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Contour integration in propagator

  1. Aug 6, 2012 #1
    Hi all,

    I've been playing around with (2.54) of Peskin and Schroeder, and I've some quick questions about it. Firstly, the part I'm stuck on:

    $$\int\frac{d^3p}{(2\pi)^3}\left[\left.\frac{1}{2E_p}e^{-ip\cdot (x - y)}\right|_{p^0=E_p} - \left.\frac{1}{-2E_p}e^{-ip\cdot (x - y)}\right|_{p^0=-E_p}\right]={x^0>y^0} \int\frac{d^3p}{(2\pi)^3}\int\frac{dp^0}{2\pi i}\ \frac{-1}{p^2 - m^2}e^{-ip\cdot (x - y)}$$

    where the contour for the last integral is shown here http://en.wikipedia.org/wiki/Propagator#Causal_propagator.

    My first question is to check that the factor of [itex]-1[/itex] in the second term is since the contour is clockwise if we close below, unlike the usual anticlockwise? I'm using the residue theorem here, not the Cauchy integral formula.

    Secondly, how does [itex]x^0>y^0[/itex] come in? PS assume it when they say that
    $$\langle|0|[\phi(x),\phi(y)]|0\rangle = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}\left(e^{-ip\cdot (x - y)} + e^{ip\cdot (x - y)}\right)$$
    but I'm not sure if the assumption [itex]x^0>y^0[/itex] is required in this equation, or in the future integration steps.

    The answer to the above might affect the next question, but PS say that, to evaluate the case for [itex]y^0>x^0[/itex], one closes the integral above, which gives zero. Why don't they just flip the integral upside down, so it's non-zero? I assume there is a physical reason such as causality that makes us want to be zero, but could someone explain this?

    Cheers
     
  2. jcsd
  3. Aug 7, 2012 #2
    I thought I'd solved the following problem, but I guess not:

    I understand that, using the residue theorem, one can get the integral around the curve that closes from below. However, I am not sure how to convert that into a [itex]dp^0[/itex] integral, as the poles lie on this line, and if I recall correctly, theorems for contour integration require the poles to lie in the interior of a region.

    I would understand displacing the poles by [itex]\pm i\epsilon[/itex] and then letting [itex]\epsilon \to 0[/itex], which seems to be what Wikipedia indicates. Is this what Peskin and Schroeder do, just without explicitly mentioning it?

    My prior questions still stand as well.

    Cheers
     
  4. Aug 8, 2012 #3

    strangerep

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    That sounds right (though I'd have to recheck the full calculation to make sure -- it's quite a few years now since I did it myself).

    To evaluate the contour integral, the contribution from the large semicircle at infinity must vanish. For a given sign of the exponent, this happens in one halfplane, but not the other.
     
  5. Aug 8, 2012 #4

    strangerep

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    Yes, the residues of only the poles interior to the contour contribute.
    That also sounds right. Sometimes people tilt the real axis by an infinitesimal amount to shorten the calculation. Others cheat (imho) by displacing the poles by an infinitesimal imaginary amount. Personally, I don't like either of these shortcuts and its more reliable to use small semicircles around the pole(s), as you said, and then evaluate the contribution on the small semicircle in the limit of zero radius. The result of the integral should then be the same regardless of which way you deform the contour, provided the contributions on the small semicircles are both evaluated correctly.

    If you can cough up a few bucks, by your own copy of this inexpensive textbook. :-)
    https://www.amazon.com/Schaums-Outl...qid=1344404847&sr=1-1&keywords=schaum+complex
    It's quite useful for checking whether integrands do/don't vanish on the large and small semicircles in these sorts of contour integrals.
     
    Last edited by a moderator: May 6, 2017
  6. Aug 8, 2012 #5
    Ah, so simple! I didn't bother checking the limits for the arc as I was confused with the following point...

    Hmmm, I agree. That feels like cheating. I am not sure how using the semicircles around the poles differs from the case of displacing the poles slightly. In the case of slightly displaced poles, at least there are no poles on the real axis. With the semicircles, there are poles on the flat edge, so one cannot evaluate the integrals...so don't the poles need to be displaced anyway?

    Anyway, thanks a heap for that clarification strangerep,

    Ianhoolihan.

    PS --- that books looks really good, as it seems to be based around solved examples.
     
