Chapter 05 Vibrational Motion

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 05: Vibrational Motion Simple Harmonic Oscillator

Simplest model for harmonic oscillator: mass attached to one end of spring while other end is held ﬁxed

-x 0 +x

Mass at x = 0 corresponds to equilibrium position x is displacement from equilibrium. Assume no friction and spring has no mass.

P. J. Grandinetti Chapter 05: Vibrational Motion Simple Harmonic Oscillator

What happens? An oscillation. Pull mass and let go.

-x 0 +x

Mass at x = 0 corresponds to equilibrium position x is displacement from equilibrium. Assume no friction and spring has no Time for 1 complete cycle is T. mass. How do we get and solve equation of motion for this system?

P. J. Grandinetti Chapter 05: Vibrational Motion Simple Harmonic Oscillator Hooke’s law 휅 For small displacements from equilibrium restoring force is F = − fx 휅 where f is force constant for spring. 휅 Use Newton’s 2nd law: F = ma = − fx

̈ 휅 to obtain diﬀerential equation of motion: mx(t) + fx(t) = 0 Propose solution

x(t) = A cos(휔t + 휙),x ̇ (t) = A휔 sin(휔t + 휙), ẍ(t) = −A휔2 cos(휔t + 휙)

( ) Substitute into diﬀerential equation: 휅 − m휔2 A cos(휔t + 휙) = 0 f √ 휔2 휅 휔 휔 휅 Satisfy solution for all t by making m = f through deﬁnition: = 0 = f∕m

P. J. Grandinetti Chapter 05: Vibrational Motion Simple Harmonic Oscillator

Solution is x(t) = A cos(휔 t + 휙) 0 √ 휔 ≡ 휔 휔 휅 where 0 natural oscillation frequency given by = 0 = f∕m Velocity of mass is ̇ 휔 휔 휙 v(t) = x(t) = − 0A sin( 0t + )

Make x(t) and ẋ (t) equations satisfy initial conditions of

x(t = 0) = A and ẋ (t = 0) = 0

by setting 휙 = 0 and A = x(0) to get ﬁnal solution

휔 휔 휔 x(t) = x(0) cos 0t and v(t) = − 0x(0) sin( 0t)

P. J. Grandinetti Chapter 05: Vibrational Motion Simple Harmonic Oscillator – Other trial solutions

휔 휙 ̈ 휅 Instead of using x(t) = A cos( t + ) as trial solution for mx(t) + fx(t) = 0 ▶ Could have used x(t) = A cos 휔t + B sin 휔t.

▶ Could also have used complex variables: ★ replace x(t) with a complex variable, z(t), ̈ 휅 mz(t) + fz(t) = 0 ★ Make an initial guess of

z(t) = Aei(휔t+휙) or z(t) = Aei휔t + Be−i휔t

★ Obtain real solution by taking real part of complex solution, z(t).

P. J. Grandinetti Chapter 05: Vibrational Motion Complex Variables Review

P. J. Grandinetti Chapter 05: Vibrational Motion Complex Variables Review Complex variables are a mathematical tool that simpliﬁes equations describing oscillations. Consider the 2D motion of this vector.

How would you describe this mathematically? You probably would suggest: x(t) = r cos 휔t and y(t) = r sin 휔t

P. J. Grandinetti Chapter 05: Vibrational Motion Complex Variables Review With complex notation we combine two equations into one Start with x(t) = r cos 휔t and y(t) = r sin 휔t First we deﬁne the square root of −1 as √ if i = −1 then i2 = −1

Second we deﬁne complex variable z as

z = x + iy

x is real part and y is imaginary part of z. Two circular motion equations become one circular motion equation

z(t) = r cos 휔t + ir sin 휔t

P. J. Grandinetti Chapter 05: Vibrational Motion Complex Variables Review Euler’s formula

In 1748 Euler showed that ei휃 = cos 휃 + i sin 휃 Euler’s formula

With Euler’s formula z(t) = r cos 휔t + ir sin 휔t becomes z(t) = rei휔t

휔 휔 휔 휔 Calculate the product (cos at + i sin at)(cos bt + i sin bt). 휔 휔 휔 휔 휔 휔 휔 휔 휔 휔 휔 휔 (cos at+i sin at)(cos bt+i sin bt) = cos at cos bt+i cos at sin bt+i sin at cos bt−sin at sin bt

2 cos 휃 cos 휙 = cos(휃 − 휙) + cos(휃 + 휙) 2 sin 휃 sin 휙 = cos(휃 − 휙) − cos(휃 + 휙) 2 sin 휃 cos 휙 = sin(휃 + 휙) + sin(휃 − 휙) 2 cos 휃 sin 휙 = sin(휃 + 휙) − sin(휃 − 휙)

휔 휔 휔 휔 휔 휔 휔 휔 (cos at + i sin at)(cos bt + i sin bt) = cos( a + b)t + i sin( a + b)t

Using Euler’s formula: 휔 휔 휔 휔 휔 휔 휔 휔 i at i bt i( a+ b)t 휔 휔 휔 휔 (cos at + i sin at)(cos bt + i sin bt) = e e = e = cos( a + b)t + i sin( a + b)t

