A particle of mass m in a harmonic oscillator potential #V(x)=(momega_0^2)/(2)x^2# has an initial #psi(x,0)=1/(sqrt2)[phi_0(x) + iphi_1(x)] # where #phi_n# are the normalized eigenstates for harmonic oscillator?
(a) Write down ψ(x,t) and |ψ(x,t)|^2
2. (For this part, you may leave the expression in
terms of φ0 and φ1.)
(b) Find the expectation value #<x># as a function of time t. Notice that it oscillates with
time. What is the amplitude of the oscillation (in terms of m, ω0 and fundamental
constants)? What is its angular frequency?
(c) Find the expectation value (p) as a function of time.
(d) Show that the probability distribution of a particle in a harmonic oscillator potential
returns to its original shape after the classical period T = 2π/ωo. You should
prove this for any harmonic oscillator state, including non-stationary states. What
feature of the harmonic oscillator makes this true?
(a) Write down ψ(x,t) and |ψ(x,t)|^2
2. (For this part, you may leave the expression in
terms of φ0 and φ1.)
(b) Find the expectation value
time. What is the amplitude of the oscillation (in terms of m, ω0 and fundamental
constants)? What is its angular frequency?
(c) Find the expectation value (p) as a function of time.
(d) Show that the probability distribution of a particle in a harmonic oscillator potential
returns to its original shape after the classical period T = 2π/ωo. You should
prove this for any harmonic oscillator state, including non-stationary states. What
feature of the harmonic oscillator makes this true?
1 Answer
Well, this depends on if
DISCLAIMER: LONG ANSWER! (obviously)
First off, I think you mean
#Psi(x,t) = 1/sqrt2[phi_0e^(-iomega_0t) + iphi_1e^(-iomega_1t)]#
Here we assume
#Psi^"*"(x,t)Psi(x,t) = [1/sqrt2[phi_0e^(-iomega_0t) + iphi_1e^(-iomega_1t)]]^"*"[1/sqrt2[phi_0e^(-iomega_0t) + iphi_1e^(-iomega_1t)]]#
#= 1/2[phi_0^"*"e^(iomega_0t) - iphi_1^"*"e^(iomega_1t)][phi_0e^(-iomega_0t) + iphi_1e^(-iomega_1t)]#
But since the
#Psi(x,t)^"*"Psi(x,t) = 1/2[phi_0e^(iomega_0t) - iphi_1e^(iomega_1t)][phi_0e^(-iomega_0t) + iphi_1e^(-iomega_1t)]#
#= 1/2[phi_0^2 + iphi_0phi_1e^(-i(omega_1-omega_0)t) - iphi_0phi_1e^(i(omega_1-omega_0)t) + phi_1^2]#
#= 1/2[phi_0^2 + phi_1^2 + 2phi_0phi_1sin[(omega_1-omega_0)t]]#
For the simple harmonic oscillator, to a first approximation,
Therefore:
#color(blue)(Psi^"*"(x,t)Psi(x,t) ~~ 1/2[phi_0^2 + phi_1^2 + 2phi_0phi_1sin(omega_0t)])#
#<< x >> = << Psi | x | Psi >> = << x Psi | Psi >>#
#= int_(-oo)^(oo) xPsi^"*"Psi dx#
#= int_(-oo)^(oo) x/2[phi_0^2 + phi_1^2 + 2phi_0phi_1sin[(omega_1-omega_0)t]]dx#
#= 1/2 int_(-oo)^(oo) x[phi_0^2 + phi_1^2]dx + sin[(omega_1-omega_0)t]int_(-oo)^(oo) xphi_0phi_1dx#
We know that
#phi_0 = (alpha_0/pi)^(1//4) e^(-alpha_0x^2//2)#
#phi_1 = (alpha_1/pi)^(1//4)sqrt(2alpha_1)xe^(-alpha_1x^2//2)# where
#alpha_n = momega_n//ℏ# . Here we again assume their angular frequencies are NOT the same.