  7. Aug 8, 2012 #6

    vanhees71

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    To precisely decide which propagator (there are many different propagators; in vacuum QFT you usually want the time-ordered propagator, which in this case is identical with the Feynman propagator; in linear-response theory you need the retareded propagator, etc.) you want to calculate, you should either give a clear contour in complex $p^0$ plane or an appropriate regularization by displacing the poles. Usually the latter form is more convenient for calculations.

    For more details on different propagators in the real-time Schwinger-Keldysh formalism of relativistic many-body theory, you can find here:

    http://fias.uni-frankfurt.de/~hees/publ/green.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Aug 8, 2012 #7

    strangerep

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    In the case where the contribution from the small semicircle vanishes in the limit of zero radius, they'll be the same. (Sorry I don't have more spare time -- it would be helpful to do all these integrals explicitly once and for all -- and post them somewhere -- since I've seen such questions re-asked here several times.)

    Actually, I might have been thinking of this one:
    https://www.amazon.com/complex-vari...ywords="the+complex+variables+problem+solver"

    which is out of print now -- but Amazon shows some used copies at reasonable price. The new edition's price is a bit excessive, though.

    Anyway, between the two of them, it should be easy to review basic contour integral stuff.
     
    Last edited by a moderator: May 6, 2017
  9. Aug 8, 2012 #8
    It doesn't sit that well with me that the propagator depends on the choice of contour, but I guess I'll have to read further before I have an informed opinion.
     
    Last edited by a moderator: May 6, 2017
  10. Aug 8, 2012 #9
    OK, since I'm doing it explicitly myself, I'll post as much as I can here, and hopefully all you have to do is say "Correct!"

    Using the figure below, we'll close the contour below with an arc C of radius R, and let the two semicircles be [itex]\alpha[/itex] and [itex]\beta[/itex] about the poles at [itex]E_p[/itex] and [itex]-E_p[/itex], of radius [itex]\epsilon[/itex] and [itex]\rho[/itex], respectively, so that the total contour is
    $$\oint_{Total} = \int_{-R}^{-E_p-\epsilon}dp^0 + \int_\alpha + \int_{-E_p+\epsilon}^{E_p-\rho}dp^0 + \int_\beta + \int_{E_p+\rho}^R + \int_C$$.
    Now, let's evaluate the separate parts for the function
    $$\frac{-1}{z^2 -E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}$$
    where we use coordinates [itex]z=p^0 + iv[/itex].

    Firstly, we are going to use the residue theorem, so note that
    $$\frac{-1}{z^2 -E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} = \frac{-1}{(z-E_p)(z+E_p)} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}$$
    so has poles at [itex]\pm E_p[/itex] with residues [itex](-1/2E_p)\ e^{-iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}[/itex] and [itex](+1/2E_p)\ e^{+iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}[/itex]. Hence, the residue theorem tells us that
    $$\oint_{Total} \frac{-1}{z^2 -E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} = -\frac{2\pi i}{2E_p}\left[-e^{-iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} + \ e^{+iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}\right].
    $$
    (Note the minus sign as we're going the wrong way round the contour.)

    Secondly, for the large arc C, we can use Jordan's lemma. Note that, for sufficiently large R,
    $$\left|\frac{-1}{z^2 - E_p^2} e^{-ip\cdot(x-y)}\right|= \frac{1}{R^2-E_p^2}\quad\forall |z|=R$$
    which clearly tends to zero as [itex]R\to \infty[/itex]. Furthermore, given that we have a lower semicircular integral of the form
    $$\int_C e^{-iaz}f(z)dz$$
    then as long as [itex]a>0[/itex], we may apply Jordan's lemma. That is,
    $$\int_C \frac{-1}{z^2 - E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} \to 0\quad\text{as }R\to\infty$$
    as long as [itex]x^0-y^0>0[/itex]. So that first piece is solved.

