Spin (physics)

 ( Rotational Symmetry ) 

Since elementary particles are point-like, self-rotation is not well-defined for them. However, spin implies that the phase of the particle depends on the angle as $e^\{i\; S\; \backslash theta\}\backslash \; ,$ for rotation of angle around the axis parallel to the spin . This is equivalent to the quantum-mechanical interpretation of momentum as phase dependence in the position, and of orbital angular momentum as phase dependence in the angular position.

For fermions, the picture is less clear: From the Ehrenfest theorem, the angular velocity is equal to the derivative of the Hamiltonian to its conjugate momentum, which is the total angular momentum operator Therefore, if the Hamiltonian has any dependence on the spin , then must be non-zero; consequently, for classical mechanics, the existence of spin in the Hamiltonian will produce an actual angular velocity, and hence an actual physical rotation – that is, a change in the phase-angle, , over time. However, whether this holds true for free electrons is ambiguous, since for an electron, ² is a constant and one might decide that since it cannot change, no partial () can exist. Therefore it is a matter of interpretation whether the Hamiltonian must include such a term, and whether this aspect of classical mechanics extends into quantum mechanics (any particle's intrinsic spin angular momentum, , is a quantum number arising from a "spinor" in the mathematical solution to the Dirac equation, rather than being a more nearly physical quantity, like orbital angular momentum ). Nevertheless, spin appears in the Dirac equation, and thus the relativistic Hamiltonian of the electron, treated as a Dirac field, can be interpreted as including a dependence in the spin .

• Spin quantum numbers may take either half-integer or integer values.
• Although the direction of its spin can be changed, the magnitude of the spin of an elementary particle cannot be changed.
• The spin of a charged particle is associated with a magnetic dipole moment with a -factor that differs from 1. (In the classical context, this would imply the internal charge and mass distributions differing for a rotating nonconductive object.)

The conventional definition of the spin quantum number is , where can be any non-negative integer. Hence the allowed values of are 0, , 1, , 2, etc. The value of for an elementary particle depends only on the type of particle and cannot be altered in any known way (in contrast to the spin direction described below). The spin angular momentum of any physical system is quantized. The allowed values of are $$S\; =\; \backslash hbar\; \backslash ,\; \backslash sqrt\{s(s\; +\; 1)\}\; =\; \backslash frac\{h\}\{2\backslash pi\}\; \backslash ,\; \backslash sqrt\{\backslash frac\{n\}\{2\}\backslash frac\{(n\; +\; 2)\}\{2\}\}\; =\; \backslash frac\{h\}\{4\backslash pi\}\; \backslash ,\; \backslash sqrt\{n(n\; +\; 2)\},$$ where is the Planck constant, and $\backslash hbar\; =\; \backslash frac\{h\}\{2\backslash pi\}$ is the reduced Planck constant. In contrast, orbital angular momentum can only take on integer values of ; i.e., even-numbered values of .

This has some profound consequences:

• Quarks and leptons (including electrons and neutrinos), which make up what is classically known as matter, are all fermions with spin . The common idea that "matter takes up space" actually comes from the Pauli exclusion principle acting on these particles to prevent the fermions from being in the same quantum state. Further compaction would require electrons to occupy the same energy states, and therefore a kind of pressure (sometimes known as degeneracy pressure of electrons) acts to resist the fermions being overly close. Elementary fermions with other spins (, , etc.) are not known to exist.
• Elementary particles which are thought of as force carrier are all bosons with spin 1. They include the photon, which carries the electromagnetic force, the gluon (strong force), and the W and Z bosons (weak force). The ability of bosons to occupy the same quantum state is used in the laser, which aligns many photons having the same quantum number (the same direction and frequency), superfluid liquid helium resulting from helium-4 atoms being bosons, and superconductivity, where Cooper pair (which individually are fermions) act as single composite bosons. Elementary bosons with other spins (0, 2, 3, etc.) were not historically known to exist, although they have received considerable theoretical treatment and are well established within their respective mainstream theories. In particular, theoreticians have proposed the graviton (predicted to exist by some quantum gravity theories) with spin 2, and the Higgs boson (explaining electroweak symmetry breaking) with spin 0. Since 2013, the Higgs boson with spin 0 has been considered proven to exist. Information about Higgs Boson in CERN's official website. It is the first scalar boson (spin 0) known to exist in nature.
• Atomic nuclei have nuclear spin which may be either half-integer or integer, so that the nuclei may be either fermions or bosons.

