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

Feynman propagator as distribution

  1. Feb 25, 2008 #1

    This question is quite relevant to some other posts at the end of the very long "very simple QFT questions" thread, but I've decided to start a new thread with a heading which is more indicative of what I wish to ask the group. As a question, it's a fairly concise, but the analysis is lengthy and I've tried to sum up what I've done, and I hope it's of interest.

    The Feynman propagator is almost exclusively used and referred to in terms of its Fourier transform. For instance, for a scalar particle we write:
    [tex] G(k) = \frac{1}{k^2 - m^2 + i\epsilon} [/tex]​
    However, I wish to evaluate the propagator in its position representation (i.e. as a function of spacetime (t,x)). Doing so may provide some useful insight. This means evaluating the Fourier transform
    [tex] G(x) = \frac{1}{(2 \pi)^4} \int d^4k \; \frac{e^{-ikx}}{k^2 - m^2 + i\epsilon} [/tex]​

    Evaluation of this integral was discussed on another recent thread of this forum ("Propagator - Hankel function - infinite? - Weinberg"), which I replied to. I've now more or less worked through the whole problem and my solution is developed below. My solution disagrees with the Wikipedia article 'Propagator', so I agree with Hans de Vries in the long 'very simple QFT' thread that the Wikipedia article needs correcting.

    My position is that this integral is formally not convergent for any x. For instance, for t > 0 we do the k_0 integral first, completing the contour in the lower half plane and collecting the pole at [tex] k_0 = \sqrt{k^2 + m^2} [/tex]; this gives
    [tex] G(x) = \frac{-i}{(2\pi)^3} \int \frac{d^3k}{2 E_k} e^{-ikx} [/tex]​
    where [tex] E_k = \sqrt{k^2 + m^2} [/tex]. This integral is clearly divergent - try to do it in polar coordinates, for instance: the denominator only goes as 1/k for large 3-momenta k, while the volume of integration goes as k^3.

    This indicates that the propagator cannot be a function, but rather a generalised function (distribution). Generalised functions are defined by their action on some suitable class of test functions, typically smooth functions with compact support. For instance, the delta function acts on test functions h according to
    [tex] \delta^4(x) : h \rightarrow h(0) . [/tex]​
    Fourier analysis for distributions is a rigorous theory, though it is not always expressible in terms of convergent integrals, and this is what we see above. Carrying out the analysis, I find that the propagator acts on test functions h according to
    [tex] G : h \rightarrow \lim_{\epsilon \rightarrow 0} \int_{|t^2-x^2|>\epsilon^2} d^4x \; B(x) h(x) [/tex]​
    [tex] B(x) = \frac{-im H_1(m\tau)}{8\pi \tau} \qquad (\tau = \sqrt{t^2 - x^2}) [/tex]
    [tex] B(x) = \frac{m K_1(m\tau)}{4\pi^2 \tau} \qquad (\tau = \sqrt{x^2 - t^2}) [/tex]​
    for timelike and spacelike x, respectively. H_1 is a Hankel function of order 1, and K_1 is a modified Bessel function of order 1. (The spacelike result is the result mentioned on page 202 of Weinberg QTF1 and discussed in the other thread.) The function B is a good indicator of the spacetime dependence of the propagator - for instance, the exponential decay of K_1 with distance means that the propagator is mostly confined to the forward and backward timelike regions, consistent with causality.

    The \epsilon -> 0 construction in the solution is necessary because the Bessel functions -iH_1 and K_1 in the integrand diverge at the light cone, one to -infinity and the other to +infinity, and both diverge too strongly to simply define the action of G on h in terms of an integral over all space time. However, nonzero epsilon excludes the light cone from the domain of integration, and a (delicate) proof shows that the limit exists as epsilon tends to zero. It's like trying to define the integral of 1/x over [-1,1]; the best we can do is define a Cauchy 'principal value' of the divergent integral.

