HW14_prob2_GPS35_G33mod

HW14_prob2_GPS35_G33mod - Goldstein 9-33 (3rd ed. 9.35)...

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Goldstein 9-33 (3 rd ed. 9.35) plus an additional part (b) (a) The Hamiltonian for this one dimensional system is H = p 2 2 m + mk x 2 . (1) We want to f nd y ( t ) ,wh e r e y is the quantity x 2 , using the Poisson bracket operator formalism. The formal solution is y ( t )= y 0 +( H y ) 0 t + 1 2 ¡ H 2 y ¢ 0 t 2 + 1 3! ¡ H 3 y ¢ 0 t 3 + ··· , (2) where the action of the operator H on y is de f ned as H y =[ y,H ] . (3) With y = x 2 , the explicit form of Eq.(3) is H x 2 =[ x 2 ,H ]= 1 2 m [ x 2 ,p 2 ]= 2 xp m . (4) The next term in the series is H 2 x 2 = H ¡ H x 2 ¢ =[ H x 2 ,H ]= 2 m [ xp,H ] . (5) Proceeding to evaluate this we f rst note that [ xp,H ]= x [ p, H ]+ p [ x, H ]= x [ p, 1 x 2 ] mk + p 2 m [ x,p 2 ] , (6) andthenthat [ p, 1 x 2 ]= x 1 x 2 = 2 x 3 , (7) so that [ xp,H ]= 2 mk x 2 + p 2 m . (8) If we compare this result with Eq.(1) we see that [ xp, H ]=2 H, (9) and Eq.(5) results in H 2 x 2 = 4 m H. (10) It should be clear from this result that H n x 2 =0 for n 3
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HW14_prob2_GPS35_G33mod - Goldstein 9-33 (3rd ed. 9.35)...

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