Normalizable free particle wave on 1D

Question

I am trying to solve a free 1D particle non-relativistic Schrodinger equation with initial wavefunction $\psi(x,0)=\delta(x)$, where $$\delta(x)=\lim_{a\to0}(a/2)|x|^{(a-1)}.$$

Here is my approach:

Set $$i\hbar \frac{\partial}{\partial t} \Psi (x,t) =E\Psi(x,t)= -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi (x,t).$$

The general solutions look like:

$$\Psi=Ae^ae^b$$

Because the sign of real number $E$ could be arbitrarily defined, I choose $E<0$ to make the solution normalizable.

$$\Psi(x,t) = A e^{-kx+i\omega t} $$

$$\Psi^*\Psi\equiv\delta(x)$$

Am I right?

How do I get the sensible solution depicting the evolution of density function of a free particle on $\mathbb R$? (like the gif video in https://en.wikipedia.org/wiki/Uncertainty_principle)

A second approach is to consider $E>0$ and set boundary conditions like a particle in a "very large" 1-D box. Could you please help me with it?

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From the top answer by Jan, I learnt three ways to deal with un-normalizable wave functions.

Normalizing the solution to free particle Schrödinger equation

One of them was "use only normalizable functions to calculate probability"

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The solution process was learnt from http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/Scheq.html#c2

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Sorry if the question is already asked.

Answer 1

... seems to me there are several problems here.

1. $\psi(x)=\delta(x)$ is not an eigenfunction of the Hamiltonian, so searching for a solution to the time-independent Schrodinger equation is not useful. The solution with your initial condition will be a superpositoin of plane waves since plane waves are eigenfunctions of the free-particle problem.
2. If $\Psi(x,t)=Ae^{-i(\omega t-kx)}$, the $\Psi(x,t)^*\Psi(x,t)=\vert A\vert^2$, not $\delta(x)$. Moreover, your $\Psi(x,t)$ is a plane wave and so certainly not concentrated at a point for any value of $x$.
3. Next, if you are solving for a free particle, then $E$ should be non-negative since the kinetic energy is non-negative.

Since the free-particle solutions are of the form $e^{ikx}$ at $t=0$, why not try $$ \psi(x)=\int_{-\infty}^{\infty}dk\phi(k) e^{ikx} $$ i.e. find $\psi(x)$ as a wave packet and look for a function $\phi(k)$ such that $\psi(x)=\delta(x)$? You might want to recall that \begin{align} \delta(x-x_0)=\langle x\vert x_0\rangle &= \int_{-\infty}^{\infty} dk \langle x\vert k\rangle\langle k\vert x_0\rangle\, ,\\ &=\int_{-\infty}^\infty dk \frac{1}{2\pi}e^{ik(x-x_0)}\, . \end{align}