Quantum Mechanics in Three Dimensions

The three-dimensional time-independent Schrodinger equation is:

where $\nabla^2$ is the Laplacian: $\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ in Cartesian coordinates, or $\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}$ in spherical coordinates.

For a separable potential $V(x,y,z) = V_x(x) + V_y(y) + V_z(z)$, the wave function factors as $\psi(x,y,z) = X(x)Y(y)Z(z)$, and the energy adds: $E = E_x + E_y + E_z$. This reduces the 3D problem to three 1D problems.

The three-dimensional infinite square well is the simplest example. For a box with sides $a$, $b$, $c$, the energy levels are $E_{n_x n_y n_z} = \frac{\hbar^2\pi^2}{2m}\left(\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2}\right)$, and the wave functions are products of sine functions.

For a central potential $V(r)$ that depends only on the radial distance $r$, the Schrodinger equation is separable in spherical coordinates. We write $\psi(r,\theta,\phi) = R(r)Y_l^m(\theta,\phi)$, where $R(r)$ is the radial wave function and $Y_l^m$ are the spherical harmonics.

The angular equation separates from the radial equation. The angular solutions $Y_l^m(\theta,\phi)$ are eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$:

The quantum number $l = 0, 1, 2, \dots$ is the orbital angular momentum quantum number. For each $l$, $m$ can take $2l+1$ values: $m = -l, -l+1, \dots, l-1, l$. These are the magnetic quantum numbers.

The spherical harmonics $Y_l^m$ form a complete orthonormal set on the unit sphere. Any function of angles can be expanded in spherical harmonics.

Substituting the separation ansatz into the 3D Schrodinger equation and dividing by $R(r)Y_l^m$ gives the radial equation:

It is conventional to define $u(r) = rR(r)$. The radial equation then becomes:

This looks exactly like the one-dimensional Schrodinger equation with an effective potential:

The centrifugal term $\hbar^2 l(l+1)/(2mr^2)$ acts like a repulsive barrier that pushes the particle away from the origin. For $l = 0$, there is no centrifugal barrier, and the radial equation reduces to the 1D form.

In classical mechanics, angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ is conserved for a central potential. In quantum mechanics, the angular momentum operators are:

These operators satisfy the angular momentum algebra:

The raising and lowering operators $\hat{L}_\pm = \hat{L}_x \pm i\hat{L}_y$ shift the $m$ quantum number:

The maximum value of $|\langle\mathbf{L}\rangle|$ for a given $l$ is $\hbar\sqrt{l(l+1)}$, achieved when the state is an eigenstate of some component perpendicular to the $z$-axis. However, it is impossible to have simultaneously definite values of all three components because $[\hat{L}_i, \hat{L}_j] = i\hbar\epsilon_{ijk}\hat{L}_k$.

Find the energy levels of a three-dimensional isotropic harmonic oscillator with potential $V(r) = \frac{1}{2}m\omega^2 r^2$.

Step 1 — In Cartesian coordinates, the potential separates: $V = \frac{1}{2}m\omega^2(x^2 + y^2 + z^2) = V_x(x) + V_y(y) + V_z(z)$.

Step 2 — The wave function factors: $\psi = X(x)Y(y)Z(z)$, with each factor being a 1D harmonic oscillator eigenfunction. The energy is $E = (n_x + n_y + n_z + \frac{3}{2})\hbar\omega$.

Step 3 — Define $n = n_x + n_y + n_z$. Then $E_n = (n + \frac{3}{2})\hbar\omega$. The degeneracy of level $n$ is the number of ways to write $n$ as a sum of three non-negative integers: $g_n = \frac{(n+1)(n+2)}{2}$.

Step 4 — In spherical coordinates, the same energy appears as $E = (2n_r + l + \frac{3}{2})\hbar\omega$, where $n_r$ is the radial quantum number and $l$ is the orbital quantum number. The two descriptions are equivalent.