CHAPTER 8 The Quantum Theory of Motion

Transcription

1 I. Translational motion. CHAPTER 8 The Quantum Theory of Motion A. Single particle in free space, 1-D. 1. Schrodinger eqn H ψ = Eψ! 2 2m d 2 dx 2 ψ = Eψ ; no boundary conditions 2. General solution: ψ = Ae ikx + Be ikx 3. Energy eigenvalues E = k2! 2 2m Note: all values of k are permitted, not just integers. Therefore the energy of a free particle is not quantized. E can take on any (+) value. 4. Note: ψ can also be written as: Since: ψ = Aʹ sin(kx) + Bʹ cos(kx) e iθ = cos θ +i sin θ e iθ = cos θ i sin θ CHAPTER 8 1

2 B. A particle in a one-dimensional box. 1. The potential function V(x). V(x) = 0 for 0 < x < L V(x) = elsewhere 2. Schrodinger equation: ˆ H ψ = Eψ where H = kinetic energy operator + V(x) =!2 2m So we can write: d 2 dx 2 + V(x)! 2 2m d 2 ψ(x) + V(x)ψ(x) = Eψ(x) 2 dx V(x) = 0 in box Above is equivalent to equation: = outside box! 2 2m d 2 ψ(x) dx 2 = Eψ(x) plus boundary condition. Boundary condition: x restricted to interval from 0 to L (in which case V(x) = 0) and ψ is required to vanish at x=0 and x=l. Boundary condition allows us to drop V(x) term. Now, need to find ψ(x) which is a solution of this differential eqn. In other words, find a function ψ such that, if we differentiate with kinetic energy operator we obtain Eψ; that is, a constant times ψ. CHAPTER 8 2

3 Solution: # ψ n = B sin nπx & % $ L ' ( n=1, 2, 3, whole series normaliz. quantum number of solutions coefficient (state number) ψ 1, ψ 2, ψ 3, & associated energy eigenvalues Problem E n = n2 h 2 Let s check to see if this is a solution:! 2 2m d 2 dx 2 ψ n =!2 2m =!2 2m =!2 2m B nπ L 8mL 2 d 2 $ nπx ' B sin 2 & ) dx % L ( d $ nπx ' B cos& ) nπ dx % L ( L ( $ nπx ' ) sin& ) nπ % L ( L = +!2 2m B $ nπ ' $ & ) sin nπx ' & ) % L ( % L ( * =!2 n 2 π 2 - $ nπx ', / B sin +, 2m L 2./ & ) % L ( 2 n 2 h 2 8mL 2 ψ n or = E n ψ n Hψ n = E n ψ n CHAPTER 8 3

4 Interpretation: Lowest energy quantum state: n=1 ψ 1 = 2 L sin πx L ; E 1 = h2 8mL 2 First excited state n=2 ψ 2 = 2 L sin2πx L ; E 2 = 4h2 8mL 2 Energy level diagram: Wavefunctions for n=1 through 5 Limit of large n: ψ *ψ uniform probability (just like classical result) CHAPTER 8 4

5 Correspondence principle: limit (Quantum Results) = Classical results n Orthogonality Condition: If ψ n is a correct, well-constructed solution to the Schrodinger equation, then the wavefunctions are mutually orthogonal: L * dx ψ i ψ j = 1 if i = j 0 and L * dx ψ i ψ j = 0 if i j 0 ψ i is said to be orthogonal to ψ j. Illustration of orthogonality. Here multiply n=1 and n=3 wave functions together and then integrate the product from 0 to L. Should get 0. CHAPTER 8 5

6 Practical Applications (particle in 1-D box): Energy levels of delocalized π electrons in long conjugated molecules: hexatriene C= C -- C = C -- C = C C -- C -- C -- C -- C -- C Delocalized π-bond system Approximately 1-D 6 electrons E n = n2 h 2 8mL 2 L Length where: E n is the energy levels for π electrons m is electron mass L 6 times C-C bond length Spectroscopic transitions in the hexatriene system: n=4 ground state xx xx xx n=3 n=2 n=1 6 electrons fill 3 levels: remember, Pauli Exclusion principle excludes more than 2 e - from occupying a given quantum state. 1st excited state x xx x n=4 n=3 n=2 ΔE = hν, where ν is the frequency of light promoting transition xx n=1 CHAPTER 8 6

