Spin (the Mystery of) Archives - Time Travel, Quantum Entanglement and Quantum Computing https://stationarystates.com/category/basic-quantum-theory/spin-the-mystery-of/ Not only is the Universe stranger than we think, it is stranger than we can think...Hiesenberg Tue, 02 Dec 2025 16:10:43 +0000 en-US hourly 1 https://wordpress.org/?v=6.9.1 Spin versus Angular Momentum https://stationarystates.com/basic-quantum-theory/spin-the-mystery-of/spin-versus-angular-momentum/?utm_source=rss&utm_medium=rss&utm_campaign=spin-versus-angular-momentum Tue, 02 Dec 2025 16:10:09 +0000 https://stationarystates.com/?p=1080 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 […]

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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.


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]]> Choosing the Axis to Measure Spin https://stationarystates.com/basic-quantum-theory/spin-the-mystery-of/choosing-the-axis-to-measure-spin/?utm_source=rss&utm_medium=rss&utm_campaign=choosing-the-axis-to-measure-spin Wed, 15 Oct 2025 20:19:47 +0000 https://stationarystates.com/?p=1051 Spin-½ measurement along a rotated axis (Pauli-matrix derivation) Spin-½ measurement along a rotated axis (Pauli-matrix derivation) We will show explicitly—using Pauli matrices—why a spin-½ measurement along any direction still yields […]

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Spin-½ measurement along a rotated axis (Pauli-matrix derivation)


Spin-½ measurement along a rotated axis (Pauli-matrix derivation)

We will show explicitly—using Pauli matrices—why a spin-½ measurement along any direction still yields exactly two outcomes, and compute the probabilities when the measurement axis is rotated by 30^\circ from the original x-axis.

1) Pauli matrices and spin operators

    \[ \sigma_x = \begin{pmatrix} 0 & 1 \\[4pt] 1 & 0 \end{pmatrix},\quad \sigma_y = \begin{pmatrix} 0 & -i \\[4pt] i & 0 \end{pmatrix},\quad \sigma_z = \begin{pmatrix} 1 & 0 \\[4pt] 0 & -1 \end{pmatrix}. \]

The spin operator along a unit direction \hat n is

    \[ \hat S_{\hat n} \;=\; \frac{\hbar}{2}\,(\vec\sigma\!\cdot\!\hat n)\,, \quad \text{with}\quad \vec\sigma\!\cdot\!\hat n \;=\; n_x \sigma_x + n_y \sigma_y + n_z \sigma_z. \]

For any direction \hat n, the eigenvalues of \hat S_{\hat n} are always \pm \frac{\hbar}{2}. Thus there are always two outcomes, independent of the axis.

2) Choose a new axis rotated by 30^\circ from the x-axis

Let the original measurement axis be \hat x. Rotate the apparatus by an angle \theta=30^\circ toward \hat z about the y-axis. The new unit vector is

    \[ \hat n' \;=\; (\cos\theta,\,0,\,\sin\theta). \]

Then

    \[ \vec\sigma\!\cdot\!\hat n' \;=\; \cos\theta\,\sigma_x \;+\; \sin\theta\,\sigma_z \;=\; \begin{pmatrix} \sin\theta & \cos\theta \\[4pt] \cos\theta & -\sin\theta \end{pmatrix}. \]

For \theta=30^\circ with \cos\theta=\tfrac{\sqrt{3}}{2} and \sin\theta=\tfrac{1}{2},

    \[ \vec\sigma\!\cdot\!\hat n' \;=\; \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\[6pt] \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}, \qquad \hat S_{\hat n'} \;=\; \frac{\hbar}{2}\,\vec\sigma\!\cdot\!\hat n'. \]

3) Eigenvalues (two outcomes only) and eigenvectors

The characteristic equation of \vec\sigma\!\cdot\!\hat n' is

    \[ \det\big(\vec\sigma\!\cdot\!\hat n' - \lambda I\big) = \lambda^2 - 1 = 0 \;\;\Rightarrow\;\; \lambda=\pm 1. \]

Therefore the measurement outcomes for \hat S_{\hat n'} are always \pm \frac{\hbar}{2}.

