Dirac Field Quantization and Propagators

The Dirac equation by itself is a relativistic wave equation for spinors. A quantum field theory of spin-$\frac12$ particles requires one more step: the spinor solution must become an operator-valued field. The modes of this field create and destroy particles and antiparticles, and the algebra of those modes must be fermionic.

This is the first place in the course where the sign rules of fermions become unavoidable. The scalar field was quantized by commutators. The Dirac field is quantized by anticommutators. That change is not cosmetic: it is what turns the negative-frequency solutions of the Dirac equation into positive-energy antiparticles, enforces the Pauli principle, and produces the minus signs in fermion loops and fermionic Wick contractions.

The practical goal is to connect three statements that are often learned separately: the $v_s(p)e^{+ip\cdot x}$ branch creates antiparticles, the equal-time anticommutator is local, and the momentum-space propagator has numerator $\gamma\cdot p+m$. They are not separate miracles; they are the same free-field construction viewed from three angles.

Required background. Dirac equation and spinor solutions supplies the plane-wave equations, Dirac adjoint, spinor normalization, and completeness relations used below.

The Dirac field as an operator

Dirac oscillator-mode normalization

Write $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$ and normalize the spinors by

$$\bar u_s(p)u_{s'}(p)=2m\delta_{ss'}, \qquad \bar v_s(p)v_{s'}(p)=-2m\delta_{ss'},$$

so that

$$\sum_s u_s(p)\bar u_s(p)=\gamma\cdot p+m, \qquad \sum_s v_s(p)\bar v_s(p)=\gamma\cdot p-m,$$

where $p^0=E_{\mathbf p}$ in both spin sums. This is the common relativistic spinor normalization.

The mode expansion below uses the $1/\sqrt{2E_{\mathbf p}}$ oscillator convention,

$$\{b_s(\mathbf p),b_{s'}^\dagger(\mathbf q)\} =(2\pi)^3\delta_{ss'}\delta^{(3)}(\mathbf p-\mathbf q),$$

and similarly for $d_s$. Covariantly normalized one-particle states are obtained by inserting the matching factor,

$$|p,s\rangle_{\rm cov}=\sqrt{2E_{\mathbf p}}\,b_s^\dagger(\mathbf p)|0\rangle,$$

and similarly for antiparticles. This is the same convention-moving discussed for scalar fields: physical amplitudes do not depend on where the $2E_{\mathbf p}$ factors are stored.

The free Dirac Lagrangian is

$$\mathcal L=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi.$$

The Euler–Lagrange equation is

$$(i\gamma^\mu\partial_\mu-m)\psi(x)=0.$$

Plane waves split into two families:

$$\psi(x)\sim u_s(p)e^{-ip\cdot x}, \qquad (\gamma\cdot p-m)u_s(p)=0,$$

and

$$\psi(x)\sim v_s(p)e^{+ip\cdot x}, \qquad (\gamma\cdot p+m)v_s(p)=0.$$

The label $s=1,2$ denotes the two spin states. In a relativistic quantum field theory both families must appear. If one tried to keep only the positive-frequency $u$ modes, the field anticommutator would fail to have the locality properties required by relativity. The $v$ modes are not optional bookkeeping; they are the antiparticle sector of the field.

The operator expansion is

$$\psi(x)=\sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}} \left( b_s(\mathbf p)u_s(p)e^{-ip\cdot x} +d_s^\dagger(\mathbf p)v_s(p)e^{+ip\cdot x} \right),$$

and

$$\bar\psi(x)=\sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}} \left( b_s^\dagger(\mathbf p)\bar u_s(p)e^{+ip\cdot x} +d_s(\mathbf p)\bar v_s(p)e^{-ip\cdot x} \right).$$

Here $b_s^\dagger(\mathbf p)$ creates a particle, $b_s(\mathbf p)$ annihilates a particle, $d_s^\dagger(\mathbf p)$ creates an antiparticle, and $d_s(\mathbf p)$ annihilates an antiparticle. This page uses the common canonical oscillator normalization, in which the factors $1/\sqrt{2E_{\mathbf p}}$ sit in the field expansion and the mode anticommutators contain no $2E_{\mathbf p}$.

The Dirac field contains a particle-annihilation term $b_s(\mathbf p)u_s(p)e^{-ip\cdot x}$ and an antiparticle-creation term $d_s^\dagger(\mathbf p)v_s(p)e^{+ip\cdot x}$. The two branches are tied together by Lorentz covariance and by locality.