7 So can use: Particle-in-box to predict ν-spectrum Pauli Excl princ ΔE=hν C. Particle in a 2-D Box (i.e. a particle confined to a surface) 1. Picture a = length in x-direction b = length in y-direction 2. ˆ H ψ = Eψ H ˆ = 2 $ 2 2m x ' & % y 2 ) partial differential equation now ( Solved by separation of variables technique (see text). 3. E nx n y = h2 " 2 n x 8m a 2 + n 2 % y $ # b 2 ' n x,n y =1,2,3... & 4 ψ nx,n y = ab sin $ n xπx' $ & ) sin n yπy' & ) % a ( % b ( 4. Special case: Energy level diagram when a=b=l (square box) Energy (3,1) (1,3) 10h 2 /8mL 2 (2,2) 8h 2 /8mL 2 (n x,n y ) (2,1) (1,2) 5h 2 /8mL 2 degeneracy (1,1) 2h 2 /8mL 2 CHAPTER 8 7

8 5. Attempt to draw pictures of ψ 2. (1,1) (1,2) (2,1) (2,2) The above is first example of degeneracy = two or more different wave functions (quantum states) having the same energy. e.g., states (1,2) and (2,1) D. Particle in 3-D Box: 1. states (1,3) and (3,1) c a ψ nx,n y,n z = b 8 abc sin $ n xπx' $ & ) sin n yπy' $ & ) sin n zπz' & ) % a ( % b ( % c ( E nx,n y,n z " 2 n x 8m a 2 + n 2 y b 2 + n z $ # = h2 2 c 2 % ' & CHAPTER 8 8

9 2. Energy level diagram for a=b=c (CUBE) II. Vibrational Motion. A. The 1-D Harmonic Oscillator. 1. An important model for bond vibration. V(x) = 1 2 kx2 where k= bond force constant (stiffness) x = displacement of bond length b from equilibrium length b o x = b - b o x = 0 represents bond at its equilibrium distance F=-kx (Force) CHAPTER 8 9

10 2. Classical trajectory. x(t) = A sin(2πνt) (if vibration starts at x=0 moving in (+) direction) A = amplitude of vibration (can be any value) ν = frequency of oscillation ν = 1 2π k µ let s also define ω 2πν = k µ µ = m 1 m 2 m 1 + m 2 reduced mass m 1 m 2 3. Quantum mechanical Schrodinger Eqn. Ĥψ = Eψ Ê K =!2 2µ ˆV = k 2 x2 d 2 dx 2 where Ĥ = Ê K + ˆV as usual S.E. is then: 2 2µ d 2 dx 2 ψ(x) + k 2 x2 ψ(x) = Eψ 4. Stationary Solutions: ψ v (x) vibrational quantum number = v = 0,1,2,3 Energies E v = ( v + 1 2)hν E v = ( v + 1 2) ω = 1 2 ω, 3 2 ω, 5 2 ω (evenly spaced levels) CHAPTER 8 10

11 where: ν = 1 2π k µ OR ω = k µ Wavefunctions: ψ ν ( x) = N v H v (y)e y2 /2 ; where y x α N v = normalization constant H v (y)= the Hermite polynomials and α ( 2 µk )1/4 The Hermite polynomials Ground state ψ o x ( ) = N o e x2 /2α 2 = Gaussian function symmetric function around x=0 (said to be an even function) CHAPTER 8 11

12 ψ o (x) = ψ o ( x) property of even function 1 st excited state 2x ψ 1 ( x) = N 2 1 α e x /2a 2 x times Gaussian an odd function ψ 1( x ) = ψ 1 ( x) ψ 1 is odd because: $ even function' $ odd function' & e x2 /2α % 2 ) & ) ( % x ( ( ) = odd function ψ 2 x % 4x ( ) 2 = N 2 ' & α 2 ( 2 * e x2 /2α 2 ) even power even power even of x of x (even function) (x 0 =1) ψ 2 is an even function CHAPTER 8 12

13 Here are the first 5 vibrational state wavefunctions: Here are the probability density functions ψ * ψ CHAPTER 8 13

14 5. Notes on Parity (symmetry of functions). ψ 0 has even parity ψ 1 has odd parity ψ 2 has even parity ψ 3 has odd parity etc. (even function) x (even function) = even function (even function) x (odd function) = odd function (odd function) x (odd function) = even function Integrals: dx f ( x ) = 2 dx f ( x ) if f(x) is even 0 dx f ( x ) = 0 if f(x) is odd Very useful properties in solving certain integrals. Some functions, however, have neither even or odd parity. Examples: 0 0 Knowing the wavefunctions we can calculate a variety of properties of the harmonic oscillator in any given quantum state v, such as average position <x>, average <x 2 >, the average potential energy <kx 2 /2>. Here < > denotes expectation value. See text for examples. CHAPTER 8 14

