5.61 F17 Lecture 7: Classical Mechanical Harmonic Oscillator

5.61 Fall, 2017 Lecture #7 Page 1 Lecture #7: Classical Mechanical Harmonic Oscillator Last time What was surprising about Quantum Mechanics? Free particle (almost an exact reprise of 1D Wave Equation) Can't be normalized to 1 over all space! Instead: Normalization to one particle between x1 and x2. What do we mean by “square integrable?” |a|2 |b|2 hp̂i = 2 |a| + |b|2 What free particle ψ(x) has this expectation value? What does this mean in a click-click experiment? Motion not present, but ψ is encoded for it. Node spacing: λ/2 (generalize this to get "semiclassical”) 1/2 h , pclassical (x) = ⎡⎣ 2m ( E − V (x)) ⎤⎦ Semiclassical: λ(x) = p(x) Particle in Infinite Box h2 2 En = n 2 8ma 1/2 ⎛ 2⎞ ⎛ nπ ⎞ ψn (x) = ⎜ ⎟ sin ⎜ x ⎝ a⎠ ⎝ a ⎟⎠ nodes, zero-point energy change: a, V0, location of left edge importance of pictures 3D box  = ĥ + ĥ + ĥ H x y z ( commuting operators ) Enxnynz = Enx + Eny + Enz ψnxnynz = ψnx (x)ψny (y)ψnz (z) revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 2 Today (and next 3+ lectures) Harmonic Oscillator 1) Classical Mechanics (“normal modes” of vibration in polyatomic molecules arise from classical mechanics). Preparation for Quantum Mechanical treatment. 2) Quantum mechanical brute force treatment — Hermite Polynomials 3) Elegant treatment with memorable selection rules: “creation/annihilation” operators. 4) Non-stationary states (i.e. moving) of Quantum Mechanical Harmonic Oscillator: wavepackets, dephasing and recurrence, and tunneling through a barrier. 5) Perturbation Theory. Harmonic Oscillator We have several kinds of potential energy functions in atoms and molecules. V (r) = − 1 r En ∝ − H atom 1 n2 En ∝ n2 particle in infinite-wall finite-length box (also particle on a ring) 0 L Level pattern tells us qualitatively what kind of system we have. Level splittings tell us quantitatively what are the properties of the class of system we have. revised 9/19/17 1:50 PM b a⇤i = ai because A 5.61 Fall, 2017 corresponds to a classically observable quantity Rigid rotor Harmonic Oscillator Lecture #7 Page 3 En ∝ n ( n + 1) V (x) = 1 2 kx 2 En ∝ ( n + 1 / 2 ) x0 The pattern of energy levels tells us which underlying microscopic structure we are dealing with. Typical interatomic potential energy: 6 Hard inner wall ... ......... ................ . ... . . . ... .... 6 . . . . ... . ... .. Bond-Breaking . ... . .. Soft outer wall ... . . ... .. ... . .... .. . . .... looks like V (R) = k2 (R Re )2 ..... ... . . . ..... . harmonic near the bottom of the well .......... ............ ...... Re We will use x rather than R here. Expand any potential energy function as a power series: X − X0 ≡ x dV V (x) = V (0) + dx d 2V x+ 2 dx x=0 x=0 x 2 d 3V x 3 + 2 dx 2 6 For small x, OK to ignore terms of higher order than x2. [What do we know about dV at the dx minimum of any V(x)?] For example, Morse Potential revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 4 ⎛ ⎞ a2 x 2 +…⎟ ⎜⎝ 1− ax + ⎠ 2 ⎛ ⎞ 4a 2 x 2 +…⎟ ⎜⎝ 1− 2ax + ⎠ 2 V (x) = De [1− e− ax ] = De [1− 2e− ax + e−2ax ] 2 some algebra = V (0) + 0 + a 2 De x 2 − a 3 De x 3 + why physically is there no linear in x term? 7 4 a De x 4 +… 12 V(∞) = De (dissociation energy), V(0) = 0. If ax ! 