In our class we said for the following fourier transformation:
$$\phi(\vec r,t)=\int_{-\infty}^{\infty}d\omega\int d^3k \hat{\phi}(\vec k,\omega)e^{i(\vec k \vec r - \omega t)}$$
We said that $\phi(\vec r,t)^*=\phi(\vec r,t)$ and therefore it's fourier transform satisfies the condition $\hat{\phi}^*(\vec k,\omega)=\hat{\phi}(-\vec k,-\omega)$
So I have two questions about it:
- How is $\phi(\vec r,t)^*=\phi(\vec r,t)$ when you have $e^{i...}$ and it's complex is $e^{-i...}$. How are they equal? + We don't know the equation of it's Fourier transform $\hat{\phi}(\vec k,\omega)$. It might contain complex numbers and therefore change when you consider it's conjugate.
2.How do we come up with this: $\hat{\phi}^*(\vec k,\omega)=\hat{\phi}(-\vec k,-\omega)$?
And if more info is needed. By solving the homogeneous wave equation one gets $\omega = \pm ck$ and you can write $$\hat{\phi}(\vec k,\omega)=2\omega f(\vec k,\omega)\delta(\omega^2- c^2k^2)$$