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 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Johnn.27
    Feb 21, 2021 at 22:11
  • 1
    $\begingroup$ Yep, that's right. $\endgroup$
    – Philip
    Feb 21, 2021 at 22:12

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