Imagine a mass $m$ anchored in a position by four springs of spring constant $k/2$ (being pushed to the center by the springs)
Or we can think of a mass anchored to a single fixed point (the origin) by a spring of spring constant $k$ and zero natural length (potential energy is zero when the mass is at the origin) (being pulled to the center by the spring)
We can show the potential function is
$$ \begin{aligned} V(x,y)&=\frac 12 k(x^2+y^2) \\ &=\frac 12 m\omega^2 (x^2+y^2)
\end{aligned} $$
The kinetic energy is
$$ T=\frac{\vec{p}^2}{2m}=\frac{p_x^2+p_y^2}{2m} $$
The corresponding quantum operator is
$$ \widehat{T}=\frac{\widehat{p}_x^2+\widehat{p}_y^2}{2m}=\frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) $$
Now we can write the TISE
$$ \frac{-\hbar^2}{2m}\left(\frac{\partial^2\psi(x,y)}{\partial x^2}+\frac{\partial^2\psi(x,y)}{\partial y^2}\right)+\frac12m\omega^2\left(x^2+y^2\right)\psi(x,y)=E\,\psi(x,y) $$
the wavefunction is two dimensional $\psi(x,y)$ and gives a two dimension probability distribution $|\psi(x,y)|^2$ where the probability of it being at a point is $|\psi(x,y)|^2\,\text dx\,\text dy$
We can write the Hamiltonian as
$$ \widehat{H}=\widehat{H}_x+\widehat{H}_y $$
where
$$ \begin{aligned} \widehat{H}_x &= \frac{-\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2} +\frac12m\omega^2\,x^2 \\ \widehat{H}_y &= \frac{-\hbar^2}{2m}\,\frac{\partial^2}{\partial y^2} +\frac12m\omega^2\,y^2\end{aligned} $$
The eigenfunction of the Hamiltonian can be written as the product functions of the eigenfunctions of $\widehat{H}_x$ and $\widehat{H}_y$
$$ \begin{aligned} \widehat{H}x\psi{n_x}(x) &= \left(n_x+{\textstyle{\frac12}}\right)\hbar\omega\,\psi_{n_x}(x) \\ \widehat{H}y\psi{n_y}(y) &= \left(n_y+{\textstyle{\frac12}}\right)\hbar\omega\,\psi_{n_y}(y)\end{aligned} $$
where $n_x,n_y\in\N$
We can show the product of the eigenfunctions of $\widehat{H}_x$ and $\widehat{H}_y$ are eigenfunctions of $\widehat{H}$
$$ \begin{aligned} \widehat{H}\Bigl(\psi_{n_x}(x)\psi_{n_y}(y)\Bigr) &= \Bigl(\widehat{H}x+\widehat{H}y\Bigr)\Bigl(\psi{n_x}(x)\psi{n_y}(y)\Bigr) \\ &= \Bigl(\widehat{H}x\psi{n_x}(x)\Bigr)\psi_{n_y}(y)+\psi_{n_x}(x)\Bigl(\widehat{H}y\psi{n_y}(y)\Bigr) \\ &= \Bigl(n_x+{\textstyle{\frac12}}\Bigr)\hbar\omega\,\psi_{n_x}(x)\psi_{n_y}(y)+\psi_{n_x}(x)\Bigl(n_y+{\textstyle{\frac12}}\Bigr)\hbar\omega\,\psi_{n_y}(y) \\ &= \Bigl(n_x+n_y+1\Bigr)\hbar\omega\,\psi_{n_x}(x)\psi_{n_y}(y)\end{aligned} $$
Thus the eigenvalues of $\widehat H$ which are the energies of the stationary states are
$$ E_{n_x,n_y}=\Bigl(n_x+n_y+1\Bigr)\hbar\omega $$
There are more than one state with the same energy
| $E_{n_x,n_y}$ | $\hbar\omega$ | $2\hbar \omega$ | $3\hbar\omega$ | |||
|---|---|---|---|---|---|---|
| $n_y$ | 0 | 1 | 0 | 2 | 1 | 0 |
| $n_x$ | 0 | 0 | 1 | 0 | 1 | 2 |
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/dcbf9043-9b67-4d37-93d3-976fbd677b89/degeneracies.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/dcbf9043-9b67-4d37-93d3-976fbd677b89/degeneracies.png" width="40px" /> Degeneracy: is the fact that different eigenfunctions of an operator can have the same eigenvalue. It is most commonly applied to the energy, where it is said that a given energy level is degenerate if there is more than one state that has that energy. Any linear combination of those states is also an eigenstate with the same energy. (example $3\hbar\omega$ has 3 energy levels)
</aside>
🗒️ Note: if $\psi_{10}(x,y)$ and $\psi_{01}(x,y)$ have the same energy $E=2\hbar\omega$ then any linear combination of them $a\psi_{10}+b\psi_{01}$ also have the same energy
$$ \widehat{H}\Bigl(a\psi_{10}+b\psi_{01}\Bigr)=2\hbar\omega\Bigl(a\psi_{10}+b\psi_{01}\Bigr) $$
In particular if $\psi_{10}$ and $\psi_{01}$ are normalized then
$$ \psi_\pm=\frac1{\sqrt2}\psi_{10}\pm\frac1{\sqrt2}\psi_{01} $$
is also normalized
We will look at the case of a central potentials
