ps5d - 4 15.4 0 t< O V t 2 36{Axcoswfie‘o‘t t> O...

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Unformatted text preview: 4. 15.4. 0 t < O V t 2 36 ( ) {Axcoswfie‘o‘t t> O ( ) Lets calculate the matrix elements of the perturbation in the eigenstate basis for t > O. (an(t)|m) 7- Acoswlte’atmlflm) (37) _a 71 = Acoswlte t‘/—— 2mw (nKA + .4le) ‘(38) = /\ cos mute—at fig;(\/fi6mm_1 + x/m + 16mm“) (39) l h : )xcosu.’1te_o‘t mh/n +15n’m_1 + fi6mm+fl (40) Born this calculation we can see clearly that7 to first order in the perturbation, the only allowed transitions are for m = n i 1. Now lets apply the first order formula for the transition amplitude. We) : 7; 0 dt'eiwmt’mwe'um) (41) M t . , = —— 1/71 + 16 m_ + 716” m / dt’ 6("‘“’m"_“>t cosw t’ 42 w( 71Y 1 \/— 7 +1) 0 1 ( l : WOOL—l— 6mm—1+\/—5nm+1)/tdtl (6(i(wmn+w1)—a)t’+e(i(wmn—w1l~a)t’> (43) + , . (wmn+w1) oz 1(Wmn—Wl)_ (1(wmn+m) cat 1 (1(wmn7w1kait71 e e «n+16nm_1+fanm+1)(—————— ————a> (44) 2v—< Where wmn = (m ~ n)w : :tw. Lets focus on the simpler case if ——> 00. Then 1 1 c 00 V71 +10 + d ———-—-———-———— + -,—-———-—- 45 m( )2 2 /—~2m _______( nm— 1 \/‘ nm+1)<. '(Wmn +001) a “Lam” —w1)—oz) ( ) 1'me —— a : 1 — n m - 46 W(Vn + 6mm 1+ x/Flci , +1)w% _w3nn _|_ a2 _ 2%}an ( l The transition probability is given by A2 W2 +a2 Pn—vm 2 ((TL + 1)6n‘mAl ’l' n6n,m+l) (47) 2mhw(w1 — £112 + a2)2 + 4w2a2 This expression, taken as a function of £411, clearly shows a resonance pattern achieving its maximum at inf = 0.12 — of. In the limit a -—+ 0 we can see a divergence in the probability for w; = w characteristic of harmonic perturbations. ...
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