P. J. Grandinetti Chapter 05: Vibrational Motion Complex Variables Review

Any complex number can be written in the form

z = x + iy = |z|ei휃

where |z| is the magnitude of the complex number √ |z| = x2 + y2

and 휃 is the argument of the complex number

tan 휃 = y∕x

P. J. Grandinetti Chapter 05: Vibrational Motion Complex Variables Review Complex Conjugate Complex conjugate, z∗, of z is obtained by changing sign of imaginary part

if z = x + iy then z∗ = x − iy

1 + 4i 1 − 4i if z = then z∗ = 4 − 5i 4 + 5i 1 + ia 1 − ia∗ if z = then z∗ = . 4 − ib 4 + ib∗

Related identities√ ▶ |z| = zz∗, since

zz∗ = (x + iy)(x − iy) = x2 + iyx − ixy + y2 = x2 + y2 = |z|2

▶ ⋯ ∗ ∗ ∗ ∗ ⋯ (z1z2z3 ) = z1z2z3

P. J. Grandinetti Chapter 05: Vibrational Motion Energy of simple harmonic oscillator

P. J. Grandinetti Chapter 05: Vibrational Motion Energy of simple harmonic oscillator Total energy of simple harmonic oscillator is sum of kinetic and potential energy of mass and spring.

Kinetic energy is given by 1 p2 K = mv2, or K = 2 2m

Potential energy is energy stored in spring and equal to work done in extending and compressing spring,

x x ′ ′ 휅 ′ ′ 1휅 2 V(x) = − ∫ F(x )dx = ∫ fx dx = fx 0 0 2

Expression above is work associated with extending spring. For work in compressing spring just change integral limits to −x to 0 (get same result).

P. J. Grandinetti Chapter 05: Vibrational Motion Energy of simple harmonic oscillator Potential energy is given by 1 V(x) = 휅 x2 2 f

V(x)

0 x

P. J. Grandinetti Chapter 05: Vibrational Motion Energy of simple harmonic oscillator

Although both K and V are time dependent during harmonic motion the total energy, E = K + V, remains constant.

1 1 p2(t) 1 E = mv2(t) + 휅 x2(t) or E = + 휅 x2(t) 2 2 f 2m 2 f

Total Energy Potential Energy Energy Kinetic Energy

P. J. Grandinetti Chapter 05: Vibrational Motion Energy of simple harmonic oscillator Substitute equation of motion into energy expression 1 1 1 1 E = mv2(t) + 휅 x2(t) = m휔2x2(0) sin2(휔 t) + 휅 x2(0) cos2 휔 t 2 2 f 2 0 0 2 f 0 휅 휔2 Recall that f = m 0 so 1 [ ] 1 E = m휔2x2(0) sin2(휔 t) + cos2 휔 t = m휔2x2(0) 2 0 0 0 2 0 Solve for initial amplitude, x(0), in terms of energy √ 1 2E x(0) = 휔 0 m and rewrite oscillation as

√ √ 2E 2E x(t) = x(0) cos 휔 t = cos 휔 t and v(t) = −휔 x(0) sin(휔 t) = − sin 휔 t 0 휔2 0 0 0 m 0 0m

P. J. Grandinetti Chapter 05: Vibrational Motion Simple Harmonic Oscillator - Phase Space Trajectory

√ √ 2E 2E x(t) = x(0) cos 휔 t = cos 휔 t and v(t) = −휔 x(0) sin(휔 t) = − sin 휔 t 0 휔2 0 0 0 m 0 0m

State of oscillator speciﬁed by point in phase space E = constant means oscillator state always lies on ellipse

x2 v2 휅 + = 1 2E∕ f 2E∕m Trajectory moves in clockwise direction. Trajectories with diﬀerent E can never cross.

P. J. Grandinetti Chapter 05: Vibrational Motion Position probability distribution for harmonic oscillator

P. J. Grandinetti Chapter 05: Vibrational Motion Position probability distribution for harmonic oscillator Scale x(t) by initial amplitude x(0), to obtain a function, y(t), that oscillates between y = −1 and y = +1 휔 y(t) = x(t)∕x(0) = cos 0t Calculate normalized probability density, p(y), for ﬁnding the mass at any scaled position between y = ±1. Probability of ﬁnding mass in interval dy at given y is proportional to time spent in dy interval, dy dt b p(y) dy = b dt = b dt = b dy = dy dy dy ẏ b ≡ proportionality constant, and ẏ ≡ speed at a given y Use derivative of y(t) ̇ 휔 휔 y(t) = − 0 sin 0t to obtain b b b b p(y) = = = = ̇ 휔 휔 휔 2 휔 1∕2 휔 2 1∕2 y(t) − 0 sin 0t − 0(1 − cos 0t) − 0(1 − y )

P. J. Grandinetti Chapter 05: Vibrational Motion Position probability distribution for harmonic oscillator

b p(y) = 8 휔 2 1∕2 − 0(1 − y ) 6 Normalizing probability distribution gives 4 1 p(y) = 휋(1 − y2)1∕2 2

0 -1.0 -0.5 0.0 0.5 1.0 m

-x 0 +x Mass spends majority of time at maximum excursions, that is, turning points where velocity is slowest and changes sign.

P. J. Grandinetti Chapter 05: Vibrational Motion Diatomic molecule vibration as Harmonic oscillator

Dashed line is harmonic oscillator potential. Solid line is Morse potential.

