- #1
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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?
[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?
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