1, V(x) ≈ V(0)+(Dea2)x2. A very good starting point for the molecular vibrational potential energy curve. Call Dea2 = k/2. Ignore the x3 and x4 terms. Let’s first focus on a simple harmonic oscillator in classical mechanics. Hooke’s Law F = −k ( X − X0 )  force is — gradient of potential F=− When X < X0 Force pulls mass back up toward X0 dV dX ∴ V (x) = When X > X0 Force pushes mass back down toward X0 1 2 k ( X − X0 ) 2 revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 5 Newton’s equation: d 2 ( X − X0 ) F = ma = m = −k ( X − X0 ) dt 2 x ≡ X − X0 substitute and rearrange 2nd order ordinary linear differential equation: solution contains two linearly independent terms, each multiplied by one of 2 constants to be determined d2 x k =− x 2 dt m ⎛ k⎞ x(t) = Asin ⎜ ⎟ ⎝ m⎠ 1/2 ⎛ k⎞ t + B cos ⎜ ⎟ ⎝ m⎠ 1/2 t It is customary to write ⎛ k⎞ ⎜⎝ ⎟⎠ m 1/2 =ω. (ω is conventionally used to specify an angular frequency: radians/second) Why? What is frequency of oscillation? τ is period of oscillation. ⎡⎛ k ⎞1/2 ⎤ ⎡⎛ k ⎞1/2 ⎤ ⎡⎛ k ⎞1/2 ⎤ ⎡⎛ k ⎞1/2 ⎤ x (t + τ ) = x(t) = Asin ⎢⎜ ⎟ t ⎥ + B cos ⎢⎜ ⎟ t ⎥ = Asin ⎢⎜ ⎟ (t + τ )⎥ + B cos ⎢⎜ ⎟ (t + τ )⎥ ⎣⎝ m ⎠ ⎦ ⎣⎝ m ⎠ ⎦ ⎣⎝ m ⎠ ⎦ ⎣⎝ m ⎠ ⎦ requires 1/2 ⎛ k⎞ ⎜⎝ ⎟⎠ τ = 2π m ν= τ= 2π 2π 1 = = as required. ω 2πν ν 1 τ period How long does one full oscillation take? 1/2 ⎛ k⎞ we have sin, cos functions of ⎜ ⎟ t = ωt ⎝ m⎠ when the argument of sin or cos goes from 0 to 2π, we have one period of oscillation. ⎛ k⎞ 2π = ⎜ ⎟ ⎝ m⎠ τ= 1/2 τ = ωτ 2π 1 = . ω ν So everything makes sense. ω is “angular frequency” v is ordinary frequency τ is period radians/sec. cycles/sec. sec revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 6 x(t) = A sin ωt + B cos ωt Need to get A,B from initial conditions: V(x)  ASK!   [e.g. starting at a turning point where E = V (x± ) = (1 / 2)kx±2 ] E– ⇓ x–(E) 0 ⎛ 2E ⎞ ±⎜ ⎝ k ⎟⎠ x+(E) 1/2 = x± 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 term is B at t = 0) ⎛ 2E ⎞ x (0) = ⎜ ⎝ k ⎟⎠ 1/2 ⎛ 2E ⎞ ⇒B=⎜ ⎝ k ⎟⎠ 1/2 Note that the frequency of oscillation does not depend on the initial amplitude. To get A for initial condition x(0) = x+, look at t = τ/4, where x(τ/4) = 0. Find A = 0. Alternatively, we can use frequency, phase form. For x(0) = x+ initial condition: ⎛ ⎛ k ⎞ 1/2 ⎞ x(t) = C sin ⎜ ⎜ ⎟ t + φ⎟ ⎝⎝ m⎠ ⎠ ⎛ 2E ⎞ if x(0) = x+ = ⎜ ⎝ k ⎟⎠ 1/2 1/2 ⎛ 2E ⎞ C=⎜ , φ = −π / 2 ⎝ k ⎟⎠ We are done. Now explore Quantum Mechanics - relevant stuff. What is: Oscillation Frequency Kinetic Energy T(t), T Potential Energy, V(t), V Period τ? revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Oscillation Frequency: ν = ω independent of E 2π T (t) = Kinetic Energy: Page 7 1 mv(t)2 2 1/2 2E x(t) = ⎡⎢ ⎤⎥ sin [ ωt + φ ] ⎣ k ⎦ take derivative of x(t) with respect to t 1/2 2E v(t) = ω ⎡⎢ ⎤⎥ cos [ ωt + φ ] ⎣ k ⎦ 1 2 ⎡ 2E ⎤ 2 T (t) = m ω  ⎢ ⎥ cos [ ωt + φ ] 2 k/m ⎣ k ⎦ = E cos 2 ( ωt + φ ) Now some time averaged quantities: T ∫ =T = E τ 0 dt cos 2 ( ωt + φ ) = E /2 τ recall τ = 2π ω 1 k ⎛ 2E ⎞ 2 V (t) = kx 2 = ⎜ ⎟ sin ( ωt + φ ) 2 2⎝ k ⎠ Calculate 〈V〉 by ∫ τ 0 dt or by simple algebra, below = E sin ( ωt + φ ) 2 E = T (t) + V (t) = T + V V = E /2 Really neat that T = V = E / 2 . τ⎞ ⎛ Energy is being exchanged between T and V. They oscillate π/2 out of phase: V (t) = T ⎜ t − ⎟ ⎝ 4⎠ V lags T. What about x(t) and p(t) when x is near the turning point? 