5.1 Classical Harmonic Oscillator

Transcription

1 Chapter 5 Harmonic Oscillator 5.1 Classical Harmonic Oscillator m l o l Hooke s Law give the force exerting on the mass as: f = k(l l o ) where l o is the equilibrium length of the spring and k is the force constant Let x = l l o then f = kx From Newton s second law: f = ma = m d2 x dt 2 m d2 x dt 2 = kx m d2 x dt 2 + kx = 0 1

2 We can solve this differential equation to obtain an equation for x as a function of t. Starting with one of the general solution forms discussed in class, x(t) = c 1 sin (ωt) + c 2 cos (ωt) where ω is the angular frequency ω = k m If at t = 0 we stretch the spring to a displacement A, we can write: x(0) = c 2 = A and since the initial velocity is zero, dx dt t=0 = c 1 ω = 0 thus c 1 = 0 c 2 = A x(t) = A cos (ωt) Since force is the negative derivative of potential energy f(x) = dv dx or V (x) = f(x)dx + c Potential Energy for a harmonic oscillator can be written as: V (x) = 1 2 kx2 + c setting V (0) = 0 gives: V (x) = 1 2 kx2 (V for a simple harmonic oscillator - SHO) The Kinetic Energy for a harmonic oscillator is: K = 1 2 mv2 = 1 dx m( 2 dt )2 using our equation for x(t) in K + V, 2

3 K = 1 2 mω2 A 2 sin 2 (ωt) V = 1 2 ka2 cos 2 (ωt) Then the total energy becomes: E = K + V = 1 2 mω2 A 2 sin 2 (ωt) ka2 cos 2 (ωt) using the fact that: ω = k m this reduces to: E = 1 2 ka2 (sin 2 (ωt) + cos 2 (ωt)) = 1 2 ka2 Thus, energy is conserved!! 5.2 Average kinetic & potential energy The period of oscillation for a SHO is: τ = 2π ω The average kinetic energy for one period is: < K >= 1 τ τ 0 1 dx m( 2 dt )2 dt The average potential energy for one period is: < U >= 1 τ τ kx2 dt If we integrate < K > by parts, Let u = 1 2 m dx dt and dv = dx dt dt then du = 1 2 m d2 x dt 2 dt and v = x < K >= 1 τ [ 1 2 m dx dt x τ 0 τ 0 Using the relations, dx dt 0 = dx dt τ 1 2 xm d2 x dt 2 dt] 3

4 x(τ) = x(0) kx = m d2 x dt 2 we obtain < K >= 1 τ τ kx2 dt This is the same as < U > from above!! Thus, for the SHO, we have E =< K > + < U >= 2 < K >= 2 < U > < K >=< U >= E Modeling a diatomic molecule as a SHO m 1 m 2 l x 1 x 2 There is one equation of motion for each mass: m 1 dx 1 dt 2 = k(x 2 x 1 l o ) m 2 dx 2 dt 2 = k(x 2 x 1 l o ) Where l o is the equilibrium length of the spring and k the force constant as before Note that these forces are equal and opposite, which satisfies Newton s third law. We can combine these 2 equations as: 4

5 d 2 dt 2 (m 1 x 1 + m 2 x 2 ) = 0 If we define a center of mass coordinate (x): X = m 1x 1 +m 2 x 2 M where M = m 1 + m 2 The combined second derivative equation becomes: M d2 X dt 2 = 0 Hence, the center of mass has no net force on it. We can also view the motion of the two bodies in terms of a relative coordinate: x = x 1 x 2 l o The two equations of motion can be combined as: d 2 dt 2 (x 2 x 1 = k( 1 m m 2 )(x 2 x 1 l o ) : where 1 m m 2 = m 1+m 2 m 1 m 2 Thus we can write: = 1 µ with µ as the reduced mass µ d2 x dt 2 + kx = 0 We have now reduced the original two-body problem into a two one-body problem, one that involves eimple translational motion of the center of mass, the other that involves harmonic oscillation of the reduced mass µ in a relative coordinate x = x 2 x 1 l o. 5.4 Harmonic Potential Approximation A simple approximation for the potential energy of a vibrating diatomic molecule is the harmonic potential. This parabolic potential is a good approximation near the equilibrium bond length. 5

6 E r o r Consider a Taylor expansion of the inter-nuclear potential V (r) around it s minimum r = r o V (r) = V (r o ) + ( dv ) dr r=r o (r r o ) + 1 ( d2 V ) 2! dr 2 r=ro (r r o ) ( d3 V ) 3! dr 3 r=ro (r r o ) where the minimum V (r o ) is a constant If we expand about r = r o, the second term (which contains ( dv ) dr r=r o ) vanishes because the first derivative ( dv ) is zero at the minimum (r = r dr o). i.e. ( dv dr ) r=r o = 0 Note that dv dr is the restoring force between the nuclei. Because ( dv bond length. dr ) r=r o = 0, there is no net force, and r = r o is our equilibrium If our displacement r r o is small, the (r r o ) n term gets smaller, and smaller... (i.e. it converges), and we can ignore the higher order terms 6

