Essentially all standard relativistic quantum field theory textbooks give the same argument for why the theory obeys causality. First, one computes the quantity $\langle 0 | \phi(\mathbf{x}) \phi(\mathbf{y}) |0 \rangle$ and finds that it is nonzero for $\mathbf{x} \neq \mathbf{y}$, naively suggesting that particles can propagate instantaneously. Then, one computes the commutator $[\phi(\mathbf{x}), \phi(\mathbf{y})]$ and finds that it is zero for $\mathbf{x} \neq \mathbf{y}$. This implies that simultaneous measurements of the field at distinct points can't affect each other, which is supposed to restore causality. (For simplicity, I'm considering simultaneous points only, though of course the arguments would be unchanged for spacelike separated points.)

There are several ways to explain why the vanishing commutator restores causality. One very common way, using particle language, is to say that the commutator represents "the amplitude of a particle to go from $\mathbf{y}$ to $\mathbf{x}$, minus the amplitude of an (anti)particle to go from $\mathbf{x}$ to $\mathbf{y}$". You need to include both terms, because ultimately you can't tell which way the particles are going for spacelike separation, and they precisely cancel. Another variant, using field language (given here and in some older textbooks), is that the vanishing of this commutator implies that turning on a source at $\mathbf{x}$ doesn't affect the expectation value of $\phi(\mathbf{y})$. And finally, there's the Weinberg way, which is that these details don't matter because in the end we're only after $S$-matrix elements.

However, I've never been satisfied by these arguments, because they don't seem to reflect how measurements occur in real life. In a typical detector in particle physics, there is no component that does anything remotely like "measuring $\phi(\mathbf{x})$". Instead, detectors absorb particles, and this is a different enough process that the arguments above don't seem to apply.

A thought experiment

Let's make a simple model of the production and detection of a particle. A quantum field couples to sources locally, $$H_{\text{int}}(t) = \int d \mathbf{x}' \, \phi(\mathbf{x}') J(\mathbf{x}', t).$$ Suppose we begin in the vacuum state $|0 \rangle$. At time $t = 0$, let's turn on an extremely weak, delta function localized source $J(\mathbf{x}') = \epsilon_s \delta(\mathbf{x} - \mathbf{x}')$. Right afterward, at time $t = 0^+$, the state is $$\exp\left(-i \int_{-\infty}^{0+} dt\, H_{\text{int}}(t)\right) |0 \rangle = |0 \rangle - i \epsilon_s \phi(\mathbf{x}) |0 \rangle + \ldots.$$ Now let's put a purely absorbing, weakly coupled detector localized at $\mathbf{y}$, which concretely could be an atom in the ground state. This detector is described by Hamiltonian $$H_{\text{det}} = \epsilon_d |e \rangle \langle g| \phi_-(\mathbf{y}) + \epsilon_d |g \rangle \langle e | \phi_+(\mathbf{y})$$ where $\phi_-$ is the part of $\phi$ containing only annihilation operators, and $\phi_+$ contains creation operators. Physically, the two terms represent absorption of a particle to excite the atom, and emission of a particle to de-excite it. Because the atom starts out in the ground state, only the first term of the Hamiltonian matters. The amplitude for the detector to be in the excited state, a tiny time after the source acts, is $$\mathcal{M} \propto \epsilon_s \epsilon_d \langle 0 | \phi_-(\mathbf{y}) \phi(\mathbf{x}) |0 \rangle \propto \langle 0 | \phi(\mathbf{y}) \phi(\mathbf{x}) |0 \rangle$$
which is nonzero! This appears to be a flagrant violation of causality, since the source can signal to the detector nearly instantaneously.


Note how this example evades all three of the arguments in the second paragraph:

  • The amplitude for a particle to go from the detector to the source doesn't contribute, because the detector starts in the ground state; it can't emit anything. There's no reason for the commutator to show up.
  • The detector isn't measuring $\phi(\mathbf{y})$, so the fact that its expectation value vanishes is irrelevant. (As an even simpler example, in the harmonic oscillator $\langle 1 | x | 1 \rangle$ also vanishes, but that doesn't mean an absorbing detector can't tell the difference between $|0 \rangle$ and $|1 \rangle$.)
  • The Weinberg argument doesn't apply because we're not considering $S$-matrix elements. Like most experimental apparatuses in the real world, the source and detector are at actual locations in space, not at abstract infinity.

