# Fourier transform of a real initial wavefunction

Consider the initial wavefunction given by:

$$\Psi (x,0) = \sin(k_0 x).$$

I've been taught that in order to time evolve a wavepacket one must first find the momentum space representation of the wavefunction, given by:

$$\Phi (k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \Psi (x,0) e^{-ikx} dx.$$ Then the time evolution is given by an integral in terms of this momentum space wavefunction.

When computing this, I got the following result:

$$\Phi (k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \sin(k_0 x) e^{-ikx} dx.$$ Using that $$\sin(k_0 x) = \Im (e^{ikx})$$ we obtain: $$\Phi (k) = \frac{1}{\sqrt{2\pi}} \Im (\int_{-\infty}^\infty e^{i(k_0-k)x} dx).$$ We can recognize the integral as a delta function: $$\Phi (k) = \frac{1}{\sqrt{2\pi}} \Im (\delta(k_0-k))$$ But since the delta function is real, we get that $$\Phi(k) = 0$$. So, what was wrong here? Of course $$\Phi$$ cannot be zero since the position wavefunction is nonzero, so what did I do wrong?

Your problem -- as far as I can see -- is a simple calculation error: while you are correct that $$\sin(k_0 x) = \mathcal{I}(e^{ik_0 x})$$, it does not follow that $$\mathcal{I}(e^{i(k_0-k) x}) = \sin(k_0 x)e^{ikx} \quad \text{(Wrong!)},$$ as you should quickly be able to see since the left-hand side should be real but the right hand side has an imaginary part.
You can, however, use a very similar trick to find the momentum-space wavefunction by realising that $$\sin(k_0 x) = \frac{e^{ik_0x}-e^{-ik_0 x}}{2i},$$ from which you can use the definition of the $$\delta-$$function to show that $$\Phi(k) = \delta(k_0 - k) + \delta(k_0 + k),$$ representing a superposition of plane-waves moving left and right with the same momentum $$k_0$$ .