7 If we ignore all terms higher than quadratic, we obtain a harmonic approximation: V (r) V (r o ) ( d2 V dr 2 ) r=ro (r r o ) 2 = C k(r r o) where k = ( d2 V dr 2 ) r=ro Another example of a good approximation to the intermolecular potential energy is the Morse Potential V (r) = D(1 e β(r ro) ) 2 dv = 2D(1 dr e β(r ro) )(βe β(r ro) ) remember for r = r o, ( dv dr ) r=r o = 0 d 2 V = 2D[β 2 e 2β(r ro) (1 e β(r ro) )β 2 e β(r ro) ] dr 2 = 2Dβ 2 [e 2β(r ro) e β(r ro) + e 2β(r ro) ] = 2Dβ 2 [2e 2β(r ro) e β(r ro) ] so ( d2 V dr 2 ) r=ro = 2Dβ 2 = k 5.5 Quantum Harmonic Oscillator Quantum mechanical treatment The Hamiltonian for a 1-d harmonic oscillator is given by: d 2 Ĥ = h2 2µ dx kx2 Inserting this into Schrödinger equation Ĥψ(x) = Eψ(x) leads to: or ( h2 d 2 2µ dx + 1 ) 2 2 kx2 ψ(x) = Eψ(x) 7

8 Let α = d 2 2µ ψ(x) + (E dx2 h 2 12 ) kx2 ψ(x) = 0 (5.1) ( kµ, so 2µ 1 k) = α 2. h 2 h 2 2 The above equation becomes, ( ) d 2 2µE dx ψ(x) + 2 h 2 α 2 x 2 ψ(x) = 0 We guess the solution to be of the form ψ(x) = f(x)e αx2 /2, since, d 2 /2 dx 2 e αx2 = d ( 2αx ) e αx2 /2 dx 2 = αe αx2 /2 + α 2 x 2 e αx2 /2 = ( α 2 x 2 α ) e αx2 /2 Using this, eqn 5.1 becomes: ( 2µE f 2αxf + h 2 The solution is of the form: ) α f = 0 with f(x) = C n x n n=0 f (x) = nc n x n 1 and f (x) = n=1 n (n 1) C n x n 2 The series solution of f(x) leads to a set of polynomial functions called Hermite polynomials, H n (x), where n=2 H 0 = 1, H 1 (x) = 2x, H 2 (x) = 4x 2, H 3 (x) = 8x 3 12x... H n (x) are even for even n and odd for odd n 8

9 5.5.2 Quantized Energies Recall for the Hamiltonian is: Ĥ = p2 x h2 + U(x) = + 1 2µ 2µ dx 2 2 kx2 and the Schrödinger equation is: h2 d 2 ψ(x) + 1 2µ dx 2 2 kx2 ψ(x) = Eψ(x) d 2 The second order differential equation to be solved is then: d 2 ψ(x) dx 2 + 2µ h 2 (E 1 2 kx2 )ψ(x) = 0 There is no simple, general technique to solve this equation. In fact, the only well-behaved, finite solutions for E are quantized (i.e. not continuous): E n = h( k µ ) 1 2 (n ) = hω(n + 1) = hν(n + 1 ) n = 0, 1, where ω = ( k µ ) 1 2 and ν = 1 2π ω Note: These energies are evenly spaced by hω (or hν) n = 3 E n = 2 n = 1 r 9

10 5.5.3 Wavefunctions of Harmonic Oscillator The wavefunctions for the one-dimensional harmonic oscillator are of the form: ψ n (x) = N n H n (α 1 2 x) e αx2 2 where α = ( kµ h 2 ) 1 2, N = 1 ( α) 1 (2 n n!) 1 4 is the normalization constant, 2 π and H n (α 1 2 x) are the Hermite polynomials The first two wavefunctions are: ψ 0 (x) = ( α) 1 4 e αx2 2 π ψ 1 (x) = ( α ) 1 4 (2α 1 2 x)e αx2 2 4π And their probability densities are: ψ 0 (x) 2 = ( α π ) 1 2 e αx2 ψ 1 (x) 2 = ( α π ) 1 2 4αx 2 e αx2 We have not solved the differential equation, but it can be shown that these wavefunctions work by plugging them into the Schrödinger equation. (Try it!) Some facts we know about these wavefunctions: 1. ψ n(x)ψ m (x)dx = ψ nψ m dτ = δ n,m (i.e. the wavefunctions are orthonormal) 2. The energy eigenvalues are non-degenerate (i.e. they have different energies) 3. ψ n has n 1 nodes (places where ψ n = 0) 4. ψ n 2 is symmetric about x = 0 5. ψ n is either symmetric (n even) or antisymmetric (n odd) 10

11 5.5.4 Vibrational Spectrum of Diatomic Molecules The vibrational energy levels for the harmonic oscillator are: E n = hν(n ) = h 2π ( k µ ) 1 2 (n ) n = 0, 1, 2... and µ = m 1m 2 m 1 +m 2 By absorbing or emitting a photon, a diatomic molecule can make an energy transition between levels with a change in energy: E = hν obs We will learn later that transitions can only occur to an adjacent energy level i.e. n = ±1 (we call this a selection rule) Thus, the change in energy for absorbing a photon is: E = E n+1 E n = hω = h 2π ( k µ ) 1 2 = hν obs To get the frequency in units of cm 1 we divide by the speed of light c: ν obs ν obs c = 1 2πc ( k µ ) 1 2 Since E is the same between all adjacent E levels, and therefore for all allowed transitions, the spectrum will consist of only one line whose frequency is called the fundamental vibrational frequency We can determine the force constant for a diatomic molecule from this fundamental vibrational frequency: k = µ (2πc ν obs ) 2 11