# 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?

Mar 7, 2018

Well, this depends on if $\alpha$ is the same for both ${\phi}_{n}$ or not... i.e. are ${\phi}_{0}$ and ${\phi}_{1}$ in phase or not?

First off, I think you mean $V \left(x\right) = \frac{m {\omega}_{0}^{2} {x}^{2}}{2}$, and $\psi \left(x , 0\right) = \frac{1}{\sqrt{2}} \left[{\phi}_{0} \left(x\right) + i {\phi}_{1} \left(x\right)\right]$.

a) Just add on the phase factor.

$\Psi \left(x , t\right) = \frac{1}{\sqrt{2}} \left[{\phi}_{0} {e}^{- i {\omega}_{0} t} + i {\phi}_{1} {e}^{- i {\omega}_{1} t}\right]$

Here we assume ${\omega}_{0} \ne {\omega}_{1}$. The probability density is then:

${\Psi}^{\text{*"(x,t)Psi(x,t) = [1/sqrt2[phi_0e^(-iomega_0t) + iphi_1e^(-iomega_1t)]]^"*}} \left[\frac{1}{\sqrt{2}} \left[{\phi}_{0} {e}^{- i {\omega}_{0} t} + i {\phi}_{1} {e}^{- i {\omega}_{1} t}\right]\right]$

$= \frac{1}{2} \left[{\phi}_{0}^{\text{*"e^(iomega_0t) - iphi_1^"*}} {e}^{i {\omega}_{1} t}\right] \left[{\phi}_{0} {e}^{- i {\omega}_{0} t} + i {\phi}_{1} {e}^{- i {\omega}_{1} t}\right]$

But since the ${\phi}_{n}$ are real, ${\phi}_{n}^{\text{*}} = {\phi}_{n}$. Thus:

$\Psi {\left(x , t\right)}^{\text{*}} \Psi \left(x , t\right) = \frac{1}{2} \left[{\phi}_{0} {e}^{i {\omega}_{0} t} - i {\phi}_{1} {e}^{i {\omega}_{1} t}\right] \left[{\phi}_{0} {e}^{- i {\omega}_{0} t} + i {\phi}_{1} {e}^{- i {\omega}_{1} t}\right]$

$= \frac{1}{2} \left[{\phi}_{0}^{2} + i {\phi}_{0} {\phi}_{1} {e}^{- i \left({\omega}_{1} - {\omega}_{0}\right) t} - i {\phi}_{0} {\phi}_{1} {e}^{i \left({\omega}_{1} - {\omega}_{0}\right) t} + {\phi}_{1}^{2}\right]$

$= \frac{1}{2} \left[{\phi}_{0}^{2} + {\phi}_{1}^{2} + 2 {\phi}_{0} {\phi}_{1} \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right]\right]$

For the simple harmonic oscillator, to a first approximation, ${\omega}_{1} \approx 2 {\omega}_{0}$. (For $\text{HBr}$ as an anharmonic oscillator, it is actually ${\omega}_{1} / {\omega}_{0} \approx 1.965$.)

Therefore:

$\textcolor{b l u e}{{\Psi}^{\text{*}} \left(x , t\right) \Psi \left(x , t\right) \approx \frac{1}{2} \left[{\phi}_{0}^{2} + {\phi}_{1}^{2} + 2 {\phi}_{0} {\phi}_{1} \sin \left({\omega}_{0} t\right)\right]}$

b) The position expectation value is then

$\left\langlex\right\rangle = \left\langle\Psi | x | \Psi\right\rangle = \left\langlex \Psi | \Psi\right\rangle$

$= {\int}_{- \infty}^{\infty} x {\Psi}^{\text{*}} \Psi \mathrm{dx}$

$= {\int}_{- \infty}^{\infty} \frac{x}{2} \left[{\phi}_{0}^{2} + {\phi}_{1}^{2} + 2 {\phi}_{0} {\phi}_{1} \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right]\right] \mathrm{dx}$

