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Residues in feynman diagrams

  1. Aug 5, 2010 #1
    Say you want to evaluate this integral:

    [tex]\int^{\infty}_{-\infty} \frac{dx}{x^2-A} [/tex]

    Since the bottom factors into [tex]\frac{1}{(x-\sqrt{A})(x+\sqrt{A})} [/tex],

    using the theorem of residues, the integral is

    [tex]\pm 2\pi i (\frac{1}{2\sqrt{A}}) [/tex]

    (where [tex]\pm [/tex] is + if [tex]Im \{\sqrt{A} \}>0 [/tex] and - otherwise, but the sign doesn't really matter for what I'm about to ask).

    In quantum field theory, we are asked to take integrals of the form above, except the A term contains an infinitismal imaginary element. Suppose that the mass of a particle is 1, then is [tex]\int^{\infty}_{-\infty} \frac{dx}{x^2-(1)^2} [/tex] undefined? I don't understand how this is mathematically different than:

    [tex]\int^{\infty}_{-\infty} \frac{dx}{x^2-(1)^2+i\epsilon} [/tex].

    For example, plugging these expressions into a math program (say Wolfram|Alpha.com):

    input: integrate 1/(x^2-(1-.025*I)) dx (x,-infinity,infinity)
    output: http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2-%281-.025*I%29%29+dx+%28x%2C-infinity%2Cinfinity%29

    agrees with the residue method:

    input: -(2*pi*I)(1/2)/sqrt(-.025*I+1)
    output: http://www.wolframalpha.com/input/?i=-(2*pi*I)(1/2)/sqrt(-.025*I+1)

    I set [tex]\epsilon[/tex] equal to .025, because if I set it less than .025, for some reason Wolfram|Alpha no longer understands what I want and won't evaluate the integral.

    I can however set [tex]\epsilon[/tex] to a very small value in the residue method and as you would expect, you get [tex]\pi*i [/tex], which is what you get when you naively use the residue method on:

    [tex] \int^{\infty}_{-\infty} \frac{dx}{x^2-(1)^2}[/tex]

    [tex]=2\pi i\frac{x-1}{(x-1)(x+1)}|_{x=1}=\pi*i [/tex]

    Is this is what is happening in QFT? That we are using the residue method, ignoring that the pole is actually on the line? I understand there are good reasons for shifting the pole off the line (to choose out the vacuum-to-vacuum expectation value in the path integral), but mathematically, by shifting pole, are we getting the same result if we ignored that the pole was on the line and just did a contour integral?

    Strangely, I don't ever remember solving a Feynman diagram with a residue theorem though. All I remember is Wick rotating, Feynman parametrization, and then all the [tex]\epsilon[/tex]'s seem to drop out or don't matter! How was it that I was able to escape using the residue theorem to evaluate the integrals? Can someone point out how the poles in a Feynman diagram integral don't require the residue theorem to evaluate the integral?
    Last edited: Aug 5, 2010
  2. jcsd
  3. Aug 5, 2010 #2
    The [tex]\epsilon[/tex] added to the denominator is actually selecting out the boundary conditions we want.
    The integral together with the prescription to go through the poles one the real line is a Green's function to the field equation.
    Different prescriptions correspond to different boundary conditions.
    And one of the prescription, i.e. going around one pole from above and round the other from below, is mathematical equivalent to the [tex]\epsilon[/tex] trick.
    This is the Feynmann propagator, and is physically relevant.

    Is there anyone who can address the relation between usual boundary condition in differential equations and the form of boundary conditions in terms of poles prescriptions?
  4. Aug 5, 2010 #3
    I vaguely recall a differential equations approach, with retarded and advanced green's functions as the reasoning for the [tex]\epsilon[/tex] prescription. It was confusing to me because it seems that QFT is inherently not a differential equations approach, but an integral approach.

    Anyways, I just remembered that Wick rotating the graph ensures that there is no pole! So you somehow can Wick rotate away the poles, because the denominator will be a Euclidean inner product instead of Minkowksi, so there is no zero in the denominator for non-negative mass! Is this valid in general mathematically? Can you just Wick rotate away your poles in a general contour integral?
  5. Aug 5, 2010 #4
    I guess you can Wick rotate:
    \int^{\infty}_{-\infty} \frac{dx}{x^2-1+i\epsilon}=-i \int^{\infty}_{-\infty} \frac{dx}{-x^2-1+i\epsilon}=i[arctan(\infty)-arctan(-\infty)]=i\pi

    Of course, after Wick rotating, you might still use the residue theorem to evaluate:

    [tex]\int^{\infty}_{-\infty} \frac{dx}{x^2+1} [/tex]

    But at least this expression doesn't have a weird infinitismal in it.

