Commutators, propagators, and measurement.

[ianhoolihan]

Hi all,

Reading through Peskin and Schroeder, I came across the following statement, with regards to propagators:

To really discuss causality, however, we should ask not whether particles can propagate over spacelike intervals, but whether a measurement performed at one point can affect a measurement at another point whose separation from the first is spacelike. The simplest thing we could try to measure is the field $\phi(x)$, so we should compute $[\phi(x),\phi(y)]$; if this commutator vanishes, one measurement cannot effect the other.

Could someone explain how the commutator is related to the measurement of the field in this context? Searching online, the only thing that crops up is the usual uncertainty principle and its relation to the commutator.

Secondly, as a somewhat aside point, PS go on to say that for a spacelike separation $x-y$, one can perform a continuous Lorentz transformation such that $x-y \longrightarrow -(x-y)$. On the other hand, if the separation is timelike, this is not possible. Is there a nice way to prove this in matrix notation? E.g. one can show that $\Lambda^2 = 1$ and so on, but I think things would then get messy with components etc to show it is not possible for $(1,0,0,0)$ but is for $(0,0,0,1)$...

 

[Bill_K]

If x-y is spacelike then you can perform a Lorentz transformation that makes it purely spacelike, i.e. makes its time component zero. Then a 180-degree rotation about a perpendicular axis will turn it into minus itself.

 

[Ilmrak]

Hi all,
[...]Could someone explain how the commutator is related to the measurement of the field in this context? Searching online, the only thing that crops up is the usual uncertainty principle and its relation to the commutator.[...]

Hello,

$[\phi (x), \phi (y)]$ is related to the causality connection between the fields in $x$ and $y$.

Suppose that $[\phi (x), \phi (y)] \neq 0$ for some $x, y$. Then a measurement of $\phi (x)$ would affect a subsequent measurement of $\phi (y)$, i.e. they are causally connected.
If $x-y$ is space-like this would violate causality.

I hope this helps,

Ilm

 

[ianhoolihan]

If x-y is spacelike then you can perform a Lorentz transformation that makes it purely spacelike, i.e. makes its time component zero. Then a 180-degree rotation about a perpendicular axis will turn it into minus itself.

I understand that. However, after your last step, you need to perform the inverse Lorentz transform (of the one that took it to zero time component). I an guessing that the spatial components will be the negative of the original, but will the time have been reversed as well?

Secondly, why can it not be done with timelike? One can still perform a 180 spatial rotation so that the spatial components become negative, and the time component left unchanged. Presumably the problem is trying to reverse the time component?

Hello,

$[\phi (x), \phi (y)]$ is related to the causality connection between the fields in $x$ and $y$.

Suppose that $[\phi (x), \phi (y)] \neq 0$ for some $x, y$. Then a measurement of $\phi (x)$ would affect a subsequent measurement of $\phi (y)$, i.e. they are causally connected.
If $x-y$ is space-like this would violate causality.

I hope this helps,

Ilm

Sorry Ilm, but that is not clear to me, and is really just stating my question again. Why does the commutator being non-zero mean subsequent measurements are affected? Is it similar to QM where if one has commuting operators A and B then one can come up with kets $|a,b\rangle$ which are eigenstates of both A and B? If they don't commute, then the state A|a> will not be an eigenket of B, so there will be uncertainty in the measurement of B. (Has A affected the measurement of B in this case, given that there'd be uncertainty anyway for B|a>?)

Oh yes, I should also remind myself that the fields $\phi(x)$ are actually operators (not the things being acted upon), so I can see how the above sort of argument of mine may apply.

 

[Bill_K]

Well ianhoolihan, You're doing a pretty good job of answering your own questions.

Presumably the problem is trying to reverse the time component?

For a timelike vector, the sign of the fourth component is an invariant under any (proper) Lorentz transformation.

Why does the commutator being non-zero mean subsequent measurements are affected?

Take any two noncommuting operators such as J_z and J_x. If you start with the system in an eigenstate, say J_z = +1/2, and measure J_x, the state will change to a superposition of J_z = +1/2 and J_z = -1/2. Then a subsequent measurement of J_z will be affected.

