# The allowed energies of 3D harmonic oscillator [closed]

I'm trying to calculate the allowed energies of each state for 3D harmonic oscillator.

$$E_n = (n_x+\textstyle\frac{1}{2})\hbar \omega_x+ (n_y+\textstyle\frac{1}{2})\hbar\omega_y+ (n_z+\textstyle\frac{1}{2})\hbar\omega_z$$
with $$n_x,n_y,n_z = 0,1,2,\ldots$$.

Unfortunately I didn't find this topic in my textbook.

Can somebody help me?

## closed as unclear what you're asking by Emilio Pisanty, Buzz, ZeroTheHero, Kyle Kanos, John RennieDec 12 '18 at 12:55

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• Who is the author of your textbook? Most QM textbooks have a robust section on 3D basic problems such as particle in a box and HO before moving on to the hydrogen atom problem. – Bill N Dec 11 '18 at 19:57
• What are you trying to calculate exactly? You already have an expression for $E_n$... – ZeroTheHero Dec 11 '18 at 20:20

There are three steps to understanding the 3-dimensional SHO.

1) Make sure you understand the 1D SHO. This will be in any quantum mechanics textbook. You should understand that if you have an equation that looks like

$$Ef(x) = -\frac{\hbar^2}{2m}\partial_x^2 f(x)+\frac{1}{2}m\omega^2x^2f(x)$$ then the solutions for the energies are $$E_n=\hbar\omega(n+\frac{1}{2})$$, with corresponding eigenfunctions $$f_n(x)=\frac{1}{\sqrt{2^n n!}}(\frac{m\omega}{\pi\hbar})^{\frac 1 4}e^{-\frac{m\omega x^2}{2\hbar}}H_n(\sqrt{\frac{m\omega}{\hbar}}x)$$.

2) Write down the Hamiltonian for the 3D SHO:

$$E\Psi(x,y,z) = -\frac{\hbar^2}{2m}\partial_x^2 \Psi(x,y,z)-\frac{\hbar^2}{2m}\partial_y^2 \Psi(x,y,z)-\frac{\hbar^2}{2m}\partial_z^2 \Psi(x,y,z)+\frac{1}{2}m\omega_x^2x^2\Psi(x,y,z)+\frac{1}{2}m\omega_y^2y^2\Psi(x,y,z)+\frac{1}{2}m\omega_z^2z^2\Psi(x,y,z)$$ Plug in the following separation of variables guess: $$\Psi(x,y,z)=X(x)Y(y)Z(z)$$, where $$X,Y,Z$$ are unknown functions to be determined. You should find that the above equation for $$\Psi$$ reduces to three decoupled equations for $$X,Y,Z$$:

$$E_xX(x) = -\frac{\hbar^2}{2m}\partial_x^2 X(x)+\frac{1}{2}m\omega^2_xx^2X(x)$$ $$E_yY(x) = -\frac{\hbar^2}{2m}\partial_y^2 Y(y)+\frac{1}{2}m\omega^2_yy^2Y(y)$$ $$E_zZ(z) = -\frac{\hbar^2}{2m}\partial_z^2 Z(z)+\frac{1}{2}m\omega^2_zz^2Z(z)$$ with the additional constraint that $$E=E_x+E_y+E_z$$.

3) Use your knowledge from (1) to solve the equations for $$X,Y,Z$$. Each of the three equations above is EXACTLY the equation for a 1D SHO, so we can immediately write down the allowed energies:

$$E_{xn_x}=\hbar\omega_x(n_x+\frac{1}{2})\qquad E_{yn_y}=\hbar\omega_y(n_y+\frac{1}{2})\qquad E_{zn_z}=\hbar\omega_z(n_z+\frac{1}{2})$$

Thus, in total, we have $$E_{n_x n_y n_z}=E_{xn_x}+E_{yn_y}+E_{zn_z}$$ which is exactly what you wanted.

• @EunbiSeo I think you can figure it out from what I've written. Certainly if $\omega_x=\omega_y=\omega_z=\omega$, then you just plug in $\omega$ everywhere I've written $\omega_{x,y,z}$, right? And from what I've written, you should be able to figure out what $X(x)$, $Y(y)$, and $Z(z)$ are for any given $(n_x,n_y,n_z)$. From $X,Y,Z$, you can write down $\Psi$. – Jahan Claes Dec 11 '18 at 19:45

Check out the section on n dimensional oscillators.

You only need one Omega if the spring constant is uniform.

https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator?wprov=sfla1

Do you understand the 1D quantum harmonic oscillator? The potential energy is $$\frac{1}{2}m\omega^2x^2$$. Now if the spring constant is uniform, you replace the $$x^2$$ with $$x^2+y^2+z^2$$ to account for the additional dimensions.

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

There are a couple ways you can proceed. Let $$\Psi(x,y,z)=A(x)B(y)C(z)$$. Using seperatin of variables, you can break this up into 3 1D Oscillator problems where : $$E=E_x+E_y+E_z=(n_x+\frac{1}{2})\hbar\omega+(n_y+\frac{1}{2})\hbar\omega+(n_z+\frac{1}{2})\hbar\omega=(\frac{3}{2}+n_x+n_y+n_z)\hbar\omega$$

Alternatively, you can do the subistitution, $$r^2=x^2+y^2+z^2$$ and solve the problem in spherical coordinates.

The energy level portion of the problem is reducible to a 1 D problem.

$$\frac{-\hbar^2}{2m}\nabla^2\psi+(\frac{1}{2}m\omega^2r^2+\frac{l(l+1)\hbar^2}{2mr^2})\psi=E\psi$$

That could be trickier. Should yield the same Eigen values for energy.

In the Cartesian solution, yep the eigenfunctions are just the products of the 1D solutions. The 1D solutions are a gaussian times a hermite polynomial.

In spherical coordinates i believe the eigenfunctions are the Generalized Laguerre Polynomials times the Associated Legendre polynomials for the angular wave functions:

Small description of solutions

• frankly, It is hard for me to understand.. Please explain more detaily. – Eunbi Seo Dec 11 '18 at 16:40