The vacuum obeys

$$b_s(\mathbf p)|0\rangle=0, \qquad d_s(\mathbf p)|0\rangle=0.$$

With the oscillator normalization above, the simple Fock states are

$$|\mathbf p,s;\mathrm{particle}\rangle=b_s^\dagger(\mathbf p)|0\rangle, \qquad |\mathbf p,s;\mathrm{antiparticle}\rangle=d_s^\dagger(\mathbf p)|0\rangle.$$

They have noncovariant delta-function normalization. When these states are used in LSZ or invariant phase space formulas, replace them by the covariant states $\sqrt{2E_{\mathbf p}}\,b_s^\dagger|0\rangle$ and $\sqrt{2E_{\mathbf p}}\,d_s^\dagger|0\rangle$. Keeping this dictionary explicit prevents the most common factor-of-$2E$ mistake in spinor amplitudes.

The conserved number current is

$$j^\mu=\bar\psi\gamma^\mu\psi.$$

For a field of charge $q$, the normal-ordered charge is

$$Q=q\sum_s\int\frac{d^3p}{(2\pi)^3} \left( b_s^\dagger(\mathbf p)b_s(\mathbf p) -d_s^\dagger(\mathbf p)d_s(\mathbf p) \right).$$

Thus the $d^\dagger$ excitation has the opposite charge. For the electron field one usually takes $q=-e$, so the $b^\dagger$ excitation is an electron and the $d^\dagger$ excitation is a positron.

Why anticommutators are forced

The momentum conjugate to $\psi_\alpha$ is

$$\pi_\alpha(x)=\frac{\partial\mathcal L}{\partial(\partial_0\psi_\alpha)} =i\psi_\alpha^\dagger(x).$$

The momentum conjugate to $\psi_\alpha^\dagger$ vanishes. Thus the first-order Dirac Lagrangian is a constrained system: $\pi-i\psi^\dagger\simeq0$ and $\pi^\dagger\simeq0$ are second-class constraints. A full Dirac-bracket analysis leads to the equal-time relation below once the classical graded bracket is promoted to an operator anticommutator. Relativistic locality and a Hamiltonian bounded below select this fermionic quantization; first-order time evolution alone is not a proof of spin–statistics.

$$\{\psi_\alpha(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y),$$

with

$$\{\psi_\alpha(t,\mathbf x),\psi_\beta(t,\mathbf y)\}=0, \qquad \{\psi_\alpha^\dagger(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\}=0.$$

Equivalently, with the $1/\sqrt{2E_{\mathbf p}}$ mode expansion above, the mode operators satisfy

$$\{b_s(\mathbf p),b_{s'}^\dagger(\mathbf q)\} =(2\pi)^3\delta_{ss'}\delta^{(3)}(\mathbf p-\mathbf q),$$ $$\{d_s(\mathbf p),d_{s'}^\dagger(\mathbf q)\} =(2\pi)^3\delta_{ss'}\delta^{(3)}(\mathbf p-\mathbf q),$$

and all other anticommutators vanish. If one wants covariantly normalized one-particle states, define $|\mathbf p,s\rangle_{\mathrm{cov}}=\sqrt{2E_{\mathbf p}}\,b_s^\dagger(\mathbf p)|0\rangle$ and similarly for antiparticles. Then

$${}_{\mathrm{cov}}\langle \mathbf p',s'|\mathbf p,s\rangle_{\mathrm{cov}} =(2\pi)^3 2E_{\mathbf p}\delta^{(3)}(\mathbf p-\mathbf p')\delta_{ss'}.$$

The Pauli principle is immediate:

$$\{b_s^\dagger(\mathbf p),b_s^\dagger(\mathbf p)\}=2b_s^\dagger(\mathbf p)b_s^\dagger(\mathbf p)=0,$$

so one cannot create two identical fermions in exactly the same quantum state.

