Dirac Propagators

The free Dirac propagator has the scalar mass-shell denominator and a matrix numerator that carries the spinor equation. The numerator alone does not determine its boundary condition. This page constructs the retarded, advanced, and Feynman kernels, checks their delta sources, and relates the pole residues to normalized spinors.

Required background. Klein–Gordon Propagators fixes the scalar signs and prescriptions; Free Dirac Spinors gives spin sums; The Dirac Hamiltonian as an Operator gives energy projectors. Helpful background. Relativistic Green Functions explains the background and source interpretation.

Factoring the free differential operator

Use $\hbar=c=1$, metric $(+,-,-,-)$, and Fourier kernel $e^{-ip\cdot x}$. Initially $m>0$. Define

$$D=i\gamma^\mu\partial_\mu-m, \qquad D_+=i\gamma^\mu\partial_\mu+m.$$

The Clifford relation and commuting partial derivatives give

$$DD_+=D_+D=-(\Box+m^2)I_4=-LI_4.$$

Let the scalar kernel satisfy $LG_X=\delta^4$. Then the Dirac unit-source inverse is

$$K_X=-D_+G_X, \qquad DK_X=\delta^4I_4.$$

The minus sign follows from the operator product. In particular, $G_F(p)=-1/(p^2-m^2+i0)$ implies

$$\widetilde K_F(p)=\frac{p\!\!\!/+m}{p^2-m^2+i0}.$$

The retarded and advanced versions replace the denominator by $(p^0\pm i0)^2-E_{\mathbf p}^2$, respectively. The numerator is unchanged. These are distributional limits: multiplying by the differential operator gives the delta identity after taking the prescribed limit, not an exact identity with a finite regulator left in the denominator.

For the conventional time-ordered spinor correlation define

$$S_F=iK_F,\qquad \widetilde S_F(p)=\frac{i(p\!\!\!/+m)}{p^2-m^2+i0}.$$

It satisfies $DS_F=+i\delta^4I_4$. The scalar convention instead has $LD_F=-i\delta^4$. The two conventions are reconciled by $S_F=D_+D_F$ and $G_F=iD_F$. The different source signs are consequences of factorization, not a disagreement between scalar and spinor Fourier transforms.

Energy projectors resolve the poles

At fixed spatial momentum write

$$\begin{aligned} H&=\boldsymbol\alpha\cdot\mathbf p+\beta m,\\ E&=\sqrt{\mathbf p^2+m^2},\\ P_\pm&=\frac12\left(I_4\pm\frac HE\right). \end{aligned}$$

Since $D=\beta(i\partial_t-H)$, the inverse in frequency space is a Hamiltonian resolvent followed on the right by $\beta$. For the Feynman boundary choice,

$$\widetilde K_F(p^0,\mathbf p)= \left[ \frac{P_+}{p^0-E+i0} +\frac{P_-}{p^0+E-i0} \right]\beta.$$

To check the rational expression away from its poles, combine the fractions: $(p^0+H)\beta=p\!\!\!/+m$. The opposing prescriptions encode the frequency selection. For $K_R$ use $+i0$ in both linear denominators; for $K_A$ use $-i0$ in both.

Using the scalar contour integrals gives

$$\begin{aligned} K_R(t,\mathbf p)&=-i\theta(t)e^{-iHt}\beta,\\ K_A(t,\mathbf p)&=+i\theta(-t)e^{-iHt}\beta, \end{aligned}$$

and

$$K_F(t,\mathbf p)= \left[-i\theta(t)e^{-iEt}P_+ +i\theta(-t)e^{iEt}P_-\right]\beta.$$

The retarded kernel contains both energy projectors at positive time. It is not a positive-energy projector followed by retarded evolution. Keeping only $P_+$ changes both its source and its causal properties.

All three kernels have the same equal-time jump:

$$K_X(0^+,\mathbf p)-K_X(0^-,\mathbf p)=-i\beta.$$

Hence $i\beta$ times the jump is $I_4$, as required by $DK_X=\delta(t)I_4$ at fixed $\mathbf p$. This checks the sign independently of residue orientation. No convention for the value of $\theta(0)$ changes the distributional jump.

Spin sums and negative-time propagation

Let $p=(E,\mathbf p)$ be future directed. With $u^\dagger u=v^\dagger v=2E$,

$$\begin{aligned} P_+(\mathbf p)\beta &=\frac1{2E}\sum_s u_s(p)\bar u_s(p),\\ P_-(\mathbf p)\beta &=\frac1{2E}\sum_s v_s(E,-\mathbf p)\bar v_s(E,-\mathbf p). \end{aligned}$$

The negative-sector spinor on the second line has future-directed label $(E,-\mathbf p)$: its mode has canonical spatial momentum $+\mathbf p$. This is the label reversal established on the free-spinor page.

