Question

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UESTION 2 A particle of mass m moves in a one- dimensional box of length Lwith boundaries at x-0 and x - L. Thus, V(x) - 0 for 0 x L and V(x) elsewhere. The normalized eigenfunctions of the Hamiltonian for the system are given by 1/2 -| sin 1-_- , with -, where the quantum number 2ml2 n can take on the values n -1, 2, 3, (i). Assuming that the particle is in arn eigenstate Vn(x), calculate the probability

Answer 1

psi_n (x) = (2/L)^1/2 sin n pi x/L i) probability (P) = integral^L/4_0 psi_n (x) psi_n^* (x) since psi_n (s) is real psi_n (x) = psi_n^* (x) so probability (P) = integral^L/4_0 squareroot 2/L sin n pi x/L squareroot 2/L sin n pi x/L dx = (2/L) integral^L/4_0 sin^2 n pi x/L dx = 2/L integral^L/4_0 (1 - cos 2n pi x/L)/2 dx = 2/L times 1/2 integral^L/4_0 1 dx - 2/L times 1/2 integral^L/4_0 cos 2n pi x/L dx = 1/L times [L/4 - 0] - L/2n pi L [sin(2n pi x/L)]_0^L/4 = 1/L times L/4 - 1/2n pi times (sin 2 times n times pi times L/4 times L) - 0 probability (P) = 1/4 - 1/2n pi times sin (n pi/2) so For n = 1, P = 1/4 - 1/2pi = 0.0908 For n = 2, p = 1/4 - 1/4pi times 0 = 0.25 For n = 3, p = 1/4 - 1/6pi times (-1) = 1/4 - 1/6pi = 0.303 For