A corresponding +1 eigenvector (for general \theta) can be taken as

    \[ |+\hat n'\rangle \;=\; \begin{pmatrix}\cos\frac{\theta}{2}\\[4pt]\sin\frac{\theta}{2}\end{pmatrix} \quad\text{and}\quad |-\hat n'\rangle \;=\; \begin{pmatrix}-\sin\frac{\theta}{2}\\[4pt]\cos\frac{\theta}{2}\end{pmatrix}, \]

when we work in the \sigma_z (i.e., |+\!z\rangle, |-\!z\rangle) basis and choose azimuth \phi=0 (the x\!-\!z plane).

4) Start in |+\!x\rangle and measure along the rotated axis \hat n'

In the \sigma_z basis, the \sigma_x eigenstates are

    \[ |+\!x\rangle \;=\; \frac{1}{\sqrt{2}} \begin{pmatrix}1\\[2pt]1\end{pmatrix}, \qquad |-\!x\rangle \;=\; \frac{1}{\sqrt{2}} \begin{pmatrix}1\\[2pt]-1\end{pmatrix}. \]

If the system is prepared in |+\!x\rangle and we measure \hat S_{\hat n'}, the probability of obtaining +\frac{\hbar}{2} is

    \[ P_{+}(\hat n' \mid +x) \;=\; \big|\langle +\hat n' \,|\, +x \rangle\big|^2 \;=\; \cos^2\!\frac{\theta}{2}, \]

and P_{-}(\hat n' \mid +x) = \sin^2\!\frac{\theta}{2}. This can be derived either by the explicit overlap of the spinors above, or by the Bloch-vector identity

    \[ P_{+}(\hat n' \mid \hat a) \;=\; \frac{1+\hat a\!\cdot\!\hat n'}{2}, \quad\text{with}\;\; \hat a=\hat x \;\Rightarrow\; P_{+}=\frac{1+\cos\theta}{2}=\cos^2\!\frac{\theta}{2}. \]

5) Plug in \theta=30^\circ

    \[ \cos\frac{\theta}{2}=\cos 15^\circ =\sqrt{\frac{1+\cos30^\circ}{2}} =\sqrt{\frac{1+\tfrac{\sqrt{3}}{2}}{2}} =\sqrt{\frac{2+\sqrt{3}}{4}} \quad\Rightarrow\quad \cos^2 15^\circ=\frac{2+\sqrt{3}}{4}\approx 0.933. \]

    \[ \sin^2 15^\circ=\frac{2-\sqrt{3}}{4}\approx 0.067. \]

Therefore, for a 30^\circ rotation of the measurement axis relative to x, a state prepared as |+\!x\rangle yields

  • Outcome +\frac{\hbar}{2} along \hat n' with probability \displaystyle \frac{2+\sqrt{3}}{4}\approx 93.3\%.
  • Outcome -\frac{\hbar}{2} along \hat n' with probability \displaystyle \frac{2-\sqrt{3}}{4}\approx 6.7\%.

6) What is special about the chosen direction?

Physically, no direction is special: quantum spin for a spin-½ particle always has two eigenvalues along any axis. What is special is that the experimentalist’s chosen direction defines the observable:

    \[ \hat S_{\hat n} \;=\; \frac{\hbar}{2}\,(\vec\sigma\!\cdot\!\hat n), \]

and distinct directions correspond to non-commuting operators (e.g. [\hat S_x,\hat S_z]\neq 0). Changing the apparatus orientation rotates the measurement basis; it does not change the two-valued spectrum.

7) One-line summary

Rotate the experiment by any angle you like (e.g. 30^\circ): the spin-½ measurement still has outcomes \pm \tfrac{\hbar}{2}; only the probabilities change according to \cos^2(\theta/2) and \sin^2(\theta/2).


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