Three-Dimensional Box

The three-dimensional box is the rectangular infinite well. It is the canonical first model where boundary quantization, product eigenfunctions, additive energies, and degeneracy all appear in one exactly solvable system.

Let the particle be confined to

$$0\lt x\lt L_x, \qquad 0\lt y\lt L_y, \qquad 0\lt z\lt L_z,$$

with infinite walls outside. Inside the box, $V=0$ and

$$\hat H =-\frac{\hbar^2}{2m}\nabla^2.$$

The wavefunction vanishes on every wall.

Boundary Conditions

The boundary conditions are

$$\psi(0,y,z)=\psi(L_x,y,z)=0,$$ $$\psi(x,0,z)=\psi(x,L_y,z)=0,$$

and

$$\psi(x,y,0)=\psi(x,y,L_z)=0.$$

These are hard-wall conditions on six surfaces. They define the allowed domain of the Hamiltonian for this model.

Product Solutions

Inside the box, the time-independent Schrödinger equation is

$$-\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right)\varphi =E\varphi.$$

Use the product ansatz

$$\varphi(x,y,z)=X(x)Y(y)Z(z).$$

Each factor satisfies the one-dimensional infinite-well equation on its own interval. The allowed quantum numbers are

$$n_x,n_y,n_z=1,2,3,\ldots$$

The normalized eigenfunctions are

$$\varphi_{n_xn_yn_z}(x,y,z) =\sqrt{\frac{8}{L_xL_yL_z}}\, \sin\frac{n_x\pi x}{L_x} \sin\frac{n_y\pi y}{L_y} \sin\frac{n_z\pi z}{L_z}.$$

The prefactor is the product of the three one-dimensional normalization constants.

Energy Spectrum

The energies are

$$E_{n_xn_yn_z} =\frac{\pi^2\hbar^2}{2m} \left( \frac{n_x^2}{L_x^2} +\frac{n_y^2}{L_y^2} +\frac{n_z^2}{L_z^2} \right).$$

The energy is additive because the Hamiltonian separates:

$$E=E_x+E_y+E_z.$$

The ground state is

$$(n_x,n_y,n_z)=(1,1,1),$$

with energy

$$E_{111} =\frac{\pi^2\hbar^2}{2m} \left( \frac{1}{L_x^2} +\frac{1}{L_y^2} +\frac{1}{L_z^2} \right).$$

As in one dimension, none of the quantum numbers can be zero for hard-wall sine modes.

Cubic Box And Degeneracy

For a cubic box,

$$L_x=L_y=L_z=L,$$

the spectrum becomes

$$E_{n_xn_yn_z} =\frac{\pi^2\hbar^2}{2mL^2} \left( n_x^2+n_y^2+n_z^2 \right).$$

Different triples can give the same sum of squares. For example,

$$(1,1,2), \qquad (1,2,1), \qquad (2,1,1)$$

have the same energy. This is a threefold degeneracy from permuting equal side lengths.

The ground state $(1,1,1)$ is nondegenerate. The first excited shell in a cube is the threefold set above. Higher shells can have degeneracies from both permutations and arithmetic coincidences among sums of squares.

The general distinction between symmetry-enforced and accidental repeated energies is developed in Degeneracy in Separable Systems.

Momentum-Space Interpretation

Inside the box, each separated sine factor is built from standing waves. The allowed wave numbers are

$$k_x=\frac{n_x\pi}{L_x}, \qquad k_y=\frac{n_y\pi}{L_y}, \qquad k_z=\frac{n_z\pi}{L_z}.$$

The energy can be written

$$E=\frac{\hbar^2}{2m} \left(k_x^2+k_y^2+k_z^2\right).$$

For a large box, allowed points in $k$-space become dense. Counting those points is the first step toward density-of-states formulas, but many-body and thermodynamic applications belong in later volumes.

Complete Set Of Labels

For a rectangular box with unequal side lengths, the triple $(n_x,n_y,n_z)$ usually labels the energy eigenstate uniquely. For a cube, energy alone may not identify the state because of degeneracy.

A complete description uses the commuting one-dimensional mode labels, not just the energy:

$$(n_x,n_y,n_z).$$

This is a concrete example of why energy eigenvalues and eigenstates are not the same data. Degenerate eigenspaces require additional labels or basis choices.

Common Mistakes

• Allowing one of the quantum numbers to be zero in the hard-wall box.
• Forgetting the normalization factor $\sqrt{8/(L_xL_yL_z)}$.
• Treating a rectangular box with unequal side lengths as if every permutation were degenerate.
• Assuming energy alone labels a unique state in the cubic box.
• Confusing hard-wall sine modes with periodic plane-wave modes.
• Forgetting that the wavefunction must vanish on all six walls.

Where This Is Used

• Infinite Square Well is the one-dimensional ancestor.
• Separation of Variables explains the product method.
• Schrödinger Equation in Three Dimensions gives the 3D equation and normalization.
• Free Particle in Three Dimensions compares hard-wall standing waves with periodic plane-wave regulators.
• Degeneracy in Separable Systems classifies cubic-box degeneracies and related repeated energies.
• Density of States: First Encounter compares hard-wall and periodic boxes in the large-volume counting limit.
• Boundary Conditions explains why hard walls impose Dirichlet conditions.
• Complete Sets of Commuting Observables explains why degeneracies require additional labels.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.

Exercises

1. Verify the normalization of $\varphi_{n_xn_yn_z}$.

Solution

The integral factors into three one-dimensional integrals:

$$\int_0^{L_x}\int_0^{L_y}\int_0^{L_z} \lvert\varphi_{n_xn_yn_z}\rvert^2 \,dz\,dy\,dx = \frac{8}{L_xL_yL_z} \prod_{i=x,y,z} \int_0^{L_i} \sin^2\frac{n_i\pi q_i}{L_i}\,dq_i.$$

Each integral is $L_i/2$, so the product gives

$$\frac{8}{L_xL_yL_z} \frac{L_x}{2} \frac{L_y}{2} \frac{L_z}{2} =1.$$

2. In a cubic box, find the degeneracy of the level with $(n_x,n_y,n_z)$ a permutation of $(1,1,2)$.

Solution

The distinct permutations are

$$(1,1,2), \qquad (1,2,1), \qquad (2,1,1).$$

They all have

$$n_x^2+n_y^2+n_z^2=1+1+4=6.$$

Thus this shell is threefold degenerate.

3. Why is $(0,1,1)$ not an allowed hard-wall box state?

Solution

The one-dimensional hard-wall solutions are proportional to $\sin(n\pi x/L)$ with $n=1,2,3,\ldots$. If $n_x=0$, then the $x$ factor is $\sin(0)=0$ everywhere, so the full wavefunction is the zero function, not a physical state.