8: Extra Credit

Last Time

• What was surprising about Quantum Mechanics?
• Free particle (almost exact reprise of 1D Wave Equation)
• Can't be normalized to 1 over all spaces! Instead: Normalize to one particle between \({x_1}\) and \({x_2}\). What do we mean by "square integrable?"

\[\langle p \rangle = \frac{\lvert a \rvert^2 - |b|^2}{|a|^2+|b|^2}\]

• What does this mean in a click-click experiment?
• Motion not present, but \(\Psi\) is encoded for it.
• Nodes spacing: \(\lambda/2\) (generalize this to get the "semiclassical")
• semiclassical : \(\lambda (x) = \frac{h}{p(x)}\), \(P_{classical}(x) = [2m(E - V(x))]^{1/2}\)

Particle in an Infinite Box

\[E_{n} = \frac{h^2}{8ma^2}n^2\]

\[\Psi_{n}(x) = \left(\frac{2}{a}\right)^{1/2} \sin \left(\frac{n\pi}{a}\right)x\]

(nodes, zero-point energy change: \(a , V_{0}\), locations of left edge

importance of pictures)

3D Box

\[\hat{H} = \hat{h}_x + \hat{h}_y + \hat{h}_z\]

(commuting operators)

\[E_{n_{x}n_{y}n_{z}} = E_{n_{x}}+ E_{n_{y}}+ E_{n_{z}}\]

\[\Psi_{n_{x}n_{y}n_{z}} = \Psi_{n_{x}}(X)\Psi_{n_{y}}(Y)\Psi_{n_{z}}(Z)\]

Today (and Next 3+ Lectures) Harmonic Oscillator

1. Classical Mechanics ("normal modes"" of vibration in polyatomic molecules arise from classical mechanics). Preparation for Quantum Mechanics treatment.
2. Quantum mechanical brute force treatment - Hermite Polynomials
3. Elegant treatment with memorable selection rules: creation/ annihilation operators.
4. Non-statinary states (i.e. moving) Quantum Mechanical Harmonic Oscillator: wavepackets, dephasing and recurrence, and tunneling through a barrier.
5. Perturbation Theory

Harmonic Oscillator

We have several kind of potential energy functions in atoms and molecules.

\(E_{n} = \propto n^{2}\)

particle in infinite-wall finite-length box (also particle on a ring)

Level pattern tells us quantitatively what kind of system we have.

Level splitting tell us quantitatively what are the properties of the class of system we have.

Rigid rotor

\(E_{n} \propto n(n +1)\)

Harmonic Oscillator

\[V(x) = \frac{1}{2}kx^{2}\]

\[E_{n} = \propto (n+\frac{1}{2})\]

The pattern of the energy levels tells us which underlying microscopic structure we are dealing with.

Typical interatomic potential energy:

We will use \(x\) rather than \(R\) here.

Expand the potential energy function as a power series:

\[X - X_{0} \equiv x\]

\[V(x) = V(0) + \frac{dV}{dx}|_{x=0} + \dfrac{d^{2}V}{dx^{2}}|_{x=0}\frac{x^{2}}{2}+\frac{d^{3}V}{dx^{2}}\frac{x^{3}}{6}\]

For small x, OK to ignore terms of higher order than \(x^{2}\). [What do we know about \(\frac{dV}{dx}\) at the minimum of any V(x)?]

Example \(\PageIndex{1}\): Morse Potential

\(V(x) = D_{e}[1-e^{-ax}]^{2} = D_{e}[1-\overbrace{2e^{-ax}}^{\scriptsize[1-ax+\frac{a^{2}x^{2}}{2}+...]}+\overbrace{e^{-2ax}}^{\scriptsize[1-2ax+\frac{4a^{2}x^{2}}{2}+...]}]\)

Some Algebra

\( = V(0)+\underbrace{0}_{\text{Why physically} \\ \text{is there} \\ \text{no linear} \\ \text{in x term?} }+a^{2}D_{e}x^{2}-a^{3}D_{e}x^{3}+\frac{7}{12} a^{4} D_{e} x^{4}+... \)

\(V(\infty = D_{e})\)(dissociation energy),\(V(0) = 0\).

