3

When discussing causality in relativistic QFT, it is often stated that the commutator between two operators determines whether a measurement of one can affect the other, e.g. Peskin and Schroeder pg. 28:

"...we should compute the commutator $[\phi(x),\phi(y)]$; if this commutator vanishes, one measurement cannot affect the other."

Now I understand that if two observables are "independent" in the sense that we can view them as acting on different factors of a tensor product Hilbert space$^\dagger$, then they should correspond to commuting operators, such as for example the momentum components $[\hat p_x,\hat p_y]=0$; measuring $p_x$ doesn't tell us anything about $p_y$.

But this is not the only case of vanishing commutators - obviously, every operator trivially commutes with itself or with a function of itself, for example $[\hat p_x,\hat p_x^2]=0$, and of course measuring $p_x$ does tells us something about $p_x^2$.

So my conclusion is that a vanishing commutator is a necessary but insufficient condition for the argument given above.

My questions are:

  1. Is the conclusion correct?
  2. If so, what is the proper mathematical way of telling if two operators are "independent" in the desired sense?
  3. How does it apply to $[\phi(x),\phi(y)]$?

Note: there seems to be quite a few questions about this passage from P&S (e.g. here, here and here) but none that I found seem to address the same question as I am raising (mostly, the answers to those questions explain why it is necessary for the commutator to vanish, i.e. why causality implies commutativity, but not the other way around)

$^\dagger$ Corrected following Valter Moretti's comment. What I mean is that if we have two operators $\hat A,\hat B$ that can be written as $\hat A = \hat A_1 \otimes \hat {\mathbb{I}}_2$, $\hat B = \hat {\mathbb{I}}_1 \otimes \hat B_2 $ (where the subscripts denote operators that act on two separate Hilbert spaces) then $[\hat A, \hat B]=0$, and we can say that the two are independent in the sense that for example $\langle \psi | \hat A \hat B | \psi \rangle = \langle \psi | \hat A| \psi \rangle \langle \psi | \hat B| \psi \rangle$ if $|\psi\rangle$ is an eigenstate of $\hat A \hat B$.

J. Delaney
  • 2,680
  • 1
  • 6
  • 15

1 Answers1

0

Maybe this perspective might help.

Consider $x,y$ be two spacetime points. Recall that the time ordered correlation function (Feynman propagator) is defined as:

\begin{equation*} T\left(\phi(x),\phi(y)\right) = \begin{cases} \phi(x)\phi(y); \hspace{0.2cm} x^{0}>y^{0}\\ \phi(y)\phi(x); \hspace{0.2cm} x^{0}<y^{0} \end{cases} \end{equation*} We can write the commutator as: \begin{equation*} \left[\phi(x),\phi(y)\right] = D_{F}(x-y) - D_{F}(y-x) \end{equation*} Since, we've established how commutators relate to correlation function, we impose that fields evaluated at two space-like separated points should be uncorrelated for a causal theory. Hence, we want the above commutator to vanish. If the fields are uncorrelated then the measurement of one will not affect the other by definition. Key point to realize is that the propagator is also the evolution operator for a particle traveling from $x$ to $y$. Having a zero commutation relation tells us that irrespective of how we measure the evolution, fields (operators) at $x$ and $y$ are independent of measuring one another. You could call them independent operators.

We require: \begin{equation*} \left[\mathcal{O}(x),\mathcal{O}(y)\right] = 0 \hspace{0.3cm} \forall \hspace{0.3cm} (x-y)^{2} < 0. \end{equation*} Unlike $p_{x}$ and its commutation relation with $p_{x}^{2}$, the commutator $[\phi(x),\phi^{2}(x)]=0$ gives us no new information about the field $\phi$ itself as it computes a 2-pt correlator evaluated at the same spacetime point. Hence, vanishing commutator is sufficient for telling us if field evaluated at 2 different points affect each other or not. I think the conclusion you draw is incorrect.

Resources: David Tong QFT notes

Venkatesh Srinivasan
  • 11