QFT: Checking Causality w/ Commutators: Examples & Criteria

  • Context: Graduate 
  • Thread starter Thread starter Phiphy
  • Start date Start date
  • Tags Tags
    Causality Qft
Click For Summary

Discussion Overview

The discussion centers on the relationship between causality and commutators in quantum field theory (QFT), particularly in the context of the Feynman propagator and its implications for space-like separations. Participants explore the criteria for causality in both classical and quantum frameworks, examining the role of commutators in determining observable properties and the nature of measurements in QFT.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the Feynman propagator is non-zero for space-like separation but argue that this does not imply a violation of causality, suggesting that commutators of field operators should be checked instead.
  • One participant explains that commuting operators correspond to observable properties, allowing for simultaneous measurements at different spatial points, while non-commuting operators complicate the interpretation of measurements across time.
  • Another participant raises a paradox regarding non-commuting operators, questioning how measurements can affect previous states and suggesting that the evolution of the system complicates the interpretation of causality.
  • Some participants discuss the nuances of propagators, noting the distinction between source propagators and self-propagators, and how these relate to causality and measurements in QFT.
  • One participant expresses a desire for a rigorous proof that a vanishing commutator for space-like separations guarantees respect for causality.
  • Another participant acknowledges confusion regarding the derivation of the self-propagating Green's function and highlights the differences in how sources are treated in relativistic versus non-relativistic theories.

Areas of Agreement / Disagreement

Participants generally agree on the importance of commutators in assessing causality, but there remains uncertainty regarding the implications of commutators being zero for space-like separations and the relationship between propagators and causality. Multiple competing views and interpretations are present, particularly concerning the nature of measurements and the implications of non-commuting operators.

Contextual Notes

Participants note limitations in understanding the derivation of certain Green's functions and the treatment of sources in relativistic theories, indicating that assumptions about the nature of measurements and the evolution of states may not be fully resolved.

Phiphy
Messages
16
Reaction score
1
The Feynman propagator in QFT is not zero for space-like separation, but we say this does not mean that causality is violated, we should check the commutator of field operators instead, and the commutators vanish for space-like separation.

My question is: why do we use commutators to check causality instead of propagators? Is there any examples that commutators are related to observation? And what is the 'causality criteria' for causality in classical relativistic field theory?
 
  • Like
Likes   Reactions: craigthone
Physics news on Phys.org
In quantum theories, the relevant criterion here is that a commuting set of operators form a set of "observable" properties. E.g. since the equal-time commutator [tex][\phi(x,t_{0}),\phi(y,t_{0})]=0[/tex] fields at different spatial points, but the same time, can be used simultaneously as physical properties of the field. More generally, if the fields commute for any 2 space-like separated points of space-time commute, then fields at all causally-separated places can be used in a single framework for computational purposes. That is, the fields at space-like separations don't fundamentally care about each other; the measurement of one is not intrinsically dependent upon the other. Of course, there can be *correlations* between such measurements (encoded in propagators).

If two operators A and B do not commute, we can still consider cases in which a system has property (corresponding to) A at one time, and property B at another time...this is consistent. But for a relativistic theory in which space-time is a single space we generally want to use operators that commute on the whole space-time to be able to work with a single framework of physical properties on the whole space-time (rather than different sets of commuting operators for each time slice).
 
  • Like
Likes   Reactions: craigthone
The case of non-commuting operators raises a curious paradox. In classical QM, if operators A and B don't commute, you can't find a set of shared eigenstates for them. So, if you measure A and get a certain value, and then you measure B, the system is no longer in the eigenstate of A.

Fast forward to QFT. We have two non-commuting operators, one in the causal future of the other. If you measure A and then you measure B, by the same logic, the second measurement kicks the system out of A's eigenstate, making the result of the first measurement invalid. In other words, your second measurement changes your past!
 
hamster143 said:
The case of non-commuting operators raises a curious paradox. In classical QM, if operators A and B don't commute, you can't find a set of shared eigenstates for them. So, if you measure A and get a certain value, and then you measure B, the system is no longer in the eigenstate of A.

Fast forward to QFT. We have two non-commuting operators, one in the causal future of the other. If you measure A and then you measure B, by the same logic, the second measurement kicks the system out of A's eigenstate, making the result of the first measurement invalid. In other words, your second measurement changes your past!

There is no paradox at all. If two operators as you say are localized at timelike separated points (A is to the past of B let's say) then A*B term in the commutator cannot be interpreted "measuring B and THEN measuring A" for obvious reasons that once can never "measure" B (which is located to the future of A) before A. The state on which the measurement is being done also evolves and moves out of the whatever operators eigenstate it was anyway (Schroedinger picture).
 
Phiphy said:
The Feynman propagator in QFT is not zero for space-like separation, but we say this does not mean that causality is violated, we should check the commutator of field operators instead, and the commutators vanish for space-like separation.

My question is: why do we use commutators to check causality instead of propagators?
Lots more can be said about the propagators:1)======================================================

The part of the Greens function outside the light-cone diminishes over time.
For an electron it's say 10-13m at t=0 but less then 10-20m after 1 microsecond

2)======================================================

There's confusion between the propagator from source and the self-propagator.
Books discuss the propagator from source as if it is the self-propagator, like:
An electron at x has and amplitude of D(y-x) to propagate to y. This should be:
A source at x has an amplitude of D(y-x) to produce an electron at y.

3)======================================================

The Klein Gordon self propagating Greens function is:

[tex] {\cal D}^{\cal S}(x^\mu) = \left \{ \qquad \begin{matrix}<br /> \frac{1}{2 \pi^2}~\frac{rt}{s}~ \frac{\partial}{\partial s}\delta(s^2) - \frac{m^2}{4 \pi}~\frac{rt}{s^2}~ H_2^{(1)}(ms) & \textrm{ if }\, s^2 \geq 0 \\ & \\<br /> -\frac{m^2}{ 2 \pi^2 }~\frac{rt}{s^2} K_2(ims) & \textrm{if }\, s^2 < 0<br /> \end{matrix} \right.[/tex]

Which is obtained by taking the 3d Fourier transform of a point at the origin
in a single time slice and then using the on the shell energy for the time -
evolution of all the individual 3-momentum components.

This Greens function is explicitly zero outside the lightcone at t=0. (!)

There is still a part of the Greens function which is non-zero outside the
light cone for t>0 (which is diminishing over time as well). Still, there is
no faster then light propagation since the wave-function which is a point
at t=0 zero is non-zero at t<0 and the part ouside the lightcone can be
considered as being a time-like propagation from the wave-function
immediately prior to t=0.Regards, Hans
 
JavierR: I think you are right. That explains why causality requires commutator=0 for space-like separations. But I am still trying to figure out a rigid proof of the inverse, ie., commutator=0 for space-like separattions => causality is respected.

Hans: Although I did not understand how you derived the self propagating Green's function, you reminded me that the "source" of a Feynman propagator is in fact not a delta function. In non-relativistic theory, Green's functions are correlation functions between point sources (delta functions in spacetime), yet in relativistic theory, we renormalize the source so that they contain spacially separated ingredients, so no wonder relativistic propagator correlates space-like separations.
 

Similar threads

Replies
18
Views
3K
  • · Replies 456 ·
16
Replies
456
Views
27K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 175 ·
6
Replies
175
Views
13K
  • · Replies 60 ·
3
Replies
60
Views
5K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 37 ·
2
Replies
37
Views
5K
  • · Replies 16 ·
Replies
16
Views
4K