Taking
#x[phi_0^2+phi_1^2] = 1/sqrtpi[sqrt(alpha_0)xe^(-alpha_0x^2) + sqrt(alpha_1)xe^(-alpha_1x^2)]#
Multiplying
#xphi_0phi_1 = ((alpha_0alpha_1)/pi^2)^(1//4)sqrt(2alpha_1)x^2 e^(-(alpha_0+alpha_1)x^2//2)#
Therefore:
#<< x >> = cancel(1/(2sqrtpi) int_(-oo)^(oo) sqrt(alpha_0)xe^(-alpha_0x^2) + sqrt(alpha_1)xe^(-alpha_1x^2) dx)^(0) + sin[(omega_1-omega_0)t] ((alpha_0alpha_1)/pi^2)^(1//4)sqrt(2alpha_1)int_(-oo)^(oo) x^2 e^(-(alpha_0+alpha_1)x^2//2)dx# where the first integral vanishes by symmetry. Odd times even equals odd function, integrated over an even interval gives zero.
The second integral is tabulated as
#int_(-oo)^(oo) x^2e^(-alphax^2)dx = sqrt(pi)/(2alpha^(3//2))#
Therefore, let
#<< x >> = sin[(omega_1-omega_0)t] ((alpha_0alpha_1)/pi^2)^(1//4)sqrt(2alpha_1) cdot (sqrtpi)/(2((alpha_0+alpha_1)/2)^(3//2))#
#= sin[(omega_1-omega_0)t] (alpha_0alpha_1)^(1//4)cdot 2sqrt(alpha_1) cdot 1/((alpha_0+alpha_1)^(3//2))#
#= (2alpha_0^(1//4)alpha_1^(3//4))/((alpha_0+alpha_1)^(3//2))sin[(omega_1-omega_0)t]#
This is indeed time-dependent... Again, we approximate
#color(blue)(<< x >>) ~~ (2alpha_0^(1//4)(2alpha_0)^(3//4))/((3alpha_0)^(3//2))sin(omega_0t)#
#= (2^(7//4))/(3^(3//2)sqrt(alpha_0))sin(omega_0t)#
#= color(blue)((2^(7//4))/(3^(3//2))sqrt(ℏ/(momega_0))sin(omega_0t))# (If we just had
#Psi(x,0) = phi_n(x)# , then#Psi^"*"Psi = psi^"*"psi# would have no time dependence and#<< x >> = 0# .)
The amplitude of the oscillation is simply the classical turning point, i.e. the maximum extension possible. That requires zero velocity, so
#0 = << v >>_("turning point")#
#= (d<< x >>)/(dt) = (2^(7//4))/(3^(3//2)sqrt(alpha_0))d/(dt)[sin(omega_0t)]#
#= (2^(7//4))/(3^(3//2)sqrt(alpha_0)) omega_0cos(omega_0t)#
Solving this for
#omega_0t = pi/2, (3pi)/2, . . . #
and that if we get a negative amplitude, it's just in the opposite direction to positive
#color(blue)(A = (2^(7//4))/(3sqrt3)sqrt(ℏ/(momega_0)))#
The angular frequency in
#<< p >> = m(d<< x >>)/(dt)# , and thus:
#color(blue)(<< p >>) = (2^(7//4))/(3^(3//2)sqrt(alpha_0)) m omega_0cos(omega_0t)#
#= color(blue)((2^(7//4))/(3^(3//2)) sqrt(momega_0ℏ) cos(omega_0t))# (If we just had
#Psi(x,0) = phi_n(x)# , then#Psi^"*"Psi = psi^"*"psi# would have no time dependence and#<< p >> = 0# .)
#Psi^"*"(x,0)Psi(x,0) = Psi^"*"(x,(2pi)/omega_0)Psi(x,(2pi)/(omega_0))#
Here we have:
#1/2[phi_0^2 + phi_1^2 + 2phi_0phi_1sin[(omega_1-omega_0)(2pi)/(omega_0)]] = 1/2[phi_0^2 + phi_1^2 + 2phi_0phi_1sin(0)]#
From this,
#sin[2picdot(omega_1-omega_0)/(omega_0)] = 0#
Under the approximation that
#sin(2pi) = 0#
And this indeed works out to be true in the harmonic approximation.
Physically we expect an oscillator to stretch and compress over time, and in one period it passes through its equilibrium position. Hence if this is a physical oscillator, its probability distribution stretches and compresses symmetrically and returns to its shape after each classical period.
