Question

`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.

Answer 1

`a)`

You just need to take `Psi^"*"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 )]`

`= [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)]`

`= 1/Lsin^2((pix)/L) + 1/L ((pix)/L)sin((2pix)/L)e^(i(omega_1-omega_2)t) + 1/L sin((pix)/L)sin((2pix)/L)e^(i(omega_2-omega_1)t) + 1/L sin^2((2pix)/L)`

`= color(blue)(1/L[sin^2((pix)/L) + sin^2((2pix)/L)] + 1/L sin((pix)/L)sin((2pix)/L)[e^(i(omega_1-omega_2)t) + e^(i(omega_2-omega_1)t)])`

`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(1/L)sin((pix)/L)` is `E_1 = (1^2pi^2ℏ^2)/(4mL^2)`, and the energy of `phi_2` is `4E_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 = (2pi)/(omega_1)` of `phi_1` is four times that of `phi_2` (`T_2 = (2pi)/(omega_2)`, and is also a period of `phi_2`.

The period is thus `color(blue)(T = (2pi)/(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 = (2pi)/(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 `[0, L/2]` is given as

`int_(0)^(L/2) Psi^"*"Psidx`

`= 1/Lint_(0)^(L/2) sin^2((pix)/L) + sin^2((2pix)/L)dx + 1/Lint_(0)^(L/2) sin((pix)/L)sin((2pix)/L)[e^(-3iomega_1t) + e^(3iomega_1t)]dx`

`= 1/Lint_(0)^(L/2) sin^2((pix)/L) + sin^2((2pix)/L)dx + 1/Lint_(0)^(L/2) 2sin((pix)/L)sin((2pix)/L)cos(3omega_1t)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 `4/(3pi)`, and `cos` is an arbitrary phase factor. Thus, the overall probability is

`= color(blue)(0.50 + 4/(3pi) cos(3omega_1t))`

`e)`

`color(blue)(<< x >>) = << Psi | x | Psi >> = << xPsi|Psi >>`

`= 1/Lint_(0)^(L/2) xsin^2((pix)/L)dx + 1/Lint_(0)^(L/2) xsin^2((2pix)/L)dx + 1/Lint_(0)^(L/2) 2xsin((pix)/L)sin((2pix)/L)cos(3omega_1t)dx`

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

`= L/(4pi^2) + L/8 + [(2L)/(3pi) - (8L)/(9pi^2)]cos(3omega_1t)`

`= color(blue)(((2 + pi^2)L)/(8pi^2) + [((6pi - 8)L)/(9pi^2)]cos(3omega_1t))`

`f)`

At `x = L/2`, the `sin` terms go to `sin(pi/2) = 1` and to `sin(pi) = 0`, respectively.

Since `sin(pi) = 0`, the time-dependent part of `Psi^"*"Psi` vanishes and the time-independent part retains `1/L` as the probability density.