There is also a deeper energy argument. To manipulate coincident modes, first use a periodic box of volume $V$ and a finite momentum set $\Lambda$. The discrete oscillators have unit CAR, $\{d_{\mathbf p s},d^\dagger_{\mathbf q r}\}=\delta_{\mathbf p\mathbf q}\delta_{sr}$, rather than a Dirac delta. Before normal ordering, the free Hamiltonian is

$$H_\Lambda=\sum_{\mathbf p\in\Lambda,s}E_{\mathbf p} \left(b_s^\dagger(\mathbf p)b_s(\mathbf p)-d_s(\mathbf p)d_s^\dagger(\mathbf p)\right).$$

Using the anticommutator,

$$d_s(\mathbf p)d_s^\dagger(\mathbf p) =1-d_s^\dagger(\mathbf p)d_s(\mathbf p),$$

so the Hamiltonian becomes

$$H_\Lambda=\sum_{\mathbf p\in\Lambda,s}E_{\mathbf p} \left(b_s^\dagger(\mathbf p)b_s(\mathbf p)+d_s^\dagger(\mathbf p)d_s(\mathbf p)\right)+E_{\mathrm{vac}}.$$

Here $E_{\mathrm{vac}}=-\sum_{\mathbf p\in\Lambda,s}E_{\mathbf p}$ is finite before the regulator is removed. Normal ordering subtracts the selected free-vacuum energy. Both particles and antiparticles then carry positive excitation energy. If the $d$ modes obeyed ordinary commutators, the sign would not flip in this way, and the theory would contain negative-energy excitations. The continuum expression restores the volume factor in the vacuum constant Schwartz 2014, § 12.5.2, pp. 217–218, Eqs. (12.65)–(12.68); the canonical Dirac Hamiltonian treatment gives the finite-box construction.

For one unit-normalized discrete mode, $dd^\dagger=1-d^\dagger d$ turns the $d$ sector into a positive-energy antiparticle excitation plus its negative vacuum constant. The schematic figure shows the finite-mode algebra, not a replacement of a continuum delta by one.

This is the free-field manifestation of the spin–statistics connection. A full spin–statistics theorem is deeper, but already here locality, positivity of energy, and the exclusion principle are all pointing to the same algebra.

Checking the equal-time algebra

The nontrivial check is that the mode expansion really gives the local canonical anticommutator. At equal time, using the mode algebra, one finds

$$\{\psi_\alpha(t,\mathbf x),\bar\psi_\beta(t,\mathbf y)\} =\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf p}} \left[ (\gamma\cdot p+m)_{\alpha\beta}e^{+i\mathbf p\cdot(\mathbf x-\mathbf y)} +(\gamma\cdot p-m)_{\alpha\beta}e^{-i\mathbf p\cdot(\mathbf x-\mathbf y)} \right].$$

In the second term change integration variable $\mathbf p\mapsto-\mathbf p$. The spatial part of $\gamma\cdot p=\gamma^0E_{\mathbf p}-\boldsymbol\gamma\cdot\mathbf p$ changes sign, while the energy and mass terms do not. The two terms combine to

$$\{\psi_\alpha(t,\mathbf x),\bar\psi_\beta(t,\mathbf y)\} =\int\frac{d^3p}{(2\pi)^3} (\gamma^0)_{\alpha\beta} e^{+i\mathbf p\cdot(\mathbf x-\mathbf y)} =(\gamma^0)_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

Multiplying by $\gamma^0$ on the right gives

$$\{\psi_\alpha(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

The antiparticle contribution is essential in this cancellation. Without the $v$ modes, the equal-time algebra would not collapse to the local delta function.

The same statement extends away from equal time. Define the scalar Pauli–Jordan distribution

$$\Delta(z)=\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf p}} \left(e^{-ip\cdot z}-e^{+ip\cdot z}\right), \qquad p^0=E_{\mathbf p}.$$

Then the mode expansion gives

$$\{\psi_\alpha(x),\bar\psi_\beta(y)\} =\left(i\gamma^\mu\partial_{x^\mu}+m\right)_{\alpha\beta} \Delta(x-y).$$

This local $\Delta$ is the raw scalar commutator kernel; in the convention $[\phi(x),\phi(y)]=i\Delta_{\mathrm{PJ}}(x-y)$ it equals $i\Delta_{\mathrm{PJ}}$. Because this distribution and its derivatives vanish on the open spacelike region, the Dirac field satisfies fermionic microcausality. Both the $u$ and $v$ sectors are needed for this cancellation.

Fermionic time ordering

For bosons, time ordering only reorders operators by time. For fermions, each odd permutation contributes a minus sign. For two spinor fields,

$$T\{\psi_\alpha(x)\bar\psi_\beta(y)\} =\theta(x^0-y^0)\psi_\alpha(x)\bar\psi_\beta(y) -\theta(y^0-x^0)\bar\psi_\beta(y)\psi_\alpha(x).$$

The minus sign is the whole story. It is the same sign that appears when exchanging two fermionic creation operators, and it will become the sign attached to closed fermion loops in perturbation theory.