V(r) has minimum at re—where restoring force is zero. < > V(r) causes repulsive force at r re and attractive force at r re. < > V(r) increases steeply at r re but levels out to constant at r re. At r → ∞ there is no attractive force as V(r) has a slope of zero.

P. J. Grandinetti Chapter 05: Vibrational Motion Diatomic molecule vibration as Harmonic oscillator Taylor series expansion of V(r) about equilibrium bond length, r = re, gives >0 dV(r) 1 d2V(r ) 1 d3V(r ) V(r) ≈ V(r ) + e (r − r ) + e (r − r )2 + e (r − r )3 + ⋯ e dr e 2! dr2 e 3! dr3 e

V(re) is the potential energy at equilibrium bond length.

1st-order term is zero since no restoring force, F = −dV(re)∕dr, at r = re Stop expansion at the 3rd-order term and deﬁne 2 constants d2V(r ) d3V(r ) 휅 = e and 훾 = e f dr2 f dr3 and write potential expansion as 1 1 V(r) − V(r ) ≈ 휅 (r − r )2 + 훾 (r − r )3 + ⋯ e 2 f e 6 f e For small displacements drop 3rd-order term and potential looks like simple harmonic oscillator. For slightly larger displacements keep 3rd-order term to account for vibration anharmonicity.

P. J. Grandinetti Chapter 05: Vibrational Motion Diatomic molecule vibration equations of motion

Make harmonic oscillator approximation taking force on m1 and m2 as 휅 휅 F1 = − f(r1 − r2 + re) and F2 = − f(r2 − r1 − re) Equations of motion are 2 coupled differential equations: d2r d2r m 1 = 휅 (r − r − r ) and m 2 = −휅 (r − r − r ) 1 dt2 f 2 1 e 2 dt2 f 2 1 e Transform to center of mass frame: M = m + m and R = 1 (m r + m r ) 1 2 M 1 1 1 2 Obtain 2 uncoupled differential equations, d2R d2Δr M = 0, and 휇 = −휅 Δr dt2 dt2 f 1 1 1 and 휇 is the reduced mass given by Δr = r2 − r1 − re 휇 = + m1 m2 P. J. Grandinetti Chapter 05: Vibrational Motion Diatomic molecule vibration equations of motion Differential equation of motion describing the vibration

dΔr2(t) 휇 + 휅 Δr(t) = 0 dt2 f Same diﬀerential equation of motion as simple harmonic oscillator. Solutions takes the same form, √ 휔 휔 휅 휇 Δr(t) = Δr(0) cos 0t where 0 = f∕

In spectroscopy vibrational frequencies are given in terms of the spectroscopic wavenumber, √ 휔 휅 ∕휇 ( ) ̃휈 0 f , rearranges to 휅 휇 휋 ̃휈 2 = 휋 = 휋 f = 2 c0 2 c0 2 c0

P. J. Grandinetti Chapter 05: Vibrational Motion Force constants for selected diatomic molecules

휅 휇 −28 ̃휈 −1 Bond f/(N/m) ∕10 kg /cm Bond length/pm H–H 570 8.367664 4401 74.1 D–D 527 16.72247 2990 74.1 H–35Cl 478 16.26652 2886 127.5 H–79Br 408 16.52430 2630 141.4 H–127I 291 16.60347 2230 160.9 35Cl–35Cl 319 290.3357 554 198.8 ( ) 79Br–79Br 240 655.2349 323 228.4 휅 휇 휋 ̃휈 2 f = 2 c0 127I–127I 170 1053.649 213 266.7 16O=16O 1142 132.8009 1556 120.7 14N≡14N 2243 116.2633 2331 109.4 12C≡16O 1857 113.8500 2143 112.8 14N=16O 1550 123.9830 1876 115.1 23Na–35Cl 117 230.3282 378 236.1 39K–35Cl 84 306.0237 278 266.7 23Na–23Na 17 190.8770 158 307.8

P. J. Grandinetti Chapter 05: Vibrational Motion Coupled Harmonic Oscillators and Normal Modes

P. J. Grandinetti Chapter 05: Vibrational Motion Two coupled oscillators constrained to move in one dimension

m m

-x1 0 +x1 -x2 0 +x2

휅 휅 2 equal masses connected with 3 springs with force constants 1 and 2.

3 springs are at equilibrium lengths when x1 = x2 = 0.

Middle spring is also at equilibrium length whenever x2 = x1. 휅 휅 Restoring force acting on left mass: F1 = − 1x1 + 2(x2 − x1) 휅 휅 Restoring force acting on right mass: F2 = − 1x2 − 2(x2 − x1) Coupled diﬀerential equations of motion for each mass:

