- #1
AndrewGRQTF
- 27
- 2
I am having trouble following a step in a book. So we are given that $$\varphi (x) = \int \frac {d^3k}{(2\pi)^3 2\omega} [a(\textbf{k})e^{ikx} + a^*(\textbf{k})e^{-ikx}] $$
where the k in the measure is the spatial (vector) part of the four-momentum k=(##\omega##,##\textbf{k}##) and the k in the exponentials is the four-momentum. Note that ##x = (t,\textbf{x})## and that ##kx## = ##k^\mu x_\mu = -\omega t + \textbf{k}\cdot \textbf{x} ##.
After applying ##\int d^3x e^{-ikx}## to both sides, my book says that we get
$$\int d^3x e^{-ikx} \varphi (x) = \frac{1}{2\omega} a(\textbf{k}) + \frac {1}{2\omega} e^{2i\omega t}a^*(-\textbf{k})$$
$$\int d^3x e^{-ikx} \frac {\partial}{\partial t}\varphi (x) = \frac{-i}{2} a(\textbf{k}) + \frac {i}{2} e^{2i\omega t}a^*(-\textbf{k})$$
My question is: where does the ##e^{2i\omega t}## on the right side of the last two equations come from? The result I get has no ##e^{2i\omega t}##.
Also, can someone tell me how the positive/negative signs work in Fourier transforms? Like, if I instead applied ##\int d^3x e^{ikx}## to both sides I would get rid of the integration on the right, so isn't that also inversion of the transform?
where the k in the measure is the spatial (vector) part of the four-momentum k=(##\omega##,##\textbf{k}##) and the k in the exponentials is the four-momentum. Note that ##x = (t,\textbf{x})## and that ##kx## = ##k^\mu x_\mu = -\omega t + \textbf{k}\cdot \textbf{x} ##.
After applying ##\int d^3x e^{-ikx}## to both sides, my book says that we get
$$\int d^3x e^{-ikx} \varphi (x) = \frac{1}{2\omega} a(\textbf{k}) + \frac {1}{2\omega} e^{2i\omega t}a^*(-\textbf{k})$$
$$\int d^3x e^{-ikx} \frac {\partial}{\partial t}\varphi (x) = \frac{-i}{2} a(\textbf{k}) + \frac {i}{2} e^{2i\omega t}a^*(-\textbf{k})$$
My question is: where does the ##e^{2i\omega t}## on the right side of the last two equations come from? The result I get has no ##e^{2i\omega t}##.
Also, can someone tell me how the positive/negative signs work in Fourier transforms? Like, if I instead applied ##\int d^3x e^{ikx}## to both sides I would get rid of the integration on the right, so isn't that also inversion of the transform?