If \(ax\ll 1\), \(V(x)\approx V(0)+ ((D_{e})a^{2})x^{2}\). A very good starting point for the molecular vibrational potential energy curve.

Call \(D_{e}a^{2} = k/2\). Ignore the \(x^{3}\) and \(x^{4}\) terms.

Hooke's Law

Let's first focus on a simple harmonic oscillator in the classical mechanics.

\[F = \underbrace {-k(X-X_{0})}_\text{force is - gradient of potential}\]

When \(X>X_{0}\),Force pushes mass back down toward \(X_{0}\)

\[F = -\frac{dV}{dX}\]

When \(X<X_{0}\), Force pulls mass back up towards \(X_{0}\)

\(\therefore V(x) = \frac{1}{2}k(X-X_{0})^2\)

Newton's Equation:

\[F = ma = m \frac{d^{2}(X-X_{0})}{dt^{2}} = -k(X-X_{0})\]

where \(x \equiv X-X_{0}\).

Substitute and rearrange

\(\frac{d^{2}x}{dt^{2}} = -\frac{k}{m}x\), \(2^{nd}\) order ordinary linear differential equation: solution contains two linearly independent terms, each multiplied by one of 2 constants to be determined

\[x(t) = A\sin \left(\frac{k}{m} \right)^{\frac{1}{2}} t + B\cos \left(\frac{k}{m} \right)^{\frac{1}{2}} t \]

It is customary to write

\[\left(\frac{k}{m} \right)^{\frac{1}{2}} = \omega.\]

(\(\omega\) is conventionally used to specify an angular frequency: radians/ second).

Why?

What is frequency of oscillation? \(\tau\) is the period of oscillation.

\[x(t+\tau)=x(t)= A\sin\left[\left(\frac{k}{m}\right)^{\frac{1}{2}} t \right] + B\cos\left[\left(\frac{k}{m} \right)^{\frac{1}{2}} t \right] = A\sin \left[\left(\frac{k}{m} \right)^{\frac{1}{2}} (t+\tau) \right] + B\cos \left[\left(\frac{k}{m} \right)^{\frac{1}{2}} (t+\tau) \right]\]

requires

\(\left(\frac{k}{m}\right)^{\frac{1}{2}}\tau = 2\pi\), \(v = \frac{1}{\tau}\) where \(\tau\) is period.

\(\tau = \frac{2\pi}{\omega}=\frac{2\pi}{2\pi v} = \frac{1}{v}\) as required.

How long does one full oscillation take?

We have sin, cos functions of \((\frac{k}{m})^{\frac{1}{2}}\) \(t = \omega t\)

when argument of sin or cos goes from 0 to \(2\pi\), we have one period of oscillation.

\[2\pi = \left(\frac{k}{m}\right)^{\frac{1}{2}}, \tau = \omega \tau\]

\[\tau = \frac{2\pi}{\omega}=\frac{1}{v}\]

So everything makes sense.

\(\omega\) is "angular frequency" (radians/sec).

\(v\) is ordinary frequency (cycles/sec).

\(\tau\) is period (sec).

\[x(t)=A\sin\omega t + B\cos\omega t\]

Need to get A, B from initial conditions:

[e.g. starting at a \(\overbrace{\text{turning point}}^{ASK!}\) where \(E=V(x_{\pm})=(1/2)kx_{\pm}^{2}\) and \( E \Rightarrow \pm(\frac{2E}{k})^{\frac{1}{2}} = x_{\pm}\)]

Initial amplitude of oscillation depends on the strength of the pluck!

If we start at \(x_{+}\) at t = 0, (the sine term is zero at t = 0, the cosine terms is B at t = 0)

\[x(0) = (\frac{2E}{k})^{\frac{1}{2}} \Rightarrow B = (\frac{2E}{k})^{\frac{1}{2}})\]

Note that the frequency of oscillation does not depend on initial amplitude. To get A for initial condition x(0)=\(x_{+}\), look at \(t=\tau /4\), where \(x(\tau /4)\) = 0. Find A = 0.

Alternatively, we can use frequency, phase form. For \(x(0) = x_{+}\) initial condition:

\[x(t) = C\sin \left( \left(\frac{k}{m} \right)^{\frac{1}{2}} t + \phi \right) \]

if \(x(0) = x_{+} = \left(\frac{2E}{k} \right)^{\frac{1}{2}}\)

C = \( \left(\frac{2E}{k} \right)^{\frac{1}{2}}\), \(\phi = -\pi /2\)

We are done. Now explore Quantum Mechanics - relevant stuff.

