The Harmonic Oscillator in Quantum Mechanics, Lecture notes of Physics

Download The Harmonic Oscillator in Quantum Mechanics and more Lecture notes Physics in PDF only on Docsity! 5.61 Fall 2007 Lectures #12-15 page 1 THE HARMONIC OSCILLATOR • Nearly any system near equilibrium can be approximated as a H.O. • One of a handful of problems that can be solved exactly in quantum mechanics examples m1 m2 B (magnetic field) A diatomic molecule µ (spin magnetic moment) E (electric field) Classical H.O. m X0 k X Hooke’s Law: f = −k X − X0 ) ≡ −kx ( (restoring force) d 2 x d 2 x ⎛ k ⎞f = ma = m dt2 = −kx ⇒ dt2 + ⎝⎜ m⎠⎟ x = 0 5.61 Fall 2007 Lectures #12-15 page 2 Solve diff. eq.: General solutions are sin and cos functions x t Asin ωt ω =( ) = ( ) + Bcos (ωt) k m or can also write as x t( ) = C sin (ωt + φ) where A and B or C and φ are determined by the initial conditions. e.g. x 0 v 0( ) = x0 ( ) = 0 spring is stretched to position x0 and released at time t = 0. Then x ( ) 0 = A sin 0 ( ) = x0 ⇒( ) + B cos 0 B = x0 v 0 = ω cos 0( ) −ω sin 0( ) = 0 ⇒ A = 0( ) = dx dt x=0 So x t( ) = x0 cos (ωt) Mass and spring oscillate with frequency: ω = k m and maximum displacement x0 from equilibrium when cos(ωt)= ±1 Energy of H.O. Kinetic energy ≡ K 1 2 1 ⎛ dx ⎞ 2 1 2 1 kx0 2 sin2 ωtK = mv = m = m −ω x0 sin ωt = 2 2 ⎝⎜ dt ⎠⎟ 2 ⎡⎣ ( )⎤⎦ 2 ( ) Potential energy ≡ U f x ⇒ U = − f x dx = kx dx = kx2 =( ) = − dU ∫ ( ) ∫ ( ) 1 1 kx0 2 cos2 (ωt)dx 2 2 5.61 Fall 2007 Lectures #12-15 page 5 Total energy of molecule in 1D m1 m2 XX1 X2XCOM xrel M = m1 + m2 total mass m1m2µ = reduced mass m1 + m2 XCOM m1 X1 + m2 X2 COM position = m1 + m2 xrel = X2 − X1 ≡ x relative position 1 ⎛ dX1 ⎞ 2 1 ⎛ dX2 ⎞ 2 1 ⎛ dXCOM ⎞ 2 + 1 ⎛ dx ⎞ 2 K = = µ 2 m1 ⎝⎜ dt ⎠⎟ + 2 m2 ⎝⎜ dt ⎠⎟ 2 M ⎝⎜ dt ⎠⎟ 2 ⎝⎜ dt ⎠⎟ 1U = kx2 2 1 ⎛ dXCOM ⎞ 2 + 1 ⎛ dx ⎞ 2 1E = K + U = µ 2 M ⎝⎜ dt ⎠⎟ 2 ⎝⎜ dt ⎠⎟ + 2 kx2 COM coordinate describes translational motion of the molecule 2 1 ⎛ dXCOM ⎞ E = trans 2 M ⎝⎜ dt ⎠⎟ QM description would be free particle or PIB with mass M We’ll concentrate on relative motion (describes vibration) 1 ⎛ dx ⎞ 2 1 = µEvib 2 ⎝⎜ dt ⎠⎟ + 2 kx2 and solve this problem quantum mechanically. ! 