    Having defined the propagator G, it is then necessary to show that it satisfies the defining relation
    [tex] (\partial^\mu \partial_\mu + m^2 ) G = \delta^4(x) [/tex]​
    This means proving the remarkable claim that
    [tex] \lim_{\epsilon \rightarrow 0} \int_{|t^2-x^2|>\epsilon^2} d^4x \;
    B(x) (\partial^\mu \partial_\mu + m^2 )h(x) = h(0) [/tex]​
    for all test functions h. This (lengthy) calculation uses Green's theorem (integration by parts); surface integral terms from the boundaries (\tau = \epsilon near the light cone) can be collected to give the h(0) as required. We use the fact that B satisfies the massive wave equation at all x except on the lightcone. It's really tricky to get the whole thing watertight because the Bessel functions are diverging as you move toward the light cone. However, it's done now and I'm fully convinced of the validity of the claim, give or take minor errors. I'm looking for references to check the analysis and answer against, but I haven't found any that do the problem in full mathematical rigour. Feynman (Theory of Fundamental Forces, Benjamin 1961) mentions a similar result but includes a delta function, which I disagree with, and doesn't indicate that the epsilon -> 0 limit is necessary.

    Has anyone ever seen this problem worked through before? I'm puzzled why none of the textbooks discuss this problem from a formal standpoint. It's such an ubiquitous quantity and I think it's crucial to know what it means and under what conditions it can be used.


  2. jcsd
  3. Feb 25, 2008 #2


    User Avatar
    Science Advisor

    Try Scharf's "Finite Quantum Electrodynamics", pp64-69. He actually calculates
    the Pauli-Jordan function, but the various related distributions are also obtained (advanced,
    retarded, +/-, etc). At a superficial glance, it doesn't look quite the same as your result,
    but I haven't checked the tricky details of all the special functions involved.

    (I also recall that the derivation is also done in one of Walter Greiner's volumes, but I
    can't now remember which one.)
  4. Feb 25, 2008 #3


    User Avatar
    Homework Helper

    clearly... well... technically, I guess you could call that integral "divergent", but it doesn't blow up, it just oscillates, and so you can fix that up by giving t a small imaginary part to make the integral converge. I.e., for small |k| the integral does not diverge, and for large |k| the integral goes like
    \frac{1}{|\vec x|}\int dk e^{-i E_k t}\sin(k|\vec x|)
    which "clearly" can be fixed right up by letting [itex]t \to t-i 0^+[\itex]. This fix up is similar to considerations of generalized functions...
  5. Feb 26, 2008 #4


    User Avatar
    Science Advisor

  6. Feb 27, 2008 #5
    I do not know if this related to the topic but watching it i remember i 'found' a curious relation between distributions and functions.

    Let be the divergent integral [tex] \int_{-\infty}^{\infty}dx f(x) e^{ipx} [/tex]

    then expanding f(x) into a power series and using the properties of Dirac delta and other distributions i came across the formula [tex] \int_{-\infty}^{\infty}dx f(x) e^{ipx}= f( -i \partial _{x}) \delta (x) [/tex]

    this means that although divergent the integral could be considered as an infinite series involving delta and its derivatives (for a more general case replace the derivative respect to x by the gradient)
  7. Feb 28, 2008 #6
    Thanks very much for the responses. I'm looking at the books now. I have General Principles of QFT by Bogolubov, Logunov, Oksak & Todorov, and I agree it has an excellent exposition on generalised functions. The various Green's functions for the scalar field are listed on pp 353-354.

    Scharf in the preface of his Finite QED makes quite a claim: that the ubiquitous UV and IR divergences in QFT can be avoided by careful and rigorous application of the mathematics. The method develops work by Epstein and Glaser in the 70s, the result of which is that 'no infinity can appear and life is beautiful'. It sounds wonderful...? I've recently been learning about different methods of regularisation (cutoffs, dimensional reg, Pauli-Villars fictitious heavy particles), all of which feel distinctly ad hoc or unphysical (or both). Scharf's promise sounds attractive indeed - but possibly too good to be true? Any comments?


  8. Feb 28, 2008 #7


    User Avatar
    Science Advisor

    The Epstein-Glaser-Scharf approach essentially boils down to smoothing the step
    function that is implicitly used in the time-ordered product. However, after reading rather
    a lot of stuff in Scharf's book, it eventually becomes clear that the smoothed function
    is being constructed perturbatively. That is, at each order of perturbation, the smoothed
    function is adjusted a bit more to cure the infinities that would otherwise emerge at
    that order.

