I am trying to generate a complex 2D field in Fourier space that will give a real-only output when computing the 2D inverse Fourier transform. Is there a way to specifically set some constraint on the input field (e.g. phase, some sort of symmetry) such that the output will have no imaginary component?
Initially, I was thinking for an inverse Fourier transform of some field $g(k)=A(k)e^{\theta(k)}$ with magnitude $A(k)$ and complex phase $\theta(k)$
$$F^{-1} \{g(k)\} = \int e^{2 \pi ix\cdot k} A(k)e^{\theta(k)} dk, $$
the imaginary part will only come from the complex exponential, which has zero imaginary component when $\sin(2\pi x \cdot k + \theta(k))=0$, i.e. when $2\pi x \cdot k + \theta(k) = n\pi$ for integer $n$. I feel like I should be able to constrain the phase to do this, but I can't wrap my head around how that $x$ plays in since I can't directly impose any value to the phase that relies on $x$ given that I'm making the field in $k$-space.
Best Answer
Yes, you can do this!
Note that $F^{-1}g$ will be real if and only if it equals its conjugate $\overline{F^{-1}g}$, which is given by $$\overline{F^{-1}g(x)} = \overline{\int_\mathbb{R} e^{2\pi i x k}g(k)dk}=\int_\mathbb R e^{-2\pi i x k}\overline{g(k)}dk = \int_\mathbb{R} e^{2\pi i x k}\overline{g(-k)}dk$$ by making the substitution $k\mapsto -k$ in the last step. Therefore, it is sufficient for $g(k)=\overline{g(-k)}$.
Can you find out whether this condition is necessary as well as sufficient?