The Feynman propagator is

$$S_{F,\alpha\beta}(x-y) =\langle0|T\{\psi_\alpha(x)\bar\psi_\beta(y)\}|0\rangle.$$

Some books instead define $iS_F$ to be the time-ordered expectation value. Here $S_F$ itself is the expectation value, so its Fourier transform and the internal-line factor are

$$\frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon}.$$

For $x^0>y^0$, only $b b^\dagger$ contributes, so

$$S_F(x-y)=\sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf p}} u_s(p)\bar u_s(p)e^{-ip\cdot(x-y)}.$$

Using the spin sum,

$$S_F(x-y)=\int\frac{d^3p}{(2\pi)^3}\frac{\gamma\cdot p+m}{2E_{\mathbf p}} e^{-ip\cdot(x-y)}, \qquad x^0>y^0.$$

For $x^0<y^0$, the time-ordering sign and the $d d^\dagger$ contraction give

$$S_F(x-y)=-\sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf p}} v_s(p)\bar v_s(p)e^{+ip\cdot(x-y)}.$$

Equivalently,

$$S_F(x-y)=-\int\frac{d^3p}{(2\pi)^3}\frac{\gamma\cdot p-m}{2E_{\mathbf p}} e^{+ip\cdot(x-y)}, \qquad x^0<y^0.$$

These two pieces combine into the compact four-dimensional expression

$$S_F(x-y)=\int\frac{d^4p}{(2\pi)^4} \frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon} e^{-ip\cdot(x-y)}.$$

In momentum space,

$$S_F(p)=\frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon}.$$

One often writes this suggestively as $i/(\gamma\cdot p-m)$ with the Feynman prescription understood. The precise formula is the fraction above: the inverse of the Dirac operator is obtained by multiplying by $\gamma\cdot p+m$, while the $i\epsilon$ prescription belongs to the scalar denominator $p^2-m^2+i\epsilon$. The pole locations are the same as for a scalar particle; the numerator is now a spinor matrix.

A fermion propagator is a matrix in spinor indices. In momentum space the line carries $S_F(p)=i(\gamma\cdot p+m)/(p^2-m^2+i\epsilon)$. The arrow tracks fermion-number flow, not necessarily momentum flow.

The propagator is the Green function of the Dirac operator:

$$(i\gamma^\mu\partial_{x^\mu}-m)S_F(x-y)=i\delta^{(4)}(x-y)\mathbf 1_4.$$

This sign matches the scalar convention $D_F(p)=i/(p^2-m^2+i\epsilon)$ because $S_F(p)=(\gamma\cdot p+m)D_F(p)$ as a matrix-valued distribution.

This is the spinor analogue of the scalar equation

$$(\partial_x^2+m^2)G_F(x-y)=-i\delta^{(4)}(x-y).$$

For the Dirac field the differential operator is first order, but the pole structure is still the mass-shell pole $p^2=m^2$.

Contact terms from differentiating time ordering

The source of the delta function in the Green-function equation is the same as in the scalar case: differentiating the step functions in the time-ordered product creates a contact term. For fermions,

$$S_{F,\alpha\beta}(x-y) =\theta(x^0-y^0)\langle0|\psi_\alpha(x)\bar\psi_\beta(y)|0\rangle -\theta(y^0-x^0)\langle0|\bar\psi_\beta(y)\psi_\alpha(x)|0\rangle.$$

When $i\gamma^0\partial_{x^0}$ acts on the step functions, it produces

$$i\gamma^0\delta(x^0-y^0) \langle0|\{\psi(x),\bar\psi(y)\}|0\rangle.$$

Using the equal-time anticommutator,

$$\{\psi_\alpha(t,\mathbf x),\bar\psi_\beta(t,\mathbf y)\} =(\gamma^0)_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y),$$

this becomes precisely the four-dimensional contact term

$$i\delta^{(4)}(x-y)\delta_{\alpha\beta}.$$

Away from $x=y$, the two Wightman functions solve the homogeneous Dirac equation, so only the contact term remains. This is why the propagator is the inverse of the Dirac operator as a distribution, not as an ordinary function.