15 III. Rotational Motion and Angular Momentum. A. Rotation in Two Dimensions (particle on a ring) 1. Physical picture: particle of mass m constrained to move on a circular path of radius r in the x-y plane. 2. Treat classically first. Energy is all kinetic, so classically E = p 2 /2m Classical angular momentum around the z-axis is J z =+ or -pr So we can write: E = J z 2 2mr 2 = J 2 z ; where I=moment of inertia 2I 3. Apply quantization condition simply using de Broglie equation since and J z = ±pr λ = h p opposite signs denote opposite directions of travel then J z = ± hr λ However, not all wavelengths λ are allowed but only those that exactly repeat themselves around the ring (they satisfy periodic b.c.). CHAPTER 8 15

16 Thus, only wavelengths obeying the following condition will work: λ = 2πr m where m = 0,±1,±2,... (here is the angular J z = m momentum quantum #) So we have the quantization of angular momentum. 4. Now what about the energies: E = J 2 z 2I = m 2I The wavefunctions: eim φ ψ m (φ) = (2π) 1/2 CHAPTER 8 16

17 B. The 2-Particle Rigid Rotor (rotation in 3-D) 1. Physical Picture. masses m 1 and m 2 moment of inertia = I = µd 2 3-dimensional problem central force problem (that is, the 2 particles are held at fixed distance by inwardly directed forces which counteract centrifugal forces.) Importance: good quantum mechanical model for rotation of molecules. 2. Full Coordinate System: 3. Simplifying Features: Since both particles rotate about a fixed center of mass (C.M.), knowledge of the position of 1 particle automatically implies position of 2 nd particle is known. Thus, really a 1-particle problem with an effective mass µ (called the reduced mass) moving on surface of sphere of radius d. µ = m 1 m 2 m 1 + m 2 CHAPTER 8 17

18 So: translational motion of system 2 particle separates as whole (motion of the C.M.) problem into internal (rotational) motion characterized by a single effective particle of mass µ confined to motion on sphere of radius d. this is the part of interest here The rigid rotor problem is mathematically homologous with a single particle of mass m=µ on a sphere of radius r=d. 4. S.E. Still 3-D problem! Ĥψ = Eψ ˆ H = 2 2µ 2 + ˆ V 2 is Laplacian operator: x 2 y 2 z 2 (in Cart. Coord.) V = 0, if the particle remains constrained to the spherical surface. 5. For convenience in central force problems, convert to spherical polar coordinates. (See The Chemist s toolkit 7B.1, page 295, Atkins 10th) x = r sin θ cos φ y = r sin θ sin φ z = r cos θ r fixed (= d) in rigid rotor CHAPTER 8 18

19 2 ψ = 2 ψ r r ψ r + 1 & 1 r 2 ( ' sin 2 θ 2 ψ φ sinθ & ψ) ) ( sinθ + θ ' θ + ** =0 radial portion (kinetic energy of radial motion = 0 since r=d rigid) angular portion ( ) = 1 d 2 Λ2 ψ where r=d Λ orbital angular momentum operator Λ 2 = the legendrian operator ˆ H ψ = Eψ!2 2I Λ2 ψ = Eψ where I = µd 2 = moment of inertia of rotor 6. Solution ψ(θ, φ) (function of the 2 polar angles) ψ(θ, φ) = Y,m ( θ, φ) where Y is special family of functions called the spherical harmonics Quantum state can be specified by 2 quantum numbers:, m = 0, 1, 2 m = -, 0 + orbital angular momentum quantum number magnetic quantum number 7. Now refer to Table 8C.1 in Atkins. (Note: H atom wave functions have these very same functions times a radial function.) 8. Energies of the rotor. ( ) 2 E = + 1 2I depend only on, not m. ( ) 2 CHAPTER 8 E = (m = =1 2I E = 1( 2) 2 2I

20 (m 9. Degeneracy =-2) =2 = (2 + 1) 10. Angular momentum J and J z. =0, m =0 E = 0 J = ( + 1) magnitude of the angular momentum J z = m angular momentum about the z-axis Table 8C.1 The spherical harmonics 11. Artist s rendering of spherical harmonics using m l =0 CHAPTER 8 20

