# Energy eigenvalues of isotropic 2D half harmonic oscillator

What are the energy eigenvalues of isotropic 2D half harmonic oscillator?

$$H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{1}{2}m\omega^2 x^2 + \frac{1}{2}m\omega^2 y^2, \quad x > 0, y > 0$$

For 1D half harmonic oscillator,

$$H = \frac{p_x^2}{2m} + \frac{1}{2}m\omega^2 x^2, \quad x > 0$$

using boundary conditions, $$\psi(0) = 0$$, so odd harmonic oscillator wave functions satisfies this conditions.

$$\psi_n(x) = \left( \frac{\alpha}{\sqrt{\pi}} \right)^{1/2} e^{-\alpha x^2/2} H_n(x), \quad n = 1, 3, 5, \dots \\ \quad \alpha^2 = \frac{m\omega}{\hbar}$$

Energy eigenvalues in case of 1D is given by

$$E_n = \left( n + \frac{1}{2} \right)\hbar \omega, \quad n = 1, 3, 5, \dots \\ = \frac{3}{2}\hbar\omega, \frac{7}{2}\hbar\omega, \frac{11}{2}\hbar\omega, \dots$$

But in case of 2D half harmonic oscillator, how do I approach this problem? These type of problems also comes under Sturm-Liouville problem.

• Just a hint: that hamiltonian is just the sum of two independent harmonic oscillators. Aug 30 at 11:14
• @DavideMorgante I already defined the Hamiltonian. Do you mean 2D half harmonic oscillator hamiltonian is sum of two 1D half harmonic oscillators? Aug 30 at 11:45
• why "half" harmonic oscillators? Aug 30 at 14:59
• @ZeroTheHero That was the question asked to me by someone, energy eigenvalues of 2D half harmonic oscillator (truncated) Aug 30 at 15:06
• @147875 Oh I see now... $x>0,y>0$. and yes it is then a sum of two "half" harmonic oscillators. Aug 30 at 15:11

## 2 Answers

What we are essentially doing is, using separation of variables to separate the half harmonic oscillator differential equation into two parts, and then solving them separately.

$$\frac{-\hbar^2}{2m}\nabla_x^2 + \frac{1}{2}m\omega^2 x^2 + \frac{-\hbar^2}{2m}\nabla_y^2 + \frac{1}{2}m\omega^2 y^2)\psi = E\psi$$

Let $$\psi = \psi_x\psi_y$$ and $$E=E_x+E_y$$, and plug this in. You'll get two separated differential equations, that you'll solve individually.

You get the following :

$$\left(\frac{-\hbar^2}{2m}\psi_y\nabla_x^2\psi_x + \frac{1}{2}m\omega^2 x^2 + \frac{-\hbar^2}{2m}\psi_x\nabla_y^2 \psi_y + \frac{1}{2}m\omega^2 y^2\right) = (E_x+E_y)\psi_x\psi_y$$

Divide by $$\psi_x\psi_y$$ on both sides, and you'll obtain

$$\left(\frac{-\hbar^2}{2m}\frac{\psi^{"}_x}{\psi_x} + \frac{1}{2}m\omega^2 x^2 \right)+ \left(\frac{-\hbar^2}{2m}\frac{\psi^{"}_y}{\psi_y} + \frac{1}{2}m\omega^2 y^2\right) = (E_x+E_y)$$

Solve these two equations separately, by solving the $$x$$ part for $$E_x$$ and $$y$$ part for $$E_y$$. You solve this exactly like two individual oscillators, and then add the energy eigenvalues.

You'll find : $$E= (n_x+ \frac{1}{2} +n_y+ \frac{1}{2})\hbar\omega$$, where both $$n_x,n_y$$ are odd.

Try solving the case for 3-d infinite well, and 3-d harmonic oscillators which are isotropic/anisotropic, to get used to this method.

• Small tip: Keep two dollar sign to enclose the equation, it'll make it look better. I've done it for you now. Aug 30 at 14:36
• ... and use \left( and \right) to automatically size the parentheses and other delimiters. Aug 30 at 14:58

Linear partial differential equations $$A\psi=0$$ in two or more variable $$x,y,\dots$$ which can be separated into a sum of differential operators, each of which only acts on a single variable, i.e. $$A(x,y,\dots) = A(x)+A(y)+\dots$$ (beware, this ambiguous usage of the "A" symbol is slightly abusive) can be solved by a product Ansatz $$\psi(x,y,\dots)=\psi(x)\cdot \psi(y)\cdots$$ It turns out, that the most general solution is a superposition of all the possible product solutions.

This should enable you to apply it to the harmonic oscillator, especially for finding the "combined" eigenvalues.