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

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  4. Bella


    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

  5. Pingback: Harmonic oscillator in 2-d and 3-d, and in polar and spherical coordinates | Physics pages

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