̈ 휅 휅 휅 ̈ 휅 휅 휅 mx1 + ( 1 + 2)x1 − 2x2 = 0 and mx2 − 2x1 + ( 1 + 2)x2 = 0

P. J. Grandinetti Chapter 05: Vibrational Motion Two coupled oscillators: Normal Modes ̈ 휅 휅 휅 ̈ 휅 휅 휅 Coupled diﬀerential equations: mx1 + ( 1 + 2)x1 − 2x2 = 0 and mx2 − 2x1 + ( 1 + 2)x2 = 0 휉 휉 Uncoupled by coordinate transformation: 1 = x1 + x2 and 2 = x1 − x2 휉 휉 1 and 2 are called normal mode coordinates 휉̈ 휅 휅 휉 휉̈ 휅 휉 Gives 2 uncoupled equations: m 1 + ( 1 + 2 2) 1 = 0 and m 2 + 1 2 = 0 휉 휔 휙 휉 휔 휙 Substitute 2 proposed solutions: 1(t) = A1 cos( 1t + 1) and 2(t) = A2 cos( 2t + 2) ( ) ( ) Obtain: 휅 휅 휔2 휔 휙 and 휅 휔2 휔 휙 ( 1 + 2 2) − m 1 A1 cos( 1t + 1) = 0 1 − m 2 A2 cos( 2t + 2) = 0

Equations true for all values of if: 휅 휅 휔2 and 휅 휔2 t ( 1 + 2 2) = m 1 1 = m 2 √ √ 휔 휅 휅 휔 휅 휔 > 휔 yields 2 normal mode frequencies: 1 = ( 1 + 2 2)∕m and 2 = 1∕m Note: 1 2

Normal mode is collective oscillation where all masses move with same oscillation frequency.

P. J. Grandinetti Chapter 05: Vibrational Motion Two coupled oscillators: Normal Modes

Imposing initial conditions:

휉 휉 , 휉̇ 휉̇ 1(0) = x1(0) + x2(0) and 2(0) = x1(0) − x2(0) and 1(0) = 2(0) = 0 휉 휔 휙 휉 휔 휙 on solutions: 1(t) = A1 cos( 1t + 1) and 2(t) = A2 cos( 2t + 2) gives: 휙 휙 휉 휉 1 = 2 = 0 and A1 = 1(0) and A2 = 2(0) 휉 휉 휔 휉 휉 휔 1(t) = 1(0) cos( 1t) and 2(t) = 2(0) cos( 2t) Converting normal mode coordinates back to original coordinates: [ ] [ ] 1 1 x (t) = 휉 (0) cos(휔 t) + 휉 (0) cos(휔 t) and x (t) = 휉 (0) cos(휔 t) − 휉 (0) cos(휔 t) 1 2 2 2 1 1 2 2 2 2 1 1

Oscillation of each mass is linear combination of two normal mode oscillations.

P. J. Grandinetti Chapter 05: Vibrational Motion Two coupled oscillators: Normal Modes Consider two cases where only one normal mode is active: 휉 휉 2(0) = 0 or 1(0) = 0 1 ⎫ 1 ⎫ x (t) = 휉 (0) cos(휔 t) x (t) = 휉 (0) cos(휔 t) 1 2 1 1 ⎪ 1 2 2 2 ⎪ ⎬ out-of-phase, ⎬ in-phase. x (t) = − 1 휉 (0) cos(휔 t) ⎪ x (t) = 1 휉 (0) cos(휔 t) ⎪ 2 2 1 1 ⎭ 2 2 2 2 ⎭

out of phase normal mode in phase normal mode m m m m

m m m m

All motion in this system can be decomposed into linear combination of these two normal modes. P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis Consider a harder problem

m1 m2

-x1 0 +x1 -x2 0 +x2

Both masses and all 3 spring constants are different. 휉 휉 Simple coordinate transformation 1 = x1 + x2 and 2 = x1 − x2 will not transform problem into uncoupled differential equations. Need systematic approach for determining coordinate transformation (normal modes) that transform coupled linear differential equations into uncoupled differential equations.

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis

m1 m2

-x1 0 +x1 -x2 0 +x2 Deﬁne mass-weighted coordinates: 1∕2 and 1∕2 q1 = m1 x1 q2 = m2 x2 Coupled diﬀerential equations are (휅 + 휅 ) 휅 휅 (휅 − 휅 ) q̈ + 1 2 q − 2 q = 0 and q̈ − 2 q − 3 2 q = 0 1 1 1∕2 2 2 1∕2 1 2 m1 (m2m1) (m2m1) m2 In matrix form: ⎡ (휅 + 휅 ) 휅 ⎤ ⎢ 1 2 − 2 ⎥ ⎡ q̈ ⎤ m (m m )1∕2 ⎡ q ⎤ ⎢ 1 ⎥ ⎢ 1 2 1 ⎥ ⎢ 1 ⎥ ⎢ ⎥ + ⎢ ⎥ ⎢ ⎥ = 0 ⎣ q̈ ⎦ ⎢ 휅 (휅 + 휅 ) ⎥ ⎣ q ⎦ 2 ⎢ − 2 3 2 ⎥ 2 ⏟⏟⏟ 1∕2 ⏟⏟⏟ ⎣ (m2m1) m2 ⎦ q̈ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ q K

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis Any coupled harmonic oscillator problem can be written in form: q̈ + Kq = 0 ̈ K is a square n × n real symmetric (i.e., K12 = K21) matrix, q and q are n × 1 matrices. Similarity transformation of real n × n symmetric matrix into diagonal matrix, 횲. 휆 ⎡ l11 l21 … ln1 ⎤ ⎡ K11 K12 … K1n ⎤ ⎡ l11 l12 … l1n ⎤ ⎡ 1 0 … 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 휆 ⎥ T l12 l22 … ln2 K21 K22 … K2n l21 l22 … l2n 0 2 … 0 횲 L KL = ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ = ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 휆 ⎦ l1n l2n … lnn Kn1 Kn2 … Knn ln1 ln2 … lnn 0 0 … n