Multiplying $K_F$ by $i$, Fourier transforming spatially, and changing $\mathbf p\to-\mathbf p$ only in the negative-time integral yields

$$\begin{aligned} S_F(x)={}&\theta(t)\sum_s\int d\Pi_p\, u_s(p)\bar u_s(p)e^{-ip\cdot x}\\ &-\theta(-t)\sum_s\int d\Pi_p\, v_s(p)\bar v_s(p)e^{ip\cdot x}. \end{aligned}$$

Here $d\Pi_p=d^3p/[(2\pi)^3\,2E]$. After free-field quantization, this is precisely

$$(S_F)_{ab}(x-y) =\langle0|T\{\widehat\psi_a(x) \widehat{\bar\psi}_b(y)\}|0\rangle.$$

For $x^0<y^0$, fermionic time ordering exchanges the two fields with a minus sign. The two spin sums therefore enter with different time-ordering signs even though their Hilbert densities are positive. Tong’s sections 5.4–5.5 give this field interpretation.

The inverse calculation can be performed with ordinary matrix-valued distributions before any quantization. Identifying the result with the displayed vacuum expectation value additionally uses the free vacuum and anticommutation relations. It is not a one-particle transition probability or a retarded signal amplitude.

Causality and the background limitation

Applying a local differential operator cannot enlarge the support of a distribution. Since $G_R$ is supported in the future light cone, $K_R=-D_+G_R$ has the same causal-support restriction. The advanced kernel has the opposite restriction. The Feynman inverse generally has spacelike correlations; the distinction is explained in Locality and Causality Warnings.

The simple scalar construction relies on commuting derivatives. In an electromagnetic background put $\mathcal D_\mu=\partial_\mu+iqA_\mu$ and $\sigma^{\mu\nu}=\tfrac i2[\gamma^\mu,\gamma^\nu]$. Then $[\mathcal D_\mu,\mathcal D_\nu]=iqF_{\mu\nu}$ gives

$$\begin{aligned} &(i\gamma\cdot\mathcal D-m)(i\gamma\cdot\mathcal D+m)\\ &\qquad=-\mathcal D_\mu\mathcal D^\mu-m^2 -\frac q2\sigma^{\mu\nu}F_{\mu\nu}. \end{aligned}$$

Thus the squared equation contains a spin-field matrix. Replacing the free scalar Green function by a minimally coupled scalar inverse while leaving the same numerator does not generally invert the Dirac operator. A second-order matrix inverse or the ordered construction on Relativistic Green Functions is required.

Exercises

1. Evaluate the three mixed time–momentum kernels at $\mathbf p=0$. Which Dirac-basis blocks occur on each time side of $K_F$?

Solution

$H=m\beta$ and $P_\pm=(I\pm\beta)/2$. For $t>0$, $K_F=-ie^{-imt}P_+$, supported in the upper block. For $t<0$, $K_F=ie^{imt}P_-\beta=-ie^{imt}P_-$, supported in the lower block. Their jump is $-i(P_+-P_-)=-i\beta$. $K_R=-i\theta(t)e^{-im\beta t}\beta$ and $K_A=i\theta(-t)e^{-im\beta t}\beta$ contain both blocks on their respective time side.

2. A proposed retarded inverse is $K_{\rm trial}=-i\theta(t)e^{-iEt}P_+\beta$. What source does it have?

Solution

Away from zero it solves the free Dirac equation. Its delta coefficient is $\beta P_+\beta$, because the derivative contributes $i\beta(-iP_+\beta)$. This is a rank-two projector, not $I_4$. It cannot respond correctly to a general four-component source.

3. Does the construction require division by $m$? State its massless limit.

Solution

No. The covariant numerator becomes $p\!\!\!/$ and the denominator $p^2$ with the same prescription. The Hamiltonian projectors use $E=|\mathbf p|$ and are defined for nonzero momentum. Their pointwise formula is undefined at $\mathbf p=0$; the full massless Green functions are understood distributionally. The construction does not use a $1/(2m)$ spinor normalization.

References

• Bjorken, James D., and Sidney D. Drell. Relativistic Quantum Mechanics. McGraw–Hill, 1964. Free-particle Green functions and propagator boundary conditions.
• Thaller, Bernd. The Dirac Equation. Springer, 1992. doi:10.1007/978-3-662-02753-0. Spectral projectors and electromagnetic Dirac operators.
• Tong, David. Lectures on Quantum Field Theory. University of Cambridge, 2006, sections 5.4–5.5. Dirac propagators.