2 dimensional quantum harmonic oscillator and the associated energy levels

Question

What I know so far:

If $V(x,y)=\frac{1}{2}m(\omega_1^2x^2+\omega_2^2y^2)$ where $V$ is the potential in Schrodinger's Equation as usual, then $E=E_n=(n_1+\frac{1}{2})\hbar\omega_1+(n_2+\frac{1}{2})\hbar\omega_2$.

Furthermore, if $w_1=w_2=\omega$ then $E=E_n=(n_1+\frac{1}{2})\hbar\omega+(n_2+\frac{1}{2})\hbar\omega=(n+1)\hbar\omega$ where $n:=n_1+n_2$.

My questions:

1). Say I have $V(x,y)=\frac{1}{2}m(3\omega^2x^2+\omega^2y^2)$, then does it mean that $E=E_n=(n_1+\frac{1}{2})\sqrt3\hbar\omega+(n_2+\frac{1}{2})\hbar\omega$

2). Then do I compile $n_1$ and $n_2$ into $n$ like this? $E=E_n=(n_1+\frac{1}{2})\sqrt3\hbar\omega+(n_2+\frac{1}{2})\hbar\omega=\hbar\omega(\sqrt3 n_1+n_2+\frac{1+\sqrt3}{2})=\hbar\omega(n+\frac{1+\sqrt3}{2})$ where$n:=\sqrt3 n_1+n_2$

3). If that is the case, then wouldn't that imply that for any $n$, $n_1$ has to be $0$ since $n_1,n_2,n$ all have to be non-negative integers?

Answer 1

You must appreciate the x variable system simply "doesn't talk" to the y variable one; that is, it is in a direct product construction with respect to it: the two Hilbert spaces are disjoint; and it is your "capricious" option to add the respective eigenvalues of the respective Hamiltonians. As a result,

1) is absolutely unimpeachable, but contains the meaningless and superfluous symbol $E_n$ for the eigenvalue of the sum Hamiltonian. It should be dropped.

2) is correct, as long as you do not imagine your scaled eigenvalue is an integer : n need not be an integer. Why do you bother introducing it? Surely, you can quickly prove there is no degeneracy in your problem, in sharp contrast to the "what you know" part of your introduction, that is, no two eigenstates may share the same eigenvalue $E$.

3) is a non-seqitur: You just proved in 2) that the pestiferous n takes both integer and irrational values, so what makes you imagine the special $n_1=0$ class is special, or preferred? After all, if you absorbed $\sqrt 3$ into the definition of $\omega$, it would be $n_2=0$ that would yield the integer class of rescaled eigenvalues instead. Surely you don't imagine the sum of two irrationally mutually scaled Hamiltonians needs have integer eigenvalues exclusively...