▶ 휆 횲 Eigenvalues, i, of K along diagonal of . ▶ T Eigenvectors, li, of K along columns of L and along rows of L . ▶ L is orthogonal matrix: product with its transpose, LT , is identity matrix, 1,

⎡ l11 l12 … l1n ⎤ ⎡ l11 l21 … ln1 ⎤ ⎡ 1 0 … 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T l21 l22 … l2n l12 l22 … ln2 0 1 … 0 LL = ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ = ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ = 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ l l … l ⎦ ⎣ l l … l ⎦ ⎣ 0 0 … 1 ⎦ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟n1 n2 nn ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟1n 2n nn ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ L LT 1

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis Insert identity matrix as 1 = LLT before and after K in matrix form of equations of motion: q̈ + Kq = q̈ + (LLT ) K (LLT ) q = q̈ + L (LT KL)(LT q) = 0 ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ 1 1 횲 Multiplying both sides by LT , regroup, and deﬁne new matrix variables: ⎛ ⎞ ⎜ ⎟ LT ⎜q̈ + L (LT KL) LT q⎟ = LT q̈ + (LT KL) (LT q) = Q̈ + 횲 Q = 0 ⎜ ⏟⏟⏟ ⎟ ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ ⎝ ⎠ 횲 Q̈ 횲 Q

Q = LT q is linear transformation of q into Q, i.e., normal mode coordinates 횲 ̈ 휆 Diﬀerential equations in Q and are uncoupled for each normal mode coordinate: Qi + iQi = 0 휔 휙 휆 휔2 휔 휙 Substitute in proposed solutions: Qi(t) = Ai cos( it + i) gives ( i − ) cos( it + i) = 0 √ i Identify the ith normal mode frequency as 휔 = 휆 i i ( ) 휔 With initial conditions, solutions become Qi(t) = Qi(0) cos it P. J. Grandinetti Chapter 05: Vibrational Motion 휆 How do we determine eigenvalues, i, and the eigenvectors, li, from K? 휆 ith row in L is li eigenvector for i eigenvalue.

⎡ K11 K12 … K1n ⎤ ⎡ li1 ⎤ ⎡ li1 ⎤ ⎡ 1 0 … 0 ⎤ ⎡ li1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ K21 K22 … K2n li2 휆 휆 li2 휆 0 1 … 0 li2 Kli = ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎢ ⋮ ⎥ = ili = i ⎢ ⋮ ⎥ = i ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ K K … K ⎦ ⎣ l ⎦ ⎣ l ⎦ ⎣ 0 0 … 1 ⎦ ⎣ l ⎦ n1 n2 nn in in ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ in 1 ⎛ ⎞ ⎜ ⎟ ⎜⎡ K11 K12 … K1n ⎤ ⎡ 1 0 … 0 ⎤⎟ ⎡ li1 ⎤ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ K21 K22 … K2n 휆 0 1 … 0 li2 ⎜⎢ ⋮ ⋮ ⋱ ⋮ ⎥ − i ⎢ ⋮ ⋮ ⋱ ⋮ ⎥⎟ ⎢ ⋮ ⎥ = 0 ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎜ Kn1 Kn2 … Knn 0 0 … 1 ⎟ lin ⎜⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⎟ ⏟⏟⏟ ⎝ ⎠ K 1 li Compactly written as 휆 (K − i1) li = 0 Encountered same problem when diagonalizing moment of inertia tensor to determine PAS components and orientation. P. J. Grandinetti Chapter 05: Vibrational Motion 휆 How do we determine eigenvalues, i, and the eigenvectors, li, from K?

Obtain eigenvalues by expanding determinant

| 휆 | K − i1 = 0

to get nth order polynomial equation in 휆 whose roots are eigenvalues.

휆 Substitute each eigenvalue back into (K − i1) li = 0 to get corresponding eigenvector.

Determining eigenvalues and eigenvectors of a matrix is usually done numerically. Software packages utilizing various algorithms for matrix diagonalization are available in various problem solving environments such as MatLab, Mathematica, or Python notebooks.

P. J. Grandinetti Chapter 05: Vibrational Motion P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis Back to our original problem

⎡ (휅 + 휅 ) 휅 ⎤ ⎢ 1 2 − 2 ⎥ ⎡ q̈ ⎤ m (m m )1∕2 ⎡ q ⎤ ⎢ 1 ⎥ ⎢ 1 2 1 ⎥ ⎢ 1 ⎥ ⎢ ⎥ + ⎢ ⎥ ⎢ ⎥ = 0 ⎣ q̈ ⎦ ⎢ 휅 (휅 + 휅 ) ⎥ ⎣ q ⎦ 2 ⎢ − 2 3 2 ⎥ 2 ⏟⏟⏟ 1∕2 ⏟⏟⏟ ⎣ (m2m1) m2 ⎦ q̈ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ q K To solve this problem we will 휆 determine eigenvalues, i and eigenvectors,√li, of K 휔 휆 obtain normal mode frequencies from i = i obtain relationship between mass-weighted and normal-mode coordinates from Q = LT q 휅 휅 To make solution more manageable, only illustrate the case where m1 = m2 = m and 3 = 1.

