# Psi(x,t) = sqrt(1/L)sin((pix)/L) e^-(iomega_1t ) + sqrt(1/L)sin((2pix)/L) e^-(iomega_2t )new question?

## a) Calculate the probability distribution P(x,t) = |ψ(x,t)|^2 b) What is the period T of this superposition – i.e., after what time T will the system return to its original configuration? c) Find the probability distribution P(x,t) evaluated at time t_∗ = π /[2(E1−E0)]. What fraction of T is t_∗? d) What’s the probability that you find the particle in the left half of the well at some arbitrary time t? e) Find the expectation value $< x >$ of the particle’s position as a function of time f) Show that the probability density P(x,t) at x = L/2 is independent of time.

Mar 4, 2018

a)

You just need to take ${\Psi}^{\text{*}} \Psi$.

color(blue)(Psi^"*"Psi) = [sqrt(1/L)sin((pix)/L) e^-(iomega_1t ) + sqrt(1/L)sin((2pix)/L) e^-(iomega_2t)]^"*" [sqrt(1/L)sin((pix)/L) e^-(iomega_1t ) + sqrt(1/L)sin((2pix)/L) e^-(iomega_2t )]

$= \left[\sqrt{\frac{1}{L}} \sin \left(\frac{\pi x}{L}\right) {e}^{i {\omega}_{1} t} + \sqrt{\frac{1}{L}} \sin \left(\frac{2 \pi x}{L}\right) {e}^{i {\omega}_{2} t}\right] \left[\sqrt{\frac{1}{L}} \sin \left(\frac{\pi x}{L}\right) {e}^{-} \left(i {\omega}_{1} t\right) + \sqrt{\frac{1}{L}} \sin \left(\frac{2 \pi x}{L}\right) {e}^{-} \left(i {\omega}_{2} t\right)\right]$

$= \frac{1}{L} {\sin}^{2} \left(\frac{\pi x}{L}\right) + \frac{1}{L} \left(\frac{\pi x}{L}\right) \sin \left(\frac{2 \pi x}{L}\right) {e}^{i \left({\omega}_{1} - {\omega}_{2}\right) t} + \frac{1}{L} \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{2 \pi x}{L}\right) {e}^{i \left({\omega}_{2} - {\omega}_{1}\right) t} + \frac{1}{L} {\sin}^{2} \left(\frac{2 \pi x}{L}\right)$

$= \textcolor{b l u e}{\frac{1}{L} \left[{\sin}^{2} \left(\frac{\pi x}{L}\right) + {\sin}^{2} \left(\frac{2 \pi x}{L}\right)\right] + \frac{1}{L} \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{2 \pi x}{L}\right) \left[{e}^{i \left({\omega}_{1} - {\omega}_{2}\right) t} + {e}^{i \left({\omega}_{2} - {\omega}_{1}\right) t}\right]}$

b)

The period can be found with minimal effort, simply by first knowing the energies, which are constants of the motion.

The energy of ${\phi}_{1} = \sqrt{\frac{1}{L}} \sin \left(\frac{\pi x}{L}\right)$ is E_1 = (1^2pi^2ℏ^2)/(4mL^2), and the energy of ${\phi}_{2}$ is $4 {E}_{1}$. Therefore, the frequency ${\omega}_{2}$ of ${\phi}_{2}$ is four times that of ${\phi}_{1}$ (${\omega}_{1}$).

As a result, the period ${T}_{1} = \frac{2 \pi}{{\omega}_{1}}$ of ${\phi}_{1}$ is four times that of ${\phi}_{2}$ (${T}_{2} = \frac{2 \pi}{{\omega}_{2}}$, and is also a period of ${\phi}_{2}$.

The period is thus $\textcolor{b l u e}{T = \frac{2 \pi}{{\omega}_{1}}}$.

c)

I'll let you plug this one in yourself as t_"*" = pi /[2(E_2−E_1)]. You don't need to do anything with it...