    Finally, let's consider the contour around [itex]\alpha[/itex], which we parameterise by [itex]z = -E_p + \epsilon e^{i\theta}[/itex]. Hence
    $$\int_\alpha \frac{-1}{z^2 -E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} = -\int_0^\pi d\theta i\epsilon e^{i\theta} \frac{-1}{(-E_p + \epsilon e^{i\theta})^2 -E_p^2} e^{-i(-E_p + \epsilon e^{i\theta})(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}= i\int_0^\pi d\theta \frac{1}{(\epsilon e^{i\theta}-2E_p)} e^{iE_p(x^0-y^0) + i\epsilon e^{i\theta}(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}$$
    (Note the extra minus as we're going the wrong way round the contour.) Now, as [itex]\epsilon\to 0[/itex] we see
    $$\lim_{\epsilon\to 0}\int_\alpha \frac{-1}{z^2 -E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} = i\pi\frac{1}{-2E_p} e^{+iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}=i\pi Res(-E_p)$$
    (Note that this result is a general one.) Similarly for [itex]\beta[/itex]
    $$\lim_{\rho\to 0}\int_\beta \frac{-1}{z^2 -E_p^2} e^{-iz(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} = i\pi\frac{1}{2E_p} e^{-iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}=i\pi Res(E_p)$$

    The last thing we need is the concept of the Cauchy principal value integral: http://mathworld.wolfram.com/CauchyPrincipalValue.html, which means that
    $$P.V.\int_{-R}^R dp^0 = \lim_{\epsilon\to 0,\rho\to 0}\left[\int_{-R}^{-E_p-\epsilon}dp^0 + \int_{-E_p+\epsilon}^{E_p-\rho}dp^0+\int_{E_p+\rho}^R dp^0\right].$$

    Finally, putting everything together in the limits [itex]\epsilon\to 0,\ \rho\to 0,\ R\to \infty[/itex] gives
    $$P.V.\int_{-\infty}^\infty dp^0 = \oint_{Total}-\int_\alpha-\int_\beta -\int_C = -\frac{2\pi i}{2E_p}\left[-e^{-iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} + \ e^{+iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}\right] - i\pi\frac{1}{-2E_p} e^{+iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} -i\pi\frac{1}{2E_p} e^{-iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} - 0$$

    That is
    $$P.V.\int dp^0\ \frac{-1}{p^2 -m^2} e^{-ip\cdot(x-y)} = \frac{\pi i}{2E_p}\left[e^{-iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})} - \ e^{+iE_p(x^0-y^0) -i\mathbf{p}\cdot(\mathbf{x-y})}\right].$$

    Unfortunately I am out by a factor of two from the correct answer of (2.54) in PS. However, I am way past tired, so need to hit the hay --- if somebody finds my error, I'd appreciate if you can let me know, and I'll edit my post tomorrow.

    Furthermore, note how I've only got the result for the principal value integral...I recall reading somewhere that it equals the usual integral if it the function is Lebesgue-integrable. I don't know what this means, but if someone does, and could confirm the function in question is, I'd be tickled pink.

    Lastly, I haven't read this yet, but the following describes how the "cheating" strangerep was referring to may actually be notational (see page 14 or so of pdf): http://hitoshi.berkeley.edu/221a/contour.pdf.
     

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    Last edited: Aug 8, 2012
  11. Aug 9, 2012 #10

    strangerep

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    At this stage, I'll just make a few suggestions...

    1) You can factor out the exponential involving spatial components.
    So they don't affect the energy integral.

    2) You can cut down on the mess enormously by using some different symbols
    for the purposes of performing the integral. E.g., you could start with
    $$ \int\!dz\; \frac{e^{ikz}}{z^2 - a^2} $$ where ##k,a## are both real.
    That should reduce your typing considerably.

    3) Your application of Jordan's lemma is a bit sloppy, since you don't
    actually evaluate the the maximum value of the denominator on the large
    semicircle.

    4) Your treatment of the small semicircles around the poles is quite wrong.
    You can't use the residue theorem here unless you've got a closed circle
    around the poles, but you've only got two little semicircles. (The treatment
    in that contour.pdf file you linked to is sloppy in this regard, since it
    assumes certain symmetry on the upper and lower sides of each pole.)
    To do it properly, you'll need to do each semicircle properly as a line integral.
    Set ##z=r e^{i\theta}## with ##r## constant, and do a ##\theta## integration
    accordingly. The integrand must be Laurent-expanded around each pole first,
    so that you can subsequently take a limit as ##r\to 0##.

    5) If you use a slightly different contour by avoiding the poles via
    semicircles in the lower half plane, you won't have to evaluate any residues
    at all, since the new contour encloses no poles. If you evaluate the small
    line integrals correctly, you'll get the right answer.

    6) I just found my old notes where I evaluated this same integral, so I can
    probably post it if you get stuck. Item 5 above can be a bit tricky if you
    haven't seen it done before, but I'll let you have a go at it first. :-)

    7) Regarding the Lebesgue-integrable thing... it's insensitive to integration
    on "sets of measure zero" (i.e., single points in this case). So you can
    exclude the isolated poles on the real axis. I hope that makes sense.
     