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis - Find the Eigenvalues

Obtain eigenvalues by solving the determinant, | 휅 + 휅 휅 | | 1 2 휆 2 | | − − | ( )( ) 2 | m m | 휅 + 휅 휅 + 휅 휅 | | 1 2 휆 1 2 휆 2 | | = − − − = 0 | 휅 휅 + 휅 | m m m2 | − 2 1 2 − 휆 | | m m | obtaining 휅 + 휅 ± 휅 휔2 = 휆 = 1 2 2 m

휆 휅 휅 휆 휅 1 = ( 1 + 2 2)∕m and 2 = 1∕m

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis - Find the Eigenvectors 휆 휅 휅 Obtain ﬁrst eigenvector, associated with 1 = ( 1 + 2 2)∕m, by expanding ⎡ 휅 휅 휆 휅 ⎤ ⎡ ⎤ ( 1 + 2)∕m − 1 − 2∕m l11 ⎢ ⎥ ⎢ ⎥ , ⎢ ⎥ ⎢ ⎥ = 0 휅 휅 휅 휆 ⎣ − 2∕m ( 1 + 2)∕m − 1 ⎦ ⎣ l12 ⎦ to get 2 identical equations: ( ) ( ) 휅 휅 휆 휅 , 휅 휅 휅 휆 , ( 1 + 2)∕m − 1 l11 − ( 2∕m)l12 = 0 and − ( 2∕m)l11 + ( 1 + 2)∕m − 1 l12 = 0

which simplify to l11 = −l12, i.e. two elements have equal magnitude and opposite sign. [ ] [ ] l11 1 l1 = = b where b is some unknown constant. l12 −1 √ √ Insisting that be normalized, i.e., 2 2 , gives and our 1st eigenvector is: l1 l11 + l12 = 1 b = 1∕ 2 [ ] [ √ ] l11 1∕ √2 l1 = = l12 −1∕ 2

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis - Find the Eigenvectors 휆 휅 Similarly, second eigenvector, associated with 2 = 1∕m, is given by √ ⎡ l ⎤ ⎡ 1∕ 2 ⎤ ⎢ 21 ⎥ ⎢ ⎥ l2 = ⎢ ⎥ = ⎢ √ ⎥ ⎣ l22 ⎦ ⎣ 1∕ 2 ⎦

Taking these two results together, we construct the eigenvector matrix √ √ [ ] ⎡ 1∕ 2 1∕ 2 ⎤ l11 l21 ⎢ ⎥ L = = √ √ l12 l22 ⎢ ⎥ ⎣ −1∕ 2 1∕ 2 ⎦

Finally, we obtain solutions, in terms of the mass-weighted coordinates, as √ √ ⎡ q ⎤ ⎡ 1∕ 2 1∕ 2 ⎤ ⎡ Q (0) cos 휔 t ⎤ ⎢ 1 ⎥ ⎢ ⎥ ⎢ 1 1 ⎥ q = LQ = ⎢ ⎥ = ⎢ √ √ ⎥ ⎢ ⎥ 휔 ⎣ q2 ⎦ ⎣ −1∕ 2 1∕ 2 ⎦ ⎣ Q2(0) cos 2t ⎦

P. J. Grandinetti Chapter 05: Vibrational Motion Web Apps by Paul Falstad

Coupled Oscillators

Web App Link: Coupled Oscillators

P. J. Grandinetti Chapter 05: Vibrational Motion Polyatomic Molecule Vibrations

P. J. Grandinetti Chapter 05: Vibrational Motion Polyatomic Molecule Vibrations Normal mode analysis of molecule with N atoms Define each atom’s displacement coordinates in PAS of molecule’s moment of inertia tensor with center of mass at origin. For example, for water molecule define 휂 , 휂 , 휂 , 1 = aO − aO,e 2 = bO − bO,e 3 = cO − cO,e 휂 = a − a , , 휂 = b − b , , 휂 = c − c , , 4 H1 H1 e 5 H1 H1 e 6 H1 H1 e 휂 = a − a , , 휂 = b − b , , 휂 = c − c , , 7 H2 H2 e 8 H2 H2 e 9 H2 H2 e subscript e represents equilibrium coordinate. 105° b ▶ , , Further define mass-weighted displacement coordinates, q1 … q3N, e.g., 1∕2휂 qi = mi i where , , . a m1 = m2 = m3 = mO m4 = m5 = m6 = mH and m7 = m8 = m9 = mH Derive equations of motion from kinetic and potential energy ( ) 휕 ̇ , , ̇ 휕 , , d K(q1 … q3N) V(q1 … q3N) 휕 ̇ + 휕 = 0 dt q qi

P. J. Grandinetti Chapter 05: Vibrational Motion Polyatomic Molecule Vibrations Kinetic and Potential Energy in mass-weighted coordinates Kinetic energy becomes ( ) [ ] ∑3N 2 1 dq 1 dq dq K = i = q̇ T q̇ where q̇ T = 1 , … , 3N and q̇ is its transpose 2 dt 2 dt dt i=1