I'm not sure why my argument fails. Some easy objections don't work:

  • Maybe I'm neglecting higher-order terms. But that probably won't fix the problem, because the $\epsilon$'s can be taken arbitrarily small.
  • Maybe it's impossible to create a perfectly localized source or detector. Probably, but there's no problem localizing particles in QFT on the scale of $1/m$, and you can make the source and detector out of very heavy particles.
  • Maybe the detector can't be made good enough to see the problem, by the energy-time uncertainty principle. But I can't see why. It doesn't matter if it's inefficient; if the detector has any chance to click at all, at $t < |\mathbf{x} - \mathbf{y}|$, then causality is violated.
  • Maybe the detector really can emit particles even when it's in the ground state. But this contradicts everything I know about atomic physics.
  • Maybe the free-field mode expansion doesn't work because the presence of the detector changes the mode structure. But I'm not sure how to make this more concrete, plus it shouldn't matter if we take $\epsilon_d$ very small.

What's going on? Why doesn't this thought experiment violate causality?


2 Answers 2


General calculation in Schrodinger picture

We can do a straightforward calculation of your setup in the Schrodinger picture. Let $x,y$ label points in space. Before $t=0$, the state is in the vacuum $|0\rangle$. Just after $t=0$, the state is $e^{-i \epsilon_s \phi(x)} |0\rangle$, for some spatial point $x$ where you place the source at $t=0$. (Really you might want to consider turning on a source smeared over a small neighborhood of $x$, to avoid singularities in the following discussion, but I'll ignore that.) At later time $t>0$, we have state $$|\psi(t)\rangle = e^{-iHt}e^{-i \epsilon_s \phi(x)} |0\rangle =e^{-iHt}e^{-i \epsilon_s \phi(x)} e^{iHt} |0\rangle = e^{-i \epsilon_s \phi(x,t)}|0\rangle$$

Then at time $t$, we make a measurement at spatial point $y$. For a moment I'll ignore your desired detector model and speak generally. Say we make a measurement in a spatial region $Y$. The observables measurable by an observer local to $Y$ are precisely the observables generated by (sums and products of) operators $\phi(y), \pi(y)$ for any $y \in Y$. Choose an observable $A_Y$ of this form, e.g. $A_Y = \phi(y)$ for some $y \in Y$, or $A_Y = \int_{y \in Y}\phi(y)\, dy$. Assume the points $(Y,t)$ are spacelike from the point $(x,t=0)$. Then the expectation of $A_Y$ in $|\psi(t)\rangle$ is $$\langle \psi(t) | A_Y |\psi(t)\rangle = \langle 0 |e^{i \epsilon_s \phi(x,t) } A_Y e^{-i \epsilon_s \phi(x,t)} | 0 \rangle = \langle 0 | A_Y |0 \rangle.$$ So the expectation of $A_Y$ is the same as if you hadn't turned on the source at $x$ at $t=0$. The second equality uses $[\phi(x,t),A_Y]=0$, by assumed spacelike separation.

Being careful with first-order expansion in $\epsilon$

Here's a possible point of confusion. Just after $t=0$, and to first order in $\epsilon_s$, we have $$ |\psi \rangle = |0\rangle - i \epsilon_s \phi(x) |0\rangle + O(\epsilon_s^2).$$ The second term $\phi(x) |0\rangle$ seems to dominate over the higher-order terms in $\epsilon_s$, and this may seem to suggest causality violation: an observable $A_Y$ at spacelike $Y$ still has nonzero expectation value in this state, i.e. $\langle 0 | \phi(x) A_Y \phi(x) |0\rangle \neq 0$. (Incidentally if you choose $A_Y=\phi(y)$ the expectation value is zero by symmetry, but you could choose e.g. $A_Y=\pi(y)$.). However, this observation is entirely unproblematic; the expectation value of $A_Y$ with respect to just the first-order term is not directly related to any measurement. If we include both the zero'th order and first-order terms in $|\psi\rangle$, then we find $\langle \psi | A_Y |\psi\rangle = \langle 0 | A_Y |0 \rangle + O(\epsilon^2)$, because the first-order contributions cancel.

Locality of your detector model

What about your detector model? There were a few problems. First, I wouldn't actually call your detector localized to the point $y$. The observables measurable at $y$ are just algebraic combinations of $\phi(y), \pi(y)$. Or again more generally, you could take a small spatial region $Y$ and consider observables $A_Y$ local to $Y$, given by algebraic combinations of $\phi(y), \pi(y)$ for $y \in Y$. If you want to imagine an external system $S$ like your atom coupled locally to $Y$, the coupling Hamiltonian for the detector should be like $$H_{det} = \sum_i O^i_S A^i_Y$$ where $O^i_S$ are some operators on the coupled system $S$, and $A^i_Y$ are operators in the QFT local to $Y$.