Actually, Peskin is being a bit careless here - the argument should not be applied to φ(x), rather to some observable like the Hamiltonian density:

[ℋ(x), ℋ(x')] = 0.

 

[ianhoolihan]

Well ianhoolihan, You're doing a pretty good job of answering your own questions.

For a timelike vector, the sign of the fourth component is an invariant under any (proper) Lorentz transformation.

Hmmm, now that I see it in those words, I recall it. Cheers

Take any two noncommuting operators such as J_z and J_x. If you start with the system in an eigenstate, say J_z = +1/2, and measure J_x, the state will change to a superposition of J_z = +1/2 and J_z = -1/2. Then a subsequent measurement of J_z will be affected.

Actually, Peskin is being a bit careless here - the argument should not be applied to φ(x), rather to some observable like the Hamiltonian density:

[ℋ(x), ℋ(x')] = 0.

Yup, that was what I was meaning. It's an unfortunate choice of labels, but for your case $[J_x(\mathbb{x}), J_z(\mathbb{y})]\neq 0$ only when $\mathbb{x}-\mathbb{y}$ is timelike. That's the causality part of it.

Thank you for also clarifying that PS should be talking about observables. That also had me a little confused.

 

[ianhoolihan]

Take any two noncommuting operators ...

Oh, and to clarify, the importance that they are noncommuting is so that eigenstates of one operator are not eigenstates of the other? (For example, with Jx and Jz in your above example.) The reason I ask is that a large part of the reason I was getting confused was that I was trying to think why a measurement required to do four operations, as such: i.e. for commutator [A,B] this is A then -B then B then A.

 

[USeptim]

Before all, I am not an expert on QM but I would like to give my opinion.

From the scattering theory, if we don't take Born approximation one can see that the scattered wave alters the incipient wave and a Green function is necessary to determinate the impact of the scattered wave on the incident one. The Green function turns the wave into a sort of operator, because now every point of the four space $\psi$(x,t) affects a region of the wave space. This region will be always confined inside the light cone, and therefore to causality.

If two points x, y are connected by a space like vector, it cannot be a casual relation between them and therefore no one of them could affect the other so they can be treated as independent variables and so [$\psi$(y), $\psi$(x)] = 0.

On the other hand [$\psi$(y), $\psi$(x)] ≠ 0 means that one of the points (the previous) acts on the other.

 

[sheaf]

There is a rather long discussion, which contains many interesting points, on this issue here.

 

[ianhoolihan]

There is a rather long discussion, which contains many interesting points, on this issue here.

Ok, that is a rather long discussion. I've had a fair read through it, and as far as I can tell, there's a lot more to this. The discussion does seem to indicate that PS's justification for a zero commutator for spacelike separations is incorrect/unnecessary. I like Hans's explanations.

 

[cygnet1]

Not everyone agrees that commutators do vanish for space-like separated fields. See the following link, if you have access to AIP Scitation. (Sorry, I don't think copyright restrictions will allow me to attach the article itself.)

 

[ianhoolihan]

Not everyone agrees that commutators do vanish for space-like separated fields. See the following link, if you have access to AIP Scitation. (Sorry, I don't think copyright restrictions will allow me to attach the article itself.)

Ah, it is nice that this paper predates OPERA by a few months, but I wonder if the author had inside information? Nonetheless, I'm sure there are many such papers as a result of OPERA, which predict tachyonic particles. I've only read the abstract (no access to AIP) but this does seem to take a more reasoned approach than throwing about every possibility for tachyons. It is cited only once, however...

 

[cygnet1]

The tachyons that paper refers to are virtual, and cannot be directly observed. They have nothing to do with Opera. Still, any tachyons are controversial.

 

[strangerep]

[...] no access to AIP [...]

Some preprints are publically available.
Search Google Scholar for:

wolf "causality is inconsistent"

[Edit: I just skimmed the paper. Imho, take it with a very large dose of salt. I'm surprised it was accepted in a conference. The author seems unaware of the fact that the Feynman propagator does not have causal support -- this has been known a long time, as elaborated in (eg) Scharf's textbook. I could complain more about Wolf's paper, but I don't want to give it any more oxygen.]