21 CHAPTER 8 21

22 Conversion of notation to the rigid diatomic rotor: Use E J = J( J + 1)!2 2I where J= rotational energy level of rotating diatomic CHAPTER 8 22

23 J=2 M= J=1 J=0 IV. Intrinsic Spin of Microscopic Particles A. Properties 1. Spin emerges when special relativity is applied to quantum mechanics. 2. Microscopic particles possess an intrinsic angular momentum about their axis called spin. 3. This is treated differently than orbital angular momentum because it has different boundary conditions. 4. The magnitude of the spin angular momentum of a particle is determined by the spin quantum number s, and is given by: { s(s +1) } 1/2 5. The value of s for a given type of particle is fixed. It cannot change. For an electron or proton, s=1/2. 6. The component of spin angular momentum along an axis z defined by an external magnetic field is dictated by the spin magnetic quantum number m s. m s can take on values from -s, -s+1, s-1, s For an electron or proton where s=1/2 m s can take on values -1/2 and +1/2 Properties of angular momentum CHAPTER 8 23

24 V. Tunneling. A. Definition: passage of a particle through an energy barrier that exceeds the energy of the particle. i.e. passage through a classically forbidden zone. B. Example. Classically, the particle of energy E trapped in the well will remain trapped forever. In Q.M., it has finite probability of escape. In fact, ψ of the particle has a small contribution through and outside barrier. CHAPTER 8 24

25 ( ) #% κ = spatial decay parameter = 2m V b E '% $ &% 2 ( )% C. Transmission probability e κl 1/2 Note that the larger κ is, the more ψ is damped as it passes through barrier. As κ Transmission Probability κ depends on V b barrier height V b κ Prob E energy of particle E κ Prob m mass of particle m κ Prob D. Exact calculation of transmission probability through square barriers can be obtained by properly piecing together solutions to S.E. in each region. Matching boundary conditions provide us simultaneous eqns to solve to obtain weighting coefficient for each region. CHAPTER 8 25

26 E. Result: Transmission Probability ( ) 2 % ' T = 1+ eκl e κl & ' 16ε 1 ε ( ( ) ) ' * ' + 1 ε = E V where E is the energy of incoming particle, V is barrier height F. Approx. result in limit of high and long barrier. κl >> 1 e κl >> e κl Simplest results: T 16ε ( 1 ε) e 2κL T depends on L, m, E, V CHAPTER 8 26

27 VI. Approximation Techniques. A. Most interesting systems cannot be solved exactly in terms of known analytic functions. Two main approximation approaches are used: 1. Variation theory (will introduce with molecular orbital theory) 2. Perturbation theory (here) B. Time-independent perturbation theory 1. Hamiltonian split into easy part (0-th order part) with known eigenfunctions and eigenvalues, and the complicating part (the perturbation): H ˆ = H ˆ (0) + H ˆ (1) 2. Energy eigenvalues are then represented by a series of corrections, with the 0-th term being the eigenvalue of the non-perturbed Hamiltonian. E = E (0) +E (1) +E (2) Wavefunctions work the same way, with the 0-th order term being the eigenfunctions of the o-th order Hamiltonian. ψ = ψ (0) + ψ (1) + ψ (2) It can be shown that the 1 st order correction to the energy (of the ground state) can be written as: E o (1) = ψ o (o)* ˆ H (1) ψ o (o) dτ Notice that the 1 st order energy correction requires only a knowledge of the 0-th order wavefunction (the unperturbed solution). It is the perturbation Hamiltonian averaged over the unperturbed probability distribution. 5. The 2 nd -order energy correction is a bit more complicated: (1) H 2 no E (2) o = n 0 E (o) (o) o E n where we have introduced the notation often used in QM: * Ω nm = ψ n ˆΩψ m dτ CHAPTER 8 27

28 C. Time-dependent perturbation theory 1. Extremely useful in spectroscopy for calculating transition probability between quantum states due to electromagnetic wave (light). 2. Perturbation Hamiltonian is now time-dependent: H ˆ = H ˆ (0) + H ˆ (1) (t) 3. Let s specifically treat the perturbing effect of an oscillating electric field of frequency ω and amplitude E interacting with a molecule s dipole moment µ z : ˆ H (1) (t) = µ z Ecosωt 4. It can be shown that the rate of transition between two quantum states i and f can be written as: w f i E 2 ψ f * µ z ψ i 2 The term in absolute value brackets is called the transition dipole moment, which is central to the understanding of spectroscopy. It is a measure of the charge redistribution that accompanies a transition. CHAPTER 8 28