  12. Aug 9, 2012 #11
    True. That would make things a lot neater. My mistake.
    OK, my notes say that as long as there is an upper bound such that the integral is less than or equal to it for all ##|z|=R## then one can use Jordan's lemma. When you say I haven't evaluated it, is it not ##1/(R^2 - E_p^2)##? (I was assuming ##E_p = \sqrt{\mathbf{p}^2 +m^2}## was constant...)

    I only used the residue theorem for the complete closed loop integral, which certainly includes the two poles. So I don't follow that point.

    Secondly, I did parameterise each semicircle exactly as you said. I'm not sure if what I did was "Laurent-expanded", but it seems right to me. So I don't really follow this point either.
    OK, true, that would be quicker. Thinking in my head, the integrals on the semicircles would change sign from what I gave in the last post, and that gives ... the same answer I originally had. I believe this is wrong, as it is a factor of two out from the top equation of (2.54) on page 30 of PS. Maybe it isn't...

    I think I've addressed your points, and I'm not sure if I'm as wrong as I felt initially... I think I did the integration on the semicircles correctly, but maybe you skimmed over it?

    Not really...does this mean that the principal value is equal to the normal integral?
     
  13. Aug 9, 2012 #12

    vanhees71

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    I haven't followed your long calculation in detail. What's for sure is that you calculate the retarded propagator if you use the contour depicted in #9. In the following I name this contour [itex]\mathcal{C}_{\text{ret}}[/itex]. To see that you get the retarded propagator, you do the integral along the contour by closing it with large semicircles in the upper (lower) [itex]p_0[/itex]-half-plane such that due to the exponential of the Fourier integral, which is the Mills representation of the free vacuum propagator of a scalar field,
    [tex]\tilde{G}_{\text{ret}}(t,\vec{p})=\int_{\mathcal{C}_{\text{ret}}} \frac{\mathrm{d} p_0}{2 \pi} \frac{\exp(-\mathrm{i} p_0 t)}{p_0^2-\omega^2}[/tex]
    the contribution of these additional circles becomes 0 in the limit of infinite radius, i.e., you have to close the contour for [itex]t<0[/itex] in the upper and for [itex]t>0[/itex] in the lower half-plain. Since there are no poles inside the loop for [itex]t<0[/itex] you have
    [tex]\tilde{G}_{\text{ret}}(t,\vec{p}) \propto \Theta(t).[/tex]
    For [itex]t>0[/itex] you pick up both poles of the integrand. Due to the residue theorem you have
    [tex]G_{\text{ret}}(t,\vec{p})=-\Theta(t) \mathrm{i} \left [\frac{\exp(-\mathrm{i} \omega t)}{2 \omega} - \frac{\exp(\mathrm{i}\omega t)}{2 \omega} \right ] =-\Theta(t) \frac{\sin(\omega t)}{\omega}.[/tex]
    Of course, in the whole calculation [itex]\omega=\sqrt{\vec{p}^2+m^2}.[/itex] The sign comes from the clockwise, i.e., negative orientation of the contour in the complex plane.

    It is easy to check that the Mills representation obeys the equation for the Green's function of the KG equation, i.e.,
    [tex](\partial_t^2+\omega^2) \tilde{G}_{\text{ret}} ( t, \vec{p} )=-\delta(t),[/tex]
    corresponding to the equation in position representation
    [tex](\Box+m^2) G(t,\vec{x})=-\delta^{(4)}(x).[/tex]
    In energy-momentum representation you can alternatively write
    [tex]\bar{G}_{\text{ret}}(p)=\frac{1}{(p_0+\mathrm{i} 0^+)^2-\omega^2} = \frac{1}{p^2-m^2+\mathrm{i} \mathrm{sign}(p_0) 0^+.}[/tex]
    This shifts the poles in the complex [itex]p_0[/itex] plane both infintesimally into the lower plane. Now you can integrate along the real axis and do the same calculation again, closing the contour as shown above. You'll get the same result as before.