Quantum Mechanics

What is:

Oscillation Frequency

Kinetic Energy \(T(t)\), \(\overline{T}\)

Potential Energy \(V(t)\), \(\overline{V}\)

Period \(\tau\)?

Oscillation Frequency:

\[v = \frac{\omega}{2\pi}\] independent of E

Kinetic Energy:

\[T(t) = \frac{1}{2}mv(t)^{2}\]

\[x(t) = \left[\frac{2E}{k} \right]^{\frac{1}{2}}\sin \left[\omega t + \phi \right]\]

take derivative of x(t) with respect to t,

\[v(t) = \omega \left[\frac{2E}{k} \right]^{\frac{1}{2}}\cos \left[\omega t + \phi \right]\]

\[T(t) = \frac{1}{2}m\underbrace{\omega ^{2}}_{\scriptsize{k/m}} \left[\frac{2E}{k} \right]\cos^2 \left[\omega t + \phi \right]\]

\[E\cos^2 (\omega t + \phi)\]

Now some time averaged quantities:

\[\langle{T}\rangle = \overline{T} = E\frac{\int_0^\tau dt \cos^{2}(\omega t + \phi)}{\tau}\]

= E/2. Recall \(\tau = \frac{2 \pi}{\omega}\).

\[V(t) = \frac{1}{2} kx^{2} = \frac{k}{2} \left(\frac{2E}{k}\right) \sin^{2} (\omega t + \phi)\]

Calculate \(\langle{V}\rangle\) by \(\int_0^\tau dt\) or by simple algebra, below

\(= Esin^{2}(\omega t + \phi)\)

\(E = T(t)+V(t)= \overline{T}+ \overline{V}\)

\(\overline{V} = E/2\)

Really neat that \(\overline{T} = \overline{V} = E/2\).

Energy is being exchanged between \(T\) and \(V\). They oscillate \(\pi / 2\) out of phase:

\[V(t) = T\left(t - \frac{\tau}{4}\right)\]

V lags T.

What about x(t) and p(t) when x is near the turning point?

\[x(t) = \left[\frac{2E}{k} \right]^{\frac{1}{2}} \cos \omega t\]

where \(x(t=0) = x_{+}\).

\(x\) changing slowly near \(x\) turning point

\(p\) changing fastest near \(x\) turning point

Insights for wavepacket dynamics. We will see that "survival probability" \(|\Psi^{*}(x,t)\Psi(x,0)|^{2}\) decays near t.p. mostly \(\hat{p}\) rather than \(\hat{x}\).

What about time-averages of \(x,x^{2},p,p^{2}\)?

\( \left. \begin{array} \\ \langle{x}\rangle = 0 \\ \langle{p}\rangle = 0 \end{array} \right\} \text{is the HO potential moving in space?}\)

\[x^{2} = V(x)/ \frac{k}{2}\]

take t-average

\[\langle{x^{2}}\rangle=\frac{2}{k}\langle{v(x)}\rangle=\frac{2}{k}\frac{E}{2}=\frac{E}{k}\]

\[p^{2}=2mT\]

\[\langle{p^{2}}\rangle=2m\frac{E}{2}= mE\]

\[\Delta x = \langle{x^{2}- \langle{x}\rangle ^{2}}\rangle ^ {\frac{1}{2}}=(E/k)^{\frac{1}{2}}\]

\[\Delta p = \langle{p^{2}-\langle{p}\rangle^{2}}\rangle^{\frac{1}{2}}=(mE)^{\frac{1}{2}}\]

\[\Delta x \Delta p = E \left(\frac{m}{k}\right)^{\frac{1}{2}} = E/ \omega \text{ Small at low E}\].

We will see an uncertainty relationship between x and p in Quantum Mechanics.

Probability of finding oscillation between x and x + dx: consider one and half period, oscillator going from left to right turning point.

\[P(x)dx = \frac{time(x,x+dx)}{\tau / 2} = \frac{\frac{distance}{velocity}}{\frac{1}{2}\frac{2 \pi}{\omega}}\]

\(= \frac{\frac{dx}{v(x)}}{\frac{2 \pi}{2 \omega}}=\frac{2 \omega}{v(x)2 \pi}dx\)

large probability at turning point. Goes to \(\infty\) at \(x_{\pm}\).

