# 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?

## 1 Answer

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 $$k_0$$ .

Side note: While this is not essential to your question, when you want to find the time evolution of a wavefunction what you really want to do is express it as a linear combination of the energy eigenstates, not the momentum eigenstates. In this case, since I assume you are speaking of a free particle, you can find a basis of momentum eigenstates that are also energy eigenstates, but in more general problems this will certainly not be possible. I'm guessing this approach is due to Griffiths, and I find it confuses a lot of students initially.

• Thanks! That was an akward error, I'm just too used to taking Im and Re out of integrals if the integrand is a product of two functions. Didn't take into account the fact that the exponential was complex! On the last part of your question, I assume that you mean finding the momentum representation this way is applied only for free particles, but for the general case energy eigenstates are the way to go, right? Feb 21, 2021 at 22:11
• Yep, that's right. Feb 21, 2021 at 22:12