 

[ianhoolihan]

Some preprints are publically available.
Edit: I just skimmed the paper. Imho, take it with a very large dose of salt. I'm surprised it was accepted in a conference. The author seems unaware of the fact that the Feynman propagator does not have causal support -- this has been known a long time, as elaborated in (eg) Scharf's textbook. I could complain more about Wolf's paper, but I don't want to give it any more oxygen.]

It is cited only once, however...

Thanks strangerep, that was what I was worried about. Given my inexperience in such matters and your advice, I think my supply of salt is too small to risk using up on this paper.

 

[cygnet1]

The author seems unaware of the fact that the Feynman propagator does not have causal support -- this has been known a long time, as elaborated in (eg) Scharf's textbook. I could complain more about Wolf's paper, but I don't want to give it any more oxygen.

Are you saying the Feynman way of computing propagators is incorrect because it leads to noncausal virtual interactions? Then how did this method manage to correctly compute the electron magnetic moment to 7 decimal places? Perhaps you're saying that it doesn't matter whether you compute propagators the Feynman way or the more conventional causal way because both methods lead to the same measurable results. If that's what you mean, then isn't it really a matter of personal preference whether you believe in virtual tachyons or not, since no experiment could ever tell the difference?

 

[strangerep]

Are you saying the Feynman way of computing propagators is incorrect because it leads to noncausal virtual interactions? [...] Perhaps you're saying that it doesn't matter whether you compute propagators the Feynman way or the more conventional causal way because both methods lead to the same measurable results. [...]

I said what I said, which was neither of these things that you're trying to put into my mouth.

 

[Hans de Vries]

The problem really is with the propagator from source of the complex Klein Gordon
field and not with, for instance, the Real valued Klein Gordon Field, see for instance
here: http://physics-quest.org/Book_Chapter_Klein_Gordon_real_propagators.pdf

There is also no problem with the time evolution of the much more important Dirac
field where, starting with ψ=δ(x) at t=0, the time evolution dψ/dt = 0 for x≠0 at
t=0 which follows directly from the Dirac equation itself.

For those people that insist on science fiction like "FTL Tachyons"... I will copy
the calculations from my book that show that the part which is initially there at
t=0 actually diminishes rapidly at t>0. See the next post.


Hans.

 

[Hans de Vries]

(from the notes of my book)


Calculation of the diminishing part outside the light cone of Feynman's
Klein Gordon propagator in configuration space

The propagator in momentum space and configuration space:$${\cal D}_4^{\cal F}(x,y) = \lim_{\epsilon \to 0} \frac{1}{(2 \pi)^4} \int d^4p \, \frac{e^{-ip(x-y)}}{p^2 - m^2 + i\epsilon}$$
$$= \left \{ \qquad \begin{matrix} -\frac{1}{8 \pi} \delta\left(\tfrac12 s^2\right) + \frac{m}{8 \pi s} H_1^{(1)}(ms) &~~~~~~ \textrm{ if }\, s^2=t^2-r^2 \geq 0 \\ & \\ -\frac{i m}{ 4 \pi^2 s} K_1(ims) &~~~~~~ \textrm{if }\, s^2=t^2-r^2 < 0 \end{matrix} \right.$$

In this section we'll turn our attention to the problematic "superluminal" part of the
Feynman propagator. The propagator (which is a propagator from source and not a
self-propagator) is non zero outside x=0 at time t=0. However, it diminishes rapidly
at t>0 and in this sense the part which is outside the light cone initially at t=0
could be considered to propagate at a speed lower than c.

