Question

A particle in an infinite square well has the initial wave function: (x,0)- A sin(x/a) (0 S a (a) (b) Determine A Find$(z,t) (Hint: You will need to break up this wavefunction into a superposition of pure states. Use orthogonality to find the coefficients.) (c) Calculate (x). Is it a function of time? (d) Calculate (H).

Answer 1

psi (x, 0) = A sin^3 (pi x/4) because sin 3 theta = 3 sin theta - 4 sin^3 theta implies sin^3 (pi x/a) = 3/4 sin(pi x/4) - 1/4 sin(3 pi x/4) a) so y (x, 0) = A sin^3 pi x/a)) = A[3/4 sin(pi x/4) - 1/4 sin(3 pi x/a)] = A[3/4 it squareroot 2/4 squareroot 9/2 sin(pi x/4) - 1/4 squareroot 2/4 squareroot a/2 sin(3 pi x/4)] = A[3/4 squareroot a/2 y_1 (x) - squareroot a/2 1/4 psi_3 (x)] A squareroot a/2[3/4 y_1 (x) - 1/4 psi_3 (x)] [psi_x (x) = squareroot x/4 sin(n pi 2/4) x - 1, 2, 3 [because integral psi_n psi_m dx = delta_nm Now integral psi^x psi dx = 1 = 1 A^2 a/2(9/4 + 1/16) = 1 implies A = 4/squareroot 5 a so A = 4/squareroot 50 b. because if psi (x_0) = psi (x) implies psi_n (x, t) = psi (x) e^- i E_n t/hwithstroke so psi (x, t_= psi (x, 0) sigma e^- i E_n t/hwithstroke