Fermionic Wick contractions

The contraction of a free Dirac field with its adjoint is defined by

$$[\psi_\alpha(x)\bar\psi_\beta(y)]_{\mathrm{contr}} =S_{F,\alpha\beta}(x-y).$$

There are no nonzero contractions between two $\psi$ fields or two $\bar\psi$ fields for a Dirac field with conserved charge:

$$[\psi_\alpha(x)\psi_\beta(y)]_{\mathrm{contr}}=0, \qquad [\bar\psi_\alpha(x)\bar\psi_\beta(y)]_{\mathrm{contr}}=0.$$

For an alternating displayed order, Wick’s theorem gives

$$\begin{aligned} &\langle0|T\{\psi_{\alpha_1}(x_1)\bar\psi_{\beta_1}(y_1) \psi_{\alpha_2}(x_2)\bar\psi_{\beta_2}(y_2)\}|0\rangle \\ &\qquad =S_{F,\alpha_1\beta_1}(x_1-y_1)S_{F,\alpha_2\beta_2}(x_2-y_2) -S_{F,\alpha_1\beta_2}(x_1-y_2)S_{F,\alpha_2\beta_1}(x_2-y_1). \end{aligned}$$

The direct pairing appears with a plus sign and the exchanged pairing with a minus sign. This alternating order is the determinant of the matrix of contractions $S_{F,\alpha_i\beta_j}(x_i-y_j)$. If the two $\psi$ fields are instead grouped before the two $\bar\psi$ fields, one odd interchange changes the overall sign:

$$\begin{aligned} &\langle0|T\{\psi_1\psi_2\bar\psi_1\bar\psi_2\}|0\rangle \\ &\qquad=-S_{11}S_{22}+S_{12}S_{21}. \end{aligned}$$

Thus the order written on the page is part of the formula; “the determinant sign” is not independent of that order.

A useful way to remember the rule is this:

$$\text{fermionic Wick sum} = \sum_{\text{pairings}}(-1)^{P}\prod_{\text{pairs}}S_F,$$

where $P$ is the parity of the permutation required to move the paired fermionic operators next to one another.

For the grouped order $\psi_1\psi_2\bar\psi_1\bar\psi_2$, the direct pairing has a minus sign and the exchanged pairing has a plus sign. Reordering the fields to $\psi_1\bar\psi_1\psi_2\bar\psi_2$ reverses the overall sign and yields the familiar determinant $S_{11}S_{22}-S_{12}S_{21}$.

For a Majorana field the distinction between particle and antiparticle is absent, and contractions between two Majorana fields are possible after inserting the charge-conjugation matrix. The sign principle is unchanged: every exchange of fermionic operators contributes a minus sign.

From operator algebra to the fermion line

The scalar propagator carried one number per momentum. The Dirac propagator carries a $4\times4$ matrix because it propagates spinor information. The matrix multiplication order is fixed by the orientation of the fermion line; reversing it is not an innocent cosmetic change. An internal fermion line in a Feynman diagram therefore contributes

$$\frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon},$$

with spinor indices contracted along the line. The arrow on the line is not decoration. It tells us how to multiply the matrices and how charge flows through the diagram.

A typical bilinear coupling to a scalar field has the form

$$\mathcal L_{\mathrm{int}}=-g\phi\bar\psi\psi.$$

A chain of fermion propagators and vertices then gives a matrix product such as

$$\bar u(p')\left[(-ig)S_F(k_1)(-ig)S_F(k_2)\cdots(-ig)\right]u(p),$$

where the order of matrices is the order along the fermion line. This is one of the first practical differences between scalar and spinor perturbation theory: spinor diagrams carry matrix algebra as well as momentum integrals.

A second difference is the closed-loop sign. Relative to the analogous bosonic contraction, closing a chain of fermionic contractions leaves one odd cyclic reordering. Consequently each independent closed fermion loop contributes a factor

$$-1.$$

The loop also involves a trace over spinor indices. Thus a closed fermion loop usually produces an expression of the schematic form

$$-\int\frac{d^4\ell}{(2\pi)^4}\mathrm{tr}\left[S_F(\ell)\Gamma_1S_F(\ell+k_1)\Gamma_2\cdots\right],$$

where the $\Gamma_i$ are the spinor matrices appearing in the vertices.

Summary

The quantized Dirac field is built from two kinds of ladder operators. The $b^\dagger$ operators create particles, while the $d^\dagger$ operators create antiparticles. Both appear in the same local spinor field.

