In Peskin and Schroeder pg. 27-28, they discuss Klein-Gordon theory and causality. For a spacelike separation $(x-y)^2 < 0$, they show that $$\langle 0| \phi(x)\phi(y) |0\rangle \neq 0$$

They go on to say that this doesn't actually break causality. Rather, what one should be looking at isn't whether particles can propagate over space-like intervals, but whether space-like separated measurements can affect one another. Hence, they argue that to understand the measurements of the field $\phi(x)$, one should be trying to understand the the commutator $[\phi(x),\phi(y)]$.

This last statement is opaque to me. In the QM setting, the ordering of the operators can be physically interpreted as one being applied first in time prior to the second. However, this doesn't make sense in the QFT context because one applies the operator at a specific point in space-time; flipping the order is not equivalent to changing the time-order of making the measurements.

  1. In the QFT context, what is meant by the "measurement of a field," in analogy to the QM measurement of some operator?
  2. Why is the commutator the object of choice when wanting to understand causality? What is the physical interpretation of the commutator here, particularly with respect to measurements of $\phi(x)$?
  • 1
    $\begingroup$ The commutator behaves as in QM: you're comparing the result of applying one operator before the other and vice-versa - here, it corresponds to the creation of a $\phi$ particle at point $x$ or $y$. If the commutator is $0$, the result is the same either way: they don't talk to each other. If it isn't, your "measurements" can affect each other. $\endgroup$ – Demosthene Apr 3 '17 at 23:23
  • $\begingroup$ Well, if I remember right from Tong's lectures, that expression decays exponentially with the spacelike distance, and he went on to explain that it's not that bad, a negative exponential, and anyway it's just an expectation value or probability, just not exactly zero. He went on with the class. I was bothered by the nonzero. Don't know if the exponential decay is quick and sharp or not, and if not zero whether one could treat the probability as a fluctuation, conceptually. But, why should the probability of creating it at x and annihilating at y with x-y spacelike NOT be zero? $\endgroup$ – Bob Bee Apr 4 '17 at 1:13
  • $\begingroup$ @Demosthene Thanks for the comment, but your answer doesn't answer the fundamental question of why should you be comparing the result of applying one operator before the other. In QM, the order of the operators corresponds to the order in which you measure in time. This is patently not true in the QFT scenario, yet causality is a statement about time-ordering, hence the meaning of the commutator with relation to causality becomes ambiguous (to me). Also, the interpretation of $\phi(x)$ as creating a particle at $x$ is only true when it acts on the vacuum state. $\endgroup$ – Aaron Apr 4 '17 at 2:06
  • $\begingroup$ "the interpretation of ϕ(x) as creating a particle at x". I don't think this is correct. $a^+(k)$ creates a particle of momentum (wave number) $k$. The interpretation would be correct if ϕ was the Fourier transform of $a$, but there is the additional factor of $(\omega_k)^{-{1/2}}$. $\endgroup$ – Keith McClary Apr 4 '17 at 5:00
  • $\begingroup$ Excellent question. It's what I've been trying to understand. If you've worked it out in the last two years, please answer your own question. The best I can come up with is that the commutator being zero means that you can choose a state $\lvert \Psi \rangle$ which is both an eigenvector of one operator and an eignevector of the other, but if it's non-zero you can't. I think of states as possible block universes, but I have no idea if this is correct. And I'm not sure what this has to do with causality. Perhaps freedom to choose mutual eigenvectors is the same as non-causality. $\endgroup$ – Alex Zeffertt Feb 15 at 12:19

According to Weinberg in his text, the components of most quantum fields are not really measurable in any obvious way, so it's best not to think in those terms.

However, the fields do have to get commuted past each other when you evaluate the S-matrix, and then the Lorentz invariance of the S-matrix depends crucially on the fields commuting at space like separations.

Lots of aspects of the physical interpretation in QFT are at best subtle, and philosophically weak but plausible-sounding heuristic arguments are not uncommon. (You can already see people disagreeing about something so basic as whether $\phi$ creates a particle in the comments!) I found the early chapters in Peskin hard going for this exact reason- it's much better when you get to phenomenology and the physics is less opaque. If you want a book you can't argue with, try Weinberg- but this does come at the price of taking twice as long to cover the material, unfortunately in a rather idiosyncratic notation that makes it hard to dip in and out of.

| cite | improve this answer | |

This answer is to add some detail to @rwold's answer.

In particle physics experiments we measure the cross section for particle interactions which depends on the S-matrix. The LSZ (Lehmann-Symanzik-Zimmermann reduction formula), which relates the Lorentz invariant S-matrix elements $ \langle f| S|i\rangle$ for $n$ asymptotic momentum eigenstates to an expression involving the quantum fields $\phi(x)$: $$ \langle f|S |i\rangle =\left[i\int d^4x_1\square \left(+m^2\right) e^{-i p_1 x_1}\right]\cdots \left[i\int d^4x_n\square \left(+m^2\right) e^{+i p_n x_n}\right]\times \langle \Omega |T\left\{\phi\left(x_1 \right) \phi \left(x_2 \right) \phi\left(x_3 \right) \cdots \phi \left(x_n\right)\right\}|\Omega \rangle $$ The $T \{\cdots \}$ refers to the time ordered product and it indicates that all operators should be ordered so that those at later times are always on the left of those at earlier times. E.g. $T \left\{ \phi \left(x_1\right) \phi \left(x_2\right)\right\}=\phi\left(x_2 \right) \phi \left(x_1 \right)$ if $t_2>t_1$ regardless of whether $\phi \left(x_1 \right)$ and $\phi \left(x_2 \right)$ commute or not. However, if $x_1$ and $x_2$ are space like separated then one can change to a different frame which reverses the time ordering. I.e. if we could have $t_2>t_1$ in one frame and $t_1>t_2$ in another. Therefore, for the S-matrix to be Lorentz-invariant (i.e frame independent) we require that $[\phi(x_1),\phi(x_2)]=0$ when $x_1$ and $x_2$ are space like separated.

| cite | improve this answer | |

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.