Potential energy expanded as Taylor-series expansion ( ) 0 ( ) ∑3N ¨¨* ∑3N ∑3N 휕V(0¨) 1 휕2V(0) V(q) = V(0) + ¨ q + q q + ⋯ ¨휕q i 2 휕q 휕q i k i=1 ¨ i i=1 k=1 i k Rewriting as 1 [ ] V(q) ≈ V(0) + qT  q where qT = q , … , q , and q is its transpose 2 1 3N where  is 3N × 3N matrix with elements given by ( ) 휕2  V(0) i,k = 휕 휕 qi qk P. J. Grandinetti Chapter 05: Vibrational Motion Polyatomic Molecule Vibrations Normal mode analysis of molecule with N atoms Derive equations of motion from kinetic and potential energy ( ) 휕 ̇ , , ̇ 휕 , , d K(q1 … q3N) V(q1 … q3N) 휕 ̇ + 휕 = 0 dt q qi and obtain a set of 3N coupled diﬀerential equations (in matrix notation), q̈ + q = 0 휆  As before, heart of the problem is determining eigenvalues, i, and eigenvectors, li of . Orthogonal eigenvector matrix, L, transforms problem into 3N uncoupled diﬀerential equations, q̈ + q = q̈ + (LLT )(LLT )q = 0 ⟶ LT q̈ + (LT L)(LT q) = Q̈ + 횲Q = 0

▶ 휔 solutions: Qi(t) = Qi(0) cos( it) √ ▶ 휔 휆 normal mode frequencies: i = i ▶ normal mode coordinates: Q = LT q Normal modes with identical frequencies are called degenerate modes.

P. J. Grandinetti Chapter 05: Vibrational Motion Polyatomic Molecule Vibrations Normal mode analysis of molecule with N atoms Since translational and rotational coordinates are included, 5 or 6 normal mode frequencies, depending on whether molecule is linear or not, respectively, will be zero. Potential and kinetic energy in terms of the vibrational normal modes coordinates are

3∑N−6 3∑N−6 1 T 횲 1 휆 2, 1 ̇ T ̇ 1 ̇ 2. V − Ve ≈ Q Q = iQi and K = Q Q = Qi 2 2 i 2 2 i

Keep in mind that approximations were made along the way. ▶ Ignored anharmonic (3rd and higher order) terms ignored in Taylor-series expansion of V(q). Can still use  matrix, but anharmonicities cause small couplings between the “normal modes.” ▶ Coordinate system fixed on molecule, i.e., PAS of moment of inertia tensor, is a rotating frame. We neglected fictitious forces of such a frame. Coriolis forces will cause small couplings between the normal modes. Primary task: obtain  matrix and find its eigenvalues and eigenvectors.

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis of diatomic molecule

b Deﬁne displacements in PAS of its moment of inertia tensor. 휂 , 휂 , 휂 , 1 = aA − aA,e 2 = bA − bA,e = 0 3 = cA − cA,e = 0 휂 , 휂 , 휂 , 4 = aB − aB,e 5 = bB − bB,e = 0 6 = cB − cB,e = 0 a From 6 displacement coordinates, deﬁne mass-weighted coordinates, 1∕2휂 , . qi = mi i, where m1 = m2 = m3 = mA and m4 = m5 = m6 = mB

Since 휂 휂 1∕2 1∕2 we obtain potential energy function Δr = r − re = 4 − 1 = q1∕m1 − q4∕m4 ( ) ( ) 2 2 2 1 1 q q 1 q 2q q q V − V ≈ 휅 (r − r )2 = 휅 1 − 4 = 휅 1 − 1 4 + 4 e f e f 1∕2 1∕2 f 1∕2 2 2 2 m1 (m1m4) m4 m1 m4 Identify non-zero matrix elements of  as ( )  휅 ,  휅 ,  휅 1∕2 11 = f∕m1 and 44 = f∕m4 and 14 = − f∕ m1m4

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis of diatomic molecule Equations of motion in matrix form is ⎡ 휅 휅 ⎤ f 0 0 − f 0 0 ⎢ ( )1∕2 ⎥ ⎡ q̈ ⎤ m1 m m ⎡ q ⎤ ⎢ 1 ⎥ ⎢ 1 4 ⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ q̈ ⎢ ⎥ q ⎢ 2 ⎥ ⎢ 0 0 0 0 0 0 ⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ̈ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ q3 ⎥ 0 0 0 0 0 0 ⎢ q3 ⎥ ⎢ ⎥ + ⎢ ⎥ ⎢ ⎥ = 0 ⎢ 휅 휅 ⎥ ⎢ ̈ ⎥ f f ⎢ ⎥ q4 ⎢ − 0 0 0 0 ⎥ q4 ⎢ ⎥ ( )1∕2 ⎢ ⎥ ⎢ m4 ⎥ ⎢ ⎥ m1m4 ⎢ ⎥ ̈ ⎢ ⎥ ⎢ q5 ⎥ ⎢ q5 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎣ q̈ ⎦ ⎢ ⎥ ⎣ q ⎦ 6 ⎢ ⎥ 6 ⏟⏟⏟ ⎣ 0 0 0 0 0 0 ⎦ ⏟⏟⏟ q̈ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ q  휆 Note: Four equations are already uncoupled and have eigenvalues of i = 0.