    What you usually need in the Feynman rules of vacuum quantum field theory is a slightly different propagator. This propagator is uniquely defined by the time ordering prescription, which comes from the Dyson series of the time-evolution operator in the interaction picture due to Wick's theorem. The corresponding propagator is the socalled causal propagator. In the vacuum it's at the same time the Feynman propagator that is defined that the particles correspond to positive energies and are propagated according to the retarded propagator and the antiparticles (or holes) are propagated according to the advanced propgator. In time-position representation the definition is
    [tex]G_{c}(t,\vec{x})=-\mathrm{i} \langle \Omega|\mathcal{T} \hat{\phi}(t,\vec{x}) \hat{\phi}^{\dagger}(0)| \Omega \rangle.[/tex]
    Using the mode decomposition of the quantum fields into creation and annihilation operators wrt. the one-particle momentum eigenstate basis, you find
    [tex]\bar{G}_c(p)=\frac{1}{p^2-m^2 + \mathrm{i} 0^+}.[/tex]
    By closing the contour for the [itex]p_0[/itex] integration you get for the Mills representation
    [tex]\tilde{G}(t,\vec{p})=-\mathrm{i} \Theta(t) \frac{\exp(-\mathrm{i} \omega t)}{2 \omega} -
    \mathrm{i} \Theta(-t) \frac{\exp(+\mathrm{i} \omega t)}{2 \omega}=-\mathrm{i}\frac{\exp(-\mathrm{i} \omega |t|)}{2 \omega}.[/tex]
     
  14. Aug 9, 2012 #13
    Thanks vanhees, the effort is much appreciated. However, at the moment I'm still stuck on the position integral, not momentum. Your post will be most helpful when I spend more time of the link between the two. (I was stuck on (2.57) and the following equation in PS for a while, especially why there is the opposite sign to normal in the Fourier transform.)

    Two points though:

    Firstly, I had problems with the following:

    $$\bar{G}_{\text{ret}}(p)=\frac{1}{(p_0+\mathrm{i} 0^+)^2-\omega^2} = \frac{1}{p^2-m^2+\mathrm{i} \mathrm{sign}(p_0) 0^+.}$$

    When you expand the LHS demominator out, there is a term ##2ip^0 0^+##. Is the reason you only put ##i\mathrm{sign}(p^0)0^+## because the size of ##p^0## is irrelevant, given the limit ##0^+##? The reason I bring it up is because every other time I've seen the equation, the ##\mathrm{sign}(p^0)## part is left out...

    Lastly, something trivial: did you miss a subscript "c" on the propagator in your last equation?
     
  15. Aug 9, 2012 #14

    strangerep

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    Unbelievable! Now it's my turn to use the "I was way past tired" excuse. Yes, I did indeed skim too fast. (Maybe I wouldn't have if there weren't so many symbols floating around, but, oh, that's a lame excuse.) I'll try to check your factors of 2 later after I wipe the egg off my face.
     
  16. Aug 10, 2012 #15

    vanhees71

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    The full transform from energy-momentum to time-position representation is not that easy. You have to calculate (take the time-ordered propagator as an example)
    [tex]G_c(t,\vec{x})=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p}{(2 \pi)^4} \frac{\exp(-\mathrm{i} p \cdot x)}{p^2-m^2+\mathrm{i} 0^+}.[/tex]
    This is pretty involved and leads to modified Bessel functions (except in the massless case, where you get [itex]\delta[/itex] distributions).

    To simplify the task you can use the fact that the Green's functions are Lorentz scalars (under proper orthochronous Lorentz transformations). Thus you can evaluate the integral for special choices of [itex]x[/itex] and then write it covariantly to get the general case. Unfortunately, there are singularities along the light cone, and you have to investigate spacelike ([itex]x^2<0[/itex]) and timelike ([itex]x^2>0[/itex]) space-time arguments separately. In the latter case you can set [itex]\vec{x}=0[/itex] in the former [itex]t=0, \quad \vec{x}=(0,0,r)[/itex].

    You find these calculations in older textbooks, where the space-time formulation is treated in more detail then in modern ones, e.g., in Bjorken Drell, Schweber, or Bogoliubov Shirkov.


    That's correct. Of course you can as well write [itex]\mathrm{i} p_0 0^+[/itex].

    Yes, I did. I added these subscript "ret" and "c" to clearly indicate the different propagators.
     
  17. Aug 10, 2012 #16

    strangerep

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    I now think the expression (2.54) in PS is wrong, and should in fact have
    the following denominator in the integrand:
    $$
    p^2 - m^2 + i\epsilon p^0
    $$
    This corresponds to Scharf, eq(2.3.22). The extra ##\epsilon## terms shifts
    both poles into the lower halfplane, so you don't need the little semicircles
    to avoid them on the real axis.

