1. The problem statement, all variables and given/known data A particle is placed in the potential (a 2 dimensional square well) V(x) = (0 for -a/2 <= x =< a/2 and -a/2 <= y =<a/2, infinity for x>a/2, x<-a/2 and y>a/2, y<-a/2) The hamiltonian commutes with the parity operator P, Pψ(x,y) = ψ(-x,-y) = λψ(x,y), where the eigenvalue λ can take two possible values +(-)1 Write down the eigenstates corresponding to the four lowest energies in such a way that they are also eigenfunctions of the parity operator P. What is the parity of these states? 2. Relevant equations I calculated the eigenfunctions and i got: ψ(x) = √(2/a)*sin(nπx/a), n=2,4,6,.. (odd) ψ(x) = √(2/a)*cos(nπx/a), n=1,3,5 (even) 3. The attempt at a solution E(n,m) = E(n) + E(m) E(n,m) = (π^2*(h-bar)^2)/(2*M*a^2)*(n^2+m^2) In the solution manual it says (n^2+m^2) = 2,5,8,10 How did they come up with those numbers ? the odd is 2,4,6 and the even is 1,3,5, so how can (n^2+m^2) = 2,5,8,10 ? and this also in the solution manual: E(1,1) = one state (n^2+m^2) = 2, odd*odd = even E(1,2)=E(2,1) = two states (n^2+m^2) = 5, odd*even = odd E(2,2) = one state (n^2+m^2) = 8, even*even = even E(1,3) = E(3,1) = two states (n^2+m^2)=10, odd*odd = even What do they mean with E(1,1), E(1,2) etc ?. How can E(1,1) be one state and E(2,2) be two states ? Where did they get odd*even from ?