Question

Exercise 10.14 A particle is initially in its ground state in an infinite one-dimensional potential box with sides at x = 0 and x a. If the wall of the box at x-a is suddenly moved to x = 10a, calculate the probability of finding the particle in (a) the fourth excited (n = 5) state of the new box and (b) the ninth (n 10) excited state of the new box.

Answer 1

a) old box ground state wave function psi_1 (x) = squareroot 2/a sin(pi x/a) and fourth excite state of new box psi_5 (x) = squareroot 2/10a sin(5 pi x/10a) Now probability = Vertical-bar < psi_5 Vertical-bar psi_1 > Vertical-bar ^2 < psi_5 Vertical-bar psi_1 > = 2/squareroot 10 a integral^a_0 sin(pi x/a) sin(5 pi x/10a) dx = 2/squareroot 10 a integral^a_0 sin(pi x/a) sin(pi x/2a) dx [use 2 sin A sin B = cos(A - B) - cos(A + B)] = 1/a squareroot 10 integral^a_0 {cos pi x/a (1 - 1/2) - cos pi x/a (1 + 1/2)} dx = 1/a squareroot 10 integral^a_0 {cos(pi x/2a) - cos(3 pi x/2a)} dx = 1/a squareroot 10 [2a/pi (sin(pi x/2a))^a_0 - 2a/3 pi (sin(3 pi x/2a))^a_0] = 1/a squareroot 10 [2a/pi + 2a/3 pi] = 1/squareroot 10 2/pi (1 + 1/3) = 8/3 squareroot 10 pi Now probability = (8/3 squareroot 10 pi)^2 = 64/9 times 10 times pi^2 = 0.07205 (b) Now ninth excited state of new box wave function psi_10 (x) = squareroot 2/10a sin