# #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.

a) Calculate the probability distribution

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

What fraction of T is

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

f) Show that the probability density P(x,t) at x = L/2 is independent of time.

##### 1 Answer

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)])#

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))# .

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))#

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))#

#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))#

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.