I Deriving the time-dependent wave equation from Energy eigenfunctions

[Dirac_Clarity_GOAT]

TL;DR

$\sum_{E} c_E \exp(- \frac{-i}{\hbar} Et) u_{E}(x)$ where $c_E = \int_{-\infty}{+\infty} u_{E}^{*}(x) \psi_{0}$

Note: My Intro QM course is structured such that we work with Fourier Transformations etc. Before Bra-Ket Notation. Sorry if the notation is cumbersome/exoteric

Hi, in my Homework I was tasked to "derive" the time dependent wavefunction $\psi(x,t)$ from the initial wave function:
I have attempted the derivation:
Please help me see if I make sense:
$$
\psi(x,0) = \frac{1}{\sqrt{ \sqrt{ 2 \pi } \sigma_{0} }} \exp\left( \frac{i}{\hbar} p_{0} x \right) \exp\left( -\frac{1}{4 \sigma_{0}^{2}} (x- x_{0})^{2} \right)
$$

I derive the energy eigenfunctions from the Potential free Schrodinger equation to get:
$$
u_{E_{n}}(x) = A_{1} \exp(i k_{n} x) + A_{2} \exp(- i k_{n} x)
$$
Where $k_{n} = \pm \sqrt{ \frac{2mE_{n}}{\hbar^{2}} }$ is the n-th wave number of the free-particle.
And $A_1 , A_2$ are Complex numbers to be determined.

Then substituting into:
$$
\psi(x,t) = \sum_{n = 1}^{\infty} c_{E_n} \exp\left( -\frac{i}{\hbar} E_nt \right) u_{E_n}(x)
$$

Although the expression is sum over the different energy eigenvalues, for the next step I have Chosen to integrate over the wave-vector. I do not know the reason. I have done so because my energy eigenfunction are expressed in k. I also have the vague notion that in the Fourier transform of from k -> x involves taking a superposition of different standing waves. For high wave-number => small wave-length => tends to be out of phase => destructive interference => 0 amplitude => localised wave packet which is what I want.
Here is the expression:
$$
\psi(x,t) = \int_{-\infty}^{\infty} [C_1 e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} + C_2 e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}}] e^{-i\frac{\hbar k^2}{2m}t} [A_1 e^{ikx} + A_2 e^{-ikx}] dk
$$

After some tedious calculations I get the wave-function as a sum of 4-components:
1. $$C_1A_1 \int_{-\infty}^{\infty} e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{ikx} dk$$
2. $$C_1A_2 \int_{-\infty}^{\infty} e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{-ikx} dk$$
3. $$C_2A_1 \int_{-\infty}^{\infty} e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{ikx} dk$$
4. $$C_2A_2 \int_{-\infty}^{\infty} e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{-ikx} dk$$


From this helpful video
http://Phase Velocity versus Group velocity
And Ehrenfest Theorem: I deduce the Group velocity of the Gaussian Packet is:
$$
v_{g} = \frac{\hbar k_{0}}{m} = \frac{p_{0}}{m}
$$


But what was the initial phase velocity of the energy eigenfunctions?

 

[Albertus Magnus]

The spectrum of the Hamiltonian for the free particle is continuous as opposed to discrete, thus there is no need to index them by a discrete label like "$n$", rather think of them as labeled by the wave number $k$:
$$E={{\hbar}^2k^2\over 2m}.$$
The solutions are:
$$\Psi(x,t)=Ae^{ik(x-{\hbar k\over 2m}t)}+Be^{-ik(x+{\hbar k\over 2m}t)},$$
These are "travelling waves" with velocity
$$v={\hbar |k|\over 2m}.$$
Thus, without loss of generality we can write the solution as:
$$\Psi(x,t)=Ae^{i(kx-{\hbar k^2\over 2m}t)}.$$
To form the general solution the trick is to take all possible linear combinations of the above solutions. In this case we do not sum over a discrete index but over the continuous index $k$. Thus the sum is replaced by integration.
The general solution to the time dependent Schrodinger equation is:
$$\Psi(x,t)={1\over\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{i(kx-{\hbar k^2\over 2m}t)}\; dk$$
Thus, the initial value is given by:
$$\Psi(x,0)={1\over\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}\; dk,$$
therefore by Plancherel's theorem we can find $\phi(k)$ which are analogous to the Fourier coefficients used to take the linear combinations when the energy eigenvalues are discrete,
$$\phi(k)={1\over\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}\; dx.$$