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8: Extra Credit

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    94075
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    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.

    Screenshot (68).png

    220px-Infinite_potential_well-en.svg.png

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

    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

    Screenshot (59).png

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

    Harmonic Oscillator

    Screenshot (61).png

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

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

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

    Typical interatomic potential energy:

    Screenshot (67).png

    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.

    Screenshot (69).png

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

    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:

    Screenshot (70).png

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

    Screenshot (65).png

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

    Screenshot (66).png

    \(p\) 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.

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

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

    Screenshot (71).png

    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

    Screenshot (72).png

    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).


    8: Extra Credit is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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