We know that $T = \frac{2 \pi}{{\omega}_{1}}$, and that (iEt)/ℏ = iomegat, so

E_n = omega_nℏ.

As a result,

pi/(2(E_2-E_1)) = pi/(2(omega_2-omega_1)ℏ)

and

color(blue)(t_"*"/T) = pi/(2(omega_2-omega_1)ℏ) cdot (omega_1)/(2pi)

= 1/(2(4omega_1-omega_1)ℏ) cdot (omega_1)/(2)

= omega_1/(4ℏ(3omega_1))

= color(blue)(1/(12ℏ))

d)

The probability of finding the particle in $\left[0 , \frac{L}{2}\right]$ is given as

${\int}_{0}^{\frac{L}{2}} {\Psi}^{\text{*}} \Psi \mathrm{dx}$

$= \frac{1}{L} {\int}_{0}^{\frac{L}{2}} {\sin}^{2} \left(\frac{\pi x}{L}\right) + {\sin}^{2} \left(\frac{2 \pi x}{L}\right) \mathrm{dx} + \frac{1}{L} {\int}_{0}^{\frac{L}{2}} \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{2 \pi x}{L}\right) \left[{e}^{- 3 i {\omega}_{1} t} + {e}^{3 i {\omega}_{1} t}\right] \mathrm{dx}$

$= \frac{1}{L} {\int}_{0}^{\frac{L}{2}} {\sin}^{2} \left(\frac{\pi x}{L}\right) + {\sin}^{2} \left(\frac{2 \pi x}{L}\right) \mathrm{dx} + \frac{1}{L} {\int}_{0}^{\frac{L}{2}} 2 \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{2 \pi x}{L}\right) \cos \left(3 {\omega}_{1} t\right) \mathrm{dx}$

The first two terms are symmetric with half the amplitude, and yield 50% overall.

The third term would have a stationary state probability of $\frac{4}{3 \pi}$, and $\cos$ is an arbitrary phase factor. Thus, the overall probability is

$= \textcolor{b l u e}{0.50 + \frac{4}{3 \pi} \cos \left(3 {\omega}_{1} t\right)}$

e)

$\textcolor{b l u e}{\left\langlex\right\rangle} = \left\langle\Psi | x | \Psi\right\rangle = \left\langlex \Psi | \Psi\right\rangle$

$= \frac{1}{L} {\int}_{0}^{\frac{L}{2}} x {\sin}^{2} \left(\frac{\pi x}{L}\right) \mathrm{dx} + \frac{1}{L} {\int}_{0}^{\frac{L}{2}} x {\sin}^{2} \left(\frac{2 \pi x}{L}\right) \mathrm{dx} + \frac{1}{L} {\int}_{0}^{\frac{L}{2}} 2 x \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{2 \pi x}{L}\right) \cos \left(3 {\omega}_{1} t\right) \mathrm{dx}$

There is no trivial solution to this... This turns out to be:

$= \frac{L}{4 {\pi}^{2}} + \frac{L}{8} + \left[\frac{2 L}{3 \pi} - \frac{8 L}{9 {\pi}^{2}}\right] \cos \left(3 {\omega}_{1} t\right)$

$= \textcolor{b l u e}{\frac{\left(2 + {\pi}^{2}\right) L}{8 {\pi}^{2}} + \left[\frac{\left(6 \pi - 8\right) L}{9 {\pi}^{2}}\right] \cos \left(3 {\omega}_{1} t\right)}$

f)

At $x = \frac{L}{2}$, the $\sin$ terms go to $\sin \left(\frac{\pi}{2}\right) = 1$ and to $\sin \left(\pi\right) = 0$, respectively.

Since $\sin \left(\pi\right) = 0$, the time-dependent part of ${\Psi}^{\text{*}} \Psi$ vanishes and the time-independent part retains $\frac{1}{L}$ as the probability density.