40
$\begingroup$

In relativistic quantum field theories (QFT),

$$[\phi(x),\phi^\dagger(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0\,.$$

On the other hand, even for space-like separation

$$\phi(x)\phi^\dagger(y)\ne0\,.$$

Many texts (e.g. Peskin and Schroeder) promise that this condition ensures causality. Why isn't the matrix element $\langle\psi| \phi(x)\phi^\dagger(y)|\psi\rangle$ of physical interest?

What is stopping me from cooking up an experiment that can measure $|\langle\psi| \phi(x)\phi^\dagger(y)|\psi\rangle|^2$? What is wrong with interpreting $\langle\psi| \phi(x)\phi^\dagger(y)|\psi\rangle \ne 0$ as the (rather small) amplitude that I can transmit information faster than the speed of light?

$\endgroup$
36
$\begingroup$

Recall that commuting observables in quantum mechanics are simultaneously observable. If I have observables A and B, and they commute, I can measure A and then B and the results will be the same as if I measured B and then A (if you insist on being precise, then by the same I mean in a statistical sense where I take averages over many identical experiments). If they don't commute, the results will not be the same: measuring A and then B will produce different results than measuring B and then A. So if I only have access to A and my friend only has access to B, by measuring A several times I can determine whether or not my friend has been measuring B or not.

Thus it is crucial that if A and B do not commute, they are not spacelike separated. Or to remove the double negatives, it is crucial that A and B must commute if they are spacelike separated. Otherwise I can tell by doing measurements of A whether or not my friend is measuring B, even though light could not have reached me from B. Then with the magic of a lorentzian spacetime I could end up traveling to my friend and arriving before he observed B and stop him from making the observation.

The correlation function you wrote down, the one without the commutator, is indeed nonzero. This represents the fact that values of the field at different points in space are correlated with one another. This is completely fine, after all there are events that are common to both in their past light cone, if you go back far enough. They have not had completely independent histories. B U T the point is that these correlations did not arise because you made measurements. You cannot access these correlations by doing local experiments at a fixed spacetime point, you can only see these correlations by measuring field values at spatial location x and then comparing notes with your friend who measured field values at spatial location y. You can only compare notes when you have had time to travel to get close to each other. The vanishing commutator guarantees that your measurements at x did not affect her measurements at y.

It is dangerous to think of fields as creating particles at spacetime locations, because you can't localize a relativistic particle in space to a greater precision than its compton wavelength. If you are thinking of fields in position space it is better to think of what you are measuring as a field and not think of particles at all.

(Actually I should say that I don't think you could actually learn that your friend was measuring B at y by only doing measurements at A. But the state of the field would change, and the evolution of the field would be acausal. I think this is a somewhat technical point, the main idea is that you don't want to be able to affect what the field is going OVER THERE outside the light cone by doing measurements RIGHT HERE because you get into trouble with causality)

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ This is a fantastic answer. Two points: 1) I don't think you can travel to my friend and stop him from making the observation because he is still space-like separated, right? 2) I think you can learn that your friend was measuring B if [A,B]=/=0, at least in principle. For example, you could imagine setting up some crazy experiment which prepared the quantum field in some eigenstate. Your friend could be doing something that brings the field out of its eigenstate. Then, if you found that your field wasn't in its original eigenstate, you know (superluminally) that your friend did something. $\endgroup$ – hwlin Feb 9 '13 at 0:13
  • 2
    $\begingroup$ ...but from what I understand, the fields in QFT are not actually observables. So it seems to me at best a heuristic to say that this is why they commute at spacelike separations...and therefore not an actual explanation. I would be interested to know if there is something I am missing here; or if not, I would be interested in another argument that does explain why locality requires $[\phi(t, \vec{x}), \phi(t, \vec{y})]=0$. $\endgroup$ – doublefelix Dec 15 '17 at 22:16
3
$\begingroup$

If you'd like to see a small computation to show why microcausality is related to the vanishing of the commutator, here is a simple exercise that one can do.