$= \frac{1}{2} {\int}_{- \infty}^{\infty} x \left[{\phi}_{0}^{2} + {\phi}_{1}^{2}\right] \mathrm{dx} + \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right] {\int}_{- \infty}^{\infty} x {\phi}_{0} {\phi}_{1} \mathrm{dx}$

We know that

${\phi}_{0} = {\left({\alpha}_{0} / \pi\right)}^{1 / 4} {e}^{- {\alpha}_{0} {x}^{2} / 2}$

${\phi}_{1} = {\left({\alpha}_{1} / \pi\right)}^{1 / 4} \sqrt{2 {\alpha}_{1}} x {e}^{- {\alpha}_{1} {x}^{2} / 2}$

where alpha_n = momega_n//ℏ. Here we again assume their angular frequencies are NOT the same.

Taking $x \left[{\phi}_{0}^{2} + {\phi}_{1}^{2}\right]$ gives:

$x \left[{\phi}_{0}^{2} + {\phi}_{1}^{2}\right] = \frac{1}{\sqrt{\pi}} \left[\sqrt{{\alpha}_{0}} x {e}^{- {\alpha}_{0} {x}^{2}} + \sqrt{{\alpha}_{1}} x {e}^{- {\alpha}_{1} {x}^{2}}\right]$

Multiplying $x {\phi}_{0} {\phi}_{1}$ gives:

$x {\phi}_{0} {\phi}_{1} = {\left(\frac{{\alpha}_{0} {\alpha}_{1}}{\pi} ^ 2\right)}^{1 / 4} \sqrt{2 {\alpha}_{1}} {x}^{2} {e}^{- \left({\alpha}_{0} + {\alpha}_{1}\right) {x}^{2} / 2}$

Therefore:

$\left\langlex\right\rangle = {\cancel{\frac{1}{2 \sqrt{\pi}} {\int}_{- \infty}^{\infty} \sqrt{{\alpha}_{0}} x {e}^{- {\alpha}_{0} {x}^{2}} + \sqrt{{\alpha}_{1}} x {e}^{- {\alpha}_{1} {x}^{2}} \mathrm{dx}}}^{0} + \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right] {\left(\frac{{\alpha}_{0} {\alpha}_{1}}{\pi} ^ 2\right)}^{1 / 4} \sqrt{2 {\alpha}_{1}} {\int}_{- \infty}^{\infty} {x}^{2} {e}^{- \left({\alpha}_{0} + {\alpha}_{1}\right) {x}^{2} / 2} \mathrm{dx}$

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

${\int}_{- \infty}^{\infty} {x}^{2} {e}^{- \alpha {x}^{2}} \mathrm{dx} = \frac{\sqrt{\pi}}{2 {\alpha}^{3 / 2}}$

Therefore, let $\alpha = \frac{{\alpha}_{0} + {\alpha}_{1}}{2}$ to get:

$\left\langlex\right\rangle = \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right] {\left(\frac{{\alpha}_{0} {\alpha}_{1}}{\pi} ^ 2\right)}^{1 / 4} \sqrt{2 {\alpha}_{1}} \cdot \frac{\sqrt{\pi}}{2 {\left(\frac{{\alpha}_{0} + {\alpha}_{1}}{2}\right)}^{3 / 2}}$

$= \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right] {\left({\alpha}_{0} {\alpha}_{1}\right)}^{1 / 4} \cdot 2 \sqrt{{\alpha}_{1}} \cdot \frac{1}{{\left({\alpha}_{0} + {\alpha}_{1}\right)}^{3 / 2}}$

$= \frac{2 {\alpha}_{0}^{1 / 4} {\alpha}_{1}^{3 / 4}}{{\left({\alpha}_{0} + {\alpha}_{1}\right)}^{3 / 2}} \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) t\right]$

This is indeed time-dependent... Again, we approximate ${\omega}_{1} \approx 2 {\omega}_{0}$, so:

$\textcolor{b l u e}{\left\langlex\right\rangle} \approx \frac{2 {\alpha}_{0}^{1 / 4} {\left(2 {\alpha}_{0}\right)}^{3 / 4}}{{\left(3 {\alpha}_{0}\right)}^{3 / 2}} \sin \left({\omega}_{0} t\right)$

$= \frac{{2}^{7 / 4}}{{3}^{3 / 2} \sqrt{{\alpha}_{0}}} \sin \left({\omega}_{0} t\right)$

= color(blue)((2^(7//4))/(3^(3//2))sqrt(ℏ/(momega_0))sin(omega_0t))

(If we just had $\Psi \left(x , 0\right) = {\phi}_{n} \left(x\right)$, then ${\Psi}^{\text{*"Psi = psi^"*}} \psi$ would have no time dependence and $\left\langlex\right\rangle = 0$.)

The amplitude of the oscillation is simply the classical turning point, i.e. the maximum extension possible. That requires zero velocity, so

$0 = {\left\langlev\right\rangle}_{\text{turning point}}$

$= \frac{d \left\langlex\right\rangle}{\mathrm{dt}} = \frac{{2}^{7 / 4}}{{3}^{3 / 2} \sqrt{{\alpha}_{0}}} \frac{d}{\mathrm{dt}} \left[\sin \left({\omega}_{0} t\right)\right]$

$= \frac{{2}^{7 / 4}}{{3}^{3 / 2} \sqrt{{\alpha}_{0}}} {\omega}_{0} \cos \left({\omega}_{0} t\right)$

Solving this for $t$, you can convince yourself that

${\omega}_{0} t = \frac{\pi}{2} , \frac{3 \pi}{2} , . . .$

and that if we get a negative amplitude, it's just in the opposite direction to positive $x$. $\sin$ maxes out at $1$, so:

color(blue)(A = (2^(7//4))/(3sqrt3)sqrt(ℏ/(momega_0)))

The angular frequency in $\text{rad/s}$ in terms of ${\omega}_{0}$ is then given as $\textcolor{b l u e}{\omega = \frac{{\omega}_{0}}{2 \pi}}$ since we've rewritten this all in terms of ${\omega}_{0}$...

c) Here's the easy way to do it... $p = m v$ for a one-body particle, so

$\left\langlep\right\rangle = m \frac{d \left\langlex\right\rangle}{\mathrm{dt}}$, and thus:

$\textcolor{b l u e}{\left\langlep\right\rangle} = \frac{{2}^{7 / 4}}{{3}^{3 / 2} \sqrt{{\alpha}_{0}}} m {\omega}_{0} \cos \left({\omega}_{0} t\right)$

= color(blue)((2^(7//4))/(3^(3//2)) sqrt(momega_0ℏ) cos(omega_0t))

(If we just had $\Psi \left(x , 0\right) = {\phi}_{n} \left(x\right)$, then ${\Psi}^{\text{*"Psi = psi^"*}} \psi$ would have no time dependence and $\left\langlep\right\rangle = 0$.)

d) To mathematically show the distribution returns to its original shape after each classical period, show that

${\Psi}^{\text{*"(x,0)Psi(x,0) = Psi^"*}} \left(x , \frac{2 \pi}{\omega} _ 0\right) \Psi \left(x , \frac{2 \pi}{{\omega}_{0}}\right)$

Here we have:

$\frac{1}{2} \left[{\phi}_{0}^{2} + {\phi}_{1}^{2} + 2 {\phi}_{0} {\phi}_{1} \sin \left[\left({\omega}_{1} - {\omega}_{0}\right) \frac{2 \pi}{{\omega}_{0}}\right]\right] = \frac{1}{2} \left[{\phi}_{0}^{2} + {\phi}_{1}^{2} + 2 {\phi}_{0} {\phi}_{1} \sin \left(0\right)\right]$

From this,

$\sin \left[2 \pi \cdot \frac{{\omega}_{1} - {\omega}_{0}}{{\omega}_{0}}\right] = 0$

Under the approximation that ${\omega}_{1} \approx 2 {\omega}_{0}$, we get:

$\sin \left(2 \pi\right) = 0$

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.