The intrinsic magnetic moment of a spin- particle with charge , mass , and spin angular momentum isPhysics of Atoms and Molecules, B. H. Bransden, C. J. Joachain, Longman, 1983, .

$\backslash boldsymbol\{\backslash mu\}\; =\; \backslash frac\{g\_\backslash text\{s\}\; q\}\{2m\}\; \backslash mathbf\{S\},$

The electron, being a charged elementary particle, possesses a nonzero magnetic moment. One of the triumphs of the theory of quantum electrodynamics is its accurate prediction of the electron -factor, which has been experimentally determined to have the value , with the digits in parentheses denoting measurement uncertainty in the last two digits at one standard deviation. The value of 2 arises from the Dirac equation, a fundamental equation connecting the electron's spin with its electromagnetic properties; and the deviation from arises from the electron's interaction with the surrounding quantum fields, including its own electromagnetic field and virtual particles.

 \hat S^2 |s, m_s\rangle &= \hbar^2 s(s + 1) |s, m_s\rangle, \\
 \hat S_z |s, m_s\rangle &= \hbar m_s |s, m_s\rangle.
     

The spin raising and lowering operators acting on these eigenvectors give $$\backslash hat\; S\_\backslash pm\; |s,\; m\_s\backslash rangle\; =\; \backslash hbar\; \backslash sqrt\{s(s\; +\; 1)\; -\; m\_s(m\_s\; \backslash pm\; 1)\}\; |s,\; m\_s\; \backslash pm\; 1\backslash rangle,$$ where $\backslash hat\; S\_\backslash pm\; =\; \backslash hat\; S\_x\; \backslash pm\; i\; \backslash hat\; S\_y$.

But unlike orbital angular momentum, the eigenvectors are not spherical harmonics. They are not functions of and . There is also no reason to exclude half-integer values of and .

All quantum-mechanical particles possess an intrinsic spin $s$ (though this value may be equal to zero). The projection of the spin $s$ on any axis is quantized in units of the reduced Planck constant, such that the state function of the particle is, say, not $\backslash psi\; =\; \backslash psi(\backslash mathbf\; r)$, but $\backslash psi\; =\; \backslash psi(\backslash mathbf\; r,s\_z)$, where $s\_z$ can take only the values of the following discrete set: $$s\_z\; \backslash in\; \backslash \{-s\backslash hbar,\; -(s\; -\; 1)\backslash hbar,\; \backslash dots,\; +(s\; -\; 1)\backslash hbar,\; +s\backslash hbar\backslash \}.$$

One distinguishes (integer spin) and (half-integer spin). The total angular momentum conserved in interaction processes is then the sum of the orbital angular momentum and the spin.

For the special case of spin- particles, , and are the three Pauli matrices: $$\backslash sigma\_x\; =\; \backslash begin\{pmatrix\}$$

0 & 1\\
1 & 0
     

0 & -i\\
i & 0
     

1 & 0\\
0 & -1
     

(-1)^{2s} \psi(\dots, \mathbf r_j, \sigma_j, \dots, \mathbf r_i, \sigma_i, \dots).
     

Thus, for the prefactor will reduce to +1, for to −1. This permutation postulate for -particle state functions has most important consequences in daily life, e.g. the periodic table of the chemical elements.

For a generic particle with spin , we would need such parameters. Since these numbers depend on the choice of the axis, they transform into each other non-trivially when this axis is rotated. It is clear that the transformation law must be linear, so we can represent it by associating a matrix with each rotation, and the product of two transformation matrices corresponding to rotations A and B must be equal (up to phase) to the matrix representing rotation AB. Further, rotations preserve the quantum-mechanical inner product, and so should our transformation matrices: $$$$

\sum_{m=-j}^j a_m^* b_m = \sum_{m=-j}^j \left(\sum_{n=-j}^j U_{nm} a_n\right)^* \left(\sum_{k=-j}^j U_{km} b_k\right),
     

\sum_{n=-j}^j \sum_{k=-j}^j U_{np}^* U_{kq} = \delta_{pq}.
     

