Spin-1/2 Particles

For a particle with spin \( \tfrac{1}{2} \), spin is not a fixed, intrinsic label. It is a property that depends on the axis along which the measurement is performed, because a quantum state is not a single configuration but a linear superposition of two basis states. This superposition is written as the spinor \( (\alpha,\ \beta) \). $$ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$ Here \( \alpha \) and \( \beta \) are the complex amplitudes associated with the "spin up" and "spin down" states along the \( z \) axis.

In simple terms, "spin up" along \( z \) is not the same physical state as "spin up" along \( x \) or \( y \).

The probabilities of different measurement outcomes depend on the magnitudes of the spinor amplitudes, and these probabilities are defined with respect to the axis in which the state is represented.

Consequently, when the measurement axis changes, the state must be re expressed in a new basis. This produces new amplitudes and therefore different probabilities.

This is why a particle prepared as "spin up" along \( z \) can yield equal probabilities of "spin up" and "spin down" when measured along \( x \).

Explanation

A spin \( \tfrac{1}{2} \) particle (such as an electron, proton, or neutron) has only two allowed values of the magnetic quantum number \( m_s \):

• \( m_s = \tfrac{1}{2} \) (spin up) 
• \( m_s = -\tfrac{1}{2} \) (spin down)

Yet a spin \( \tfrac{1}{2} \) system can be oriented along any direction in three dimensional space.

The quantum state is not restricted to the two eigenstates of \( S_z \). Instead, it is described by a linear superposition of them, encoded by two complex amplitudes.

$$ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \\ 1
\end{pmatrix} $$

This two component column vector is called a spinor, the standard representation for the state space of a spin 1/2 particle.

Note. The two fundamental spinors are the eigenstates of the spin operator along the \( z \) axis. $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \left| \tfrac{1}{2}, \tfrac{1}{2} \right\rangle $$ $$ \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \left| \tfrac{1}{2}, -\tfrac{1}{2} \right\rangle $$ Applying the operator $$ S_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ yields the expected eigenvalues: $$ S_z \begin{pmatrix}1 \\ 0\end{pmatrix} = +\tfrac{1}{2}\hbar $$ $$ S_z \begin{pmatrix}0 \\ 1\end{pmatrix} = -\tfrac{1}{2}\hbar $$ Thus, the spinor encodes the state, while \( S_z \) returns the two allowed values of spin angular momentum along the \( z \) axis: \( +\tfrac{1}{2}\hbar \) and \( -\tfrac{1}{2}\hbar \).

The amplitudes \( \alpha \) and \( \beta \) are not merely labels for "up" and "down". They represent the relative weights of the two \( S_z \) eigenstates in the superposition.

Their probabilistic interpretation is straightforward:

• \( |\alpha|^2 \) gives the probability of measuring spin up along \( z \)
• \( |\beta|^2 \) gives the probability of measuring spin down along \( z \)

Since these are the only possible outcomes, normalization requires:

$$ |\alpha|^2 + |\beta|^2 = 1 $$

What changes when measuring along the x axis?

The eigenstates of \( S_x \) are not the same as those of \( S_z \).

To compute the outcome of a measurement of \( S_x \), the state \( (\alpha,\ \beta) \) must be expressed in the eigenbasis of \( S_x \).

This requires a change of basis, which produces two new amplitudes:

$$ a = \frac{\alpha + \beta}{\sqrt{2}} $$

$$ b = \frac{\alpha - \beta}{\sqrt{2}} $$

These coefficients represent the state in the \( S_x \) eigenbasis.

Their interpretation parallels that of \( \alpha \) and \( \beta \):

• \( |a|^2 \) is the probability of measuring spin up along \( x \)
• \( |b|^2 \) is the probability of measuring spin down along \( x \)

This follows from the fact that the eigenstates of \( S_x \) are specific linear combinations of the eigenstates of \( S_z \).

