- #1
NanakiXIII
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This may be more of a maths question, but because I may actually just be interpreting the expression wrong, I think I'd better post it here.
I'm reading Quantum Field Theory in a Nutshell by A. Zee and I'm stuck on a bit of maths he does. He provides an expression for the free propagator for a particles described by the Klein-Gordon equation,
[tex]
D(x-y) = \int \frac{d^4 k}{(2 \pi)^4} \frac{e^{i k (x-y)}}{k^2 - m^2 + i \epsilon}.
[/tex]
Now, if I am not mistaken, the integral over four counts of [itex]k[/itex] means integrating over [itex]k^0, k^1, k^2, k^3[/itex], each with integration limits [itex]-\infty[/itex] and [itex]\infty[/itex].
He goes on to perform the integral over [itex]k^0[/itex], and he describes this as a contour integral in the complex plain. He takes this contour to be the real axis and an infinite semicircle to get back to [itex]-\infty[/itex]. My question is, why does he add that semicircle? Once you've integrated over the real line, since the integration limits are [itex]-\infty[/itex] and [itex]\infty[/itex], aren't you done? Or have I perhaps misinterpreted what he means to integrate over?
I'm reading Quantum Field Theory in a Nutshell by A. Zee and I'm stuck on a bit of maths he does. He provides an expression for the free propagator for a particles described by the Klein-Gordon equation,
[tex]
D(x-y) = \int \frac{d^4 k}{(2 \pi)^4} \frac{e^{i k (x-y)}}{k^2 - m^2 + i \epsilon}.
[/tex]
Now, if I am not mistaken, the integral over four counts of [itex]k[/itex] means integrating over [itex]k^0, k^1, k^2, k^3[/itex], each with integration limits [itex]-\infty[/itex] and [itex]\infty[/itex].
He goes on to perform the integral over [itex]k^0[/itex], and he describes this as a contour integral in the complex plain. He takes this contour to be the real axis and an infinite semicircle to get back to [itex]-\infty[/itex]. My question is, why does he add that semicircle? Once you've integrated over the real line, since the integration limits are [itex]-\infty[/itex] and [itex]\infty[/itex], aren't you done? Or have I perhaps misinterpreted what he means to integrate over?