    So a Wick rotation is just a nice way of getting rid of the weird [tex]\epsilon[/tex] (and as a side-effect usually allows a simple integral over a hypersphere instead of having to use a residue theorem). It would be useless if there wasn't an [tex]\epsilon[/tex]. Only physicists can come up with expressions like:

    [tex]\int^{\infty}_{-\infty} \frac{dx}{x^2-1}[/tex]


    [tex]\int^{\infty}_{-\infty} \frac{dx}{x^2-1+i\epsilon}[/tex] ,

    so only physicists need a Wick rotation to make that look nicer. So a Wick rotation is not a general mathematical method to solve contour integrals.

    addendum: maybe all these types of integrals give results of the order of unity. In this example, the integral gave [tex]i\pi [/tex]. If you add another integral/loop, maybe you get that squared which is ~10, which is much smaller than an extra fine structure constant ~1/137 you have to add. So I guess for magnitude of order estimates, the fine structure constant is the best guide, and you don't have to worry that maybe the extra integrals can give you a result greater than 137, canceling the extra fine structure constant when you add more loops! So perturbation theory works! But of course you have to prove that all these integrals are of order unity, so that they don't cancel the effect of extra fine structure constants! Is there a proof of this? It shouldn't be too hard, since all the integrals after Wick rotation are of the form:

    \int^{\infty}_{-\infty} \frac{dx}{x^2+1}*
    \int^{\infty}_{-\infty} \frac{dy}{y^2+1} *
    \int^{\infty}_{-\infty} \frac{dz}{z^2+1} *...

    All the denominators are a lot bigger than the numerator (i.e., 1), so these integrals can't get too big, at least compared to the fine structure constants. Actually, now I remember that there is a theorem called Weinberg's theorem that says something about this, but I never read his book. But this is assuming the mass is still 1. So if the mass is very tiny, then maybe these integrals can get really big! If set close to zero you probably get your infrared divergence:

    \int^{\infty}_{-\infty} \frac{dx}{x^2+.000001}*
    \int^{\infty}_{-\infty} \frac{dy}{y^2+.000001} *
    \int^{\infty}_{-\infty} \frac{dz}{z^2+.000001} *...

    So this is a little more complicated than I initially thought. Fortunately we are saved and perturbation theory still works by saying that photons get brehmed off in a stream. Still, it's hard to see how brehming off lots of photons takes care of these very large higher-loop integrals just from looking at the equations above. But evidently when we take a cross-section, we multiply all these large integrals by some very small quantities. But the problem is that the higher order we go in loops, the larger that the product of the integrals gets, so we have to calculate the amplitude of the whole process to all orders before multiplying by a cross-sectional term. So something is wrong, since we obviously don't calculate amplitudes to all orders.
    Last edited: Aug 5, 2010
  6. Aug 5, 2010 #5

    Ben Niehoff

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    Wick rotation is equivalent to the [itex]+i\varepsilon[/itex] pole prescription. There are two poles: one on the positive real axis, and one on the negative real axis. The pole prescription moves the positive pole down, and the negative pole up.

    Wick rotation instead takes the entire contour and rotates it 90 degrees counter-clockwise. There is only one complex infinity, so this contour is equivalent to the original contour + pole prescription. The only thing that matters is which side of the contour each pole is on; since the contour is no longer on the real line, the [itex]+i\varepsilon[/itex] does not matter.
  7. Aug 5, 2010 #6
    Is this true though? Take the following integral:

    [tex]\Sigma(p^2)=\int d^6k \mbox{ } \frac{1}{k^2-m^2+i\epsilon}* \frac{1}{(k-p)^2-m^2+i\epsilon} =\int d^6k \mbox{ } \frac{1}{k_{o}^2-E_{k}^2+i\epsilon}* \frac{1}{(k_{o}-p_{o})^2-E_{kp}^2+i\epsilon} [/tex]

    where [tex]E_k=\sqrt{\vec{k}^2+m^2} [/tex] and [tex]E_{kp}=\sqrt{(\vec{k}-\vec{p})^2+m^2}[/tex]

    This can for example be the self-energy in a [tex]\phi^3 [/tex] theory (renormalizeable in 6 dimensions). The first fraction in the integrand has a pole on the negative and positive real axis (the [tex]k_0 [/tex]-axis). The second one would have a pole at the negative and positive real axis, except that it is shifted by [tex]p_o [/tex]! So if [tex]p_o [/tex] is really large, can you shift both poles to the positive real axis? If this is the case, are you still allowed to Wick rotate? Evidently you still can, because the textbooks always use Wick rotations, but it doesn't really make sense to me, since you've moved a pole to above the positive real axis!
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