Mathematically speaking, these matrices furnish a unitary projective representation of the rotation group SO(3). Each such representation corresponds to a representation of the covering group of SO(3), which is SU(2). There is one -dimensional irreducible representation of SU(2) for each dimension, though this representation is -dimensional real for odd and -dimensional complex for even (hence of real dimension ). For a rotation by angle in the plane with normal vector $\backslash hat\{\backslash boldsymbol\{\backslash theta\}\}$, $$U\; =\; e^\{-\backslash frac\{i\}\{\backslash hbar\}\; \backslash boldsymbol\{\backslash theta\}\; \backslash cdot\; \backslash mathbf\{S\}\},$$ where $\backslash boldsymbol\{\backslash theta\}\; =\; \backslash theta\; \backslash hat\{\backslash boldsymbol\{\backslash theta\}\}$, and is the vector of spin operators.

A generic rotation in 3-dimensional space can be built by compounding operators of this type using Euler angles: $$\backslash mathcal\{R\}(\backslash alpha,\; \backslash beta,\; \backslash gamma)\; =\; e^\{-i\backslash alpha\; S\_x\}\; e^\{-i\backslash beta\; S\_y\}\; e^\{-i\backslash gamma\; S\_z\}.$$

An irreducible representation of this group of operators is furnished by the Wigner D-matrix: $$$$

 D^s_{m'm}(\alpha, \beta, \gamma) \equiv
 \langle sm' | \mathcal{R}(\alpha, \beta, \gamma) | sm \rangle =
 e^{-im'\alpha} d^s_{m'm}(\beta)e^{-i m\gamma},
     

Recalling that a generic spin state can be written as a superposition of states with definite , we see that if is an integer, the values of are all integers, and this matrix corresponds to the identity operator. However, if is a half-integer, the values of are also all half-integers, giving for all , and hence upon rotation by 2 the state picks up a minus sign. This fact is a crucial element of the proof of the spin–statistics theorem.

In case of spin- particles, it is possible to find a construction that includes both a finite-dimensional representation and a scalar product that is preserved by this representation. We associate a 4-component Dirac spinor with each particle. These spinors transform under Lorentz transformations according to the law $$\backslash psi\text{'}\; =\; \backslash exp\{\backslash left(\backslash tfrac\{1\}\{8\}\; \backslash omega\_\{\backslash mu\backslash nu\}\; \backslash gamma\_\{\backslash mu\},\backslash right)\}\; \backslash psi,$$ where are gamma matrices, and is an antisymmetric 4 × 4 matrix parametrizing the transformation. It can be shown that the scalar product $$\backslash langle\backslash psi|\backslash phi\backslash rangle\; =\; \backslash bar\{\backslash psi\}\backslash phi\; =\; \backslash psi^\backslash dagger\; \backslash gamma\_0\; \backslash phi$$ is preserved. It is not, however, positive-definite, so the representation is not unitary.

 \psi_{x+} = \left|\frac{1}{2}, \frac{+1}{2}\right\rangle_x =
   \displaystyle\frac{1}{\sqrt{2}} \!\!\!\!\! & \begin{pmatrix}{1}\\{1}\end{pmatrix},  &
 \psi_{x-} = \left|\frac{1}{2}, \frac{-1}{2}\right\rangle_x =
   \displaystyle\frac{1}{\sqrt{2}} \!\!\!\!\! & \begin{pmatrix}{1}\\{-1}\end{pmatrix}, \\
 \psi_{y+} = \left|\frac{1}{2}, \frac{+1}{2}\right\rangle_y =
   \displaystyle\frac{1}{\sqrt{2}} \!\!\!\!\! & \begin{pmatrix}{1}\\{i}\end{pmatrix},  &
 \psi_{y-} = \left|\frac{1}{2}, \frac{-1}{2}\right\rangle_y =
   \displaystyle\frac{1}{\sqrt{2}} \!\!\!\!\! & \begin{pmatrix}{1}\\{-i}\end{pmatrix}, \\
 \psi_{z+} = \left|\frac{1}{2}, \frac{+1}{2}\right\rangle_z =                          &
   \begin{pmatrix} 1 \\ 0 \end{pmatrix}, &
 \psi_{z-} = \left|\frac{1}{2}, \frac{-1}{2}\right\rangle_z =                          &
   \begin{pmatrix} 0 \\ 1 \end{pmatrix}.
     

(Because any eigenvector multiplied by a constant is still an eigenvector, there is ambiguity about the overall sign. In this article, the convention is chosen to make the first element imaginary and negative if there is a sign ambiguity. The present convention is used by software such as SymPy; while many physics textbooks, such as Sakurai and Griffiths, prefer to make it real and positive.)

