# Harmonic oscillator in 3-d – rectangular coordinates

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.38.

The 3-d harmonic oscillator can be solved in rectangular coordinates by separation of variables. The Schrödinger equation to be solved for the 3-d harmonic oscillator is

$\displaystyle -\frac{\hbar}{2m}\nabla^{2}\psi+\frac{1}{2}m\omega^{2}(x^{2}+y^{2}+z^{2})\psi=E\psi \ \ \ \ \ (1)$

To use separation of variables we define

$\displaystyle \psi(x,y,z)=\xi(x)\eta(y)\zeta(z) \ \ \ \ \ (2)$

Dividing 1 through by this product we get

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{\xi''}{\xi}+\frac{1}{2}m\omega^{2}x^{2}-\frac{\hbar^{2}}{2m}\frac{\eta''}{\eta}+\frac{1}{2}m\omega^{2}y^{2}-\frac{\hbar^{2}}{2m}\frac{\zeta''}{\zeta}+\frac{1}{2}m\omega^{2}z^{2}=E \ \ \ \ \ (3)$

where the double prime notation indicates the second derivative of a function with respect to its independent variable, so ${\xi''=d^{2}\xi/dx^{2}}$, etc.

We now have three groups of two terms each of which depends on only one of the variables ${x,\, y}$ and ${z}$, and the sum of all these terms is the constant ${E}$. We can therefore use the usual argument that each group of two terms must be a constant on its own, so the 3-d equation reduces to the sum of three 1-d harmonic oscillators. From the analysis of the 1-d harmonic oscillator, we know that each of these will contribute ${(n+1/2)\hbar\omega}$ to the total energy, with the ground state at ${n=0}$. Thus the ground state for the 3-d oscillator will have energy ${3\hbar\omega/2}$, and the general energy level will increase in steps of ${\hbar\omega}$ so the energy levels are given by

$\displaystyle E_{n}=\left(n+\frac{3}{2}\right)\hbar\omega \ \ \ \ \ (4)$

Unlike the 1-d case, the energies of the 3-d oscillator are degenerate. A given value of ${n}$ is composed of the sum of 3 quantum numbers: ${n=n_{x}+n_{y}+n_{z}}$ where all numbers are non-negative integers. Suppose we choose a value for ${n_{x}}$ so that ${n_{y}+n_{z}=n-n_{x}}$. The number of pairs of integers that can be used for ${n_{y}+n_{z}}$ is ${n-n_{x}+1}$ (since ${n_{y}}$ can be anything between 0 and ${n-n_{x}}$). Since ${n_{x}}$ itself can range between 0 and ${n}$, the total number of combinations of quantum states that can make up state ${n}$ is

 $\displaystyle d(n)$ $\displaystyle =$ $\displaystyle \sum_{n_{x}=0}^{n}(n-n_{x}+1)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle (n+1)\sum_{n_{x}=0}^{n}1-\sum_{n_{x}=0}^{n}n_{x}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle (n+1)^{2}-\frac{1}{2}n(n+1)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}(n+1)(n+2) \ \ \ \ \ (8)$

## 7 thoughts on “Harmonic oscillator in 3-d – rectangular coordinates”

1. Bella

Hi,

Great information. I have been looking for something like this. I was wondering tho, how will the solution look like for an harmonic oscillator that has different values for x, y and z? What happens to the energy? The degeneration?

Best regards