Spin (physics)

Author: Leandro Alegsa Creation date: November 13, 2022

In physics, spin is the constant rotation of an object.

For large visible objects like the Earth, spin is the angular momentum of the turning of the Earth around its axis. This tells the amount of rotation that it has. Angular momentum changes with the mass and shape of the object, and with how fast it is turning.

Spin operator, eigenvalues and quantum numbers

The spin operator ${\hat {{\vec {s}}}}=({\hat {s}}_{x},\,{\hat {s}}_{y},\,{\hat {s}}_{z})$ obeys the same three interchange relations as orbital angular momentum operator and total angular momentum:

$[{\hat {s}}_{x},{\hat {s}}_{y}]=i\hbar {\hat s}_{z}$ (also for x,y,z z} cyclically interchanged).

Therefore, all other general rules of quantum mechanical angular momentum also apply here. While for the orbital angular momentum due to ${\hat {{\vec {l}}}}$ $\hat{\vec{l}} \cdot \hat{\vec{p}} =0$ only integer multiples of the quantum of action can occur as eigenvalues, half-integer multiples are also possible as eigenvalues for the spin.

Since the three components are not interchangeable, one chooses as the maximum possible set of interchangeable operators, analogous to orbital angular momentum, the square of magnitude, ${\hat {{\vec {s}}}}^{2}$ , and its -component, ${\hat {s}}_{z}$ (the projection on the axis). An eigenstate of the particle at ${\hat {{\vec {s}}}}^{2}$ has the eigenvalue $s{\mathord {(}}{\mathord {s}}+{\mathord {1}})\,\hbar ^{2}$ ; the set of values for the spin quantum number $\,s$ is thereby $s=0,\,{\tfrac {1}{2}},\,1,\,{\tfrac {3}{2}}\;\dots$ . For abbreviation purposes, a particle with spin quantum number $\,s$ often referred to as "particle with spin $\,s$ ".

The eigenvalues for ${\hat {s}}_{z}$ $\,m_{s}\hbar$ denoted by In this, the magnetic spin quantum number has one of the $\,({\mathord {2}}{\mathord {s}}+{\mathord {1}})$ Values which are $\,m_{s}=-s,\,-(s-{\mathord {1}}),\,\dots ,\,+s$ all together $\,s$ either only half-integer (then in even number) or only integer (then in odd number) depending on the value

Observed values for the spin quantum number of elementary particles are

$\,s=\tfrac{1}{2}$ for all elementary particles of the fermion type, e.g. electron, neutrino, quarks.
$\,s=1$ for the exchange bosons: Photon, Gluon, W-Boson and Z-Boson.
$\,s=0$ for the Higgs boson.

The rules for the addition of two angular momentums apply in exactly the same way to orbital angular momentum and spin. Therefore, the addition of two half-integer angular momentums results in an integer one (as is also the case with two integer ones), while a half-integer and an integer angular momentum add up to a half-integer angular momentum. A system of bosons and fermions therefore has a half-integer total angular momentum exactly when it contains an odd number of fermions.

In the colloquial language of physics, the angular momentum around the centre of gravity is also referred to as spin for many composite particles and quasiparticles (e.g. proton, neutron, atomic nucleus, atom, ...). Here it can also have different values for the same type of particle depending on the excited state of the particle. In these composite systems, the angular momentum is formed from the spins and orbital angular momentums of their fundamental components according to the generally valid rules of quantum mechanical addition. They are not considered further here.

Boson, fermion, particle number conservation

Spin leads to the fundamental and invariable classification of elementary particles into bosons (spin integer) and fermions (spin half-integer). This is a basis of the Standard Model. Thus, the total angular momentum of a fermion in every conceivable state is also half-integer, and that of a boson is integer. It follows further that a system that contains an odd number of fermions in addition to any number of bosons can only have a half-integer total angular momentum, and with an even number of fermions only an integer total angular momentum.