The crucial new ingredient is the fermionic algebra. Equal-time anticommutators give a local delta function, enforce the exclusion principle, and convert the negative-frequency branch of the classical Dirac equation into positive-energy antiparticles. This is why the normal-ordered Hamiltonian contains

$$E_{\mathbf p}\left(b^\dagger b+d^\dagger d\right)$$

rather than a negative-energy antiparticle sector.

The Dirac propagator is the inverse of the first-order operator $i\gamma^\mu\partial_\mu-m$:

$$S_F(p)=\frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon}.$$

It has the same mass-shell denominator as the scalar propagator but a matrix numerator that carries spinor information. Fermionic time ordering introduces minus signs, and fermionic Wick contractions therefore have determinant-like signs rather than purely permanent-like bosonic signs.

Common pitfalls

A frequent mistake is to say that $v_s(p)$ is simply a negative-energy particle spinor. In the field operator, the $v_s(p)e^{+ip\cdot x}$ term multiplies $d_s^\dagger$, so it creates a positive-energy antiparticle.

Another common mistake is to forget the minus sign in fermionic time ordering. Without it, the Dirac propagator would not be the correct inverse of the Dirac operator, and Wick’s theorem would give wrong relative signs.

It is also easy to confuse momentum flow with arrow flow. The arrow on a fermion line tracks fermion-number flow and spinor-index contraction. Momentum may be assigned along or against the arrow, as long as it is done consistently.

Do not mix the oscillator normalization used in the mode expansion with the covariant state normalization used in cross sections. The conversion factor is $\sqrt{2E_{\mathbf p}}$ per external one-particle state.

Finally, the expression

$$\frac{i}{\gamma\cdot p-m}$$

is only shorthand for a matrix inverse with the scalar Feynman pole prescription understood. The explicit propagator is

$$\frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon}.$$

Exercises

Exercise 1: recovering the equal-time anticommutator

Starting from the mode expansion of $\psi(x)$ and the spin sums

$$\sum_s u_s(p)\bar u_s(p)=\gamma\cdot p+m, \qquad \sum_s v_s(p)\bar v_s(p)=\gamma\cdot p-m,$$

prove that

$$\{\psi_\alpha(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

Solution

At equal time,

$$\begin{aligned} \{\psi_\alpha(t,\mathbf x),\bar\psi_\beta(t,\mathbf y)\} &=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf p}} \left[ (\gamma\cdot p+m)_{\alpha\beta}e^{+i\mathbf p\cdot(\mathbf x-\mathbf y)} +(\gamma\cdot p-m)_{\alpha\beta}e^{-i\mathbf p\cdot(\mathbf x-\mathbf y)} \right]. \end{aligned}$$

In the second term let $\mathbf p\mapsto-\mathbf p$. Then

$$\gamma\cdot p=\gamma^0E_{\mathbf p}-\boldsymbol\gamma\cdot\mathbf p$$

becomes

$$\gamma^0E_{\mathbf p}+\boldsymbol\gamma\cdot\mathbf p.$$

Adding the two numerators gives $2E_{\mathbf p}\gamma^0$. Therefore

$$\{\psi_\alpha(t,\mathbf x),\bar\psi_\beta(t,\mathbf y)\} =(\gamma^0)_{\alpha\beta}\int\frac{d^3p}{(2\pi)^3} e^{+i\mathbf p\cdot(\mathbf x-\mathbf y)} =(\gamma^0)_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

Since $\bar\psi_\beta=\psi_\rho^\dagger(\gamma^0)_{\rho\beta}$, multiply the last equation by $\gamma^0$ on the right:

$$\{\psi_\alpha(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

Exercise 2: the Dirac propagator as a Green function

Show that

$$S_F(x-y)=\int\frac{d^4p}{(2\pi)^4} \frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon} e^{-ip\cdot(x-y)}$$

satisfies

$$(i\gamma^\mu\partial_{x^\mu}-m)S_F(x-y)=i\delta^{(4)}(x-y)\mathbf 1_4.$$

Solution

Acting on the exponential gives

$$i\partial_{x^\mu}e^{-ip\cdot(x-y)}=p_\mu e^{-ip\cdot(x-y)}.$$

Therefore

$$(i\gamma^\mu\partial_{x^\mu}-m)S_F(x-y) =\int\frac{d^4p}{(2\pi)^4} (\gamma\cdot p-m) \frac{i(\gamma\cdot p+m)}{p^2-m^2+i\epsilon} e^{-ip\cdot(x-y)}.$$

