Definitions Intrinsic angular momentum A particle is said to possess intrinsic spin s if the quantum-mechanical states of that particle in its own rest frame are eigenstates of the operator
\mathbf{J}^2 with eigenvalue \hbar^2 s(s+1).
Relativistic quantum theory In relativistic quantum theory, spin is defined as an invariant label of a time-like unitary irreducible representation of the
Poincaré group. For time-like representations, corresponding to massive particles, the unitary irreducible representations are characterized by the mass M and the spin s, where s labels an irreducible
projective representation of the rotation group SO(3).
Operator Spin obeys
commutation relations analogous to those of the
orbital angular momentum: \left[\hat S_j, \hat S_k\right] = i \hbar \varepsilon_{jkl} \hat S_l, where is the
Levi-Civita symbol. It follows (as with
angular momentum) that the
eigenvectors of \hat S^2 and \hat S_z (expressed as
kets in the total
basis) are 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 \hat{\boldsymbol{\theta}}, U = e^{-\frac{i}{\hbar} \boldsymbol{\theta} \cdot \mathbf{S}}, where \boldsymbol{\theta} = \theta \hat{\boldsymbol{\theta}}, and is the vector of
spin operators. {{math proof Working in the coordinate system where \hat{\theta} = \hat{z}, we would like to show that and are rotated into each other by the angle . Starting with . Using units where : \begin{align} S_x \rightarrow U^\dagger S_x U &= e^{i \theta S_z} S_x e^{-i \theta S_z} \\ &= S_x + (i \theta) \left[S_z, S_x\right] + \left(\frac{1}{2!}\right) (i \theta)^2 \left[S_z, \left[S_z, S_x\right]\right] + \left(\frac{1}{3!}\right) (i \theta)^3 \left[S_z, \left[S_z, \left[S_z, S_x\right]\right]\right] + \cdots \end{align} Using the
spin operator commutation relations, we see that the commutators evaluate to for the odd terms in the series, and to for all of the even terms. Thus: \begin{align} U^\dagger S_x U &= S_x \left[ 1 - \frac{\theta^2}{2!} + \cdots \right] - S_y \left[ \theta - \frac{\theta^3}{3!} \cdots \right] \\ &= S_x \cos\theta - S_y \sin\theta, \end{align} as expected. Note that since we only relied on the spin operator commutation relations, this proof holds for any dimension (i.e., for any principal spin quantum number ) }} A generic rotation in 3-dimensional space can be built by compounding operators of this type using
Euler angles: \mathcal{R}(\alpha, \beta, \gamma) = e^{-i\alpha S_x} e^{-i\beta S_y} e^{-i\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}, where d^s_{m'm}(\beta) = \langle sm' | e^{-i\beta s_y} | sm \rangle is
Wigner's small d-matrix. Note that for and ; i.e., a full rotation about the axis, the Wigner D-matrix elements become D^s_{m'm}(0, 0, 2\pi) = d^s_{m'm}(0) e^{-i m 2 \pi} = \delta_{m'm} (-1)^{2m}. 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.
Lorentz transformations We could try the same approach to determine the behavior of spin under general
Lorentz transformations, but we would immediately discover a major obstacle. Unlike SO(3), the group of Lorentz transformations
SO(3,1) is
non-compact and therefore does not have any faithful, unitary, finite-dimensional representations. 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 \psi' = \exp{\left(\tfrac{1}{8} \omega_{\mu\nu} [\gamma_{\mu}, \gamma_{\nu}]\right)} \psi, where are
gamma matrices, and is an antisymmetric 4 × 4 matrix parametrizing the transformation. It can be shown that the scalar product \langle\psi|\phi\rangle = \bar{\psi}\phi = \psi^\dagger \gamma_0 \phi is preserved. It is not, however, positive-definite, so the representation is not unitary.
Measurement of spin along the , , or axes Each of the (
Hermitian) Pauli matrices of spin- particles has two
eigenvalues, +1 and −1. The corresponding
normalized eigenvectors are \begin{array}{lclc} \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}. \end{array} (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: \psi = \begin{pmatrix} a + bi \\ c + di \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 \big|\langle \psi_{x+}|\psi\rangle\big|^2. Correspondingly, the probability that its spin will be measured as is just \big|\langle\psi_{x-}|\psi\rangle\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 \big|\langle\psi_{x+} | \psi_{x+}\rangle\big|^2 = 1, etc.), provided that no measurements of the spin are made along other axes.