From the theorem of the conservation of the total angular momentum of a system in all possible processes follows the restriction - consistent with observation - that fermions can only be created or annihilated in pairs, never individually, because otherwise the total angular momentum would have to change from an integer to a half-integer value or vice versa. Bosons, on the other hand, can also be created or annihilated individually.

Swap symmetry, statistics, Pauli principle

The classification into bosons (spin integer) and fermions (spin half-integer) has strong implications for the possible states and processes of a system in which several particles of the same kind are present. Since, due to the indistinguishability of particles of the same kind, swapping two of them produces the same physical state of the system, the state vector (or wave function) can also only remain the same or change its sign when this swap takes place. All observations show that for bosons, the first case always applies (symmetry of the wave function in case of interchange), but for fermions, the second case always applies (antisymmetry of the wave function in case of interchange). The direct consequence of the antisymmetry is the Pauli principle, according to which there can be no system that contains two identical fermions in the same one-particle state. This principle determines, for example, the structure of the atomic shell and is thus one of the foundations for the physical explanation of the properties of macroscopic matter (e.g. in the chemical behaviour of the elements in the periodic table and in the (approximate) incompressibility of liquids and solid bodies). The fact that there are two different exchange symmetries explains the great differences between many-particle systems of fermions or bosons. Examples are the electron gas in metal (fermions) or the photons in cavity radiation (bosons), but also the entire astrophysics. In the treatment with statistical methods, fermions follow the Fermi-Dirac statistics, bosons the Bose-Einstein statistics. A profound justification for this connection is provided by the spin-statistics theorem. Although the forces emanating from the spins are mostly negligible (magnetic dipole interaction!) and are usually completely neglected in the theoretical description, the mere property of the particles to have a half- or integer spin thus shows far-reaching consequences in the macroscopically experienceable world.

Spin operator and base states for spin ½

The spin operator ${\hat {{\vec {s}}}}=({\hat {s}}_{x},\,{\hat {s}}_{y},\,{\hat {s}}_{z})$ has for $s={\tfrac {1}{2}}$ three components, each of which has exactly two eigenvalues $\pm {\tfrac {\hbar }{2}}$ However, since the three components satisfy the same interchange relations as any angular momentum operator, no common eigenstates exist. If one chooses (as usual) the alignment along the axis, then the two eigenstates of ${\hat {s}}_{z}$ with the quantum numbers $m_{s}=\pm {\tfrac {1}{2}}$ "parallel" and "antiparallel" to the respectively.} ${\hat {s}}_{x}$ and ${\hat {s}}_{y}$ then have the expected values zero.

Beyond the general properties of quantum mechanical angular momentum, spin ${\tfrac {1}{2}}$ additional special properties. They are based on the fact that ${\hat {s}}_{z}$ has only two eigenvalues. Therefore, the double application of the ascent or descent operator ${\hat {s}}_{{\pm }}={\hat {s}}_{{x}}\pm i{\hat {s}}_{{y}}$ always results in zero: ${\hat {s}}_{{\pm }}^{2}=0$ .

In order to simplify the formulas, Wolfgang Pauli used

${\hat {s}}_{{i}}={\tfrac {\hbar }{2}}{\hat {\sigma }}_{{i}}$ (for $i=x,\,y,\,z$ )

the three Paulian spin operators σ $\sigma _{{x}},\sigma _{{y}},\sigma _{{z}}$ introduced. From ${\hat {s}}_{{\pm }}^{2}=0$ then follows (for $i,j=x,y,z;\ \ i\neq j$ )

${\hat {\sigma }}_{{i}}^{2}=1\ ,\quad {\hat {\sigma }}_{{j}}{\hat {\sigma }}_{{i}}=-{\hat {\sigma }}_{{i}}{\hat {\sigma }}_{{j}}\;,\quad ({\hat {{\vec {\sigma }}}}\cdot {\hat {{\vec {p}}}})^{2}={\hat {{\vec {p}}}}\;^{2}$ .