By the postulates of quantum mechanics, an experiment designed to measure the electron spin on the , , or  axis can only yield an eigenvalue of the corresponding spin operator (, or ) on that axis, i.e. or . The quantum state of a particle (with respect to spin), can be represented by a two-component spinor: $$\backslash psi\; =\; \backslash begin\{pmatrix\}\; a\; +\; bi\; \backslash \backslash \; c\; +\; di\; \backslash end\{pmatrix\}.$$

When the spin of this particle is measured with respect to a given axis (in this example, the  axis), the probability that its spin will be measured as is just $\backslash big|\backslash langle\; \backslash psi\_\{x+\}|\backslash psi\backslash rangle\backslash big|^2$. Correspondingly, the probability that its spin will be measured as is just $\backslash big|\backslash langle\backslash psi\_\{x-\}|\backslash psi\backslash rangle\backslash big|^2$. Following the measurement, the spin state of the particle collapses into the corresponding eigenstate. As a result, if the particle's spin along a given axis has been measured to have a given eigenvalue, all measurements will yield the same eigenvalue (since $\backslash big|\backslash langle\backslash psi\_\{x+\}\; |\; \backslash psi\_\{x+\}\backslash rangle\backslash big|^2\; =\; 1$, etc.), provided that no measurements of the spin are made along other axes.

The operator has eigenvalues of , just like the usual spin matrices. This method of finding the operator for spin in an arbitrary direction generalizes to higher spin states, one takes the dot product of the direction with a vector of the three operators for the three -, -, -axis directions.

A normalized spinor for spin- in the direction (which works for all spin states except spin down, where it will give ) is $$\backslash frac\{1\}\{\backslash sqrt\{2\; +\; 2u\_z\}\}\; \backslash begin\{pmatrix\}\; 1\; +\; u\_z\; \backslash \backslash \; u\_x\; +\; iu\_y\; \backslash end\{pmatrix\}.$$

The above spinor is obtained in the usual way by diagonalizing the matrix and finding the eigenstates corresponding to the eigenvalues. In quantum mechanics, vectors are termed "normalized" when multiplied by a normalizing factor, which results in the vector having a length of unity.

\big| \langle \psi_{x\pm} | \psi_{z\pm} \rangle \big|^2 =
\big| \langle \psi_{y\pm} | \psi_{z\pm} \rangle \big|^2 = \tfrac{1}{2}.
     

So when measure the spin of a particle along the  axis as, for example, , the particle's spin state collapses into the eigenstate $|\backslash psi\_\{x+\}\backslash rangle$. When we then subsequently measure the particle's spin along the  axis, the spin state will now collapse into either $|\backslash psi\_\{y+\}\backslash rangle$ or $|\backslash psi\_\{y-\}\backslash rangle$, each with probability . Let us say, in our example, that we measure . When we now return to measure the particle's spin along the  axis again, the probabilities that we will measure or are each (i.e. they are $\backslash big|\; \backslash langle\; \backslash psi\_\{x+\}\; |\; \backslash psi\_\{y-\}\; \backslash rangle\; \backslash big|^2$ and $\backslash big|\; \backslash langle\; \backslash psi\_\{x-\}\; |\; \backslash psi\_\{y-\}\; \backslash rangle\; \backslash big|^2$ respectively). This implies that the original measurement of the spin along the  axis is no longer valid, since the spin along the  axis will now be measured to have either eigenvalue with equal probability.

The resulting irreducible representations yield the following spin matrices and eigenvalues in the z-basis:

                          \begin{pmatrix}
                            0 & 1 & 0 \\
                            1 & 0 & 1 \\
                            0 & 1 & 0
                          \end{pmatrix}, &
 \left|1, +1\right\rangle_x &= \frac{1}{2} \begin{pmatrix} 1 \\{\sqrt{2}}\\ 1 \end{pmatrix},  &
 \left|1,  0\right\rangle_x &= \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, &
 \left|1, -1\right\rangle_x &= \frac{1}{2} \begin{pmatrix} 1 \\{-\sqrt{2}}\\ 1 \end{pmatrix}  \\
                        S_y &= \frac{\hbar}{\sqrt{2}}
                          \begin{pmatrix}
                            0 & -i  &  0 \\
                            i &  0  & -i \\
                            0 &  i  &  0
                          \end{pmatrix}, &
 \left|1, +1\right\rangle_y &= \frac{1}{2} \begin{pmatrix} -1 \\ -i\sqrt{2} \\ 1 \end{pmatrix}, &
 \left|1,  0\right\rangle_y &= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix},    &
 \left|1, -1\right\rangle_y &= \frac{1}{2} \begin{pmatrix} -1 \\  i\sqrt{2} \\ 1 \end{pmatrix}  \\
                        S_z &= \hbar
                          \begin{pmatrix}
                            1 & 0 &  0 \\
                            0 & 0 &  0 \\
                            0 & 0 & -1
                          \end{pmatrix}, &
 \left|1, +1\right\rangle_z &= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, &
 \left|1,  0\right\rangle_z &= \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, &
 \left|1, -1\right\rangle_z &= \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}  \\
     

