1. The problem statement, all variables and given/known data A particle of mass m, is in an infinite square well of width L, V(x)=0 for 0<x<L, and V(x)=∞, elsewhere. At time t=0,Ψ(x,0) = C[((1+i)/2)*√(2/L)*sin(πx/L) + (1/√L)*sin(2πx/L) in, 0<x<L a) Find C b) Find Ψ(x,t) c) Find <E> as a function of t. d) Find the probability as a function of t, that measurement of energy will yield ħ2π2/2mL2 e) Find <x> as a function of t 2. Relevant equations Nonrelativistic quantum mechanical model for the infinite square well. Currently using the Griffith's textbook "Intro to QM" 3. The attempt at a solution So I'm at part e) and feeling odd about it. For part a) I got C=1 For b) Ψ(x,t) = c1Ψ1(x)e-iE1t/ħ + c2Ψ2(x)e-iE2t/ħ where c1=(1+i)/2 & c2=√2/2. Now for part E, here is what I have so far. <x>=<Ψ(x,t)|xΨ(x,t)> = <c1Ψ1+c2Ψ2|xc1Ψ1+xc2Ψ2>=(1/2)<Ψ1|xΨ1>+(1/2)<Ψ2|xΨ2>+c1*c2<Ψ1|xΨ2>+c2*c1<Ψ2|xΨ1> Assuming all this is fine and dandy my main problem seems to stem from the terms c1*c2<Ψ1|xΨ2>&c2*c1<Ψ2|xΨ1> Ignoring the constants with respect to the integral that comes out of the inner product I think the integral becomes ∫xsin(πx/L)sin(2πx/L)dx Perhaps this is wrong? If it isn't wrong I've been trying to look at the integrand as xsinAsinB then using a trig identity sinAsinB=(1/2)[cos(A-B)-cos(A+B)]. Now I have (1/2)[∫xcos(-πx/l)dx-∫xcos(3πx/L)dx] then of course I take advantage of cos(-x)=cos(x) to get rid of that negative sign in the first integral. At this point, I wonder if I'm alright as far as my steps go. This integral in particular has been giving me fits. I've worked it out and have answers, would be happy to keep sharing steps if so far it's all good.