In addition to last equation ${\hat {\vec {p}}}$ also applies to any other vector operator whose components ${\hat {{\vec {s}}}}$ are interchangeable with each other and with

The inconclusive conclusions:

Because σ ${\hat {\sigma }}_{i}^{2}=1$ ${\hat {s}}_{x}^{2}={\hat {s}}_{y}^{2}={\hat {s}}_{z}^{2}=({\tfrac {\hbar }{2}})^{2}$ . That is, in every conceivable state, a spin- ${\tfrac {1}{2}}$ particle squared to the component of its spin in any direction has a well-determined and always equal value, the largest possible. In the two states of "(anti-)parallel" alignment to the -axis, the two components perpendicular to it are twice as large as the component along the alignment axis. A normal vector with these properties is not parallel to the axis, but even closer to the perpendicular xy-plane.
The component of the vector ${\hat {\vec {p}}}$ in the direction of the spin always has the same magnitude as the vector itself.

The two states $|m_{s}\rangle =\left|\pm {\tfrac {1}{2}}\right\rangle$ (in linguistic usage "spin parallel or antiparallel to the axis", often also referred to $\left|\downarrow \right\rangle$ by the descriptive symbols $\left|\uparrow \right\rangle$ or ) form a basis in the two-dimensional complex state space ${\mathbb C}^{2}$ for the spin degree of freedom of a spin ${\tfrac {1}{2}}$ particle. Also the state with spin parallel to any other direction is a linear combination of these two basis vectors with certain complex coefficients. For the state with spin parallel to the axis, for example, both coefficients have the same magnitude, for the state parallel to the axis also, but with a different complex phase. Even if the spatial directions and are perpendicular to each other, the correspondingly aligned states are not orthogonal (the only $\left|{+{\tfrac {1}{2}}}\right\rangle$ state orthogonal to ∈ $\in {\mathbb C}^{2}$ is $\left|{-{\tfrac {1}{2}}}\right\rangle$ ).

Note: The matrix representation of the Paulian spin operators are the Pauli matrices. Mathematically, the smallest representations of the spin algebra are the spinors.

Spin ½ and three-dimensional vector

The expected value of the angular momentum vector ⟨ $\langle {\hat {{\vec s}}}\rangle =(\langle {\hat s}_{x}\rangle ,\,\langle {\hat s}_{y}\rangle ,\,\langle {\hat s}_{z}\rangle )$ has among all possible values of the angular momentum quantum number (0, 1/2, 1, 3/2, ...) only for spin ½ has the two properties that are vividly associated with a vector in three-dimensional space: In every possible state, it always has the same length $\vert \langle {\hat {{\vec s}}}\rangle \vert ={\tfrac {1}{2}}\hbar$ and always a well-determined direction.

For at any spin state $\vert \chi \rangle =\alpha \left|\uparrow \right\rangle +\beta \left|\downarrow \right\rangle$ (normalised by $\vert \alpha \vert ^{2}+\vert \beta \vert ^{2}=1$ ) is

$\vert \langle {\hat {{\vec s}}}\rangle \vert ^{2}=\langle \chi \vert {\hat s}_{x}\vert \chi \rangle ^{2}+\langle \chi \vert {\hat s}_{y}\vert \chi \rangle ^{2}+\langle \chi \vert {\hat s}_{z}\vert \chi \rangle ^{2}={\tfrac {1}{4}}\hbar ^{2}(\vert \alpha \vert ^{2}+\vert \beta \vert ^{2})^{2}\equiv ({\tfrac {1}{2}}\hbar )^{2}\ .$

Furthermore, it holds that for any spin state (i.e. for any linear combination of $\left|{+{\tfrac {1}{2}}}\right\rangle$ and $\left|{-{\tfrac {1}{2}}}\right\rangle$ ) there is exactly one direction in three-dimensional space, to which the spin is then parallel as in the state $\left|{+{\tfrac {1}{2}}}\right\rangle$ to the -axis. For the linear combination $\left\vert \chi \right\rangle =\alpha \left|\uparrow \right\rangle +\beta \left|\downarrow \right\rangle$ polar angles θ $\theta$ and azimuth angle ϕ $\phi$ of the orientation direction from the equation α}}}}. ${\tfrac {\alpha }{\beta }}={\tfrac {\cos(\theta /2)}{\exp {(i\phi )\,\sin(\theta /2)}}}$ axis.

