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# ps5c - is given by 15 age A dt’eiw21t(11V1(t’)|2(25 n...

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Unformatted text preview: is given by - 15 age) : _% A dt’eiw21t/(11V1(t’)|2) (25) , n : gﬁzjs/ dt’ em”, sinwt’ (26) i 7r 0 8M; t . I . , : (it, eL(u121+w)t __ 61(w21—wﬂ ) (27) 9795/0 ( "5(cuzi‘tw)t _ i(w21~w)t n : 8A: (6. 1_e" 1) (28) 971' h 1(0121 + w) 1(w21 — w) : 16)“); e§(w21+w)t81n(ﬂ1§ﬂt) __ €§(w21-—w)t Sin (2115:3175) (29) 97T2h, U121 + w UJ21 — w where E 2h E — 3 (4)21 = —2—1 ’* 7T (30) h _ 2ma,2 The transition probability at a given it is 256A2a2 sin2 (L get) 277r’1h2 + 2 _ 2 2 (W21 + cu)? (L021 — (0)2 wgl — 0.12 . 2 w 7w t w +0.} , w *w Icz<t>l2— (Lt) W) cm (b) The transition probability between the ground state and the second excited state is zero since (1W1 l3) 2 0. This is easy to see using a symmetry argument. The wave functions corresponding to these two eigenstates are both symmetric under reﬂexion about m = % whereas the perturbation is antisymmetric. Since the calculation of the matrix element involves an integration over a symmetric interval around g- of the product of these three factors, the result must be zero. (0) It is easy to check that the probability calculated in part a) goes to zero as w —> 0. This corresponds to the fact that in the absence of the perturbation there are no transitions. 3. 15.3 ms, t) : /\ (a; — 5%) erg/T2 (32) ' Given that in this case the perturbation goes to zero rapidly as t —> :too we will assume that at t 2 —00 the system is in the ground state and we will be interested in the transition probability for t : 00. The calculation of the matrix elements of the perturbation is the same as in the previous problem, the only diﬁerence in the calculation of the probability is in the time integral. The relevant time integral for this problem is oo -oo / dtl ei'nzglt’e—t’Q/v‘2 = e—m3172/4/ _ dtl 6A(t'7iw2172/2)/-r2 (33) sec —00 : ﬁre—“51”” (34) Therefore the transition probability from the ground state to the ﬁrst excited state is 256A2a272 €_w§172/2 817t3ﬁ2 ' As 7' ~—> 00 the probability goes to zero. Moreover, in this case we see an exponential suppression which is a clear illustration of the adiabatic theorem. P1—>2’ = (35) ...
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