5.61 Fall 2007 Lectures #12-15 page 6 THE QUANTUM MECHANICAL HARMONIC OSCILLATOR Ĥψ x ⎡ − !2 d 2 1 ⎤ ⎥ψ ( ) = Eψ ( ) + kx2 x x( ) = ⎢ ⎣ 2m dx2 2 ⎦ K U Note: replace m with µ (reduced mass) if Goal: Find eigenvalues En and eigenfunctions ψn(x) Rewrite as: m1 m2 d 2ψ x( ) + 2m dx2 !2 ⎡ ⎢ ⎣ E − 1 2 kx2 ⎤ ⎥ ⎦ ψ x( ) = 0 This is not a constant, as it was for P-I-B, so sin and cos functions won’t work. TRY: f ( ) = e−α x2 2 (gaussian function) x d 2 f x 2 2 −α x2 2 ( ) + α 2 x ( ) = −αe−α x 2 + α 2 x e = −α f x x2 f ( ) dx2 d 2 f x( ) + α f x x2 f xor rewriting, ( ) −α 2 ( ) = 0w dx2 which matches our original diff. eq. if 2mE mk α = and α 2 = !2 !2 ∴ E = 2 k m 5.61 Fall 2007 Lectures #12-15 page 7 We have found one eigenvalue and eigenfunction m k 1 or ν =Recall ω = 2π k m 1 1 ∴ E = !ω = hν 2 2 This turns out to be the lowest energy: the “ground” state For the wavefunction, we need to normalize: where N is the normalization constant ψ x x 2 2( ) = Nf ( ) = Ne−α x 1 4 ∞ N 2 −α x∫−∞ ψ ( ) x 2 dx = 1 ⇒ ∫−∞ ∞ e 2 = 1 ⇒ N = ⎝⎜ ⎛ α π ⎠⎟ ⎞ π α ψ 0 ( ) = ⎝⎜ ⎛ α ⎞ 1 4 −α x2 2x π ⎠⎟ e x∴ ψ 0 ( ) 11 1 E0 = !ωE0 = !ω = hν 22 2 x Note ψ 0 x x( ) is symmetric. It is an even function: ψ 0 ( ) = ψ 0 (−x) There are no nodes, & the most likely value for the oscillator displacement is 0. So far we have just one eigenvalue and eigenstate. What about the others? 5.61 Fall 2007 Lectures #12-15 page 10 Symmetry properties of ψ’s ψ 0,2,4,6,.... are even functions ψ −x) = ψ ( ) ( x ψ 1,3,5,7,.... are odd functions ψ −x) = −ψ ( ) ( x Useful properties: (even) ⋅(even) = even (odd) ⋅(odd) = even (odd) ⋅(even) = odd d (odd) = (even) d (even) = (odd)dx dx ∫ ∞ (odd) dx = 0 ∫ ∞ (even) dx = 2∫ ∞ (even) dx −∞ −∞ 0 Just from symmetry: ∞ ∞ ⎛ d ⎞ ( ) ∗ x n = ∫−∞ ψ n ( ) xψ n ( ) dx = 0x x p = ∫ ψ ∗ ⎝⎜ −ih ⎠⎟ ψ n x dx = 0 n −∞ n dx odd odd Average displacement & average momentum = 0 IR spectroscopy ⇒ H.O. selection rules Intensity of vibrational absorption features n’ = 1 Vibrational transition hν n = 0 δ+ δ- 5.61 Fall 2007 Lectures #12-15 page 11 2 Intensity Inn′ ∝ dµ ∞ ∫−∞ ψ n ∗ xψ n ' dx dx 1) Dipole moment of molecule must change as molecule vibrates ⇒ HCl can absorb IR radiation, but N2, O2, H2 cannot. 2) Only transitions with n′ = n ± 1 allowed (selection rule). (Prove for homework.) QUANTUM MECHANICAL HARMONIC OSCILLATOR & TUNNELING Classical turning points Classical H.O.: Total energy ET = 1 kx0 2 2 EToscillates between K and U. Maximum displacement x0 occurs when all the energy is potential. E x -x0 x0 x0 = 2ET k is the “classical turning point” The classical oscillator with energy ET can never exceed this displacement, since if it did it would have more potential energy than the total energy. 212xψ 212kx 5.61 Fall 2007 Lectures #12-15 page 12 Quantum Mechanical Harmonic Oscillator. x( ) 2 x( ) 2 x( ) 2 x( ) 2 x ψ 3 ψ 2 At high n, probability ψ 1 density begins to look classical, peaking at turning points. ψ 0 ψ 12 x( ) 2 () 1 2 kx2 xψ 3 ( ) 2 ψ 2 ψ 1 ψ 0 x( ) 2 x( ) 2 x( ) 2 Non-zero probability at x > x0! Prob. of (x > x0, x < -x0): 1 ∞ 2∫ ψ 0 2 ( ) x dx = 2 ⎝⎜ ⎛ α π ⎞ 2 ∞ −α x2 e dx α−1 2 α−1 2 ⎠⎟ ∫ = 2 π 1 2 ∞ e− y2 dy = erfc 1∫1 ( ) “Complementary error function” tabulated or calculated numerically Prob. of (x > x0, x < -x0) = erfc(1) = 0.16 Significant probability! x ()01costψψω 20xψ 21xψ ()012costψψω 5.61 Fall 2007 Lectures #12-15 page 15 ωt = 0, π 4, π 2, 3π 4, π Ψ (x,t) 2 x ψ 1 x( ) 2 () ψ 0 x( ) 2 () x 2ψ 0ψ1 cos (ωt) ωt = 0, π 4, π 2, 3π 4, π What happens to the expectation value <x>? 5.61 Fall 2007 Lectures #12-15 page 16 ∞ Ψ∗ x = ∫−∞ (x,t) x̂Ψ (x,t) dx 1 ∞ = ∫−∞ ⎣ ⎡ψ 0 ∗ x iω0t + ψ 1 ∗ ( )eiω1t ⎤ ⎣⎡ψ 0 ( )e− iω0t x − iω1t ⎦⎤dx( )e x ⎦ x x + ψ 1 ( )e2 ∞ ∞ ∞ ∞ = 1 ⎣⎢ ⎡∫−∞ ψ 0 ∗ xψ 0 dx + ∫−∞ ψ 1 ∗ xψ 1dx + ∫−∞ ψ 1 ∗ xψ 0 e i(ω1 −ω0 )t dx + ∫−∞ ψ 0 ∗ xψ 1e − i(ω1 −ω0 )t dx ⎦⎥ ⎤ 2 <x>0 = 0 <x>1 = 0 = cos (ω vibt) ∫ ∞ ψ 0 xψ 1dx −∞ <x>(t) oscillates at the vibrational frequency, like the classical H.O.! ∞ Vibrational amplitude is ∫ ψ 0 xψ 1dx −∞ 1 1 2 2 ψ 0 x ⎝⎜ π ⎠⎟ 1 2 ⎝⎜ ⎛ α π ⎠⎟ ⎞ 4 (2α1 2 x)e−α x 2e( ) = ⎛ α ⎞ 4 −α x 2 ψ 1 ( ) x = 1 ⇒ xψ 0 ( ) = ⎝⎜ ⎛ α π ⎠⎟ ⎞ 4 xe−α x2 2 = (2α )−1 2 xx ψ 1 ( ) ∞ )−1 2 ∞ )−1 2 2∴ dx = dx = x ( ) t = (2α )−1 2 cos (ω vibt)∫−∞ ψ 0 xψ 1 (2α ∫−∞ ψ 0 (2α Relations among Hermite polynomials Recall H.O. wavefunctions 1 2 ψ n x 1 1/2 ⎝⎜ ⎛ α π ⎠⎟ ⎞ 4 H (α1 2 x)e−α x 2 n = 0,1,2,... ( ) = n 2n n!( ) Normalization Gaussian Hermite polynomial ! ! 5.61 Fall 2007 Lectures #12-15 page 17 y even n = 0 H1 ( ) = 2 y ( ) H0 ( ) = 1 ( ) y odd n = 1 H2 ( ) = 4 y ( )y 2 − 2 even n = 2 H3 ( ) = 8y ( )y 3 − 12 y odd n = 3 H4 ( ) = 16 y ( )y 4 − 48y2 + 12 even n = 4 Generating formula for all the Hn: H y −1 ey2 en ( ) = ( )n d n − y2 dyn A useful derivative formula is: n ( ) = −1 2 yey2 e− y2 + −1 ey2dH y d n d n+1 e− y2 = 2 yH ( ) − H ( ) ( )n ( )n dy dyn dyn+1 n y n+1 y Another useful relation among the Hn’s is the recursion formula: ( ) − 2 yH ( ) + 2nH ( ) = 0Hn+1 y n y n−1 y Substituting 2 yH y ( ) above gives ( ) = H ( ) + 2nH n y n+1 n−1 y dH y dy = 2nHn−1 yn ( ) ( ) Use these relations to solve for momentum <p>(t) p = ∞ Ψ∗ (x,t) p̂Ψ (x,t) dx∫−∞ = 1 ∞ ⎡ ∗ ( )eiω0t ∗ x iω1t ⎤⎦ ˆ ⎡ψ 0 ( )e− iω0t x − iω1t ⎤ ∫−∞ ⎣ψ 0 x + ψ 1 ( )e p ⎣ x + ψ 1 ( )e ⎦ dx 2 ∞ ∞ ∞ ∞ = 2 ⎣⎢∫−∞ ψ 0 pψ 0 ∫−∞ ψ 1 pψ 1 ∫−∞ ψ 1 pψ 0 ∫−∞ ψ 0 pψ 1 ⎦⎥ 1 ⎡ ∗ ˆ dx + ∗ ˆ dx + ∗ ˆ ei(ω1 −ω0 )t dx + ∗ ˆ e − i(ω1 −ω0 )t dx⎤ <p>0 = 0 <p>1 = 0