Among all quantum mechanically possible angular momenta, both are only valid for the quantum number $s={\tfrac {1}{2}}$ . In this respect, among all quantum mechanical angular momenta, the spin ${\tfrac {1}{2}}$ comes closest to the idea of a vector. The vector operator ${\hat {{\vec s}}}=({\hat s}_{x},\,{\hat s}_{y},\,{\hat s}_{z})$ on the other hand has some highly unusual properties (see previous section).

Spin ½ as equivalent of all 2-state systems

If a physical system has only two basic states (at least in approximate terms, e.g. with two neighbouring energy levels, while the other, more distant ones are neglected), it is formally an exact image of the 2-state system for spin ${\tfrac {1}{2}}$ . Three operators can be defined for this system without regard to their physical meaning: An ascend operator and a descend operator transforms the second base state into the first and vice versa, respectively, and otherwise yields zero. The third operator gives the first base state the eigenvalue $+{\tfrac {1}{2}}$ and the second $-{\tfrac {1}{2}}$ . Calling these operators in order ${\hat s}_{+},\,{\hat s}_{-},{\hat s}_{z}$ , they satisfy the same equations as the operators of the same name for spin ${\tfrac {1}{2}}$ . They can also be ${\hat {\vec {s}}}=({\hat {s}}_{x},\,{\hat {s}}_{y},{\hat {s}}_{z})$ rewritten into the vector operator which, like any angular momentum operator, describes the infinitesimal rotations in an (abstract) three-dimensional space due to its interchange relations.

The mathematical background of this equivalence is the fact that the basis transformations in two-dimensional complex Hilbert space form a representation of the group SU(2), which is "twice as large" as the group SO(3) of rotations in real three-dimensional space. The difference to the "normal" rotations in three-dimensional space is that the rotation generated by the spin operator with the angle of rotation 360° is not represented by the unit matrix $\mathbf{1}$ but by $-{\mathbf {1}}$ . Here the physical state merges into itself, but the state vector merges into its negative. One is compatible with the other because state vectors that differ only by a complex factor describe the same state. Only a 720° rotation produces the same state vector again.

If one takes different elementary particles for the two basic states, for example proton and neutron, or electron and electron-neutrino, the physical quantity defined by this procedure is called the isospin of the particle. This also proves useful for multi-particle systems, i.e. their states can be classified according to how the isospins of their individual particles add up to the total isospin, whereby the rules of addition of quantum mechanical angular momentum are fully valid. This isospin concept has played a significant role in the development of elementary particle physics.

Two particles with spin ½

The total spin here can $\,S=0$ have the values $\,S=1$ and With the designation $\left|\uparrow \right\rangle \ ,\left|\downarrow \right\rangle$ for the ground states of each of the particles, the two-particle states with the quantum numbers and $M_{S}$ are thus formed:

$\{\,\left|{\uparrow \uparrow }\right\rangle \ ,\ {\tfrac {1}{\sqrt {2}}}(\left|{\uparrow \downarrow }\right\rangle +\left|{\downarrow \uparrow }\right\rangle )\ ,\ \left|{\downarrow \downarrow }\right\rangle \,\}$ for $\,S=1\;,\ M_{S}=+1,\,0,\,-1$ (triplet)

${\tfrac {1}{\sqrt {2}}}(\left|\uparrow \downarrow \right\rangle -\left|\downarrow \uparrow \right\rangle )$ for $\,S=0,\;M_{S}=0$ (singlet)

