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

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Unformatted text preview: CHAPTER 15 1. With the perturbing potential given, we get C (1 s → 2 p ) = eE i h 〈 φ 210 | z | φ 100 〉 dte i ϖ t ∞ ∫ 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 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 2 to complete the calculation. 2. Here we need to calculate the absolute square of 1 i h dt T ∫ e i ϖ 21 t sin ϖ t × 2 a λ dx a ∫ sin 2 π x a ( x- a 2 )sin 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 dte i ϖ 21 t ∞ ∫ sin ϖ t = 1 2 i dt { e i ( ϖ 21 + ϖ ) t ∞ ∫- 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 a ∫ sin π x a ( x- a 2 ) = 1 a cos π x a- cos 3 π x a 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 a ∫ = 1 a a 2 π 2 cos π x a- a 2 9 π 2 cos 3 π x a a = - 2 a π 2 8 9 The probability is therefore 1 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....
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This note was uploaded on 03/09/2008 for the course PHYS 401 taught by Professor Mokhtari during the Spring '05 term at UCLA.

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

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