# Causality and Operators

## Main Question or Discussion Point

Hi. I'm reading David Tong's notes (Causality at page 36: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf ) on QFT and I'm currently trying to understand the causality requirement that

$$[O_1(x), O_2(y)] = 0 \ \ \forall \ \ (x-y)^2 < 0.$$

For two operators O1 and O2. He then states that this ensures that a measurement at x can not affect a measurement at y when x and y are not causally connected.

I know that in classical relativity, a spacelike separation between events, imply that the event at the first location could not have been the cause of the event at the second location because it would require a signal to go faster than the speed of light.
However exactly how this statement is related to measurements and operators in QFT is not clear to me. Could someone enlighten me?

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The concept behind this requirement is that if two observables do not commute then they follow the Heisenber's indetermination principle. Hypothetically speaking you could use this kind of correlation in order to transmit informations from x to y. So if x-y is a spacelike vector then you have to require the commutation of the two operators.
I hope I was clear and, obviously, not to be wrong!

The concept behind this requirement is that if two observables do not commute then they follow the Heisenber's indetermination principle. Hypothetically speaking you could use this kind of correlation in order to transmit informations from x to y. So if x-y is a spacelike vector then you have to require the commutation of the two operators.
I hope I was clear and, obviously, not to be wrong!
So since they obey Heisenberg's uncertainty relation which states that

$$\Delta O_1(x) \Delta O_2 (y)$$

is less than or equal to some function of the commutator between the two, one could in principle make O1 sharp (by measuring it?) at the event x, and then O2 would have to become very indeterminate at y, such that you have in some sence sent a signal to y that you measured O1 at x? Am I right?

That's right. At least that's the way I understood that too. Maybe there are other ways of using indetermination's relation to transmit information but I don't know it for sure.

The re is an excelent explanation of this in the first 200 pages of weinberg qft. Much more profound than HUP. I really recommend to take a look at that book.

Which volume?

Bill_K
Yes, it seems to take Weinberg 200 pages to explain anything. He would not do well on Twitter.

In fact he requires the vanishing of the commutator/anticommutator of the fields not to guarantee causality but rather to guarantee Lorentz invariance.

Do you think is my previous interpretation wrong or is just the problem deeper? I'd like to know if my current understanding is correct.

Bill_K
I think the interpretation of [O1(x), O2(y)] = 0 where x - y is spacelike and O1, O2 are any two observables is clear. If this were not the case, a measurement at point x would influence the outcome of a measurejment at point y.

Weinberg just wants to skirt the details of what constitutes an observable and what constitutes a measurement, especially for spinor fields, and so he derives it an easier way, from Lorentz invariance.

Understood. Thank you.

I do not know if it is not enough. I found more satisfying the explanation of Weinberg. In fact what I dont like of the usual explanation is that, although

[O(x);O(y)] = 0 for x-y spacelike, (1)

One also finds that

<0|O(x);O(y)|0> ≠ 0 for creation operators. (2)

So there is some apparent incongruence between the first equality and the the second because in the first eq one interpretates that what happens in x does not affect what happens in y, but in the second eq one finds that what happens in "x" kind of affect what happens in "y".

In fact, what I never understood and perhaps some of you can enlight me is why it is usually said from (1) that O(x) cant affect the probability distribution of O(y) (ensuring in this way the locality property of any theory). To me, it only says that we can know both the value of O(x) and O(y) which is weaker I think. Am I loosing something?

Thanks!

Bill_K
[O(x);O(y)] = 0 for x-y spacelike, (1)
One also finds that
<0|O(x);O(y)|0> ≠ 0 for creation operators. (2)
(1) holds only for observables. In (2), creation operators are not observables and so their commutator is not required to vanish. What saves causality is that the field operator is constructed from both particle and antiparticle. One term is creation, the other term annihilation, and the nonzero commutator from the first term cancels the commutator from the other.

The Heisenberg uncertainty principle, by the way, holds only in the special case that the two operators involved are canonical conjugates of each other. Thus it holds for [x, p] = iħ, but not for Jx, Jy which also don't commute.

The re is an excelent explanation of this in the first 200 pages of weinberg qft. Much more profound than HUP. I really recommend to take a look at that book.
Thanks for all the anwers! I will check out Weinberg's book for sure. However I've heard it is not suitable for an introduction to QFT. What do you think?

dextercioby
Homework Helper
Thanks for all the anwers! I will check out Weinberg's book for sure. However I've heard it is not suitable for an introduction to QFT. What do you think?
It most certainly depends on the level of math you possess. If you come from QM learnt from the 2-volume set of Galindo & Pascual, then Weinberg Vol.1 is the proper exposure to QFT. If you come from Griffiths, Liboff & perhaps with a touch from Sakurai, then a text on QFT by Ryder or Peskin & Schroeder would be appropriate.