ch15 - 1 CHAPTER 15 1 With the perturbing potential given...

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CHAPTER 15 1. With the perturbing potential given, we get C (1 s 2 p ) = eE 0 i h φ 210 | z | φ 100 d te i ϖ t 0 e - γ t where ϖ = ( E 21 E 10 ). The integral yields 1 /( γ - i ϖ ) so that the absolute square of C (1 s 2 p ) is P (1 s 2 p ) = e 2 E 0 2 | φ 210 | z | φ 100 | 2 h 2 ( ϖ 2 + γ 2 ) We may use | φ 210 | z | φ 100 | 2 = 2 15 3 10 a 0 2 to complete the calculation. 2. Here we need to calculate the absolute square of 1 i h d t 0 T e i ϖ 21 t s in ϖ t × 2 a λ dx 0 a s in 2 π x a ( x - a 2 )s in π x a Let us first consider the time integral. We will assume that at t = 0 the system starts in the ground state. The time integral then becomes d te i ϖ 21 t 0 s in ϖ t = 1 2 i d t { e i ( ϖ 21 + ϖ ) t 0 - e i ( ϖ 21 - ϖ ) t } = ϖ ϖ 2 - ϖ 21 2 We have used the fact that an finitely rapidly oscillating function is zero on the average. In the special case that ϖ matches the transition frequency, one must deal with this integral in a more delicate manner. We shall exclude this possibility. The spatial integral involves 2 a dx sin 2 π x a 0 a sin π x a ( x - a 2 ) = 1 a cos π x a - cos 3 π x a 0 a ( x - a 2 ) = 1 a dx d dx a π sin π x a - a 3 π sin 3 π x a ( x - a 2 ) - a π sin π x a - a 3 π sin 3 π x a 0 a = 1 a a 2 π 2 cos π x a - a 2 9 π 2 cos 3 π x a 0 a =- 2 a π 2 8 9 The probability is therefore 1
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P 12 = λ h 2 16 a 9 π 2 2 ϖ 2 ( ϖ 21 2 - ϖ 2 ) 2 (b) The transition from the n = 1 state to the n = 3 state is zero . The reason is that the eigenfunctions for all the odd values of n are all symmetric about x = a /2, while the potential ( x a /2) is antisymmetric about that axis, so that the integral vanishes. In fact, quite generally all transition probabilities (even even) and (odd odd) vanish. (c) The probability goes to zero as ϖ 0. 3. The only change occurs in the absolute square of the time integral. The relevant one is d te i ϖ 21 t -∞ e - t 2 / τ 2 = π e - ϖ 2 τ 2 /4 which has to be squared.
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