I'm interested in the retarded propagator for a free massless Dirac fermion, i.e. solutions $ψ$ to the inhomogeneous PDE

$$ (∂_t- \nabla·\vec σ) ψ(x,t) = f(x,t) $$

with boundary conditions

$$\quad ψ(x,t) \to 0 \text{ for } t \to -∞$$

where $\vec σ = (σ_1,σ_2,σ_3)^T$ are the three Pauli matrices. (The boundary conditions can be even more restrictive, I just want the solution to decay sufficiently quickly at infinity so that it becomes unique and has a well-defined Fourier transform.)

Now, solving the Dirac equation is a standard exercise in virtually every QFT book, but all the books I've looked at only consider the Fourier transform of the propagator.

However, I am interested in the real space formula for the retarded propagator

Using the retarded propagator for the wave equation in $3+1$ dimensions, we can write

$$ ψ(x,t) = (∂_t + \nabla·\vec σ)(∂_t^2 - \nabla^2)^{-1} f(x,t) $$

$$ = (∂_t + \nabla·\vec σ) \frac1{4π·\text{something}}∫d^3x'dt' \frac1{|x-x'|}\delta(|x-x'|-|t-t'|) f(x',t')$$

but this formula strikes me as seriously weird: carrying out the differentiation with respect to $x$ and $t$ will differentiate the $\delta$-function in the integral, which means that the solution depends on the derivatives of the function $f$. This goes against my intuition that a linear first-order PDE should depend on the initial values directly, and not on their time and space derivatives!

Is there a reference where I can find a discussion of the retarded propagator of the (massless) Dirac equation in real space?

  • $\begingroup$ You are summing on all $t'$, so your expression does not depends on a value of the time derivative of $f$ at one precise initial time $t_0$, but for all $t'$. $\endgroup$ – Trimok Oct 17 '13 at 10:50
  • $\begingroup$ @Trimok: We can choose $f(x',t') = g(x')\delta(t'-t_0)$ to model initial conditions at a time $t_0$. In general, the propagator for a driving force $f$ can always be used to construct the propagator for initial conditions $ψ(x,t_0)$. $\endgroup$ – Greg Graviton Oct 18 '13 at 8:08
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    $\begingroup$ "This goes against my intuition that a linear first-order PDE should depend on the initial values directly, and not on their time and space derivatives!" . In fact, one may always rewrite an expression with explicit first derivatives. Take for instance the differential equation $\frac{d\psi(x)}{dx} = f(x)$. The solution is $\psi(x) = \psi(0)+\int_0^x du f(u)$. However, with an integration by parts, this could be written $\psi(x) = \psi(0)+ xf(x) -\int_0^x du ~u ~f'(u)$, or even $\psi(x) = \psi(0)+ xf(0) + x\int_0^x du ~f'(u) -\int_0^x du ~u ~f'(u)$ $\endgroup$ – Trimok Oct 26 '13 at 19:01
  • $\begingroup$ Your formula is correct. You may also use an alternative form of the massless scalar retarded propagator, this will give you : $\psi(x) = (\sigma.\partial)\int d^4x' \frac{1}{2 \pi} \delta((x-x')^2)\theta(x_0 - x'_0)f(x')$ $\endgroup$ – Trimok Oct 29 '13 at 6:58
  • $\begingroup$ @Trimok: Moving the differentiation under the integral sign would mean that the propagator involves the derivative of a delta function, though, which I find odd. You make the argument that you "involving a derivative" is not an invariant property of the propagator, which I can't deny. One way to formalize it is to ask whether the solution changes when we add a constant to the function $f \to f + C$. (But this may be forbidden due to boundary conditions.) $\endgroup$ – Greg Graviton Oct 29 '13 at 14:15

The real space propagator for the massive Dirac fermion in $3+1$ dimensions is calculated in R. Feynman's book Quantum Electrodynamics (Lecture 17, page 84 in the edition linked to).

The result is very much as indicated in the question: first solve the wave equation, then differentiate the solution with the Dirac operator again. In particular, Feynman calculates the propagator of the Klein-Gordon equation in real space:

$$ I_+(t,x) = ∫ \frac{d^4p}{(2π)^4} \frac{\exp[-i(p\cdot x)]}{p^2 - m^2 + i\varepsilon} = -(4π)^{-1} \delta(s^2) + \frac{m}{8πs}H_1^{(2)}(ms)$$

Here, $s = +(t^2-x^2)^{1/2}$ for $t>|x|$ and $s = -i(x^2-t^2)^{1/2}$ for $t < |x|$. Moreover, $\delta(s^2)$ is a delta function and $H^{(2)}_1(ms)$ is a Hankel function. Then, you have to differentiate the delta function, indeed.

However, note that to be physically meaningful, the propagator $G(x_2,t_2;x_1,t_1)$ for the Dirac equation should only take into account the positive energy states for the retarded time frame $t_2 - t_1 > 0$, while the advanced portion $t_2 - t_1 < 0$ should only take into account the negative energy eigenstates (holes).


maybe you already thought about it, but I'll ask anyway: why don't you try to tackle directly the differential operator: $\partial_t -\vec{\sigma}\cdot\nabla$ ? It's a first order PDE so you would get something of the kind: $$ \psi(x,t) = \psi_0(x,t) + \int G(x,x',t,t')f(x',t')\, dx'dt' \qquad \text{(1)} $$ Where $\psi_0(x,t)$ is a solution of the homogeneous equation and $G(x,x',t,t')$ is the solution of $$ (\partial_t -\vec{\sigma}\cdot\nabla)G(x,x',t,t') = \delta(x-x')\delta(t-t') $$

Now, it can be shown that equation 1 can be cast into the form (see Ref.): $$ \psi(x,t) = \psi_0(x,t) + \int dx' \int_{-\infty}^t G_1(x,x',t,t')f(x',t')\, dt' \qquad \text{(2)} $$

where $G_1(x,x',t,t')$ is an object that satisfy: $$ (\partial_t -\vec{\sigma}\cdot\nabla)G_1(x,x',t,t') = 0 $$

and therefore should be less clumsy to solve.

Reference: Mathematical Method of Classical and Quantum Physics, F. W. Byron and R. W. Fuller. Chapter 7: Time-dependent Green's functions: First Order

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    $\begingroup$ I think there is a problem in your equations. (1) and (2) are exactly the same (just the index $_1$ changes), so they can't be both true. I think you forget the $'$ of $f(x',t')$ in equation (1). $\endgroup$ – Adam Oct 24 '13 at 14:10
  • $\begingroup$ You're absolutely right for what concerns the primes, you're also right for what concerns the integrals, I should have specified the integration extremes! Please see the Edit $\endgroup$ – Mattia Oct 24 '13 at 14:13
  • $\begingroup$ Is $G_1(x,x',t,t)=\delta(x-x')$ ? Otherwise, I don't understand how eq. (2) can work... $\endgroup$ – Adam Oct 24 '13 at 14:22
  • $\begingroup$ More precisely: $\lim_{t'\rightarrow t} G_1(x,x',t,t') = \delta(x-x')$ $\endgroup$ – Mattia Oct 24 '13 at 14:35
  • $\begingroup$ I understand that, what I'm looking for is an explicit formula for the Greens function $G_1(x,x;t,t')$. I have the impression that it involves the derivative of a $\delta$-function, and I just want to make sure that this is correct by comparing to an explicit reference. (All textbooks that I've looked at only calculate the Fourier transform $G_1(k;t,t')$). $\endgroup$ – Greg Graviton Oct 25 '13 at 12:39

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