Question

A particle is completely confined to one-dimensional region along the x-axis between the points x = ± L The wave function that describes its state is: SP 10 elsewhere where a and b are (as yet) unknown constants that can be expressed in terms of L Use the fact that the wave function must be continuous everywhere to solve for the constant b. The square of the wave function is a probability density, which means that the area under that curve between two positions equals the probability of finding the particle there. Use this fact to solve for the constant a. a. b. c. Compute the probability that the particle will be found in a region of length L centered at the middle of the box

Answer 1

Please like. Thank you

Psi(x) = {a(x^2 - b): -L < x < + L 0, elsewhere At x = plusminus L, psi(plusminus L) = 0 (continuity condition for wave function) a(L^2 - b) = 0 therefore b = L^2 integral psi*(x) psi(x) dx = 1 a^2 integral^+ L_-L (x^2 - L^2)^2 dx = > integral^+ L_-L a^2 (x^4 + L^4 - 2 x^2 L^2) dx = > a^2[x^5/5 + L^4 x - 2 x^3/3 L^2]_-L^+ L = 1 = > 2a^2[L^5/5 + L^5 - 2/3 L^5] = 1 = > 2a^2 (3 + 15 - 10)L^5/15 = 1 16 a^2 L^5/15 = 1 = > a^2 = 15/16 L^5 a = 1/4 L^2 squareroot 15/L probability for -1/2 lessthanorequalto x + 1/2 p = a^2 integral^+ 1/2_-1/2 (x^2 - L^2) dx = a^2 [x^5/5 + L^4 x - 2 x^3 L^2/3]_-1/2^+ 1/2 p = 2a^2 [L^5/2^5 times 5 + L^5/2 - 2/3 L^2 times L^3/8] p = 2 times 15/16 L^5 [L^5/2 (1/80 + 1 - 1/6)] = 0.793 p = 0.793