    However, this also means that there's some other fudges floating around, since
    the 3D integral should then involve
    $$
    \frac{1}{E_p - i\epsilon}
    $$
    and
    $$
    \frac{1}{-E_p - i\epsilon}
    $$
    and blithely ignoring the ##\epsilon## terms at this point is unjustified.

    BTW, note that the above from the Feynman prescription where we use the
    denominator
    $$
    p^2 - m^2 + i\epsilon
    $$
    which moves one pole up and the other down.

    OTOH, the factor of 2 can probably be corrected in a better way by starting
    the whole relativistic field formalism using an appropriately normalized measure
    on the +ve energy mass shell. But that's a story for another day.
     
  18. Aug 10, 2012 #17
    I was actually starting from the time-position basis as in Peskin and Schroeder. I'd then transform to energy-momentum. Nonetheless, you've given me a lot to think about!

    That makes my late night feel a little less of a waste!

    OK, you've given me a lot to read up on, which I'll get to soon. I'm especially interested that you think PS is wrong. Also, this "appropriately normalized measure on the +ve energy mass shell..." is new to me, as I thought things were already nicely normalised? (Is what you are referring to a mainstream idea?)

    Oh, would you say my process of evaluating the integral was correct, and the factor of two was not an error in my calculations?

    Cheers strangerep.
     
  19. Aug 10, 2012 #18

    strangerep

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    My recent boo-boo makes me too shy to utter any strong statements for a while.
    But if you've still got energy, you could check my results by doing
    the following exercise:

    Start with the simpler integral I mentioned earlier, i.e.,
    $$I_1 ~:=~ \int_{-\infty}^{+\infty} \!\! dz \; \frac{e^{ikz}}{z^2 - a^2}$$
    where ##k,a## are real nonzero constants, with ##k>0##.
    (But don't refer to your previous work while doing this, so as not to bias the calculation.)
    Do it two ways:

    1) with a contour that takes little detours around the poles in the upper halfplane (and hence encloses both poles),

    2) with a contour that takes little detours around the poles in the lower halfplane (and hence encloses no poles).

    Do you get the same answer in both cases, and is the answer compatible with what you had before?

    Then do it twice more with modified denominators:

    (i) ##z^2 - a^2 + i\epsilon z##

    (ii) ##z^2 - a^2 + i\epsilon##

    Check where each of these moves the poles, and what the contour integral results are in both cases.


    It arises if one constructs QFT by causal +ve energy Poincare representations. One needs an integration measure over the +ve energy mass shell of the form
    $$
    \int\! d^4p \; \delta(p^2-m^2) \; \Theta(p_0)
    $$
    except that some constant factors are needed in order to end up with the usual 3D integral formulas and covariant commutation relations. Scharf also talks about integrals of this form -- which are closely related to all the stuff we've been talking about here.
     
  20. Aug 11, 2012 #19
    I get
    $$I_1 = -\frac{i\pi}{2a}\left(e^{ika}-e^{-ika}\right) = \frac{\pi}{a}\sin{ka}$$
    in both cases, which is compatible with what I previously had.

    OK, I will be pedantic for now, such that, the poles are at
    $$z= \frac{-i\epsilon}{2}\pm \sqrt{a^2-\frac{\epsilon^2}{4}},$$
    i.e. they're shifted into lower half plane (and displaced slightly on real axis as well).

    The residue theorem (with the assumption that the arc of semicircle goes to zero, which I haven't checked) then gives
    $$\int_{-\infty}^{+\infty} \!\! dz \; \frac{e^{ikz}}{z^2 - a^2+i\epsilon z} = -\frac{2\pi i}{\alpha_1 - \alpha_2}\left(e^{ik\alpha_1} - e^{ik\alpha_2}\right)\quad\text{where}\quad \alpha_1 = -\frac{i\epsilon}{2}+\sqrt{a^2-\frac{\epsilon^2}{4}},\ \ \alpha_2 = -\frac{i\epsilon}{2}-\sqrt{a^2-\frac{\epsilon^2}{4}}.$$
    Now, in the limit ##\epsilon \to 0##, ##\alpha_1 \to a## and ##\alpha_2 \to -a##, and hence
    $$\lim_{\epsilon \to 0}\int_{-\infty}^{+\infty} \!\! dz \; \frac{e^{ikz}}{z^2 - a^2+i\epsilon z} = -\frac{\pi i}{a}\left(e^{ika} - e^{-ika}\right)$$
    which (hey presto!) is as before, but with the factor of two problem fixed, so it will agree with PS. So, it looks like we agree PS may be wrong...