Consider a particle prepared in the "spin up" state along \( z \):

$$ (\alpha,\ \beta) = (1,0) $$

The basis transformation yields:

$$ a = \frac{\alpha + \beta}{\sqrt{2}} = \frac{1+0}{\sqrt{2}} = \frac{1}{\sqrt{2}} $$

$$ b = \frac{\alpha - \beta}{\sqrt{2}} = \frac{1-0}{\sqrt{2}} = \frac{1}{\sqrt{2}} $$

The resulting probabilities are:

$$ |a|^2 = \tfrac{1}{2} $$

$$ |b|^2 = \tfrac{1}{2} $$

Therefore, even when the particle is prepared as "spin up" along \( z \), a measurement along \( x \) produces:

• 50 percent spin up along \( x \)
• 50 percent spin down along \( x \)

This behavior is consistently observed in Stern-Gerlach experiments.

In essence, spin up along one axis does not imply spin up along a different axis.

Understanding this requires recomputing the probabilities using the amplitudes \( a \) and \( b \).

The spin operators \( S_x \), \( S_y \), and \( S_z \) each possess distinct eigenvectors.

$$
\hat S_x = \frac{\hbar}{2}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} $$

$$ \hat S_y = \frac{\hbar}{2}
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix} $$

$$ \hat S_z = \frac{\hbar}{2}
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} $$

Note. In a spin \( \tfrac{1}{2} \) system, a full \( 360^\circ \) rotation does not return the quantum state to itself but to its negative: \[ U(2\pi)\,|\psi\rangle = -|\psi\rangle \] This is a hallmark of spin \( \tfrac{1}{2} \) particles. The physical orientation of the spin remains unchanged, but the spinor that represents the state acquires an overall minus sign. To recover the exact same quantum state, a \( 720^\circ \) rotation is required, meaning two complete turns. Mathematically, this demands a 2×2 representation of rotations satisfying \[ U(2\pi) = -I \] where \( I \) is the identity matrix and \( U \) is the unitary rotation operator. This constraint forces the use of SU(2) instead of SO(3). The matrices that generate these transformations are the Pauli matrices. They are the only 2×2 matrices with the required properties: zero trace, determinant -1, correct commutation relations for rotations, anticommutation structure, and eigenvalues \( \pm 1 \). These Pauli matrices, denoted by \( \sigma \), are: $$ \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ $$ \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$ $$ \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ As an example, consider a generic spinor: \[ \psi = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix} \] Multiplying by \( -I = \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix} \) gives the effect of a \( 360^\circ \) rotation: \[ (-I)\psi = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix} = \begin{pmatrix} -0.6 \\ -0.8 \end{pmatrix}. \] The resulting spinor points in the same physical direction but carries the opposite phase: \[ \psi' = -\psi \] The physical state is unchanged because measurable probabilities depend on \( |\alpha|^2 \) and \( |\beta|^2 \), not on the overall sign. Originally: $$ |0.6|^2 = 0.36 $$ $$ |0.8|^2 = 0.64 $$ After the rotation: $$ |-0.6|^2 = 0.36 $$ $$ |-0.8|^2 = 0.64 $$ Thus the probabilities remain identical.

This also implies that the state \( \begin{pmatrix}1 \\ 0\end{pmatrix} \), which represents "spin up" along \( z \), is not an eigenstate of either \( \hat S_x \) or \( \hat S_y \).

To measure spin along the \( x \) axis, the state must be rewritten as a linear combination of the eigenvectors of \( \hat S_x \). The probabilities for the outcomes \( \pm \tfrac{1}{2}\hbar \) then follow from the new coefficients.

Therefore, each spatial axis has its own up and down eigenstates, and these do not coincide with those of any other axis.

This is why spin measurements performed along different axes yield different outcomes, even for the same initial spinor.

The Eigenvectors of \( S_x \) and Basis Change in Spin-1/2 Systems

Any spinor \( (\alpha,\beta) \) can be decomposed into the eigenvectors of the operator \( S_x \):

$$ \begin{pmatrix}\alpha \\ \beta\end{pmatrix} = a \begin{pmatrix} \tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} \end{pmatrix} + b \begin{pmatrix} \tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}}\end{pmatrix} $$

The eigenvectors are:

• $ \begin{pmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}}\end{pmatrix} $
• $ \begin{pmatrix}\tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}}\end{pmatrix} $

and the coefficients \( a \) and \( b \) are:

• $ a = \dfrac{\alpha + \beta}{\sqrt{2}} $
• $ b = \dfrac{\alpha - \beta}{\sqrt{2}} $

And so on.