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Quantum Field Theory: Field Operators and Lorentz invariance

  1. Mar 31, 2008 #1
    [SOLVED] Quantum Field Theory: Field Operators and Lorentz invariance

    Hi there,

    I am currently working my way through a book an QFT (Aitchison/Hey) and am a bit stuck on an important step in the derivation of the Feynman Propagator. My problem is obviously that I am not a hard core expert of relativitiy :)

    Actually, I have TWO questions on the same matter.
    The central quantity is the Feynman Propagator

    [tex] <0|T([\hat{\phi}(x_1) \hat{\phi}(x_2)|0> [/tex]

    where the [tex] \hat{\phi} [/tex] are scalar field operators and T is the time-ordering operator and the x are 4-vectors.

    The point of interest is now this quantity's Lorentz Invariance.

    The book says: "If the two points [tex]x_1[/tex] and [tex]x_2[/tex] are separated by a time-like interval ([tex] (x_1 - x_2)^2 > 0)[/tex] then the time ordering is Lorentz invariant; this is because no proper (doesn't change the sense of time) Lorentz transformation can alter the time-ordering of time-like separated events."
    It goes on:
    "The fact that time-ordering is invariant for time-like separated events is what guarantees that we cannot influence our past, only our future "

    First question: The first (italic) part sounds suspiciously self-evident, but how can that be quickly shown mathematically? And for the second part: I would say: the fact that it is Lorentz invariant means that one can not think of a coordinate frame where the events change their order of time. Is that right?

    Now the book goes on and treats the case of space-like ([tex] (x_1 - x_2)^2 < 0)[/tex] separated events. The book says it can be shown that it can be shown that the two field operators always commute in this case. I tried to show that following a hint:
    Commutator of 2 scalar Field Operators of the same kind:

    [tex] D(x_1, x_2) = [\hat{\phi}(x_1, t_1), \hat{\phi}(x_2, t_2)] [/tex]
    .. with 3-vectors and time component treated seperatly.

    I could show that this can be written as

    [tex] D(x_1, x_2) = \int \frac{d^3 k}{(2 \pi)^3 2E} [ e^{-ik\cdot(x_1-x_2)} - e^{ik\cdot(x_1-x_2)}][/tex]
    ... with x and k being again 4-vectors.

    The right side is obviously Lorentz invariant. The book now hints that this fact is enough to show that in this case [tex] D(x_1, x_2)[/tex] actually always vanishes.

    Second question: How is Lorentz invariance enough to show that?
  2. jcsd
  3. Mar 31, 2008 #2


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    edit: actually I take all that back. Note that the integral is D(x-y) - D(y-x), so each of those two terms is lorentz invariant and you can continously transform the second term from (x-y) --> - (x-y) and the two terms will cancel for (x-y)^2 < 0
    Last edited: Mar 31, 2008
  4. Mar 31, 2008 #3
    The book suggests it should be possible without further calculations.
    I mean you can nicely show that for equal times the integral vanishes, and surely it must be possible to show that it vanishes in general, but somehow it must be possible to show in a very easy way that it is sufficient that the integral is Lorentz invariant and x_1 and x_2 are space-like separated... I certainly don't see it at the moment...
    Last edited: Mar 31, 2008
  5. Mar 31, 2008 #4


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    Have you calculated D(x-y) for spacelike seperation? eg x - y = r? You should see that the statement I have written down in the edited post makes sense.
  6. Mar 31, 2008 #5
    edit: where are my manners :) Thanks so much for helping me out so far! I see your comment is certainly going into the right direction, all I need now is my humble mind to catch on :)

    well I am working on it. The quantities in D(x_1, x_2) as I wrote it are 4 vectors. I know something about the square of (x_1 - x_2), but nothing about the quantity k(x_1 - x_2).

    As I said, I somehow always avoided delving deeper into special relativity (which I regret A LOT lately...), so I probably just fail to see the obvious.

    Let me summarize: In the commutator I have those 2 exponentials of which I know that x_1 and x_2 are space-like separated. I don't understand what you mean by D(x-y) - D(y-x) to be honest. 2 commutators? or are you just referring to the exponential functions under the integral?
    Last edited: Mar 31, 2008
  7. Mar 31, 2008 #6


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    This is straight out of P&S if you want to look at a copy.. But D(x-y) is simply
    <0|phi(x)phi(y)|0> which will be proportional to something like int (1/2E) exp (-ip(x-y))
  8. Mar 31, 2008 #7
    Ah ok I just found the (as expected: obvious) solution.

    As I stated I already showed that the integral vanishes for t_1 = t_2 and the integral is also Lorentz invariant. So I can transform without changing its value.
    Now when x_1 and x_2 are space-like separated I can always transform to a system where t_1' = t_2' and the integral vanishes as shown -> the integral must be zero.

    Thanks for hinting me into the right direction! I definetely should look deeper into relativistics :)
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