- #1
SuperStringboy
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Homework Statement
I am facing problem to derive the 2nd expression from the first one. My problem is the 2nd term of the 2nd expression.
Homework Equations
[tex]
\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}[\exp(-ip\cdot(x - y)) - \exp(ip\cdot (x - y))]=\int\ \frac{d^3p} {(2\pi)^3}\ \{ \frac {1}{2E_p}\ e^{-ip.(x-y)}\left|_{p^0 = E_p}\ +\ \frac {1}{-2E_p}\ e^{-ip.(x-y)}\left|_{p^0 = -E_p}\ \}
[/tex]
The Attempt at a Solution
[tex]
p\cdot (x - y)= p^0(x^0 - y^0) - \textbf p\cdot(x-y)
[/tex]
For Po = - Ep we can take
[tex]
p\cdot (x - y)= - p^0(x^0 - y^0) - \textbf p\cdot(x-y)
[/tex]
If i am not wrong yet, then what now?
should i change the dummy variable as p = - p? But if do it then i think another change comes d3p becomes -d3p for the 2nd term and i loose the minus sign before the 2nd term.
i don't know how much wrong i am but i am expecting good solution from you guys.
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