    OK, now the the poles are at
    $$z=\pm re^{i\theta}\quad\text{where}\quad r=(a^4 + \epsilon^2)^{\frac{1}{4}},\ \ \theta=-\frac{1}{2}\tan^{-1}\left(\frac{\epsilon}{a^2}\right).$$
    That is, the one with positive real component is in the lower half plane, and the one with negative real component is in the upper half plane. That means there will be only one residue inside the contour (semicircle in lower half plane), so
    $$\int_{-\infty}^{+\infty} \!\! dz \; \frac{e^{ikz}}{z^2 - a^2+i\epsilon} = -2\pi i \frac{e^{ik re^{i\theta}}}{2 r e^{i\theta}}.$$
    As ##\epsilon \to 0,\ r\to a,\ \theta \to 0## so
    $$\lim_{\epsilon \to 0}\int_{-\infty}^{+\infty} \!\! dz \; \frac{e^{ikz}}{z^2 - a^2+i\epsilon} = -\frac{\pi i}{a} e^{ika}$$
    which, unless I've made an error, is the same apart from the missing negative exponential term.

    So, to vanhees71 --- it would seem one should include the ##p^0## explicitly.

    Where do we go from here? Conclude PS (2.54) is incorrect?

    Ok, all I know was the PS introduced the ##\sqrt{2E_p}## factor to make things Lorentz invariant. Is this what you are talking about?
     
  21. Aug 11, 2012 #20

    vanhees71

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    But your solution to the first problem cannot be completely right. If you enclose no poles, the result is 0.

    Note that you must close the contour in the upper [itex]z[/itex] half-plane. So for contour 1 you get 0, because in that case there are no poles inside the contour (contrary to what strangerep said in his question). If you close the contour in the lower half-plane the exponential blows up along the semi circle there!

    In the second case both poles are within the contour, and then you have according to the residue theorem (I take [itex]a>0[/itex] to make the notation simple)
    [tex]\int_{\mathcal{C}_w} \mathrm{d} z \frac{\exp(\mathrm{i} z k)}{k^2-a^2}=2 \pi \mathrm{i} \left [\frac{\exp(\mathrm{i} k a)}{2a}-\frac{\exp(-\mathrm{i} k a)}{2 a} \right ]=-\frac{2 \pi}{a} \sin(k a).[/tex]
    So here you have a factor 2 wrong.

    The other cases you should do carefully in the same way. I must say, I do not understand the idea behind your calculations at all.

    Now to Peskin-Schroeder (2.54). I'm always very sceptical about formulae in this book, because it's full of typos and little misconceptions. So it's always good to check his stuff. To check Eq. (2.54) we only have to do the [itex]p^0[/itex] integral at the very end, for [itex]t=x^0-y^0>0[/itex]:
    [tex]f(\vec{p},m)=-\int_{\mathrm{C}_{\text{PS}}} \frac{\mathrm{d} p^0}{2 \pi \mathrm{i}} \frac{\exp(-\mathrm{i} p^0 t)}{p^2-m^2}.[/tex]
    Now we have to close the contour in the lower half-plane (note the additional minus sign in the exponential compared to the integral before). Now the contour is clockwise and we have to take into account an additional sign in the residue theorem, which gives (since again we enclose both poles and [itex]E_{\vec{p}}=+\sqrt{\vec{p}^2+m^2}[/itex]):
    [tex]f(\vec{p},m)=+ \left [\frac{\exp(-\mathrm i E_{\vec{p}}t)}{2 E_{\vec{p}}}-\frac{\exp(+\mathrm{i} E_{\vec{p}}t)}{2 E_{\vec{p}}} \right ]=-\frac{\mathrm{i}}{E_{\vec{p}}} \sin(E_{\vec{p}} t).[/tex]
    But this is identical to the line before in Peskin/Schroeder. This is indeed also identical with one further line up since you come to it by substituting [itex]\vec{p} \rightarrow -\vec{p}[/itex] in the second part of the [itex]\vec{p}[/itex] integral. So in this case PS is correct.
     
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