# Why do we take two different approaches to the “meaning of causality” in quantum mechanics and quantum field theory?

In non-relativistic quantum mechanics, causality is violated by saying that the amplitude of propagation of a particle $$A=\langle \textbf{x}|\exp{\Big(\frac{-i\textbf{p}^2t}{2m\hslash}}\Big)|\textbf{x}_0\rangle$$ between any two points $\textbf{x},\textbf{x}_0$ is non-zero for any time $t$, however small.

But this is also true in quantum field theory. The amplitude of propagation of a particle from a spacetime point $x$ to another spacetime point $y$, given by $$A(x,y)=\langle 0|\phi(x)\phi(y)|0\rangle\neq 0$$ even for space-like separations.

Therefore, isn't by the previous argument, here too, causality violated?

This is "resolved'' in QFT by saying that we should not ask whether particles can propagate over spacelike intervals, but whether a measurement at $x$ can affect a measurement $y$ if their separation is spacelike, and when one computes $[\phi(x),\phi(y)]$ for $(x-y)^2<0$, it turns out that the commutator vanishes. So causality is preserved.

So my question is why do we take two different approaches to the "meaning of causality" in quantum mechanics and quantum field theory?

In quantum mechanics, we work with a single-particle theory. That is, if we detect a particle, we know it is our one and only particle and nothing else. If this unique particle shows up at two events at a space-like separation, it means that it has travelled at superluminal distance and our theory violates causality.

However, later on we realize that there is no "quantum mechanics + relativity" without the introduction of a possibly infinite number of particles - this compells us to throw away single-particle quantum mechanics and the foundations of quantum field theory are laid instead. The operator $\phi(x)$ then does not correspond to the detection of any unique particle, but only to the detection of some particle from the virtually infinite ensemble of indistinguishable particles of the QFT.

In other words, finding an electron in the matter around you on Earth and at the same time in the matter on the Moon corresponds to $\langle 0| \psi (x) \psi(y) | 0 \rangle$ with $x-y$ spacelike. But that does not mean that the electron has jumped to moon superluminally, it is just a different electron.

Consequently, we must relax our conditions on causality to something different: If I make any measurement at event $x$, it will not influence the measurement at $y$ if $x-y$ is space-like; finding an electron on Earth does not influence finding it on the Moon at the same moment. The measurements which are mutually non-influencing are commuting in quantum mechanics, so we end up requiring $[\psi(x),\psi(y)] =0$ for $x-y$ space-like.

• "In other words, finding an electron in the matter around you on Earth and at the same time in the matter on the Moon corresponds to with spacelike. But that does not mean that the electron has jumped to moon superluminally, it is just a different electron." Doesn't that violate one-electron theory? – Mockingbird Oct 7 '17 at 16:14
• Well, consider asking yourself if there is a way to recognize, on the fundamental level, whether the electron you have found on the Moon is different from the one on Earth. There is no way to do that, and it also shows up as a fundamental feature in the theory. Of course, the theory shows us that for massive particles and energies well below the rest-mass/energy of the particle we can use single-particle theories as an effective description and there a more conventional notion of causality arises from the QFT. – Void Oct 7 '17 at 16:46
• On the other hand, such a description turns out not to be admissible case for e.g. massless particles even at moderate conditions such as photons in everyday situations. In fact, this "semi-causal" property of QFT can be in some sense understood as giving rise to the possibility of electrostatic effects, charged particles repelling or attracting particles at space-like separations even in the very classical-physics mode. – Void Oct 7 '17 at 16:49
• I just want to say this is a wonderful answer! I have nothing constructive to add. – franz g. Oct 12 '17 at 17:53

There are several misconceptions in this post. Let me address them one by one.

In non-relativistic quantum mechanics, causality is violated by saying that the amplitude of propagation of a particle $$A=\langle \textbf{x}|\exp{\Big(\frac{-i\textbf{p}^2t}{2m\hslash}}\Big)|\textbf{x}_0\rangle$$ between any two points $\textbf{x},\textbf{x}_0$ is non-zero for any time $t$, however small.

This is wrong on several different levels. First of all, in non-relativistic mechanics a process is acausal if and only if its amplitude is non-zero for $t<0$. Whether it vanishes or not for $t>0$ is irrelevant. Indeed, if $c\equiv \infty$ perturbations can propagate infinitely fast (cf. the heat equation), and this is not considered a violation of causality. The latter is violated if and only if effects are observed before their cause, i.e., if the amplitude is non-zero for negative times.

Second of all, your expression for $A$ is not a propagation amplitude; you forgot the step function (cf. wikipedia) $$A=\langle \textbf{x}|\exp{\Big(\frac{-i\textbf{p}^2t}{2m\hslash}}\Big)|\textbf{x}_0\rangle\Theta(t)$$

Now you can see that $A$ vanishes for $t<0$, so the amplitude is causal, as required. So far so good.

