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ramparts
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In David Tong's QFT notes (http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf p. 43, eqn. 2.89) he shows how the commutator of a scalar field [tex]\phi(x)[/tex] and [tex]\phi(y)[/tex] vanishes for spacelike-separated 4-vectors x and y, establishing that the theory is causal. For equal time, [tex]x^0=y^0[/tex], the commutator is given by:
[tex] [\phi(x),\phi(y)] = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{\vec{p}^2+m^2}} (e^{i \vec{p}\cdot(\vec{x}-\vec{y})} - e^{-i \vec{p}\cdot(\vec{x}-\vec{y})}) [/tex]
He says that this vanishes because "we can flip the sign of [tex]\vec{p}[/tex] in the last exponent as it is an integration variable." What does he mean here?
I think I see how something equivalent can be done by Lorentz transforming the (x-y) in the second exponent to -(x-y), since it's just a change of coordinates, but I'm not sure why you can do that in one exponent without doing it in the other, and why Tong says you can also flip the sign of p because it's being integrated over.
Any help would be appreciated!
[tex] [\phi(x),\phi(y)] = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{\vec{p}^2+m^2}} (e^{i \vec{p}\cdot(\vec{x}-\vec{y})} - e^{-i \vec{p}\cdot(\vec{x}-\vec{y})}) [/tex]
He says that this vanishes because "we can flip the sign of [tex]\vec{p}[/tex] in the last exponent as it is an integration variable." What does he mean here?
I think I see how something equivalent can be done by Lorentz transforming the (x-y) in the second exponent to -(x-y), since it's just a change of coordinates, but I'm not sure why you can do that in one exponent without doing it in the other, and why Tong says you can also flip the sign of p because it's being integrated over.
Any help would be appreciated!