The two cases of $M_{S}=0$ (i.e. the component of the total spin is zero) are the simplest examples of an entangled state consisting of two summands each. Here, already in each of the two summands $\left|\uparrow \downarrow \right\rangle$ and $\left|\downarrow \uparrow \right\rangle$ the -components of the two individual spins together are zero. This is no longer true if, instead of the (equally sized) spins, one considers other vector operators that have different sizes for the two particles. E.g. the magnetic moments of electron and proton in the H-atom differ by a factor of about 700. If for the electron with its large magnetic moment for clarification $\left|\Uparrow \right\rangle$ resp. $\left|\Downarrow \right\rangle$ is written, the two $(M_{S}=0)$ - states are calledstates ${\tfrac {1}{{\sqrt {2}}}}(\left|\Uparrow \downarrow \right\rangle \pm \left|\Downarrow \uparrow \right\rangle )$ . While each of the summands here shows a magnetic moment of almost the same magnitude as the electron, aligned in the ( -direction or in the ( )-direction, the total magnetic moment of the atom in such an entangled state has the component zero. From this it can be seen that both summands $\left|\Uparrow \downarrow \right\rangle$ and must be present simultaneously for this to occur. $\left|\Downarrow \uparrow \right\rangle$

Two identical particles with spin ½

Swapping symmetry in spin and location coordinates

The triplet state is symmetrical, the singlet state antisymmetrical with respect to the spins, because swapping the two particles here means writing the two arrows for their spin state in the above formulae in reverse order. Since the complete state vector of two identical fermions changes sign when all their coordinates are swapped, the spin-dependent part existing besides the spin part must $|\psi ({\vec r}_{1},{\vec r}_{2})\rangle$ also have a defined symmetry, antisymmetric in the triplet, symmetric in the singlet. If the spatial coordinates are swapped, the charge distributions of both electrons are simply exchanged, but remain exactly the same in form as before. Nevertheless, if the charge distributions overlap, two different values result for the electrostatic repulsion energy: In the antisymmetrically entangled local state, the amount of energy is smaller than in the symmetrical one, because the probability of both electrons staying at the same place is certainly zero in the antisymmetrical local state, but not in the symmetrical one (in the overlap region). This purely quantum mechanical effect is called the exchange interaction. It accounts for the strong influence of the total spin of the electrons on the energy levels of their atom, although no electrostatic and only slight magnetic interaction emanates from the spins themselves.

The spherically symmetric singlet state

If one does not form the state vector for the singlet state with the spin states aligned in direction $\left|\uparrow \right\rangle \ ,\left|\downarrow \right\rangle$ but with the direction $\left|\leftarrow \right\rangle \ ,\left|\rightarrow \right\rangle$ the state is still one and the same (because there is only one):

${\tfrac {1}{\sqrt {2}}}\;(\,\left|\uparrow \downarrow \right\rangle -\left|\downarrow \uparrow \right\rangle \,)\quad \equiv \quad {\tfrac {1}{\sqrt {2}}}\;(\,\left|\leftarrow \,\rightarrow \right\rangle -\left|\rightarrow \,\leftarrow \right\rangle \,)\cdot$

Formally, this is a sequence of $\left|{\rightarrow }\right\rangle ={\tfrac {1}{\sqrt {2}}}(\,\left|{\uparrow }\right\rangle +\left|{\downarrow }\right\rangle \,)$ and $\left|{\leftarrow }\right\rangle ={\tfrac {1}{\sqrt {2}}}(\,\left|{\uparrow }\right\rangle -\left|{\downarrow }\right\rangle \,)$ .

