The Feynman propagator for the Klein-Gordon field is D_F(x-y) = <0|T{phi(x)phi(y)}|0>, where T denotes time ordering. Why is time ordering essential rather than just using <0|phi(x)phi(y)|0>?
ATime ordering ensures the propagator is real-valued
BWithout time ordering, the propagator would not be Lorentz invariant
CTime ordering ensures that positive-frequency modes propagate forward in time and negative-frequency modes propagate backward — the correct causal structure for a relativistic theory where antiparticles propagate backward in time
DTime ordering is merely a convention with no physical significance
The time-ordered product T{phi(x)phi(y)} places the later-time operator to the left. This ensures the correct boundary conditions: particles propagate forward in time (positive-energy poles contribute for t_x > t_y) and antiparticles propagate backward in time (negative-energy poles contribute for t_y > t_x). In momentum space, this corresponds to the Feynman i-epsilon prescription: D_F(p) = i/(p^2 - m^2 + i epsilon), where the small imaginary part shifts the poles off the real axis in the way that produces causal propagation. Different time orderings give different Green's functions (retarded, advanced) with different physical interpretations.
Question 2 Multiple Choice
In momentum space, the Feynman propagator for a scalar field is D_F(p) = i/(p^2 - m^2 + i epsilon). The pole at p^2 = m^2 corresponds to what physical situation?
AA virtual particle that violates energy-momentum conservation
BAn on-shell particle satisfying the relativistic energy-momentum relation — the propagator diverges when the intermediate particle is real
CA bound state of the field
DAn ultraviolet divergence that must be regularized
The condition p^2 = m^2 is exactly the on-shell condition for a real particle of mass m. When an intermediate particle in a Feynman diagram is on-shell, the propagator diverges (the denominator vanishes). Off-shell (virtual) particles have p^2 != m^2 and the propagator is finite. The i epsilon prescription tells you how to handle the pole — it determines the causal boundary conditions. In scattering calculations, the on-shell poles correspond to physical intermediate states and are handled by the optical theorem and cutting rules.
Question 3 True / False
The propagator for a massive field falls off exponentially at spacelike separations with a characteristic length scale of 1/m (the Compton wavelength). This means that virtual particles cannot propagate farther than their Compton wavelength.
TTrue
FFalse
Answer: False
The propagator does fall off as e^{-m|x-y|} for spacelike separations, and this sets the natural range scale. However, saying virtual particles 'cannot propagate farther' is misleading. The propagator is nonzero at all separations — it is exponentially suppressed but never exactly zero. In Feynman diagrams, virtual particles with any momentum (including very small momenta, corresponding to long distances) contribute. The exponential suppression means that long-range effects from massive fields are strongly suppressed, which is why the nuclear forces (mediated by massive mesons or W/Z bosons) are short-ranged while electromagnetism (massless photon propagator, falling off as 1/|x-y|^2) is long-ranged.
Question 4 Short Answer
Explain the physical meaning of the Feynman propagator and why it serves as the building block for all perturbative calculations in QFT.
Think about your answer, then reveal below.
Model answer: The Feynman propagator D_F(x-y) = <0|T{phi(x)phi(y)}|0> gives the amplitude for a field disturbance to propagate from y to x (or equivalently, for a particle created at y to be detected at x, with antiparticle contributions when x^0 < y^0). In perturbation theory, interactions are treated as small corrections to free propagation. Every internal line in a Feynman diagram represents a free propagator — the amplitude for a virtual particle to travel between two interaction vertices. The full scattering amplitude is built by multiplying propagators (internal lines) with vertex factors (interaction couplings) and integrating over all intermediate momenta. The propagator is the Green's function of the free field equation, so it encodes the response of the field to a point-like disturbance.
The connection between propagators and Green's functions is exact: (partial^2 + m^2)D_F(x-y) = -i delta^4(x-y). This means the propagator tells you how the field responds to a localized source at point y. The entire Feynman diagram expansion is an iterative solution of the full interacting field equation, using the free propagator as the kernel.