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis of diatomic molecule Remaining equations of motion are ( )1∕2 [ ] ⎡ 휅 ∕m −휅 ∕ m m ⎤ [ ] q̈ f 1 f 1 4 q 1 + ⎢ ⎥ 1 = 0 q̈ ⎢ ( ) ⎥ q 4 ⎣ 휅 1∕2 휅 ⎦ 4 − f∕ m1m4 f∕m4

Get eigenvalues from solving ( ) | 1∕2 | | 휅 m 휆 휅 m m | | f∕ 1 − − f∕ 1 4 | ( ) | | = (휅 ∕m − 휆)(휅 ∕m − 휆) − 휅2∕ m m = 0 | ( ) | f 1 f 4 f 1 4 | 휅 1∕2 휅 휆 | | − f∕ m1m4 f∕m4 − | 휅 1 1 1 m + m M gives 휆 and f where 1 4 , or 휇 , = 0 휇 휇 = + = = = (m1m4)∕M m1 m4 m1m4 m1m4 휇 is the reduced mass, and M = m1 + m4 is the total mass. 5 out of 6 eigenvalues are zero—associated with 3 translational and 2 rotational degrees of freedom. √ 휔 휅 휇 1 out of 6 degrees of freedom( is) associated with vibration with 1 = f∕ , and normal mode 휔 solution Q1(t) = Q1(0) cos 1t P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis of diatomic molecule

휆 휅 휇 Turning to normal mode eigenvectors. If we substitute the eigenvalue 1 = f∕ into ( ) ⎡ 휅 휆 휅 1∕2 ⎤ ⎡ ⎤ ⎢ f∕m1 − 1 − f∕ m1m4 ⎥ l11  휆 ⎢ ⎥ , ( − 11)l1 = ⎢ ( ) ⎥ ⎢ ⎥ = 0 ⎢ 휅 1∕2 휅 휆 ⎥ ⎣ l ⎦ ⎣ − f∕ m1m4 f∕m4 − 1 ⎦ 14 ( ) 1∕2 we obtain two equivalent equations, which simplify to l11 = − m4∕m1 l14 Write eigenvector as Same approach gives ( ) ( ) ⎡ ⎤ ⎡ 1∕2 ⎤ ⎡ 휇 1∕2 ⎤ l11 − m4∕m1 ⎢ ∕m1 ⎥ ( ) ⎢ ⎥ ⎢ ⎥ ⟶ ⎡ 휇 1∕2 ⎤ l1 = ⎢ ⎥ = b ⎢ ⎥ l1 = ⎢ ( ) ⎥ ⎢ ∕m4 ⎥ ⎣ l ⎦ ⎣ ⎦ ⎢ 휇 1∕2 ⎥ l = ⎢ ⎥ 14 1 ⎣ − ∕m4 ⎦ 2 ( ) ⎢ 1∕2 ⎥ 휇∕m ( ) ⎣ 1 ⎦ 1∕2 where b = m4∕M after normalization.

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis of diatomic molecule Bring eigenvectors together to create transformation between mass-weighted and normal coordinates: ( ) ( ) ⎡ 휇 1∕2 휇 1∕2 ⎤ ⎡ ⎤ ⎢ − ⎥ ⎡ ⎤ Q1 ⎢ m1 m4 ⎥ q1 T ⎢ ⎥ ⎢ ⎥ Q = L q = ⎢ ⎥ = ⎢ ( ) ( ) ⎥ ⎢ ⎥ ⎣ Q ⎦ ⎢ 휇 1∕2 휇 1∕2 ⎥ ⎣ q ⎦ 4 ⎢ ⎥ 4 ⎣ ⎦ m4 m1

The Q1 coordinate expands to ( ) ( ) 휇 1∕2 1∕2 1∕2 휇 1∕2 휂 휂 휇 1∕2 Q1 = ( ) q1∕m − q4∕m = ( ) 1 − 4 = ( ) Δr ⏟⏟⏟1 ⏟⏟⏟4 ⏟⏟⏟ 휂 휂 1 4 Δr Rearranging gives vibrational motion solution: ( ) 휇 −1∕2 휇 −1∕2 휔 Δr(t) = ( ) Q1(t) = ( ) Q1(0) cos 1t

P. J. Grandinetti Chapter 05: Vibrational Motion Normal mode analysis of diatomic molecule 휆 Q4 normal mode (with 4 = 0) can be expanded as ( ) ( ) ( ) 1∕2 1∕2 1∕2 1 1 Q = (휇) q ∕m + q ∕m = m 휂 + m 휂 = m (a − a , ) + m (a − a , ) 4 1 4 4 1 M1∕2 1 1 4 4 M1∕2 A A A e B B B e Recalling center of mass along the a axis,

aCOM = (mAaA + mBaB)∕M and aCOM,e = (mAaA,e + mBaB,e)∕M = 0 where center of mass when atoms are at equilibrium coordinates is zero, gives ( ) mAaA + mBaB Q = M1∕2 = M1∕2a 4 M COM

휆 ̈ Since 4 = 0, equation of motion is Q4 = 0 ̇ ̇ initial condition of aCOM = Q4 = 0 means aCOM cannot change in time. 휆 i = 0 for 5 normal modes associated with translation and rotation, so no change in center of mass or PAS orientation as molecule vibrates.

P. J. Grandinetti Chapter 05: Vibrational Motion Web Videos

Web Video Link: Vibration of Polyatomic Molecules Web Links: Crash Course Physics: Simple Harmonic Motion

P. J. Grandinetti Chapter 05: Vibrational Motion