 S_x = \frac\hbar2
   \begin{pmatrix}
     0        &\sqrt{3} &0        &0\\
     \sqrt{3} &0        &2        &0\\
     0        &2        &0        &\sqrt{3}\\
     0        &0        &\sqrt{3} &0
   \end{pmatrix}, \!\!\! &
     

 S_y = \frac\hbar2
   \begin{pmatrix}
     0         &-i\sqrt{3} &0         &0\\
     i\sqrt{3} &0          &-2i       &0\\
     0         &2i         &0         &-i\sqrt{3}\\
     0         &0          &i\sqrt{3} &0
   \end{pmatrix}, \!\!\! &
     

 S_z = \frac\hbar2
   \begin{pmatrix}
     3 &0 &0  &0\\
     0 &1 &0  &0\\
     0 &0 &-1 &0\\
     0 &0 &0  &-3
    \end{pmatrix}, \!\!\! &
     

|For spin they are

\begin{pmatrix}
0 &\sqrt{5} &0 &0 &0 &0 \\
\sqrt{5} &0 &2\sqrt{2} &0 &0 &0 \\
0 &2\sqrt{2} &0 &3 &0 &0 \\
0 &0 &3 &0 &2\sqrt{2} &0 \\
0 &0 &0 &2\sqrt{2} &0 &\sqrt{5} \\
0 &0 &0 &0 &\sqrt{5} &0
\end{pmatrix}, \\
     

\begin{pmatrix}
0 &-i\sqrt{5} &0 &0 &0 &0 \\
i\sqrt{5} &0 &-2i\sqrt{2} &0 &0 &0 \\
0 &2i\sqrt{2} &0 &-3i &0 &0 \\
0 &0 &3i &0 &-2i\sqrt{2} &0 \\
0 &0 &0 &2i\sqrt{2} &0 &-i\sqrt{5} \\
0 &0 &0 &0 &i\sqrt{5} &0
\end{pmatrix}, \\
     

\begin{pmatrix}
5 &0 &0 &0 &0 &0 \\
0 &3 &0 &0 &0 &0 \\
0 &0 &1 &0 &0 &0 \\
0 &0 &0 &-1 &0 &0 \\
0 &0 &0 &0 &-3 &0 \\
0 &0 &0 &0 &0 &-5
\end{pmatrix}.
     

 \left(S_x\right)_{ab} & = \frac{\hbar}{2}  \left(\delta_{a,b+1} + \delta_{a+1,b}\right) \sqrt{(s + 1)(a + b - 1) - ab}, \\
 \left(S_y\right)_{ab} & = \frac{i\hbar}{2} \left(\delta_{a,b+1} - \delta_{a+1,b}\right) \sqrt{(s + 1)(a + b - 1) - ab}, \\
 \left(S_z\right)_{ab} & = \hbar (s + 1 - a) \delta_{a,b} = \hbar (s + 1 - b) \delta_{a,b},
     

Also useful in the quantum mechanics of multiparticle systems, the general Pauli group is defined to consist of all -fold tensor products of Pauli matrices.

The analog formula of Euler's formula in terms of the Pauli matrices $$$$

 \hat{R}(\theta, \hat{\mathbf{n}}) = e^{i \frac{\theta}{2} \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}} =
   I \cos \frac{\theta}{2} + i \left(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}\right) \sin \frac{\theta}{2}
     

The spin of pions, a type of elementary particle, was determined by the principle of detailed balance applied to those collisions of protons that produced charged pions and deuterium. $$p\; +\; p\; \backslash rightarrow\; \backslash pi^+\; +d$$ The known spin values for protons and deuterium allows analysis of the collision cross-section to show that $\backslash pi^+$ has spin $s\_\backslash pi\; =\; 0$. A different approach is needed for neutral pions. In that case the decay produced two gamma ray photons with spin one: $$\backslash pi^0\; \backslash rightarrow\; 2\backslash gamma$$ This result supplemented with additional analysis leads to the conclusion that the neutral pion also has spin zero.