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.

  • 4
    $\begingroup$ Just a hint: that hamiltonian is just the sum of two independent harmonic oscillators. $\endgroup$ Aug 30, 2021 at 11:14
  • $\begingroup$ @DavideMorgante I already defined the Hamiltonian. Do you mean 2D half harmonic oscillator hamiltonian is sum of two 1D half harmonic oscillators? $\endgroup$
    – 147875
    Aug 30, 2021 at 11:45
  • $\begingroup$ why "half" harmonic oscillators? $\endgroup$ Aug 30, 2021 at 14:59
  • $\begingroup$ @ZeroTheHero That was the question asked to me by someone, energy eigenvalues of 2D half harmonic oscillator (truncated) $\endgroup$
    – 147875
    Aug 30, 2021 at 15:06
  • $\begingroup$ @147875 Oh I see now... $x>0,y>0$. and yes it is then a sum of two "half" harmonic oscillators. $\endgroup$ Aug 30, 2021 at 15:11

2 Answers 2


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.

  • 1
    $\begingroup$ Small tip: Keep two dollar sign to enclose the equation, it'll make it look better. I've done it for you now. $\endgroup$ Aug 30, 2021 at 14:36
  • 2
    $\begingroup$ ... and use \left( and \right) to automatically size the parentheses and other delimiters. $\endgroup$ Aug 30, 2021 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.


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