Ok,so check this situation out. We have a one-dimensional box with walls at (-a/2,a/2). We know that the particle is in a state with energy probabilities P(E1)=1/3, P(E2)=1/3, and P(E3)=1/3 while P(En)=0 for all n not equal to 1,2,3. The parity is measured ideally and -1 is found. If some time later E is measured, what value is found? What is the answer if the original measurement found the parity to be 1? I don't understand how if the parity of the state is measured ideally that -1 is found, since we know Psi(x,0)= (square root of (2/3a))(cos(n(pi)x/a) + sin(2n(pi)x/a) + cos(3n(pi)x/a) And when we take P, where P is the parity operator of our function, we note that only the sin function will become negative. solving for <P> then, <P>= 1/3 + (-1/3) + 1/3=1/3, correct? This does not equal 1. Anyway, given the initial state, now I am pretty sure that Psi(x,t)= (square root of (2/3a))(cos(n(pi)x/a)e^-iE1t/hbar + sin(2n(pi)x/a)e^-iE2t/hbar + cos(3n(pi)x/a)e^-iE3t/hbar) How do I determine energy from this? And then how will the energy change if the original parity happened to be -1?