1
$\begingroup$

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.

$\endgroup$
5
  • 4
    $\begingroup$ Just a hint: that hamiltonian is just the sum of two independent harmonic oscillators. $\endgroup$ Commented 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
    Commented Aug 30, 2021 at 11:45
  • $\begingroup$ why "half" harmonic oscillators? $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented Aug 30, 2021 at 15:11

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
2
  • 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. $\endgroup$ Commented Aug 30, 2021 at 14:36
  • 2
    $\begingroup$ ... and use \left( and \right) to automatically size the parentheses and other delimiters. $\endgroup$ Commented Aug 30, 2021 at 14:58
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.