The problematic part is given by an expression characterized by the Bessel K function
of first order.

$$-\frac{i m}{ 4 \pi^2 s} K_1(ims) ~~~~=~~~~ \frac{1}{4 \pi^2}~\frac{r_c}{\sqrt{r^2-c^2t^2} }~ K_1\left(~\frac{\sqrt{r^2-c^2t^2}}{r_c}~\right)$$

Where $r_c$ is the Compton wavelength corresponding to the mass $m$ of the particle.
The Bessel function can be approximated as follows.

$$K_1(z)~\approx~1/z ~~~~\mbox{for}~~ z<1$$$$K_1(z)~=~\sqrt{\frac{\pi}{2}}~\frac{e^{-z}}{\sqrt{z}}~\Big( 1 + \frac{3}{2^3}z^{-1}-\frac{30}{2^8}z^{-2}+\frac{840}{2^{13}}z^{-3}-\frac{37800}{2^{18}}z^{-4}+... \Big) ~~~\mbox{for}~ z>1$$

For arguments smaller as 1 and larger as 1 respectively. The part outside the light-cone
falls off exponentially outside the light-cone at a certain distance. This cut-off instance
actually becomes smaller when t increases. That is, the violation is largest at t=0.
The dominant cut-off factor is.

$$e^{-\sqrt{r^2-c^2t^2}/r_c}$$

We want to examine the part just outside the light cone, directly beyond the point
where r=t. Let $\delta r$ be the distance outside the light-cone. We then get.$$\left.~\sqrt{(r+\delta r)^2-c^2t^2} ~\right|_{(r=ct)} ~~=~~ \sqrt{2r\delta r + (\delta r)^2}$$

At r=t=0 the contribution outside the light-cone is exponentially cut-off outside the
Compton radius.

$$e^{-\delta r/r_c}$$

After a short while $r$ becomes larger as $r_c$ and the contribution outside the light-cone
becomes dominated by the other term.

$$e^{-\sqrt{2r\delta r }/r_c}$$

After a propagation over a distance of $r$ we obtain for the distance $\delta r$ outside the light-
cone where the cut-off becomes comparable to the cut-off at the Compton radius at t=0

$$\delta r ~=~ \frac{1}{2}\frac{r_c^2}{r}$$

Which decreases linear in time. So, if the decay sets in at $10^{-13}m$ at the start, which
is approximately the Compton Wavelength of the electron, then after having propagated
over $1~\mu m$ we see the outside the light-cone contribution confined to a much smaller
range of circa $10^{-20}m$

===============================================================

Hans.

 

[ianhoolihan]

(from the notes of my book)...

Would you be able to provide an eager reader the title?

Also, I've been reading through the post https://www.physicsforums.com/showthread.php?t=161235, specifically #6, where you state:

However, the concise derivation of the
Green's function does produce the Heaviside step function which eliminates
the propagation outside the lightcone.

I'm probably applying things to the wrong situation, but in one case you imply zero propagator outside the lightcone, whereas in the excerpt from your book, you imply it is non zero...?

 

[Hans de Vries]

Would you be able to provide an eager reader the title?

http://www.physics-quest.org

Also, I've been reading through the post https://www.physicsforums.com/showthread.php?t=161235, specifically #6, where you state:

I'm probably applying things to the wrong situation, but in one case you imply zero propagator outside the lightcone, whereas in the excerpt from your book, you imply it is non zero...?

The difference is the use of the iε prescription which makes D(x-y) complex instead
of real. Note that the requirement for the Greens function.

$$(\Box-m^2)D(x-y)=\delta(x-y)$$

is a problem when D(x-y) is complex because $\delta(x-y)$ is real and $(\Box-m^2)$ treats
the real and imaginary parts of D(x-y) independently so you get the requirement.

$$(\Box-m^2) Im\{D(x-y)\}=0$$

Another problem is that when you want the source to consist out of positive energy
functions only. This will give $\delta(x-y)$ an imaginary companion which is the Hilbert
transform of the Dirac function $\delta(x-y)$ which is the Fourier transform of the sign
function. (All Fourier components of the delta function get a 90 degrees phase shift
like in cos(t) ---> -i sin(t) ). The result is a function proportional to i/t and this is
incompatible with the Greens function requirements above.Hans.

 

[ianhoolihan]

http://www.physics-quest.org

Oh goody! Thanks!

