What blocks these causality-violating observables in QFT?

Question

Some questions have been asked on this site [1] [2] about causality violation in relativistic quantum mechanics of a single particle (RQM) vs quantum field theory (QFT), but I have a very specific question that builds off these. In a nutshell: It seems like certain causality-violating observables in RQM (to be described) are still present in QFT, in principle—so do we have to reject these observables by fiat as a basic assumption of QFT?

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Let's consider the QFT of a free real scalar field. One way to construct (or think about) such a theory is as the second-quantized theory of a single relativistic particle. The single-particle Hilbert space $\mathcal H$ is the usual $L^2(\mathbb R^3)$, and the Hamiltonian in momentum space is just multiplication by $\sqrt{|\textbf{p}|^2+m^2}$; the corresponding QFT has the symmetric Fock space $\mathcal F_s(\mathcal H)$ as its Hilbert space and the canonically induced Hamiltonian on this Fock space. (See, e.g., chapter 5 of Talagrand's What is a QFT for more on this approach to scalar QFT and its equivalence to the standard approach.)

The single-particle theory described above famously has problems with causality. Following the description on p. 157–8 in Nik Weaver's Mathematical Quantization, let $K_1$ and $K_2$ be compact regions of space that are distant from each other, let $P_1$ denote the orthogonal projector onto the subspace $L^2(K_1)$ at $t = 0$, and let $P_2$ denote the projector onto $L^2(K_2)$ at time $t = t_0$ (Heisenberg picture), where $t_0$ is very small so that light cannot propagate between $K_1$ and $K_2$ in this time. Then (as Weaver argues quite rigorously) $P_1$ and $P_2$ cannot commute, which violates causality in the sense that this non-commutation could be exploited for superluminal signaling.

Somehow QFT is supposed to resolve this issue, and the microlocality condition "$[\phi(x),\phi(y)] = 0$ if $x$ and $y$ are spacelike separated" is cited as justification. My question is, don't (canonical extensions of) $P_1$ and $P_2$ still exist on the Fock space, projecting down to the one-particle sector? Aren't they still non-commuting, leading to a causality violation? Is QFT's "solution" to this problem just to forbid measurements corresponding to these operators, instead insisting on observables that can be built from the operator fields? This seems to me like the only possible conclusion, but at the same time extremely weird, because $P_1$ and $P_2$ are perfectly well defined (even bounded) self-adjoint operators on the Fock space.

Answer 1

Consider a free, spin zero (scalar), neutral (free) particle in $d$ dimensions. In the context of canonical quantisation, it has the Hamiltonian: $$ H = \int\left(\frac12\pi^2+\frac12\nabla\phi^2+\frac12m^2\phi^2\right)d^d\vec x $$ with the canonical commutation relations satisfied at a certain instant: $$ [\phi(\vec x),\pi(\vec y)] = i\delta(\vec x-\vec y). $$ In the Heisenberg picture, this results in the equations of motion (the Klein-Gordon equation): $$ \partial_t\phi = \pi \quad \partial_t\pi = \Delta\phi-m^2\phi. $$ You can solve them: $$ \phi(t,\vec x) = \int \left(\partial_tG_>(t-s,\vec x-\vec y)\phi(s,\vec y)+G_>(t-s,\vec x-\vec y)\pi(s,\vec y)\right)d^d\vec y \\ \pi(t,\vec x) = \int \left(\partial_t^2G_>(t-s,\vec x-\vec y)\phi(s,\vec y)+\partial_tG_>(t-s,\vec x-\vec y)\pi(s,\vec y)\right)d^d\vec y $$ with $G_>$ the causal Green's function: $$ (\partial_t^2-\Delta +m^2)G_> = \delta. $$ You can explicitly check that causality is conserved since $G_>$ is supported on the future light cone: for space-like separation $(x-y)^2<0$: $$ [\phi(x),\phi(y)] = 0 \quad [\phi(x),\pi(y)] = 0 \quad [\pi(x),\pi(y)] = 0 . $$ More generally, for arbitrary time intervals: $$ [\phi(x),\phi(y)] = -iG_>(x-y)\quad [\phi(x),\pi(y)] = i\partial_tG_>(x-y) \quad [\pi(x),\pi(y)] = i\partial_t^2G_>(x-y) $$ with $x=(t,\vec x),y = (s,\vec y)$. Causality is therefore not an assumption, but is rather a necessary consequence of the equations of motion.

This is still compatible with the first quantisation point of view. You obtain it converting to an annihilation/creation basis. Your second quantised Hamiltonian is: $$ H = \int \mathcal H(\vec x-\vec y)a(\vec x)^\dagger a(\vec y)d^d\vec xd^d\vec y $$ with canonical commutation relations: $$ [a(\vec x),a(\vec y)^\dagger ] = \delta(\vec x-\vec y). $$ You can identify $\mathcal H$ as the first quantised Hamilontian (generalised Bessel potential): $$ \mathcal H = \sqrt{m^2-\Delta} := \int \sqrt{m^2+k^2}e^{i\vec k\cdot \vec x}\frac{d^d\vec k}{(2\pi)^d}. $$ The first quantised wavefunctions seem to violate causality. Indeed, the Schrödinger retarded Green's functions are not supported on the future lightcone: $$ (\partial_t+i\mathcal H)G = \delta $$ and represent wavefunctions when the particle starts at a specific location. In particular, if you consider the position projectors in first quantization: $$ P(\vec x) = |\vec x\rangle\langle \vec x| $$ they become in second quantisation: $$ P(\vec x) = a(\vec x)^\dagger a(\vec x). $$ Your projection operators do not commute in the Heisenberg picture even if the events are not causally linked: $$ [P(x),P(y)]\neq 0 $$ even if $(x-y)^2<0$.

When properly formalising the two approach on a common footing, the reconciliation is pretty obvious. $a$ is a non local operator when expressed in terms of the original $\phi,\pi$. Explicitly: $$ a(\vec x) = \int \left(A(\vec x-\vec y)\phi(\vec y)+iB(\vec x-\vec y)\pi(\vec y)\right)d^d\vec y $$ with: $$ A = \sqrt{\frac{\mathcal H}2} := \frac1{\sqrt2}\int \sqrt[4]{m^2+k^2}e^{i\vec k\cdot \vec x}\frac{d^d\vec k}{(2\pi)^d} \\ B = \frac1{\sqrt{2\mathcal H}} := \frac1{\sqrt2}\int \frac1{\sqrt[4]{m^2+k^2}}e^{i\vec k\cdot \vec x}\frac{d^d\vec k}{(2\pi)^d} = \frac1{\sqrt2}\frac{K_{(d-1/4)/2}\left(\frac{|\vec x|}m\right)|\vec x|^{(1/4-d)/2}}{2^{1/8-1}(2\pi)^d\Gamma(1/8)}. $$ Therefore, in terms of the elementary fields $\phi,\pi$, the projectors $P$ are causally connected and do not commute as expected. It is all about physical interpretation. Note that it is completely valid to focus on the first quantised theory in the free case. Only the second quantisation is more transparent physically by focusing on the causal operators.

Hope this helps.