Two dimensional Square well and parity

[Firben]

Homework Statement

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?

Homework 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)

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 ?

 

[physicus]

The eigenfunctions you have written down in your relevant equations are those in 1D but we have a 2D system, so an eigenfunction of the full problem has the form
$\psi_{mn}(x,y)=\psi_{m}(x)\psi_{n}(y)$
where m and n label the solutions you have given.
Now, the energy of such a solution is just the sum of the energies in 1D and they are labeled by the two integers m and n. The energies are quadratic in n and m, so the full energy depends on the sum of the squares $n^2+m^2$. The smallest sums you can get this way are 1+1=2, 1+4=5, 4+4=8 and 1+9=10.