Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutators, propagators, and measurement.

  1. Aug 6, 2012 #1
    Hi all,

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

    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 [itex]x-y[/itex], one can perform a continuous Lorentz transformation such that [itex]x-y \longrightarrow -(x-y)[/itex]. 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 [itex]\Lambda^2 = 1[/itex] and so on, but I think things would then get messy with components etc to show it is not possible for [itex](1,0,0,0)[/itex] but is for [itex](0,0,0,1)[/itex]...
     
  2. jcsd
  3. Aug 6, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    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.
     
  4. Aug 6, 2012 #3
    Hello,

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

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

    I hope this helps,

    Ilm
     
  5. Aug 6, 2012 #4
    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?

    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 [itex]|a,b\rangle[/itex] 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 [itex]\phi(x)[/itex] are actually operators (not the things being acted upon), so I can see how the above sort of argument of mine may apply.
     
  6. Aug 6, 2012 #5

    Bill_K

    User Avatar
    Science Advisor

    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.
    Take any two noncommuting operators such as Jz and Jx. If you start with the system in an eigenstate, say Jz = +1/2, and measure Jx, the state will change to a superposition of Jz = +1/2 and Jz = -1/2. Then a subsequent measurement of Jz 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.
     
  7. Aug 6, 2012 #6
    Hmmm, now that I see it in those words, I recall it. Cheers

    Yup, that was what I was meaning. It's an unfortunate choice of labels, but for your case [itex][J_x(\mathbb{x}), J_z(\mathbb{y})]\neq 0[/itex] only when [itex]\mathbb{x}-\mathbb{y}[/itex] 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.
     
  8. Aug 6, 2012 #7
    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.
     
  9. Aug 7, 2012 #8
    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 [itex]\psi[/itex](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 [[itex]\psi[/itex](y), [itex]\psi[/itex](x)] = 0.

    On the other hand [[itex]\psi[/itex](y), [itex]\psi[/itex](x)] ≠ 0 means that one of the points (the previous) acts on the other.
     
  10. Aug 7, 2012 #9
    There is a rather long discussion, which contains many interesting points, on this issue here.
     
  11. Aug 7, 2012 #10
    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.
     
  12. Aug 10, 2012 #11
    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.)
     
  13. Aug 10, 2012 #12
    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...
     
  14. Aug 10, 2012 #13
    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.
     
  15. Aug 10, 2012 #14

    strangerep

    User Avatar
    Science Advisor

    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.]
     
    Last edited: Aug 10, 2012
  16. Aug 10, 2012 #15
    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.
     
  17. Aug 11, 2012 #16
    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?
     
  18. Aug 12, 2012 #17

    strangerep

    User Avatar
    Science Advisor

    I said what I said, which was neither of these things that you're trying to put into my mouth.
     
  19. Aug 12, 2012 #18

    Hans de Vries

    User Avatar
    Science Advisor

    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.
     
  20. Aug 12, 2012 #19

    Hans de Vries

    User Avatar
    Science Advisor

    (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:


    [tex]
    {\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}
    [/tex]
    [tex]
    = \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.
    [/tex]

    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.

    [tex]
    -\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)
    [/tex]

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

    [tex]
    K_1(z)~\approx~1/z ~~~~\mbox{for}~~ z<1
    [/tex][tex]
    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
    [/tex]

    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.

    [tex]
    e^{-\sqrt{r^2-c^2t^2}/r_c}
    [/tex]

    We want to examine the part just outside the light cone, directly beyond the point
    where r=t. Let [itex]\delta r[/itex] be the distance outside the light-cone. We then get.


    [tex]
    \left.~\sqrt{(r+\delta r)^2-c^2t^2} ~\right|_{(r=ct)} ~~=~~ \sqrt{2r\delta r + (\delta r)^2}
    [/tex]

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

    [tex]
    e^{-\delta r/r_c}
    [/tex]

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

    [tex]
    e^{-\sqrt{2r\delta r }/r_c}
    [/tex]

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

    [tex]
    \delta r ~=~ \frac{1}{2}\frac{r_c^2}{r}
    [/tex]

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

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

    Hans.
     

    Attached Files:

    Last edited: Aug 12, 2012
  21. Aug 12, 2012 #20
    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:
    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...?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutators, propagators, and measurement.
  1. The propagator (Replies: 2)

Loading...