Your desired Hamiltonian $H_{det}$ may look like it takes this form, but your operators $\phi_{\pm}(x)$, by which I assume you mean something like $$\phi_{-}(x) \equiv \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{E_p}} a_p e^{ipx}$$ $$\phi_{+}(x) \equiv \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{E_p}} a_p^\dagger e^{-ipx},$$ are not strictly local to $y$. One way to see this is that they have nonzero commutator with any $\phi(x)$. (Incidentally there's some discussion about this in Section 6 and Eq. 81 here.). The operators $\phi_+(y)$ and $\phi_-(y)$ may look like they are local to $y$, by the way they are written, but if you actually re-write $\phi_{\pm}(x)$ in terms of the genuinely local $\phi(x)$ and $\pi(x)$ operators, you will find the $\phi_{\pm}(x)$ are not local.

Moreover, regardless of your detector model, I think you're a bit too quick when you say "the amplitude for the detector to be in the excited state$\dots$." You should actually think about what a measurement of the detector subsystem would yield. The analysis will then go similarly to the general discussion at the beginning of the answer.

Finally, what if we insist on using your particular detector $H_{det}$, using $\phi_{\pm}(y)$ couplings? First we must admit it's not truly local to $y$, and that it's really only approximately "localized" to a region of radius $\approx \frac{1}{m}$ around $y$ (for a massive theory). We must further admit that it's not even strictly localized to that neighborhood, or any finite region: it's really only "local" to a neighborhood of radius $r$ around $y$ with error $e^{-\frac{r}{m}}$, due to the nonzero commutators $[\phi_{\pm}(y), \phi(x)]$, or the expression of $\phi_{\pm}(y)$ in terms of genuinely local field operators $\phi(x), \pi(x)$. So you shouldn't be surprised if the detector has a $e^{-\frac{r}{m}}$ probability of registering a causality-violating signal in this model.

  • $\begingroup$ Your comment on the locality of my detector Hamiltonian is really interesting! You can find Hamiltonians just like $H_{\text{det}}$ in any atomic physics or quantum optics textbook. They're meant to be simple models for, e.g. an atom in its ground state that can be excited by the electromagnetic field, and that interaction is certainly local, $\mathcal{L} \supset \bar{e}(x) \gamma^\mu D_\mu e(x)$. So perhaps replacing this interaction with the simplified $H_{\text{det}}$ is not quite correct, and violates locality? Interesting that I've never seen this mentioned anywhere! $\endgroup$
    – knzhou
    Commented Feb 14, 2022 at 18:24
  • $\begingroup$ I would say yes, your detector doesn't actually represent a measurement that can be exactly implemented at the desired spacetime point, and it can't even be exactly implemented by acting within a small spatial neighborhood. I've seen this issue come up before, but I'm not sure about a good reference. One related keyword is Newton-Wigner localization. The paper linked in the answer (I'm a coauthor) also has some discussion. $\endgroup$ Commented Feb 15, 2022 at 4:09
  • $\begingroup$ After some thought, I'm accepting this answer because I think it identified the key point: fundamentally detectors have to couple to fields $\phi$, not $\phi_\pm$, to be local. As a reference, I found this nice paper which explicitly shows how you can model an ideal absorbing detector, using separate fields for a "ground" and "excited" state. The model is perfectly local, but it also admits an effective description written in terms of $\phi_\pm$, which naively looks slightly nonlocal. It's nice to have this bridge between QFT and "practical" detectors. $\endgroup$
    – knzhou
    Commented Mar 1, 2022 at 0:19

The $i \epsilon$ prescription / time-ordered exponential / Feynman propagator is used in scattering theory, where we measure the amplitude to go from an in state at asymptotic past infinity, to an out state at asymptotic future infinity. This is the "in-out" formalism, because the bra and the ket are in different Hilbert spaces (in and out spaces).

To ask questions like the one in your setup, which involve looking at how some observable evolves in time, we should instead use the "in-in" formalism, where the bra and the ket are in the same Hilbert space. The correct propagator is the retarded propagator, which obeys causality. To look at how an operator like $H_{\rm det}$ evolves in time, rather than multiplying by the interaction Hamiltonian, we need the commutator with the interaction Hamiltonian. (See Eq 2 of https://arxiv.org/abs/hep-th/0506236).

To give a bit more detail, let's consider the average value of the energy of the detector, $\langle H_{det, 0} \rangle$.

Let's say the initial state of the field and detector is $| \psi(0)\rangle$. The expectation value is \begin{eqnarray} \langle \psi(t) | H_{det,0} | \psi(t) \rangle &=& \langle \psi(0) | U^\dagger(t, 0) H_{det,0} U(t, 0) | \psi(0) \rangle \\ &=& \langle \psi(0) | \left(1 + i H_{int} t \right) H_{det, 0} \left(1 - i H_{int} t \right) | \psi(0)\rangle + O(t^2) \\ &=& E_g + i t \langle \psi(0) | [H_{int}, H_{det, 0}] | \psi(0) \rangle + O(t^2) \end{eqnarray} where $E_g$ is the ground state energy, and here $H_{int}$ includes the evolution of the field and the interaction of the detector and the field.