But this is also true in quantum field theory. The amplitude of propagation of a particle from a spacetime point $x$ to another spacetime point $y$, given by $$A(x,y)=\langle 0|\phi(x)\phi(y)|0\rangle\neq 0$$ even for space-like separations.

No, this is not correct. The object $\langle 0|\phi(x)\phi(y)|0\rangle$ has nothing to do with a propagation amplitude, because the object $\phi(y)|0\rangle$ does not represent a particle localised at $y$. You already know that. A (non-) vanishing $A(x,y)$ has nothing to do with causality.

Therefore, isn't by the previous argument, here too, causality violated?

No, because the first $A$ represents a completely different object from the second $A$. Use the same letter if you want, but you are dealing with two different objects.

This is "resolved'' in QFT by saying that we should not ask whether particles can propagate over spacelike intervals, but whether a measurement at $x$ can affect a measurement $y$ if their separation is spacelike, and when one computes $[\phi(x),\phi(y)]$ for $(x-y)^2<0$, it turns out that the commutator vanishes. So causality is preserved.

This is somewhat correct, but far from being the whole story. Let me quote a paragraph from Weinberg's QFT (page 198):

The condition $(5.1.32)$ is often described as a causality condition, because if $x-y$ is space-like then no signal can reach $y$ from $x$, so that a measurement of $\psi_\ell$ at a point $x$ should not be able to interfere with a measurement of $\psi_{\ell'}$ or $\psi^\dagger_{\ell'}$ at point $y$. Such a consideration of causality are plausible for the electromagnetic field, any of whose components may be measured at a given spacetime point, as shown in a classic paper of Bohr and Rosenfeld. However, we will be dealing here with fields like the Dirac field of the electron that do not see in any sense measurable. The point of view taken here is that Eq.$(5.1.32)$ is needed for the Lorentz invariance of the $S$-matrix, without any ancillary assumptions about measurability or causality.

where $(5.1.32)\equiv [\psi_\ell(x),\psi_{\ell'}(y)]=[\psi_\ell(x),\psi^\dagger_{\ell'}(y)]=0$ for $x-y$ space-like.

So my question is why do we take two different approaches to the "meaning of causality" in quantum mechanics and quantum field theory?

We do not. Causality is just the statement that the effect comes after the cause. In non-relativistic mechanics every interval is time-like so you only have to worry about the step function. In relativistic mechanics, the propagation velocity of everything is finite so you have to worry about space-like intervals as well. But that's it.

• About your first point: If the amplitude is nonzero arbitrarily short time between two points it certainly violates the fact that $c$ is finite. Do you disagree? @AccidentalFourierTransform – SRS Oct 11 '17 at 13:14
• @SRS Note that my first point is about non-relativistic mechanics, so $c$ is indeed infinite. So we agree, right? – AccidentalFourierTransform Oct 11 '17 at 13:20
• I liked your answer very much. But Peskin and Schroeder writes "This expression is nonzero for all $\textbf{x}$ and t, indicating that a particle can propagate between any two points in an arbitrarily short time. In a relativistic theory, this conclusion would signal a violation of causality." He does a non-relativistic calculation and claims what would it mean in a relativistic theory. Please see page 14. I don't know what they wanted to convey through this non-relativistic calculation. @AccidentalFourierTransform – SRS May 4 '18 at 11:14
• @SRS Yet another reason to toss P&S and get a better book. In any case, page 14 is still the "hand-wavy introduction", where the authors make no attempt at rigour/correctness. Regardless of this "intuitive interpretation" of P&S, a careful analysis reveals that the propagator cannot be thought of as a propagation amplitude. See e.g. arxiv.org/abs/1712.06605 – AccidentalFourierTransform May 4 '18 at 13:43

If the events $x$ and $y$ are at spacelike separation, then there is no Lorentz invariant notion of whether one event is earlier than the other, since there are transformations that reverse their temporal order.

To therefore speak of $\langle \phi(x)\phi(y)\rangle$ as the "amplitude for a particle to propagate from $x$ to $y$" is somewhat misleading - at least in some frames this sentence doesn't make any sense because $y$ is earlier than $x$, and they would tend to rather look at $\langle \phi(y)\phi(x)\rangle$. So both $\langle \phi(x)\phi(y)\rangle$ and $\langle \phi(y)\phi(x)\rangle$ essentially describe the same process - the propagation of a particle between the points $x$ and $y$ - and only the commutator is meaningful as looking at the sum/difference of these amplitudes.

Looking at only one of the amplitudes is like looking only at s-channel diagrams for some process and drawing all sorts of (wrong) conclusions from that.

• "s-channel diagrams for some process and drawing all sorts of (wrong) conclusions from that." Could you elaborate it? – Mockingbird Oct 7 '17 at 16:53