The difference is the use of the iε prescription which makes D(x-y) complex instead
of real. Note that the requirement for the Greens function.

$$(\Box-m^2)D(x-y)=\delta(x-y)$$

is a problem when D(x-y) is complex because $\delta(x-y)$ is real and $(\Box-m^2)$ treats
the real and imaginary parts of D(x-y independently so you get the requirement.

$$(\Box-m^2) Im\{D(x-y)\}=0$$

Another problem is that when you want the source to consist out of positive energy
functions only. This will give $\delta(x-y)$ an imaginary companion which is the Hilbert
transform of the Dirac function $\delta(x-y)$ which is the Fourier transform of the sign
function. This is a function proportional to i/t and is incompatible with the Greens
function requirements above.Hans.

OK, that'll take some further reading on my part. However, in layman's terms, is it that if one ignores the (unnecessary) $i\epsilon$ prescription, then there's no propagation outside the lightcone? If one does use it, then things just get messy, giving an exponentially decaying propagation. If this is the case, it makes me wonder why the $i\epsilon$ prescription is even used...(I'm stuck on these sorts of questions in this current thread...https://www.physicsforums.com/showthread.php?p=4031174).

(Oh, and in that excerpt you had "examen" which should be "examine". I'm not being rude --- I prefer people to point out errors in my work rather than leave them there.)

 

[Hans de Vries]

Oh goody! Thanks!



OK, that'll take some further reading on my part. However, in layman's terms, is it that if one ignores the (unnecessary) $i\epsilon$ prescription, then there's no propagation outside the lightcone? If one does use it, then things just get messy, giving an exponentially decaying propagation. If this is the case, it makes me wonder why the $i\epsilon$ prescription is even used...(I'm stuck on these sorts of questions in this current thread...https://www.physicsforums.com/showthread.php?p=4031174).

The $i\epsilon$ trick is used by Feynman to make positive energy fields propagate
forward in time and negative energy fields propagate backwards in time.

Treating the two differently is what spoils the causality of the Greens function...


You can do all kinds of tricks with $i\epsilon$ and contour integration but you can do
the same tricks with the Heaviside step function and its Fourier transform.
The latter method is more transparent for me.

So instead of $i\epsilon$'s in your propagators you get extra delta functions as defined
by the Sokhatsky Weierstrass theorem:

If $f(x)$ is a complex-valued function which is defined and continuous on the real line,
and the limits of the integral a and b are real constants with $a < 0 < b$, then

$$\lim_{\varepsilon\rightarrow 0^+} \int_a^b \left\{\,\frac{1}{x\pm i \varepsilon}\,\right\}f(x)\,dx ~~=~~ \int_a^b \left\{\,\frac{1}{x}~\mp~ i\pi \delta(x)\,\right\} f(x) \, dx$$

The latter expression between curly brackets is just the Fourier transform of the
Heaviside stepfunction. There's more about this here in sections 1.13 through 1.16

Hans

 

[Physics Monkey]

One standard interpretation of the commutator is as the change in one operator generated by another. For example, unitary transformations like $U = e^{i p a}$ that are generated by the momentum change the position operator because position and momentum don't commute. Hence $U x U^\dagger = x + i a [p,x] + ...$. This is a very general intuition.

In the context of time evolution one can ask the question, when does an expectation value change if a I add a perturbation to the Hamiltonian? The role of the commutator is visible in perturbation theory (called linear response theory http://webusers.physics.illinois.edu/~efradkin/phys582/LRT.pdf ). The key equation is 3.4 which shows that the expectation value doesn't change unless the perturbation and the operator you're measuring don't commute.

Thus if you're measuring the expectation value of $\phi(x)$ and you perturb the Hamiltonian by $\phi(y)$ then nothing will happen until $\phi(x,t)$ and $\phi(y,t)$ don't commute (the time evolution of these operators is governed by the unperturbed Hamiltonian since the calculation is done in the interaction picture). The commutator vanishing outside the light cone is hence the same as requring that a perturbation at $y$ not effect a measurement at $x$ until $c t > |x-y|$.

Hope this helps.

 

[ianhoolihan]

The $i\epsilon$ trick is used by Feynman to make positive energy fields propagate
forward in time and negative energy fields propagate backwards in time.

Treating the two differently is what spoils the causality of the Greens function...


