Solving seperable wavefunction in 2D infintie square well using parity operator

[Qbit42]

Homework Statement

You are given in a earlier stage of this problem that the wavefunction is separable, ie.)

$$\Psi(x,y) = X(x)Y(y)$$

The problem asks you to solve for the wavefunction of a particle trapped in a 2D infinite square well using Parity. ie.) solve

$$\Psi(-x,-y) = \Psi(x,y)$$ and $$\Psi(-x,-y) = -\Psi(x,y)$$.

The potential is defined by $$U(x,y) = 0$$ if $$-a \leq x \leq a$$ and $$-b \leq y \leq b$$ and infinite everywhere else.

Homework Equations

Since the wavefunction is separable I just solved the 1D infinite square potential for both X(x) and Y(y). The results are:

$$X(x) = ACos(\frac{n\pi x}{2a})$$ for n odd and $$X(x) = ASin(\frac{n\pi x}{2a})$$ for n even

and Likewise for Y(y)

My question is can I just combine these two results together to get the parity solutions for the total wavefunction?

$$\Psi(-x,-y) = \Psi(x,y)$$

$$\Psi(x,y) = ACos(\frac{n_{x}\pi x}{2a})Cos(\frac{n_{y}\pi y}{2b})$$ $$n_{x}$$ odd, $$n_{y}$$ odd

$$\Psi(x,y) = ASin(\frac{n_{x}\pi x}{2a})Sin(\frac{n_{y}\pi y}{2b})$$ $$n_{x}$$ even, $$n_{y}$$ even

$$\Psi(-x,-y) = -\Psi(x,y)$$

$$\Psi(x,y) = ACos(\frac{n_{x}\pi x}{2a})Sin(\frac{n_{y}\pi y}{2b})$$ $$n_{x}$$ odd, $$n_{y}$$ even

$$\Psi(x,y) = ASin(\frac{n_{x}\pi x}{2a})Cos(\frac{n_{y}\pi y}{2b})$$ $$n_{x}$$ even, $$n_{y}$$ odd

Is this correct?

 

[fzero]

Those are correct, but you can simplify them by considering n = 2k + 1 when n is odd and n =2k when n is even. It cuts down on the number of separate cases you need to list.