Minimum probability at x = 0.

In Quantum Mechanics, we will see that \(P(x_{\pm})\) does not blow u and there is some probability outside the classically allowed region. Tunneling.

NON Lecture: Spring with Mass (Non Lecture)

Next we want to go from one mass on an anchored spring to two masses connected by a spring

F = ma for each mass

\[m_{1} \frac{d^{2}x_{1}}{dt^{2}} = k(x_{2}-x_{1} - {\it{l}_{0}} )\]

Where \(\it{l}_{0}\) is the length of the spring at rest, i.e. when \(x_{2} - x_{1} = \it{l}_{0}\)

\[m_{2} \frac{d^{2}x_{2}}{dt^{2}} = -k(x_{2}-x_{1} - {\it{l}_{0}} )\]

2 coupled differential equations.

Uncouple them easily, as follows:

Add the 2 equations:

\(m_{1} \frac{d^{2}x_{1}}{dt^{2}} + m_{2} \frac{d^{2}x_{2}}{dt^{2}}=\frac{d^{2}}{dt^{2}} \underbrace{(m_{1}x_{1}+m_{2}x_{2})}_{\text{we will see that}\\ \text{this is at worst}\\ \text{proportional}\\ \text{to t}}=0\)

Define a center of mass coordinate.

\[\frac{m_{1}x_{1} + m_{2}x_{2}}{M} = X\]

\[M = m_{1} + m_{2}\]

replace \(m_{1}x_{1} + m_{2}x_{2}\) by MX

\(M\frac{d^{2}X}{dt^{2}}=0\)

integrate once with respect to t

\(\frac{dX}{dt}(t) = \text{const}\).

The center of mass is moving at constant velocity — no force acting.

Next find a new differential equation expressed in terms of the relative coordinate.

\[x = x_{2} - x_{1} - \it{l}_{0}\]

Divide the first differential equation by \(m_{1}\), the second by \(m_{2}\), and subtract the first from the second:

\(\frac{d^{2}x_{2}}{dt^{2}}-\frac{d^{2}x_{1}}{dt^{2}} = -\frac{k}{m_{2}}(x_{2} - x_{1} - {\it{l}_{0}} )-\frac{k}{m_{1}}(x_{2} - x_{1} - {\it {l}_{0}} )\)

\(\frac{d^{2}}{dt^2}(x_{2}-x_{1})=-k\left(\frac{1}{m_{2}}+\frac{1}{m_{1}}\right)(x_{2} - x_{1} - {\it{l}_{0}} )\)

\(= -k \left(\frac{m_{1}+m_{2}}{m_{1}m_{2}}\right)(x_{2} - x_{1} - {\it{l}_{0}} )\)

\[\mu =\frac{m_{1}m_{2}}{m_{1}+m_{2}}\]

\(\frac{d^{2}}{dt^2}\underbrace{(x_{2}-x_{1})}_{\text{killed by}\\ \text{derivative}\\=x+{\it{l}_0}}=-\frac{k}{\mu}\underbrace{(x_{2} - x_{1} - \it{l}_{0}) = -\frac{k}{\mu}x}_{\text{x is displacement} \\ \text{from equilibrium}}\)

We get a familiar looking equation fro the intermolecular displacement from equilibrium.

\[\mu\frac{d^{2}x}{dt^{2}}+kx = 0\]

Everything is the same as the one-mass-on-a-spring problem except m → μ.

Next time: Quantum Mechanical Harmonic Oscillator

\[H = \frac{\hat{p}^{2}}{2 \mu}+\frac{1}{2}k \hat{x}^{2}\]

note that this differential operator does not have time in it!

We will see particle-like motion for harmonic oscillator when we consider the Time Dependent Schrödinger equation (Lecture #10) and Ψ(x,t) is a particle-like state.

\(\Psi(x,t)\) where \(\Psi(x,0) = \sum \limits_{v=0}^{\infty} C_{v}\Psi_{v}\)

in the \(4^{th}\) lecture on Harmonic Oscillators (Lecture #11).