There is a thought experiment on this that sheds light on the difficulties of the view in understanding the superposition of indivisible particles:

In a He+ ion with the one 1s electron in the state $\left|{\leftarrow }\right\rangle$ , measure the yield with which an electron in the state $\left|{\uparrow }\right\rangle$ can be extracted. Answer: 50 %.
The He+ ion now captures a second electron into the 1s state. Because of the same spatial wave functions of both electrons, the state is symmetrical with regard to the location and antisymmetrical with regard to the spin. The new electron does not simply set its spin opposite to the existing one ( $\left|{\leftarrow \,\rightarrow }\right\rangle$ ), but automatically forms the correct entanglement for the singlet (according to the formula above). This singlet state is (although the vector looks different) the same that would have formed $\left|\uparrow \right\rangle ,\left|\downarrow \right\rangle$ from two electrons in the states
Consequently, now (i.e. after step 2.) the same measurement as in no. 1 (extraction of $\left|{\uparrow }\right\rangle$ ) shows a yield of 100 %. This apparent contradiction "per se" is only compatible with the view trained on macroscopic conditions if both electrons could have "split" and reassembled with the respective correct halves crosswise.

Spin and Dirac equation, anomalous magnetic moment

The theoretical justification of spin ${\tfrac {1}{2}}$ is based on the Dirac equation discovered by Paul Dirac in 1928, which replaces the non-relativistic Schrödinger equation as a relativistically correct wave equation. A condition for relativistic invariance of the associated equation for energy is that energy and momentum occur linearly in it. This is not the case with the Schrödinger equation, because according to classical mechanics it is based on $E={\tfrac {p^{2}}{2m}}$ , in operators: ${\hat {H}}={\tfrac {{\hat {p}}^{2}}{2m}}$ . Dirac found in

${\hat {{\vec \sigma }}}\cdot {\hat {{\vec p}}}={\hat {|{\vec p}|}}$ .

the linear operator we were looking for for the magnitude of the momentum. In the further formulation of this approach, the Paulian $2{\times }2$ matrices σ according to ${\hat {{\vec \sigma }}}$

${\hat {\vec {\alpha }}}={\begin{pmatrix}0&{\hat {\vec {\sigma }}}\\{\hat {\vec {\sigma }}}&0\end{pmatrix}}$

can be extended to $4{\times }4$ matrices. This showed that for a free particle, for which one must therefore assume conservation of angular momentum, it is not the orbital angular momentum ${\hat {\vec {l}}}={\hat {\vec {r}}}\times {\hat {\vec {p}}}$ is a constant of motion, but the quantity identified as total angular momentum ${\hat {\vec {j}}}={\tfrac {\hbar }{2}}{\hat {\vec {\sigma }}}+{\hat {\vec {r}}}\times {\hat {\vec {p}}}$ . The constant additional element ${\hat {\vec {s}}}={\tfrac {\hbar }{2}}{\hat {\vec {\sigma }}}$ is the spin.

If one adds the effect of a static magnetic field to the Dirac equation, the result is an additional energy like that of a magnetic dipole. This dipole is parallel to the spin, just as the magnetic dipole of a circular current is parallel to its orbital angular momentum. However, it has exactly twice the strength compared to the orbital angular momentum of the circular current. The anomalous magnetic moment of the Dirac particle is thus larger than classically understandable by the anomalous spin factor g $g_{s}=2$

Experimentally, however, the electron shows a value of about 2.00232. This deviation of the spin factor of the electron g $g_{e}$ explained by quantum electrodynamics.

Questions and Answers

Q: What is spin in physics?

A: Spin is the constant rotation of an object in physics.

Q: How is spin defined for large visible objects like the Earth?

A: For large visible objects like the Earth, spin is defined as the angular momentum of the turning of the Earth around its axis.

Q: What does the amount of rotation of an object tell us?

A: The amount of rotation of an object tells us its angular momentum.

Q: What factors affect the angular momentum of an object?

A: The angular momentum of an object is affected by its mass, shape, and how fast it is turning.

Q: How does spin apply to objects besides the Earth?

A: Spin applies to all objects, not just the Earth.

Q: Can the amount of spin of an object change?

A: Yes, the amount of spin of an object can change depending on external factors.

Q: How does the mass and shape of an object affect its angular momentum?

A: The mass and shape of an object affect its angular momentum because they determine how much force is required to alter the object's rotation.