You can do all kinds of tricks with $i\epsilon$ and contour integration but you can do
the same tricks with the Heaviside step function and its Fourier transform.
The latter method is more transparent for me.

So instead of $i\epsilon$'s in your propagators you get extra delta functions as defined
by the Sokhatsky Weierstrass theorem:

If $f(x)$ is a complex-valued function which is defined and continuous on the real line,
and the limits of the integral a and b are real constants with $a < 0 < b$, then

$$\lim_{\varepsilon\rightarrow 0^+} \int_a^b \left\{\,\frac{1}{x\pm i \varepsilon}\,\right\}f(x)\,dx ~~=~~ \int_a^b \left\{\,\frac{1}{x}~\mp~ i\pi \delta(x)\,\right\} f(x) \, dx$$

The latter expression between curly brackets is just the Fourier transform of the
Heaviside stepfunction. There's more about this here in sections 1.13 through 1.16

Hans

Ah, I had glanced over that before, but thought it a notational thing. I agree that Heaviside functions are more transparent. Anyway, I will look more into your book.

Thanks a bunch.

One standard interpretation of the commutator is as the change in one operator generated by another. For example, unitary transformations like $U = e^{i p a}$ that are generated by the momentum change the position operator because position and momentum don't commute. Hence $U x U^\dagger = x + i a [p,x] + ...$. This is a very general intuition.

In the context of time evolution one can ask the question, when does an expectation value change if a I add a perturbation to the Hamiltonian? The role of the commutator is visible in perturbation theory (called linear response theory http://webusers.physics.illinois.edu/~efradkin/phys582/LRT.pdf ). The key equation is 3.4 which shows that the expectation value doesn't change unless the perturbation and the operator you're measuring don't commute.

Thus if you're measuring the expectation value of $\phi(x)$ and you perturb the Hamiltonian by $\phi(y)$ then nothing will happen until $\phi(x,t)$ and $\phi(y,t)$ don't commute (the time evolution of these operators is governed by the unperturbed Hamiltonian since the calculation is done in the interaction picture). The commutator vanishing outside the light cone is hence the same as requring that a perturbation at $y$ not effect a measurement at $x$ until $c t > |x-y|$.

Hope this helps.

That first bit is useful, but I guess $U x U^\dagger = x + i a [p,x] + ...$ is just another way of writing $[p,x]\neq 0$, in the sense that I can't see how $U x U^\dagger$ is interpreted physically...

I like the second bit about the Hamiltonian perturbation, but to clarify: if $[\phi(x),\phi(y)]=0$ then "nothing happens". Assuming $[\phi(x),\phi(y)]=0$ inside lightcone, then "nothing happens" for $x-y$ is timelike. On the other hand, something happens for $[\phi(x),\phi(y)]\neq 0$ outside the light cone, with $x-y$ spacelike. My problem is that I thought there was no propagation outside the light cone, so "nothing happens" outside the light cone...?

Cheers

 

[cygnet1]

The iε trick is used by Feynman to make positive energy fields propagate
forward in time and negative energy fields propagate backwards in time.

Treating the two differently is what spoils the causality of the Greens function...

Thanks for your book excerpts, Hans De Vries. It looks like a book I'd like to buy when it comes out. Do you have an estimated publishing date and publisher?

I think I understand your point about the Feynman propagators leading to FTL (space-like) interactions, which decay exponentially as their time difference increases. This seems very analogous to evanescent waves in a waveguide or waves reflected from a plane reflector beyond the critical angle. They exist in the near field only, and cannot propagate energy or information. So then why do you label them "problematic?"

Please understand that I'm not trying to "put words in the mouth" of anyone. I'm just trying to ask probing questions to help clarify these issues in my mind, and hopefully other's as well.

My question is whether the "iε" method of Feynman, whereby he perturbs the positive and negative poles of the Green's function in opposite imaginary directions before he does his contour integration, is now considered to be "wrong" in some sense. Does this method lead to discrepancies from any observed measurement? Is it objectionable on philosophical grounds? If so, why? Is there another method which is considered to be more "acceptable" today, conceptually, philosophically or experimentally?