27 Jul 2025

Linear Harmonic Oscillator in One and Three Dimensions

harmonic-oscillator hermite-polynomials zero-point-energy three-dimensional-oscillator

One-dimensional harmonic oscillator

For

\[V(x)=\frac12m\omega^2x^2,\]

the stationary Schrödinger equation is

\[-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} +\frac12m\omega^2x^2\psi=E\psi.\]

Introduce the dimensionless coordinate and energy

\[\xi=\sqrt{\frac{m\omega}{\hbar}}x, \qquad \epsilon=\frac{E}{\hbar\omega}.\]

The factor $m\omega/\hbar$ has units $\mathrm{m^{-2}}$. Since $d^2/dx^2=(m\omega/\hbar)d^2/d\xi^2$, division by $\hbar\omega/2$ gives

\[\boxed{ \frac{d^2\psi}{d\xi^2}+(2\epsilon-\xi^2)\psi=0 }.\]

For $\lvert\xi\rvert\gg1$, the $\xi^2$ term dominates and the equation is approximately $\psi’’-\xi^2\psi\simeq0$. Its leading behaviours are $e^{-\xi^2/2}$ and $e^{+\xi^2/2}$; normalizability excludes the growing sign. Write

\[\psi(\xi)=e^{-\xi^2/2}H(\xi).\]

Differentiation gives

\[\psi'=e^{-\xi^2/2}(H'-\xi H),\] \[\psi''=e^{-\xi^2/2} \left[H''-2\xi H'+(\xi^2-1)H\right].\]

Substitution cancels the $\xi^2H$ terms and leaves

\[\boxed{H''-2\xi H'+(2\epsilon-1)H=0}.\]

Let $H=\sum_{j=0}^{\infty}a_j\xi^j$. Equating the coefficient of $\xi^j$ gives

\[(j+2)(j+1)a_{j+2} +(2\epsilon-1-2j)a_j=0,\]

or

\[\boxed{ a_{j+2}=\frac{2j+1-2\epsilon}{(j+2)(j+1)}a_j }.\]

If the series does not terminate, its large-$j$ behaviour reconstructs $e^{+\xi^2}$, making $\psi$ non-normalizable. Termination at degree $n$ requires the numerator to vanish at $j=n$:

\[2n+1-2\epsilon=0.\]

Hence

\[\boxed{E_n=\left(n+\frac12\right)\hbar\omega}, \qquad n=0,1,2,\ldots .\]

The terminating polynomials are Hermite polynomials $H_n(\xi)$. With the standard convention $H_n(\xi)=(-1)^ne^{\xi^2}d^n(e^{-\xi^2})/d\xi^n$, the normalized eigenfunctions are

\[\boxed{ \psi_n(x)= \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{H_n(\xi)}{\sqrt{2^n n!}} e^{-\xi^2/2} }.\]

The normalization follows from

\[\int_{-\infty}^{\infty}e^{-\xi^2}H_n(\xi)H_m(\xi)d\xi =\sqrt\pi\,2^n n!\,\delta_{nm},\]

and $dx=\sqrt{\hbar/(m\omega)}d\xi$. The eigenfunction parity is $(-1)^n$. The ground state has $E_0=\hbar\omega/2$, the zero-point energy required by the uncertainty principle.

Three-dimensional isotropic oscillator

For

\[V(x,y,z)=\frac12m\omega^2(x^2+y^2+z^2),\]

the Hamiltonian separates exactly:

\[\hat H=\hat H_x+\hat H_y+\hat H_z,\] \[\hat H_x=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} +\frac12m\omega^2x^2,\]

with analogous expressions for $y$ and $z$. Put

\[\Psi(x,y,z)=X(x)Y(y)Z(z).\]

After substitution and division by $XYZ$,

\[\frac{\hat H_xX}{X} +\frac{\hat H_yY}{Y} +\frac{\hat H_zZ}{Z}=E.\]

Each term depends on a different coordinate, so each equals a constant $E_x,E_y,E_z$ with $E=E_x+E_y+E_z$. Applying the one-dimensional result to every factor gives

\[\boxed{ \Psi_{n_xn_yn_z}(x,y,z) =\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z) },\] \[\boxed{ E_{n_xn_yn_z} =\left(n_x+n_y+n_z+\frac32\right)\hbar\omega }.\]

Here $n_x,n_y,n_z$ are independent non-negative integers. For total quantum number $N=n_x+n_y+n_z$, the energy is $(N+3/2)\hbar\omega$. The number of non-negative integer triples with that sum is

\[g_N=\frac{(N+1)(N+2)}2.\]

The normalized ground state is

\[\boxed{ \Psi_{000}(\mathbf r)= \left(\frac{m\omega}{\pi\hbar}\right)^{3/4} e^{-m\omega r^2/(2\hbar)} },\]

with energy $3\hbar\omega/2$, one zero-point contribution from each Cartesian direction.

The editable TikZ source generates the figure. The dimensionless differential equation, Hermite-polynomial residuals, and 3D energy sum are checked in the Maxima worksheet.