Measurement of spin along an arbitrary axis The operator to measure spin along an arbitrary axis direction is easily obtained from the Pauli spin matrices. Let be an arbitrary unit vector. Then the operator for spin in this direction is simply S_u = \frac{\hbar}{2} (u_x \sigma_x + u_y \sigma_y + u_z \sigma_z). 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 \frac{1}{\sqrt{2 + 2u_z}} \begin{pmatrix} 1 + u_z \\ u_x + iu_y \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.
Compatibility of spin measurements Since the Pauli matrices do not
commute, measurements of spin along the different axes are incompatible. This means that if, for example, we know the spin along the axis, and we then measure the spin along the axis, we have invalidated our previous knowledge of the axis spin. This can be seen from the property of the eigenvectors (i.e. eigenstates) of the Pauli matrices that \big| \langle \psi_{x\pm} | \psi_{y\pm} \rangle \big|^2 = \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
physicists measure the spin of a particle along the axis as, for example, , the particle's spin state
collapses into the eigenstate |\psi_{x+}\rangle. When we then subsequently measure the particle's spin along the axis, the spin state will now collapse into either |\psi_{y+}\rangle or |\psi_{y-}\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 \big| \langle \psi_{x+} | \psi_{y-} \rangle \big|^2 and \big| \langle \psi_{x-} | \psi_{y-} \rangle \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.
Higher spins The spin- operator forms the
fundamental representation of
SU(2). By taking
Kronecker products of this representation with itself repeatedly, one may construct all higher irreducible representations. That is, the resulting
spin operators for higher-spin systems in three spatial dimensions can be calculated for arbitrarily large using this
spin operator and
ladder operators. For example, taking the Kronecker product of two spin- yields a four-dimensional representation, which is separable into a 3-dimensional spin-1 (
triplet states) and a 1-dimensional spin-0 representation (
singlet state). The resulting irreducible representations yield the following spin matrices and eigenvalues in the z-basis: {{ordered list \begin{align} S_x &= \frac{\hbar}{\sqrt{2}} \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} \\ \end{align} \begin{array}{lclc} 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}, \!\!\! & \left|\frac{3}{2}, \frac{+3}{2}\right\rangle_x =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix} 1 \\{\sqrt{3}}\\{\sqrt{3}}\\ 1 \end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{+1}{2}\right\rangle_x =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix}{-\sqrt{3}}\\ -1 \\ 1 \\{\sqrt{3}}\end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{-1}{2}\right\rangle_x =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix}{\sqrt{3}}\\ -1 \\ -1 \\{\sqrt{3}}\end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{-3}{2}\right\rangle_x =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix} -1 \\{\sqrt{3}}\\{-\sqrt{3}}\\ 1 \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}, \!\!\! & \left|\frac{3}{2}, \frac{+3}{2}\right\rangle_y =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix}{i}\\{-\sqrt{3}}\\{-i\sqrt{3}}\\ 1 \end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{+1}{2}\right\rangle_y =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix}{-i\sqrt{3}}\\ 1 \\{-i}\\{\sqrt{3}}\end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{-1}{2}\right\rangle_y =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix}{i\sqrt{3}}\\ 1 \\{i}\\{\sqrt{3}}\end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{-3}{2}\right\rangle_y =\!\!\! & \frac{1}{2\sqrt{2}} \begin{pmatrix}{-i}\\{-\sqrt{3}}\\{i\sqrt{3}}\\ 1 \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}, \!\!\! & \left|\frac{3}{2}, \frac{+3}{2}\right\rangle_z =\!\!\! & \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{+1}{2}\right\rangle_z =\!\!\! & \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{-1}{2}\right\rangle_z =\!\!\! & \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \!\!\! & \left|\frac{3}{2}, \frac{-3}{2}\right\rangle_z =\!\!\! & \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \\ \end{array} \begin{align} \boldsymbol{S}_x &= \frac{\hbar}{2} \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}, \\ \boldsymbol{S}_y &= \frac{\hbar}{2} \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}, \\ \boldsymbol{S}_z &= \frac{\hbar}{2} \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}. \end{align} \begin{align} \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}, \end{align} where indices a, b are integer numbers such that 1 \le a \le 2s + 1, \quad 1 \le b \le 2s + 1. }} 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} for higher spins is tractable, but less simple. == Parity ==