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Unformatted text preview: Physics 731 Assignment #4, Solutions 1. The odd parity eigenfunctions of the finite square well potential, V = , | x | < a V , | x | > a, (1) for regions I ( x <- a ), region II (- a < x < a ), and region III ( x > a ) take the form ψ I =- Fe κx , ψ II = B sin kx, ψ III ( x ) = Fe- κx , (2) in which k = √ 2 mE/ ¯ h and κ = p 2 m ( V- E ) / ¯ h . Continuity of ψ and dψ/dx at x = a leads to the conditions F = B sin ka e κa (3)- κa = ka cot ka. (4) Defining as usual ξ = ka and η = κa , we have the conditions η =- ξ cot ξ, η 2 + ξ 2 = R 2 , (5) in which R = √ 2 mV a/ ¯ h . Using either graphical or numerical methods, it is straightforward to see that for there are no solutions for R < π/ 2 ( i.e. , V < π 2 ¯ h 2 / (8 ma 2 ) , there is one solution for π/ 2 < R < 3 π/ 2 , and there are two solutions for 3 π/ 2 < R < 5 π/ 2 , and so on. Therefore, as V → , there are no odd parity bound states, and for V → ∞ , there are an infinite number of bound states (as expected), with k = nπ/ (2 a ) for even values of n . Together with Eq. ( 4 ), the normalization condition for the eigenfunctions, Z a- a | B | 2 sin 2 kxdx + 2 Z ∞ a | F | 2 e- 2 κx dx = 1 , (6) yields the following result for B : | B | = 1 p a (1 + 1 / ( κa ) , (7) which reduces to the infinite square well case for V → ∞ . The complete eigenfunctions for the odd parity states thus are given by ψ I =- 1 p a + 1 /κ e κ ( x + a ) , ψ II = 1 p a + 1 /κ sin kx, ψ III ( x ) = 1 p a + 1 /κ e- κ ( x- a ) . (8) 2. We are asked to analyze the bound states of the above finite square well potential with 2 mV a 2 ¯ h 2 ! 1 2 = 2 . (9) The bound states, which have E < V , can be classified by even/odd parity. For the even parity states, the conditions which determines the bound state energies are ξ tan ξ = η, ξ 2 + η 2 = R 2 = 2 mV a 2 ¯ h 2 , (10) 1 where ξ = ka = √ 2 mEa/ ¯ h and η = κa = p 2 m ( V- E ) a/ ¯ h . Here we have R = 2 . To determine the bound state energies, one way to proceed is as we did in class, e.g. plot the intersection of ξ tan ξ and the circle ξ 2 + η 2 = R 2 . Another option is to form the ratio f ( ξ ) = p 4- ξ 2 ξ tan ξ , (11) in which case the allowed bound states are given by...
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- Fall '09
- Trigraph, Error detection and correction, Parity bit, odd parity, Hn, bound state