# Preserving Causality in QFT

1. Dec 1, 2011

### Matterwave

In Peskin and Schroeder, we calculate that the propagator $D(x-y)=\langle 0|\phi(x)\phi(y)|0\rangle$ to be non zero even if x and y are space-like separated (although it's exponentially decaying). This suggests (given our interpretation of this quantity) that a particle CAN propagate between these two points (albeit with vanishingly small probability).

Peskin argues, however, that this is not a violation of causality, because causality should only prevent measurements from affecting each other over space-like separations. He goes on to say that, we really should calculate:
$$\langle 0|[\phi(x),\phi(y)]|0\rangle$$

And that since this is 0, causality is preserved. Can someone elaborate this point for me? I'm not sure I quite understand what the commutator has to do with measurements, and why it is this quantity that is important and not the D(x-y) which is important.

2. Dec 1, 2011

3. Dec 1, 2011

### kof9595995

I was confused with this before, and I have accumulated some thought so far and I'll try to say something about it here, but it's in no way a complete answer:
Firstly, consider a generic Alice Bob experiment, and they have identically prepared some particles in some entangled state, and they know what state they have prepared is(in principle it's possible). Now say Bob is going to travel to a place very far from Alice, bringing part of the entangled particles with him, and they make the agreement that if Bob finds Alice did some measurement on the particles, then Bob will immediately start traveling back. Say Alice can measure an observable A at her position and Bob can measure B at his position, suppose this entangled state expanded in eigenstates of A is $$\frac{1}{\sqrt2}(|1\rangle+|2\rangle)$$. When Bob decides to measure B, if Alice hasn't measured A, Bob will get $$\frac{1}{2}(\langle1|+\langle2|)B(|1\rangle+|2\rangle)=\frac{1}{2}(\langle1|B|1\rangle+\langle2|B|2\rangle+\langle1|B|2\rangle+\langle2|B|1\rangle).....(1)$$; If Alice has measured A, the pure state collapes to a mixed ensemble $$\frac{1}{2}|1\rangle\langle1|+\frac{1}{2}|2\rangle\langle2|$$, and Bob will get $$\frac{1}{2}(\langle1|B|1\rangle+\langle2|B|2\rangle)......(2)$$
Now if $[A,B]=0$, $|1\rangle$ and $|2\rangle$ would also be eigenstates of $B$, so $\langle1|B|2\rangle=\langle2|B|1\rangle=0$ and $(1)=(2)$, meaning no way can Bob find out if the state has been measured. No violation of causality in this case.
On the other hand, if $[A,B]\neq0$, $|1\rangle$ and $|2\rangle$ are not eigenstates of $B$, so $\langle1|B|2\rangle\neq0$ thus $(1)\neq(2)$ and this means Bob can determine if Alice has measured A. In this way a message is sent to Bob instantly.
Secondly, from canonical perturbation theory, we can see $[\phi(x),\phi(y)]=0$ would be one of the sufficient conditions for the S-matrix to be Lorentz invariant, this is discussed in details in Weinberg's QFT,Vol 1, section 3.3 and 3.5.
I think my first point is more generic, but there is a problem when applied to fields, that is, although $[\phi(x),\phi(y)]=0$, it's possible that polynomials of the field operators do not commute. So if every Hermitian operator corresponds to an observable, doesn't this mean we can do a superluminal communication using Alice-Bob experiment I described?

EDIT:"it's possible that polynomials of the field operators do not commute",it seems I was wrong, polynomials of finite degrees of the fundamental fields have to commute with each other. But in Walter Thirring's "elementary QFT" chapter 5, he defined a number density operator $N(\vec x,t)=-i[\partial_0\phi^{(-)}\phi^{(+)}-\phi^{(-)}\partial_0\phi^{(+)}]$(5.27), where plus minus sign denotes the positive-frequency and negative-frequency part, and this doesn't commute in a spacelike region. It's not a perfect example for me since it's not constructed from the fundamental field, but this operator does look like somewhat physically meaningful, so shall we avoid discussing operators just because it violates causality?(Anyway this deviates from OP's question already)

Last edited: Dec 1, 2011
4. Dec 1, 2011

### kof9595995

Emm, what's wrong with \rangle?

5. Dec 1, 2011

### A. Neumaier

Causality requires that two observables at spacelike related arguments are statistically independent. Now independence implies that the operators are commuting, hence their commutator vanishes. In particular, the expectation of the commutator has to vanish.

On the other hand, I haven't seen any positive argument for why causality should force $\langle 0|\phi(x)\phi(y)|0\rangle$ to be zero. Hence the fact that it is nonzero says nothing about causality or its violation.

6. Dec 1, 2011

### kof9595995

Hey, nice to see you back Neumaier!

7. Dec 1, 2011

### A. Neumaier

If A and B commute then any two algebraic expressions in A and B commute. But a derivative is not an algebraic expression.

On the other hand, causality requires that _any_ local expression in the fields and/or their deriviatives evaluated at x commutes with _any_ (same or different) local expression in the fields and/or their deriviatives evaluated at y, as long as x and y are spacelike related. In particular, this also holds for the number density.

8. Dec 1, 2011

### Hans de Vries

The propagator treated in P&S is the "propagator from source" of the
complex Klein Gordon field. It's a somewhat unlucky example because
this particular propagator is non-zero outside the light cone while the
really important propagators don't have this problem.

