Ah okay, thank you for the tip!
Okay, I'm still a little bit confused. So you're saying that I can use the equation ##\langle{X}\rangle = \langle\psi|X|\psi\rangle## to find X at t=0, but somehow relate the time evolution expectation values to those at t=0?
I've seen the equation ##<X> = <\psi|X|\psi>##, but wouldn't I need to have ##\psi## as a function of time to find ##<X>## as a function of time? ##\psi(x,t)## instead of ##\psi(x,0)##?
Going off of that, I have to find both ##<X>## and ##\dot{<X>}##.
Ehrenfest's theorem is ##\frac{d}{dt}<X>...
I realized I miscalculated on my integral, and the actual answer is ##\frac{92}{15} - 6.1333.## So my normalized equation would now be
$$ \psi(x,0) = \sqrt{\frac{1}{6.1333}} \cdot e^{ip_ox/\hbar} \cdot
\begin{cases}
x^2 & 0 \leq x < 1,\\
-x^2 + 4x -2 & 1 \leq x < 3,\\
x^2 -8x +16 & 3 \leq x \leq...
So I just end up at 70.1 without a constant? So by multiplying this times the equation is it normalized?
Also, after this am I ready to find <X> and <P> as functions of time?
Okay, I think I'm getting it more, thank you! So when normalizing, do I follow this equation
$$\int_{a}^{b} \psi \psi^* dx$$
and set my piecewise accordingly? It appears that the complex conjugate causes $$\bigg(e^{\frac{ip_0x}{\hbar}}\bigg)^2 =...
So there's a free particle with mass m.
\begin{equation}
\psi(x,0) = e^{ip_ox/\hbar}\cdot\begin{cases}
x^2 & 0 \leq x < 1,\\
-x^2 + 4x -2 & 1 \leq x < 3,\\
x^2 -8x +16 & 3 \leq x \leq 4, \\
0 & \text{otherwise}.
\end{cases}
\end{equation}
What does each part of the piecewise represent...