In particular, seeing how detector observables evolve, we don't just look at $\langle \psi(0) | H_{int} H_{det, 0} | \psi(0) \rangle$. We need the commutator $[H_{int}, H_{det, 0}]$, which will involve a field commutator.

To do an analysis of this system, let's break the calculation into two stages.

First, let's compute the expectation value of the field $\phi$ at the position of the detector. We can think, physically, that we have a $\phi$-\probe that measures the $\phi$ value.

We will work in the Heisenberg picture.

We will find it useful to evaluate a few commutators. First, the interaction Hamiltonian $H_{int} = \int dx J(x) \phi(x)$ has a commutator with $\phi(y)$ given by \begin{eqnarray} \langle 0| [H_{int}, \phi(y)] | 0 \rangle &=& \int dx J(x) \langle 0 | [\phi(x), \phi(y)] | 0 \rangle \\ &=& \int dx J(x) G_R(x, y) \end{eqnarray} where $G_R = \langle 0 | [\phi(x), \phi(y)] | 0 \rangle$ is the retarded propagator, which vanishes when $x$ and $y$ are spacelike separated (ie, vanishes outside the light cone). As we'll spell out below, causality is really enforced by this property of the commutator / retarded propagator.

Eventually we will want to compute the commutator of an exponential of $H_{int}$ with $\phi(y)$. Before doing that, let's consider just a power of $H_{int}$. (There should really be a time ordering symbol, which I am ignoring to keep the answer simpler -- you can justify this since $\phi(y)$ will be after all the times in the integrals in $H_{int}$) \begin{eqnarray} \langle 0 | [H_{int}^n, \phi(y)] | 0 \rangle &=& n \langle 0 | [H_{int}, \phi(y)] H_{int}^{n-1} | 0 \rangle \end{eqnarray} So then \begin{eqnarray} \langle 0 | [e^{i H_{int} t}, \phi(y)] | 0 \rangle &=& \langle 0 | \sum_{n=0}^\infty \frac{(it)^n}{n!} [H_{int}^n, \phi(y)] | 0 \rangle \\ &=& \langle 0 | \int dx [\phi(x), \phi(y)] J(x) \sum_{n=0}^\infty \left(\frac{(i t)^n}{n!}\right) n H_{int}^{n-1} | 0 \rangle \\ &=& i t \langle 0 | \int dx [\phi(x), \phi(y)] J(x) \left[\sum_{n=0}^{\infty} \frac{(i t)^n}{n!} H_{int}^n \right] | 0 \rangle \\ &=& i t \langle 0 | \int dx [\phi(x), \phi(y)] J(x) e^{i H_{int} t} | 0 \rangle \\ \end{eqnarray} To get from the second line to the third, you have to think a bit about the case $n=0$, where the commutator vanishes. So the sum is only non-zero starting from $n=1$. But since the summand is a function of $n-1$, you can relabel the index to start at $n=0$, which is done in the third line. Note I've also used $n/n!=1/(n-1)!$.

With this result, we can evaluate the expectation value $\langle \phi(t) \rangle$ in the Heisenberg picture

\begin{eqnarray} \langle \phi(t) \rangle &=& \langle 0 | e^{i H_{int} t} \phi(y) e^{-i H_{int} t} |0 \rangle \\ &=& i t \langle 0 | \int dx [\phi(x), \phi(y)] J(x) e^{i H_{int} t} e^{- i H_{int} t} | 0 \rangle \\ &=& i t \int dx J(x) \langle 0 | [\phi(x), \phi(y)] | 0 \rangle \\ &=& i t \int dx G_R(x,y) J(x) \end{eqnarray}

I am not 100% sure I got the factors of $i$ right, and probably that $t$ isn't supposed to be there, but I am typing this up fairly quickly. Note that this answer makes complete sense -- the expectation value of $\phi$ is just the classical solution of the equations of motion (using the retarded propagator, which is what we would use in classical physics).

Now let's slightly modify your detector model, so instead of directly depending on the field operator $\phi$, it only depends on the expectation value of the field, $\langle \phi(t) \rangle$. This simplifies the calculation. It amounts to ignoring backreaction of your detector on the field.

Then the part of the detector Hamiltonian describing transitions from the ground state to the excited state is \begin{equation} H_{det} \supset \epsilon_d \langle \phi(t) \rangle |e\rangle \langle g | =\left( i t \epsilon_d \int dx G_R(x,y) J(x) \right)|e\rangle \langle g | \end{equation} which is only nonzero when the detector is on or within the light cone of the source, because of the properties of the retarded propagator / field commutators. Hence, everything is causal

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Commented Feb 15, 2022 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.