It's also a propagator from a source (rather than a self propagator) just
like $A^\mu$ is propagated from $j^\mu$

The propagator from source of the real valued Klein Gordon field is zero
outside the lightcone. You can read this back in the P&S text if you read
between the lines where they talk about the requirement of having to take
positive and negative charged particles together in order to eliminate the
part outside the lightcone.

The propagator from source of the Real Klein Gordon field.

$$\tilde{\mathcal{K}}(x^i,t) ~=~ \tfrac{1}{4\pi}\left[~\delta\left(\tfrac12 s^2\right)-\frac{m}{s}J_1(ms)~\theta\left(\tfrac12 s^2\right)\right] ~~~~~~~~\mbox{with} ~~ s^2=t^2-x^2 \gt 0$$

The most relevant propagator is the Dirac self propagator, as used in the
Feynman diagrams,which is strictly zero outside the lightcone. The easiest
way to derive it is with the help of the real Klein Gordon propagator ${\cal K}$

The self-propagator of the Dirac field

$$\tilde{\mathbb{D}}(x^i,t) = \left[ \left(\begin{array}{ll} 1 ~~ 0 \\ 0 ~~ 1 \end{array}\right) \frac{\partial }{\partial t} - c\left(\begin{array}{ll} \sigma^i~~~~0\\0\,-\sigma^i\end{array}\right) \frac{\partial}{\partial x^i} - i\left(\begin{array}{ll} 0 ~~ 1 \\ 1 ~~ 0 \end{array}\right) \frac{mc^2}{\hbar} \right] \tilde{\mathcal{K}}(x^i,t)$$

The space and time derivatives work first on ${\cal K}$ and the result is the
appropriate Green's function which is has to be convoluted with the Dirac
field to obtain the time evolution of the field. The time and space derivatives
of ${\cal K}$ are the second and third function in this table.

$$\begin{array}{|c|c|} \hline & \\ & \\ \mbox{Space-Time domain } & \mbox{Momentum-Time domain} \\ & \\ & \\ & \\ \tfrac{1}{4\pi}\left[~\delta\left(\tfrac12 s^2\right)-\frac{m}{s}J_1(ms)~\theta\left(\tfrac12 s^2\right)\right] & \frac{\sin\left(t\sqrt{p^2+m^2}~\right)}{\sqrt{p^2+m^2}} \\ & \\ \tfrac{1}{4\pi}~t\,\left[~\delta'\left(\tfrac12 s^2\right)- \frac{m^2}{2}\delta\left(\tfrac12 s^2\right)+ \frac{m^2}{s^2}J_2(ms)~\theta\left(\tfrac12 s^2\right)\right] & \cos\left(t\sqrt{p^2+m^2}~\right) \\ & \\ \tfrac{1}{4\pi} x^i \left[~\delta'\left(\tfrac12 s^2\right)- \frac{m^2}{2}\delta\left(\tfrac12 s^2\right)+ \frac{m^2}{s^2}J_2(ms)~\theta\left(\tfrac12 s^2\right)\right] & -\frac{ip^i \sin\left(t\sqrt{p^2+m^2}~\right)}{\sqrt{p^2+m^2}} \\ & \\ \hline \end{array}$$

With $\delta' (z)= \partial\delta(z)/\partial z$

The causality requirement

The requirement for the propagator to be causal (inside the light-cone)
is that it does not contain the non-local operator in the space-time domain:

$$\sqrt{-\frac{\partial^2}{\partial x^2} + m^2}$$

Or as expressed in the momentum-time domain

$$\sqrt{p^2 + m^2}$$

This operator is a non local convolution with a Bessel K function which
causes the part outside the light cone. You can check the momentum-
time propagators in the table. If you expand the sine and cosine functions
then you'll find that all the square roots are eliminated because there are
only even powers of the square root and no odd powers.

This is the reason the propagators are local, which is not the case with
the propagator treated in P&S which contains $\exp(-i\sqrt{p^2+m^2}\,t)$

Regard, Hans

Last edited: Dec 1, 2011
9. Dec 1, 2011

### kof9595995

But Thirring showed that the equal-time commutator of two number density operators do not commute, because it's constructed out of positive and negative frequency part of the fundamental field, instead of the fundamental field itself.

10. Dec 2, 2011

### A. Neumaier

This is because the positive and negative frequency part of the fundamental field is a _nonlocal_ expression in the field (involving not only fields at a single point but an integral over all of space. Thus my statement does not apply.

On the other hand, the fact that the number density field lacks the causal commutation property,
it means that it cannot be prepared, since independent preparation ad different points would lead to
acausal phenomena.

11. Dec 2, 2011

### kof9595995

Yes i understand, so does this mean it's not physical? It puzzled me because Thirring seemed to take this operator quite seriously and used it to discuss the concept of the size of a particle.

12. Dec 3, 2011

### Matterwave

I'm busy trying to do this Final right now, I'll read this thread in more detail soon. Thanks for the input guys.

13. Dec 4, 2011

### A. Neumaier

It is not ''real'' in the sense of prepareable. To get something interpretable in terms of physical position, one needs to translate it in terms of the Foldy-Wouthuysen transformation, just as in the treatment of the Dirac equation. I don't have Thirring's book, hence cannot judge what he is doing.