1/2 ⎡ 2E ⎤ x ( t ) = ⎢ ⎥ cosωt ⎣ k ⎦ x(t = 0) = x+ revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 8 x+ x changing slowly near x turning point x τ t p p changing fastest near x turning point 0 τ/4 τ/2 t Insights for wavepacket dynamics. We will see (in Lecture #11) that “survival probability” Ψ* (x,t)Ψ(x,0) 2 decays near t.p. mostly because of p̂ rather than x̂ . What about time-averages of x, x2, p, p2? x = 0⎫ ⎬ is the HO potential moving in space? p = 0 ⎪⎭ x2 = V (x) ( k / 2 ) take t-average 2 2E x2 = V (x) = = E/k k k2 p2 = 2mT p2 = 2m E = mE 2 revised 9/19/17 1:50 PM 5.61 Fall, 2017 ∆ x = x2 − x ∆ p = p2 − p Lecture #7 = (E / k) 2 1/2 1/2 = ( mE ) 2 1/2 ⎛ m⎞ ∆ x∆ p = E ⎜ ⎟ ⎝ k⎠ Page 9 1/2 1/2 = E / ω . small at low E We will see an uncertainty relationship between x and p in Quantum Mechanics. Probability of finding oscillator between x and x + dx: consider one half period, oscillator going from left to right turning point. distance time ( x, x + dx ) velocity P(x)dx = = 1 ⎛ 2π ⎞ τ/2 ⎜ ⎟ 2⎝ ω ⎠ dx 2ω v(x) = = dx 2π v(x)2π 2ω ( v(x) small at x = x ± ) P(x) x- x+ large probability at turning points. Goes to ∞ at x±. minimum probability at x = 0 In Quantum Mechanics, we will see that P(x±) does not blow up and also that there is some probability outside the classically allowed region. Tunneling. revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 10 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 d 2 x1 m1 2 = k ( x2 − x1 −  0 ) dt length of spring at rest, i.e. when x – x = "0 2 1 d2x m2 22 = −k ( x2 − x1 −  0 ) dt 2 coupled differential equations. Uncouple them easily, as follows: Add the 2 equations m1 d 2 x1 d 2 x2 d 2 + m = 2 ( m1 x1 + m2 x2 ) = 0 2  dt 2 dt 2 dt  we will see that this is at worst proportional to t Define a center of mass coordinate. revised 9/19/17 1:50 PM 5.61 Fall, 2017 Lecture #7 Page 11 m1 x1 + m2 x2 =X M replace m1 x1 + m2 x2 by MX M = m1 + m2 d2X =0 dt 2 integrate once with respect to t M dX (t) = const. dt 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 = x2 – x1 – "0. Divide the first differential equation (located at the top of page 10) by m1, the second by m2, and subtract the first from the second: d 2 x2 d 2 x1 k k − 2 =− ( x2 − x1 −  0 ) − ( x2 − x1 −  0 ) 2 dt dt m2 m1 1⎞ d2 ⎛ 1 + ⎟ ( x2 − x1 −  0 ) x − x1 ) = −k ⎜ 2 ( 2 ⎝ m2 m1 ⎠ dt ⎛ m + m2 ⎞ = −k ⎜ 1 (x − x −  ) ⎝ m1m2 ⎟⎠ 2 1 0 µ≡ killed by derivative m1m2 m1 + m2 d2 k k x − x x − x −  = − = − x ( 1) 2 1 0) 2 ( 2  dt  µ µ x+ 0  x is displacement from equilibrium We get a familiar looking equation for the intramolecular displacement from equilibrium. µ d2x + kx = 0 dt 2 Everything is the same as the one-mass-on-an-anchored-spring problem except m → µ. revised 9/19/17 1:50 PM 5.61 Fall, 2017 Next time: Lecture #7 Page 12 Quantum Mechanical Harmonic Oscillator 2  = p̂ + 1 kx̂ 2 H 2µ 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 constructed to be a particle-like state. Ψ(x,t) ∞ where Ψ(x,0) = ∑ cvψ v v=0 th in the 4 lecture on Harmonic Oscillators (Lecture #11). revised 9/19/17 1:50 PM