Using the Clifford algebra,

$$(\gamma\cdot p-m)(\gamma\cdot p+m)=p^2-m^2.$$

Thus

$$(i\gamma^\mu\partial_{x^\mu}-m)S_F(x-y) =\int\frac{d^4p}{(2\pi)^4} i\frac{p^2-m^2}{p^2-m^2+i\epsilon}e^{-ip\cdot(x-y)}.$$

As a distribution, the ratio tends to $1$ in the $\epsilon\to0^+$ prescription, so

$$(i\gamma^\mu\partial_{x^\mu}-m)S_F(x-y) =i\int\frac{d^4p}{(2\pi)^4}e^{-ip\cdot(x-y)} =i\delta^{(4)}(x-y)\mathbf 1_4.$$

Exercise 3: the first fermionic Wick sign

Use fermionic Wick’s theorem to compute

$$\langle0|T\{\psi_{\alpha}(x)\bar\psi_{\beta}(y)\psi_{\gamma}(z)\bar\psi_{\delta}(w)\}|0\rangle$$

in the free Dirac theory.

Solution

Only $\psi$–$\bar\psi$ contractions are nonzero. There are two pairings:

$$(\psi_\alpha(x),\bar\psi_\beta(y))(\psi_\gamma(z),\bar\psi_\delta(w))$$

and

$$(\psi_\alpha(x),\bar\psi_\delta(w))(\psi_\gamma(z),\bar\psi_\beta(y)).$$

The first gives

$$S_{F,\alpha\beta}(x-y)S_{F,\gamma\delta}(z-w).$$

For the second pairing, moving $\bar\psi_\delta(w)$ past $\psi_\gamma(z)$ costs one minus sign. Hence

$$\langle0|T\{\psi_{\alpha}(x)\bar\psi_{\beta}(y)\psi_{\gamma}(z)\bar\psi_{\delta}(w)\}|0\rangle = S_{F,\alpha\beta}(x-y)S_{F,\gamma\delta}(z-w) - S_{F,\alpha\delta}(x-w)S_{F,\gamma\beta}(z-y).$$

This is the two-by-two determinant sign pattern for charged fermions.

Exercise 4: positive antiparticle energy

Show how the anticommutator turns

$$H_d=-\sum_s\int\frac{d^3p}{(2\pi)^3}E_{\mathbf p} d_s(\mathbf p)d_s^\dagger(\mathbf p)$$

into a positive antiparticle Hamiltonian plus a constant.

Solution

Using

$$\{d_s(\mathbf p),d_{s'}^\dagger(\mathbf q)\} =(2\pi)^3\delta_{ss'}\delta^{(3)}(\mathbf p-\mathbf q),$$

first replace the continuum by the same finite box and momentum set as in the Hamiltonian derivation. For each discrete mode,

$$dd^\dagger=1-d^\dagger d.$$

Therefore

$$H_{d,\Lambda}=-\sum_{\mathbf p\in\Lambda,s}E_{\mathbf p} \left(1-d_s^\dagger(\mathbf p)d_s(\mathbf p)\right).$$

Thus

$$H_{d,\Lambda}=\sum_{\mathbf p\in\Lambda,s}E_{\mathbf p} d_s^\dagger(\mathbf p)d_s(\mathbf p)+E_{\mathrm{vac},\Lambda},$$

where

$$E_{\mathrm{vac},\Lambda}=-\sum_{\mathbf p\in\Lambda,s}E_{\mathbf p}.$$

At fixed UV cutoff, the large-volume limit gives

$$\frac{E_{\mathrm{vac},\Lambda}}{V} \longrightarrow-\sum_s\int_{\Lambda}\frac{d^3p}{(2\pi)^3}E_{\mathbf p}.$$

Thus the integral is an energy density, with mass dimension four; the total energy has the extra $V=(2\pi)^3\delta^{(3)}(0)$ in continuum notation. Removing the UV cutoff makes this free-vacuum constant divergent. Normal ordering leaves the positive excitation spectrum; it is not a calculation of gravitational vacuum energy.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. DOI.

Further reading

• Coleman, Sidney. Lectures of Sidney Coleman on Quantum Field Theory. Edited by Bryan Gin-ge Chen et al. World Scientific, 2019, chapters 20–21.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007, sections 37, 39, and 42.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, chapters 5–6.
• Zee, Anthony. Quantum Field Theory in a Nutshell. 2nd ed., Princeton University Press, 2010, chapter II.2.