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I was reading commutation relation of canonical momentum in KG Field from Lectures of Quantum Field Theory by Ashok Das. In page 179, He has used Integration to derive the result where he expressed integration as follow

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I was unable to understand how the exponential value changed from $k$ to $k^0$.

Can anyone please explain me?

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    $\begingroup$ Expand out the contraction of the four vectors in the exponential. The spatial part combined with the integral gives the Dirac delta. The time part gives the exponential on the rhs $\endgroup$ – innisfree May 7 '19 at 10:10
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    $\begingroup$ $k\cdot x = k^{0}x^{0} - \mathbf{k}\cdot \mathbf{x}\equiv \omega t - \mathbf{k}\cdot \mathbf{x}$ in "east-coast" Minkowski metric. $\endgroup$ – Frederic Thomas May 7 '19 at 10:11
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    $\begingroup$ ...but time like is called “west coast”, no? $\endgroup$ – Cosmas Zachos May 7 '19 at 10:24
  • $\begingroup$ @Cosmas Zachos: oh, you are right, it the other way around. $\endgroup$ – Frederic Thomas May 7 '19 at 11:37
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As already pointed out in the comments $(k-k^\prime)\cdot x$ is the four vector product with $k = \begin{pmatrix}k^0 & k^1 & k^2 & k^3\end{pmatrix}^T$. But the integral is only done in three dimensions. Therefore

$$\int \frac{\mathrm{d}^3 x}{(2\pi)^3} e^{i(k-k^\prime)\cdot x}=\int \frac{\mathrm{d}^3 x}{(2\pi)^3} e^{i\left[(k^0-k^{\prime 0})x^0-(\vec{k}-\vec{k^\prime})\cdot\vec{x}\right]}= e^{i\left[(k^0-k^{\prime 0})x^0\right]}\int\frac{\mathrm{d}^3 x}{(2\pi)^3} e^{-i\left[(\vec{k}-\vec{k^\prime})\cdot\vec{x}\right]}$$

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