Do relativistic propagators give probability amplitudes?

Question

I find myself confused as to whether relativistic propagators can be understood as probability amplitudes or not. This Wikipedia article for example, under 'Relativistic propagators', states that:

In relativistic quantum mechanics and quantum field theory the propagators are Lorentz-invariant. They give the amplitude for a particle to travel between two spacetime events.

Peskin and Schroeder similarly write on p.27 above eq. (2.50) that

$D(x-y)=\langle0|\phi(x)\phi(y)|0\rangle$ is the amplitude for a particle to propagate from $y$ to $x$.

On the other hand the answer to this question claims essentially that the propagator in relativistic QFT does not produce a probability because it is not conserved.

The problem is that in order to make sense as a probability, we should have something like (where $x=(t,\vec x)$) $$\int d^3 \vec x |D(x-y)|^2 = constant,$$

i.e. the total probability should not change over time (otherwise the particle might vanish), but this does not seem to be satisfied by the propagator (it is not even clear if this expression is Lorentz-invariant).

For example looking at the explicit expression for the Feynman propagator for timelike intervals ($\tau^2_{xy} = (x-y)^2 > 0$) $$ D(x-y) = -\frac{1}{4 \pi} \delta(\tau_{xy}^2) + \frac{m}{8 \pi \tau_{xy}} H_1^{(1)}(m \tau_{xy}) $$

trying to square this will yield terms having $\delta$-function squared, which I can't imagine can be properly integrated over.

So which one is correct? is the argument about probability conservation wrong and if so why, or is it actually satisfied but just hidden somewhere in the math, or is the statement in Wikipedia wrong?

Answer 1

There is an unambiguous and straightforward sense in which $\langle 0\vert \phi(x)^\dagger \phi(y)\lvert 0\rangle$ is a probability amplitude: It's the probability amplitude for the state $\phi(y)\lvert 0\rangle$ to be found in the state $\phi(x)\lvert 0\rangle$. The rules of QFT are just the rules of QM, and this is just the Born rule. This is what people mean when they describe the propagator as "the probability of the particle to travel from $y$ to $x$" - they handwave $\phi(y)\lvert 0\rangle$ to be in some sense localized at $\vec y$ at time $y^0$ and $\phi(x)\lvert 0\rangle$ to be in some sense localized at $\vec x$ at time $x^0$ and then this just very straightforwardly is what that overlap of states means by the Born rule.

The actual problematic part of the statement isn't that it's a probability amplitude, it's the idea that those states are "localized" in the sense we want. Localization of relativistic states is a subtle topic (starting with the non-existence of "proper" position operators and the deficiencies of their common substitutes, the Newton-Wigner operators, see also e.g. this answer by Valter Moretti) and in the end this statement of the states being localized is, outside of the non-relativistic limit, to be treated with suspicion.

The "non-conservation" of the "probability amplitude" in the sense of the question I would simply interpret as another sign of this ill-definedness of the localization. Once you accept it simply as a probability amplitude between two abstract states and not states strictly localized at $\vec x$ and $\vec y$, the integrals varying in time also cease to have the worrisome meaning of the particle "vanishing" - but at the same time of course the nice picture of the particle going from $x$ to $y$ has also gone away. Such is life.

Answer 2

First of all, notice that you (and the quoted references) are using two different notions of propagators, one is the temporally ordered one, the Feynman propagator, the other is the non-temporally ordered one, the two-point Wightman function. They coincide only if the first argument is in the causal future of the latter. I will refer to the second case below.

You can view the propagator $$\langle 0|\phi(x)\phi(y)|0\rangle$$ as the kernel for computing a probability amplitude of a pair of one-particle states.

The space of vectors $$|f\rangle :=\int f(x) \phi(x) d^4x|0\rangle = \phi(f)|0\rangle,$$ where $f$ varies in the space of Schwartz functions on Minkowski spacetime is dense in the one-particle Fock sub space. Therefore

$$\langle 0|\phi(f)\phi(g)|0\rangle =\langle f|g\rangle$$

Is a bona fide probability amplitude of a pair of states represented by vectors in the said Hilbert subspace.

From this perspective, there is no difference with the usual probability amplitude notion of standard QM.

In a sense, probability is conserved in time: if $f_t$ represent the temporally translated function $f(x-(t,0,0,0))$ then $$\langle f_t|g_t\rangle = \langle 0| U_t \phi(f) U_t^\dagger U_t\phi(g) U^\dagger_t|0\rangle = \langle 0|\phi(f)\phi(g) |0\rangle =\langle f|g\rangle$$ just because $U_t^\dagger|0\rangle =U_{-t}|0\rangle=|0\rangle$. Above $U_t$ is the time evolutor in the used Minkowski reference frame.

What it is strongly disputable (see also ACuriousMind’s answer) is the localisation notion involved in states $|f\rangle$. It does not mean that the particle is localized in spacetime in the support of $f$ in the sense of standard quantum mechanics.

According to this wiepoint, Wikipedia is wrong if strictly interpreted, even if this misinterpretation is a sort of shared folklore in QFT.

There is some literature on these issues with different opinions. What is certain is that the notion of position as presented in non relativistic QM cannot be generalised to relativistic quantum theory.

Answer 3

Perhaps to understand the process, one can look at the analogous situation in classical field theory. The quantum scenario adds the notion of self-energy corrections which do not affect the question posed by the OP.

In classical field theory, one considers the propagation of a classical field $f(x,y)$, which leads to a convolution integral $$ g(u,v,z)=\int f(x,y) K(u-x,v-y,z) dxdy . $$ Here $K(u-x,v-y,z)$ is the kernel function that is analogous to the propagator in quantum theory. It is also called an impulse response because if $f(x,y)$ is a Dirac delta function, the output would just be $K(u,v,z)$. If we set the propagation distance $z$ equal to zero, we must get $K(u-x,v-y,0)=\delta(u-x,v-y)$, because without any propagation the field must remain the same. This Dirac delta corresponds to the Dirac delta found in the quantum scenario.

One difference between the classical scenario and the quantum scenario is that we don't talk the probability, but there is an equivalent property, namely the energy. Just like probability needs to be conserved in quantum scenarios, the energy must be conserved in the classical scenario. Therefore, this propagation process must have the same property that it has in the quantum scenario: it must be unitary.

These properties therefore correspond to the same situation in the quantum scenario. As a result, we can think conceptually about the propagator as the kernel of the unitary propagation process. No physical quantum state every occupies a single point in space. Therefore the propagator by itself will never represent the propagation of a physical quantum state. (The statement is P&S makes some tacit assumptions.) Therefore, it would not make sense to consider the modulus square of the propagator by itself. Instead, the conservation of probability in the quantum scenario is provided by the unitary nature of the propagation process.