    So, unfortunately, Scharf does not give a non-perturbative answer. It is not known
    what the limit of the smoothed function might be (or even whether it exists).
  9. Mar 1, 2008 #8


    User Avatar
    Gold Member

  10. Mar 3, 2008 #9

    Hans de Vries

    User Avatar
    Science Advisor

    It looks like the he expression from the t,p domain to the t,r domain can
    be simplified substantially:

    For the Feynman propagator we have, see for instance Zee I3, eq: 23

    D^{\cal F}(t,r)\ =\ \frac{1}{(2\pi)^3}\ \frac{1}{2i}\ \int_{-\infty}^\infty dp^3\ \frac{\exp\left( -i\sqrt{p^2+m^2}\ |t| \right)}{\sqrt{p^2+m^2}}\ e^{i\vec{r}\cdot\vec{p}}

    If we go from the 3d spatial to the radial Fourier transform in 3d this becomes:

    D^{\cal F}(t,r)\ =\ \frac{4\pi}{(2\pi)^3}\ \frac{1}{2ir}\ \int_0^\infty dp\ \frac{p\ \exp\left( -i\sqrt{p^2+m^2}\ |t| \right)}{\sqrt{p^2+m^2}}\ \sin(rp)

    The argument is anti-symmetric in p so we can turn this into a 1d Fourier transform:

    D^{\cal F}(t,r)\ =\ \frac{4\pi}{(2\pi)^3}\ \frac{1}{4r}\ \int_{-\infty}^\infty dp\ \frac{p\ \exp\left( -i\sqrt{p^2+m^2}\ |t| \right)}{\sqrt{p^2+m^2}}\ e^{-irp}

    The argument can be expressed as a derivative like this.

    D^{\cal F}(t,r)\ =\ \frac{\pi}{(2\pi)^3}\ \frac{i}{r|t|}\ \int_{-\infty}^\infty dp\ \frac{\partial}{\partial p}\ \bigg\{ \exp\left( -i\sqrt{p^2+m^2}\ |t| \right)\bigg\}\ e^{-irp}

    A derivative in p is equal to a multiplication with -ir in the t-r domain, so

    D^{\cal F}(t,r)\ =\ \frac{\pi}{(2\pi)^3}\ \frac{1}{|t|}\ \int_{-\infty}^\infty dp\ \exp\left( -i\sqrt{p^2+m^2}\ |t| \right)\ e^{-irp}

    We are left with a substantially simpler expression. Using [itex]E_p = \sqrt{p^2+m^2}[/itex] we can write:

    D^{\cal F}(t,r)\ =\ \frac{\pi}{(2\pi)^3}\ \frac{1}{|t|}\ \int_{-\infty}^\infty dp\ e^{ -iE_p|t|}\ e^{-irp}

    For the massless case we get:

    D^{\cal F}(t,r)\ =\ \frac{\pi}{(2\pi)^3}\ \frac{1}{|t|}\ \int_{-\infty}^\infty dp\ e^{-i |p| |t|}\ e^{-irp}

    Which is correct as far as I can see.

    Regards, Hans
    Last edited: Mar 3, 2008
  11. Mar 4, 2008 #10

    Hans de Vries

    User Avatar
    Science Advisor

    and the correct answer is.....

    For the Green's function of the Feynman propagator:

    D^{\cal F}(E,p)\ =\ \frac{-1}{E^2-p^2-m^2+i\epsilon}\

    We get in position space: (congratulations to the starter of this thread)

    [tex] D^{\cal F}(t,r)\ =\ \frac{m}{8\pi s}\ H^{(1)}_1(ms)[/tex]

    where [itex]s^2=t^2-r^2[/itex] and [itex]H^{(1)}_1(z)[/itex] is the complex Hankel function of the first kind
    and first order. It is the Bessel equivalent of [itex]\exp(iz)=\cos(z)+i\sin(z)[/itex]


    Where J and Y are the Bessel J and bessel Y functions. As one could expect:
    The complex conjungate of the propagator looks like a positive energy
    exponential for (t>0) and like a negative energy exponential for (T<0).

    The trick was to do a Fast Fourier Transform on the function minus the function
    for the massless particle:

    \frac{\pi}{(2\pi)^3}\ \frac{1}{|t|}\ \int_{-\infty}^\infty dp\ \bigg[\ e^{ -i\sqrt{p^2+m^2}|t|}\ ~~-~~e^{-i |p| |t|}\ \bigg]\ e^{-irp}

    The argument now goes to zero when p increases and we can calculate the
    massless case:

    [tex] D^{\cal F}(t,r)\ =\ \frac{1}{4\pi}\ \delta(s^2)\ \ +\frac{i}{8\pi^3}\ \frac{1}{s^2}[/tex]

    The result of the mentioned FFT compared with the mathematical result
    can be seen in the image.

