# 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?

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(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 ...

(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 ...

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

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)])#

**position expectation value** is then

#<< 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)i)/(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)i)/(3sqrt3)sqrt(ℏ/(momega_0)))#

The **angular frequency** in

#<< p >> = m(d<< x >>)/(dt)# , and thus:

#color(blue)(<< p >>) = (2^(7//4)i)/(3^(3//2)sqrt(alpha_0)) m omega_0cos(omega_0t)#

#= color(blue)((2^(7//4)i)/(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.*

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