- #1

SuperStringboy

- 74

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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 d

^{3}p becomes -d

^{3}p 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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