Consider some operator $A(\vec{x},t)$ of which I want to measure the vacuum expectation value in some state $\psi$ $$ \mathcal{E}_A(\vec{x},t) := \langle \psi|A(\vec{x},t)|\psi\rangle\,. $$ Now give a "kick" to the Hamiltonian at a certain time $t_0$ (let's assume $t_0 = 0$). By that I mean that we perturbe the Hamiltonian by some operator that is non zero only for $t > 0$. Namely $$ H = H_0 + \theta(t)\, V(t)\,. $$ How does the expectation value of $A$ change after this perturbation? It seems that the most convenient approach would be the interaction picture, so let's do that. Without reviewing the details, we define the state $|\psi\rangle$ and the operators $\mathcal{O}$ as the time evolution operator $\exp(i H_0 t)$ applied on the Schrödinger picture $$ \psi_{\mathrm{int}}(t) = e^{i H_0 t} \psi_{\mathrm{S}}(t)\,,\qquad H_{\mathrm{int}}(t) = e^{i H_0 t}H e^{-i H_0 t}\,. $$ The time evolution operator $U(t,t_0)$ must satisfy $$ i \frac{\mathrm{d}}{\mathrm{d}t} \psi_{\mathrm{int}}(t) := i \frac{\mathrm{d}}{\mathrm{d}t} U(t,t_0) \psi_{\mathrm{int}}(t_0) = \theta(t) V(t)\,U(t,t_0) \psi_{\mathrm{int}}(t_0)\,, $$ where the first equality is a definition of $U$ and the second its differential equation. To first order in the perturbation $V$ the solution is $$ U(t,t_0) = \mathbb{1} - i \int_{t_0}^t\mathrm{d}t' \,V(t') + O(V^2)\,,\qquad \forall\;t_0 > 0\,. $$ So far all standard. The expectation value can be then seen to transform as $$ \begin{aligned} \mathcal{E}_A(\vec{x},t) &= \langle\psi(t)|A(\vec{x},t)|\psi(t)\rangle \\&= \langle \psi|U^\dagger(t,0) e^{i H_0 t}\, e^{-i H_0 t} A(\vec{x},0)e^{i H_0 t}\, e^{-i H_0 t} U(t,0) |\psi\rangle \\& \simeq \mathcal{E}_A(\vec{x},0) - i\int_0^t \mathrm dt'\langle \psi| A(\vec{x},t) V(t) - V(t) A(\vec{x},t) | \psi \rangle\,. \end{aligned} $$ Here I simply used all the definitions of the interaction picture and expanded to first order in $V$. Now let us make a physical assumption. This is similar to what one does in linear response theory. See the Kubo formula for instance.

The perturbation $V$ that I defined to be a "kick" happens not only at some specific time, but also at a specific location. Therefore it will modify the Hamiltonian as the integral of some local operator $B$. Namely $$ V(t) = \int \mathrm{d}^{d-1} x B(\vec{x},t)\,. $$ From this one has $$ \mathcal{E}_A(\vec{x},t) - \mathcal{E}_A(\vec{x},0) = \int_0^t\mathrm{d}t'\int \mathrm{d}^{d-1} x'\,\langle\psi|\big[A(\vec{x},t)\,, B(\vec{x}{}',t')\big]|\psi\rangle\,. $$ Here you see immediately that microcausality must imply that the correlator has to vanish outside the light cone. Suppose that $B$ creates a perturbation in some location in spacetime, it is impossible that $A$ knows about it if they are space-like separated. You would have to wait at least the time it takes for the light to get there in order to have a change in the expectation value. Therefore the only way to preserve causality is to require $$ \big[A(\vec{x},t)\,, B(\vec{x}{}',t')\big] = 0\quad \mathrm{if}\; (x-x')^2 < 0\,. $$ A simple contradiction that one might cook up is the following: tell a friend to make a perturbation to the Hamiltonian at time $t = 0$, or to not do it. Then you set yourself spacelike separated from your friend. If $A$ and $B$ do not commute you can infer whether your friend has decided to perturb be Hamiltonian or not by just measuring $\mathcal{E}_A$. And as you might know this leads to all sort of paradoxes in special relativity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I believe there is a mistake in your derivation. Your perturbative definition of the evolution operator $U$ is in the interaction picture (i.e. $U_I$). When you expand out $\mathcal{E}_A(\vec{x},t)$ you are evolving $|\psi\rangle$ with $U_I$, as required in the interaction picture: $|\psi_I(t)\rangle= U_I(t,0)|\psi\rangle$. However, the expression for the operator $A_I(t)$ should read $e^{i H_0 t} A(\vec{x}) e^{-i H_0 t}$ by definition, instead of $e^{i H_0 t}\, e^{-i H_0 t} A(\vec{x})e^{i H_0 t}\, e^{-i H_0 t}$, right? $\endgroup$ – FriendlyLagrangian Sep 21 at 14:28
  • $\begingroup$ Shouldn't you end up with: $\mathcal{E}_A(\vec{x},t) = \langle \psi | A_I(t) | \psi \rangle - i \int_0^t...$ ? (where I also corrected a small sign and $i$ mistake.) $\endgroup$ – FriendlyLagrangian Sep 21 at 16:01
  • $\begingroup$ Either way, the interpretation still remains, as $\langle \psi | A_I(t) | \psi \rangle$ is simply the expectation value of operator $A$ evolved in time $t$ according to the unperturbed Hamiltonian (i.e. think of the Heisenberg picture with $H=H_0$). $\endgroup$ – FriendlyLagrangian Sep 23 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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