We start with a function $${\Delta(x) = \displaystyle \int \dfrac{d^3k}{(2\pi)^3 2k^0}}\left( e^{ik^\mu x_\mu} - e^{-ik^\mu x_\mu} \right).$$ It is obvious to me that for $t = 0$ the above expression vanishes.

However, I want to show that $$\dot{\Delta}(\mathbf{x},0) = -i\delta^{3}(\mathbf{x}).$$

One simple derivative and then setting $t = 0$ gives me ${\dot{\Delta}(\mathbf{x},0) = \displaystyle -i\int \dfrac{d^3k}{(2\pi)^3}}\left( e^{i\mathbf{k \cdot x}} - e^{-i\mathbf{k \cdot x}} \right)$. This is antisymmetric with respect to $\mathbf{x}$, how can it equal the Dirac delta?

Metric convention is $\mathrm{diag}(-1,1,1,1)$.

  • 1
    $\begingroup$ Is not antisymmetric w.r.t $x$. You can see it by replacing $k$ with $-k$. The fact that the expression to be integrate is antisymmetric w.r.t. $k$ for $x\neq 0$ is what shows that it vanishes far all such $x$'s. For show that it is indeed the $\delta$-function you are looking for, try to use Fourier. $\endgroup$ – Snaporaz Oct 14 '15 at 12:24
  • $\begingroup$ I know that $\int d^3 k/(2\pi)^3 e^{i\mathbf{kx}} = \delta (\mathbf{x})$. However here I am missing a factor of 2, when I try to derive the above result. $\endgroup$ – EEEB Oct 14 '15 at 13:09
  • 3
    $\begingroup$ @user208272 That missing factor of 2, or correctly $\frac{1}{2}$ you dropped while taking the derivative. $\endgroup$ – Clever Oct 14 '15 at 13:14
  • $\begingroup$ Didn't that cancel the $2k^0$ in the denominator? $\endgroup$ – EEEB Oct 14 '15 at 13:15
  • $\begingroup$ I got your point, my bad. You're absolutely right. $\endgroup$ – EEEB Oct 14 '15 at 13:28

$${\dot{\Delta}(\mathbf{x},0) = - \int \frac{d^3k}{(2\pi)^3 2k^0}}ik^0 \left( e^{i\mathbf{k \cdot x}} + e^{-i\mathbf{k \cdot x}} \right)=-i \int \frac{d^3k}{(2\pi)^3 } e^{i\mathbf{k \cdot x}} =-i\delta^{3}(\mathbf{x})~. $$ This, naturally, is the cornerstone of the canonical quantization commutation relation.


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