Angular Momentum, Spin, and the Particle in a Box


Clarifying Angular Momentum in a Particle in a Box

The statement “a particle in a box has no angular momentum” refers specifically to the absence of orbital angular momentum as a good quantum number. This does not apply to spin, which is an intrinsic form of angular momentum independent of geometry.

1. Two Kinds of Angular Momentum

(A) Orbital Angular Momentum

The operator is:

    \[ \hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}} \]

It depends on spatial coordinates and exists only when the system has rotational symmetry.

(B) Spin Angular Momentum

    \[ \hat{\mathbf{S}} \]

Spin is an internal degree of freedom and does not depend on the potential’s shape or boundary conditions.

2. What “No Angular Momentum” Really Means

A rectangular (1D or 3D) infinite potential well has no rotational symmetry. Therefore:

    \[ [H, L^2] \neq 0, \qquad [H, L_z] \neq 0. \]

This means neither L^2 nor L_z are conserved or define good quantum numbers. The particle’s orbital motion is described only by the quantum numbers n_x, n_y, n_z.

Thus: a particle in a box has no conserved orbital angular momentum.

3. But the Particle Can Still Have Spin

The spin degree of freedom is completely unaffected by placing the particle in a box.

The full Hilbert space becomes:

    \[ \mathcal{H} = \mathcal{H}_\text{spatial} \otimes \mathcal{H}_\text{spin}. \]

The spatial eigenfunctions are the usual box states:

    \[ \psi_{n_x,n_y,n_z}(x,y,z). \]

The spin state is independent:

    \[ |\uparrow\rangle,\qquad |\downarrow\rangle. \]

The complete state is therefore:

    \[ \Psi(x,y,z) = \psi_{n_x,n_y,n_z}(x,y,z)\,|\uparrow\rangle \]

or

    \[ \Psi(x,y,z) = \psi_{n_x,n_y,n_z}(x,y,z)\,|\downarrow\rangle. \]

If no magnetic fields or spin–orbit coupling are present:

    \[ [H, S_i] = 0. \]

Spin angular momentum is fully conserved inside the box.

4. What If the Particle Is “Already Rotating”?

This phrase has two possible interpretations:

Case 1 — The particle has orbital angular momentum before entering the box

The initial state might be something like:

    \[ \psi(r,\theta,\phi) \propto Y_{\ell m}(\theta,\phi). \]

When placed in a rectangular box:

  • rotational symmetry is lost,
  • orbital angular momentum is no longer conserved,
  • the state becomes a superposition of box eigenstates.

The angular momentum “scrambles” because the box does not allow rotations.

Case 2 — The particle has spin

Spin remains a well-defined, conserved quantum number. The full state can be:

    \[ \Psi(x,y,z) = \psi_{n_x,n_y,n_z}(x,y,z) \left(a|\uparrow\rangle + b|\downarrow\rangle\right). \]

Spin exists independently of the geometry and survives unchanged inside the box.

5. Summary Table

Quantity Exists in a Box? Depends on Geometry? Conserved?
Orbital Angular Momentum ❌ Not a good quantum number Yes No
Spin Angular Momentum ✔ Always No Yes (unless external fields)
Total J = L + S ❌ Not conserved (because L is not) Yes Only for spherical potentials

In Summary

A particle in a rectangular/1D box does not have conserved orbital angular momentum because the geometry breaks rotational symmetry.
But the particle’s spin remains fully intact and unaffected by the boundary conditions.