\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf 1\) with \(\eta=\mathrm{diag}(+,-,-,-)\), show that \((\gamma^0)^2=\mathbf 1\) and \((\gamma^i)^2=-\mathbf 1\), and that \(\gamma^0\gamma^i=-\gamma^i\gamma^0\).
Solution
Set \(\mu=\nu=0\): \(\{\gamma^0,\gamma^0\}=2(\gamma^0)^2=2\eta^{00}\mathbf 1=2\mathbf 1\), so \((\gamma^0)^2=\mathbf 1\). Set \(\mu=\nu=i\) (no sum): \(2(\gamma^i)^2=2\eta^{ii}\mathbf 1=-2\mathbf 1\), so \((\gamma^i)^2=-\mathbf 1\). For \(\mu=0,\nu=i\) with \(\eta^{0i}=0\): \(\gamma^0\gamma^i+\gamma^i\gamma^0=0\), i.e. they anticommute.
By varying \(\psi\) in \(\mathcal L=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi\), derive the adjoint Dirac equation \(i\,\partial_\mu\bar\psi\,\gamma^\mu+m\bar\psi=0\).
Solution
Compute \(\partial\mathcal L/\partial\psi=-m\bar\psi\) (only the mass term contains bare \(\psi\)) and \(\partial\mathcal L/\partial(\partial_\mu\psi)=i\bar\psi\gamma^\mu\) (from the kinetic term). The Euler–Lagrange equation \(\partial\mathcal L/\partial\psi-\partial_\mu[\partial\mathcal L/\partial(\partial_\mu\psi)]=0\) gives \(-m\bar\psi-\partial_\mu(i\bar\psi\gamma^\mu)=0\). Since \(\gamma^\mu\) is constant, \(-m\bar\psi-i(\partial_\mu\bar\psi)\gamma^\mu=0\); multiply by \(-1\) to obtain \(i\,\partial_\mu\bar\psi\,\gamma^\mu+m\bar\psi=0\).
Using the Dirac equation and its adjoint, show explicitly that \(\partial_\mu(\bar\psi\gamma^\mu\psi)=0\).
Solution
\(\partial_\mu j^\mu=(\partial_\mu\bar\psi)\gamma^\mu\psi+\bar\psi\gamma^\mu(\partial_\mu\psi)\). From the adjoint equation, \((\partial_\mu\bar\psi)\gamma^\mu=im\bar\psi\); from the Dirac equation, \(\gamma^\mu\partial_\mu\psi=-im\psi\). Substituting, \(\partial_\mu j^\mu=(im\bar\psi)\psi+\bar\psi(-im\psi)=im\bar\psi\psi-im\bar\psi\psi=0\).
A free electron (\(mc^2=0.511\ \text{MeV}\)) has momentum \(|\mathbf p|c=0.300\ \text{MeV}\). Find its energy and speed \(v/c\).
Solution
\(E=\sqrt{(pc)^2+(mc^2)^2}=\sqrt{0.300^2+0.511^2}=\sqrt{0.0900+0.2611}=\sqrt{0.3511}=0.593\ \text{MeV}\). Speed \(v/c=|\mathbf p|c/E=0.300/0.593=0.506\), i.e. \(v=0.506c=1.52\times10^8\ \text{m/s}\).
Prove the trace identity \(\mathrm{Tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu}\) for \(4\times4\) gamma matrices, using only the Clifford algebra and cyclicity of the trace.
Solution
From the algebra, \(\gamma^\mu\gamma^\nu=2\eta^{\mu\nu}\mathbf 1-\gamma^\nu\gamma^\mu\). Take the trace: \(\mathrm{Tr}(\gamma^\mu\gamma^\nu)=2\eta^{\mu\nu}\mathrm{Tr}(\mathbf 1)-\mathrm{Tr}(\gamma^\nu\gamma^\mu)\). By cyclicity \(\mathrm{Tr}(\gamma^\nu\gamma^\mu)=\mathrm{Tr}(\gamma^\mu\gamma^\nu)\), so \(2\,\mathrm{Tr}(\gamma^\mu\gamma^\nu)=2\eta^{\mu\nu}\cdot 4\), giving \(\mathrm{Tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu}\) (using \(\mathrm{Tr}\,\mathbf 1=4\) in four dimensions).