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.

Helpful background. Separation of Variables to turn a separable wave equation into ordinary differential equations.

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

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

These $k_i$ are standing-wave mode labels, not definite Cartesian momentum components. A momentum measurement after releasing a hard-wall state has a continuous Fourier distribution. For a large box, the mode 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

Each triple $(n_x,n_y,n_z)$ specifies one member of this product basis, for any side lengths. Generic unequal lengths give nondegenerate energies, whereas equal lengths and special length ratios allow distinct triples to have the same energy.

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.

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.

References

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