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lecture12to15 - 5.61 F all 2007 Lectures#12-15 page 1 THE...

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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 m 1 m 2 B (magnetic field) A diatomic molecule µ (spin magnetic moment) E (electric field) Classical H.O. m X 0 k X Hooke’s Law: f = k X X 0 ) ≡ − kx ( (restoring force) d 2 x d 2 x k f = ma = m dt 2 = kx dt 2 + m x = 0
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5.61 Fall 2007 Lectures #12-15 page 2 Solve diff. eq.: General solutions are sin and cos functions x t A sin ω t ω = ( ) = ( ) + B cos ( ω t ) k m or can also write as x t ( ) = C sin ( ω t + φ ) where A and B or C and φ are determined by the initial conditions. e.g. x 0 v 0 ( ) = x 0 ( ) = 0 spring is stretched to position x 0 and released at time t = 0. Then x ( ) 0 = A sin 0 ( ) = x 0 ( ) + B cos 0 B = x 0 v 0 = ω cos 0 ( ) ω sin 0 ( ) = 0 A = 0 ( ) = dx dt x = 0 So x t ( ) = x 0 cos ( ω t ) Mass and spring oscillate with frequency: ω = k m and maximum displacement x 0 from equilibrium when cos( ω t)= ±1 Energy of H.O. Kinetic energy K 1 2 1 dx 2 1 2 1 kx 0 2 sin 2 ω t K = mv = m = m ω x 0 sin ω t = 2 2 dt 2 ( ) 2 ( ) Potential energy U f x U = f x dx = kx dx = kx 2 = ( ) = dU ( ) ( ) 1 1 kx 0 2 cos 2 ( ω t ) dx 2 2
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5.61 Fall 2007 Lectures #12-15 page 3 Total energy = K + U = E E = 1 kx 0 2 sin 2 ( ω t ) + cos 2 ( ω t ) 2 1 2 E = kx 0 2 x ( t ) x 0 ( t ) 0 t -x 0 ( t ) 1 2 U K kx 0 E 2 0 t Most real systems near equilibrium can be approximated as H.O. e.g. Diatomic molecular bond A B X U X X 0 A + B separated atoms equilibrium bond length
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5.61 Fall 2007 Lectures #12-15 page 4 U X ( X X 0 ) 2 + 1 d 3 U ( X X 0 ) 3 + ! ( ) = U ( X 0 ) + dU 1 d 2 U ( X X 0 ) + dX 2 dX 2 3! dX 3 X = X 0 X = X 0 X = X 0 Redefine x = X X 0 and U ( X = X 0 ) = U ( x = 0 ) = 0 ( ) = dU 1 d 2 U 1 d 3 U U x x + x 2 + x 3 + ! dx x = 0 2 dx 2 x = 0 3! dx 3 x = 0 U x real potential H.O. approximation dU At eq. = 0 dx x = 0 For small deviations from eq. x 3 << x 2 U x 1 d 2 U 2 1 kx 2 x ( ) 2 dx 2 2 x = 0
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5.61 Fall 2007 Lectures #12-15 page 5 Total energy of molecule in 1D m 1 m 2 X X 1 X 2 X COM x rel M = m 1 + m 2 total mass m 1 m 2 µ = reduced mass m 1 + m 2 X COM m 1 X 1 + m 2 X 2 COM position = m 1 + m 2 x rel = X 2 X 1 x relative position 1 dX 1 2 1 dX 2 2 1 dX COM 2 + 1 dx 2 K = = µ 2 m 1 dt + 2 m 2 dt 2 M dt 2 dt 1 U = kx 2 2 1 dX COM 2 + 1 dx 2 1 E = K + U = µ 2 M dt 2 dt + 2 kx 2 COM coordinate describes translational motion of the molecule 2 1 dX COM 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 = µ E vib 2 dt + 2 kx 2 and solve this problem quantum mechanically.
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