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# p09fl13 - Lecture 13 Action constraints(2 Oct 09 A Action...

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Lecture 13: Action, constraints (2 Oct 09) A. Action integral exs 1. Evaluate S = Z t 2 t 1 L [ q, ˙ q ] dt subject to q ( t 1 ) = q 1 ; q ( t 2 ) = q 2 for 3 potentials. 2. Free particle in 1D ¨ x = 0 x ( t ) = x 1 + ˙ x 1 ( t - t 1 ) and set the initial velocity by the match at t 2 : ˙ x 1 = ( x 2 - x 1 ) / ( t 2 - t 1 ) S = Z dt m ˙ x 2 / 2 = [ m ˙ x 2 1 / 2] Z dt = m ( x 2 - x 1 ) 2 2( t 2 - t 1 ) 3. Particle in 1D with constant force. x ( t 1 ) x 1 ; x ( t 2 ) x 2 Φ = - fx m ¨ x = f x ( t ) = x 1 + v 1 ( t - t 1 ) + f ( t - t 1 ) 2 / 2 x 2 - x 1 = v 1 ( t 2 - t 1 ) + ( f/ 2)( t 2 - t 1 ) 2 v 1 = x 2 - x 1 t 2 - t 1 - ( f/ 2)( t 2 - t 1 ) Calculate the action (let t 2 - t 1 t ). Algebra leads to: S = Z t 2 t 1 [ 1 2 m ˙ x 2 + fx ] dt = 1 2 t ( x 2 - x 1 ) 2 + 1 2 ft ( x 2 + x 1 ) - t 3 24 f 2 4. Particle in 1D oscillator potential, angular frequency ω ; measure times from t 1 = 0 and define τ = t 2 : x ( t ) = 1 sin ωτ [ x 1 sin ω ( τ - t ) + x 2 sin ωt ] S = 2 sin ωτ [( x 2 1 + x 2 2 ) cos ωτ - 2 x 1 x 2 ] 5. These are classical mechanics starting points in Feynman and Hibbs, “Quantum Mechanics and Path Integrals.” 1

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B.path integral in 1D 1. Solve for propagator -
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p09fl13 - Lecture 13 Action constraints(2 Oct 09 A Action...

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