    Yellow: Real value of the FFT
    White: Imaginary value of the FFT
    Blue: Real value calculated
    Red: Imaginary value calculated

    Regards, Hans

    Attached Files:

    Last edited: Mar 4, 2008
  12. Mar 5, 2008 #11
    Unfortunately, there was after all a minor error in my calculation and result presented in the header for this thread. For posterity, I had better correct it... Having checked with the two references (Scharf, and Bogolubov et al mentioned in earlier posts), I think I quote them correctly that the correct spacetime expression for the Feynman propagator is
    [tex] D(x) = \frac{1}{4\pi} \delta(\tau^2) + B(x) [/tex]​
    where [tex]\delta(\tau^2)[/tex] is defined to be the (everywhere-non-negative) distribution
    [tex]\delta(\tau^2) = \frac{1}{2r} \delta(t-r) + \frac{1}{2r} \delta(t+r)[/tex]​
    and the regular part B(x) is
    [tex]B(x) = \frac{-1}{8\pi\tau} H_1(m\tau) [/tex]
    [tex]B(x) = \frac{-i}{4\pi^2\tau} K_1(m\tau) [/tex]​
    for timelike and spacelike x, respectively. In my original post, I lost the delta function part because my notes were in a mess, having taken about a dozen attempts at the problem. But it should definitely be there, and in the last-but-one equation in Hans de Vries' post, it corresponds to the Green's function for the massless particle that was subtracted to make the integral converge.

    I write the solution in this form because D takes a fundamentally different form at spacelike and timelike x. As mentioned in previous posts, at timelike x, both types of Bessel function J_1 and Y_1 are present, in the combination H_1 = J_1 + iY_1. However, at spacelike x there is no comparable combination of the modified bessel functions K_1and I_1, because I_1 diverges exponentially at large distances and therefore we don't want any of it. Only K_1 is present.

    I think the symmetries of D are nicely summed up by saying that D is even under parity, even under time reversal.

    However, I still wish to advocate the 'principal part' idea I presented in the header post, which is discussed by Scharf, and possibly discussed amid Bogolubov's technical presentation of distributions. (I'm not 100% sure.) The problem is that leaving D in its unqualified form above is technically incorrect, because an integral like
    [tex] \int d^4x \; D(x) h(x) [/tex]​
    is not well defined (h is an arbitrary smooth, well-behaved function). This is because B(x) diverges too strongly near the light cone, so the integral converges neither in the sense of Riemann nor of Lebesgue. Yet we need this type of integral (in particular the convolution of D with h), because the ability to do so permits solution of inhomogeneous equations like
    [tex] (\partial_\mu \partial^\mu + m^2 ) \psi(x) = h(x) [/tex]​
    and the possibility of doing so is one of the principal uses of a Green's function or propagator in perturbation theory. So we define it using the lim(\epsilon -> 0) construction, according to the first post.
    Last edited: Mar 5, 2008
  13. Mar 7, 2008 #12

    Hans de Vries

    User Avatar
    Science Advisor

    Actually, it is one and the same function.

    [tex]H^{(1)}_1(iz)\ =\ J_1(iz)+iY_1(iz)\ =\ \frac{2}{i\pi}\ K_1(z)[/tex]


    The Dirac function is there. It stems from the cosine term in.

    [tex] D^{\cal F}(t,r)\ =\ \frac{\pi}{(2\pi)^3}\ \frac{1}{|t|}\ \int_{-\infty}^\infty dp\ e^{ -iE_p|t|}\ e^{-irp}[/tex]

    [tex] =\ \frac{\pi}{(2\pi)^3}\ \frac{1}{|t|}\ \int_{-\infty}^\infty dp\ \bigg(\ \cos(-E_p|t|})~~+~~i\sin(-E_p|t|})\bigg)\ e^{-irp} [/tex]

    The cosine term corresponds with the Bessel J function while the Bessel Y
    function corresponds with the sine term. The cosine term is symmetrical in
    p corresponding with two Dirac functions in the Fourier domain if [itex]E_p \rightarrow |p|[/itex].

    Regards, Hans
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook