Abstract
The spin of the electron is 1/2. Why 1/2, and not 1/3 or 0.3? This paper counts the phase acquired when the direction of a state with spin is moved over the sphere, and checks that 1/2 comes out as the ratio of “the 2π of one turn of phase” to “the solid angle 4π of the whole sphere”. No new theorem or law is claimed.
First, carrying the direction around a loop shifts the phase by s times the enclosed solid angle. States of spin s equal to 1/2, 1, 3/2 and 2 were carried around triangles on the sphere. For a triangle of solid angle 5.636735 the phases were, in order, $\mathbf{-2.818368}$, $\mathbf{+0.646450}$, $\mathbf{-2.171918}$ and $\mathbf{+1.292900}$, each equal to −s times the solid angle folded into 2π. Over four triangles and four spins the largest departure was 5.151×10⁻¹⁴.
Second, and this is the core of the paper: for the phase of one loop to be single-valued, s×4π must be an integer multiple of 2π. A closed loop on the sphere can be said to enclose the northern cap or the southern cap. The solid angles of the two caps add up to 4π, and the ratio of the phases counted in the two ways is the factor of a turn through s×4π. That factor is 1 only when s×4π is an integer multiple of 2π. The smallest positive s is 2π÷4π = $\mathbf{0.500000}$; for s equal to 1/3 the factor is $\mathbf{1.732051}$ away from 1, and for 1/4 it is $\mathbf{2.000000}$ away.
Third, a rotation by 2π changes the sign and a rotation by 4π restores it. Rotating the state by 2π about an axis multiplies it by $\mathbf{-1.000000}$ for s equal to 1/2 and 3/2 and by $\mathbf{+1.000000}$ for 1 and 2; rotating by 4π restores all of them. That a particle of spin 1/2 changes sign under a rotation by 2π has been measured in neutron interferometry (Rauch et al. 1975).
Fourth, the 1/2 stays 1/2 when the dimension is raised. Spinors in three to six dimensions, rotated by 2π within one plane, are multiplied by −1 in every dimension and restored by 4π. Under the same rotation a vector is restored. In two dimensions a rotation has a single angle and there is no sphere, so any real s satisfies the multiplication rule.
Fifth, there are three “halves of ħ”, and changing the dimension separates them. The 1/2 of the uncertainty bound is 1/2 in every dimension (Paper 129). The 1/2 of spin is 1/2 in three or more dimensions and undetermined in two. The ħ/2 obtained by dividing h by the mass (Paper 1083) moves with the dimension: 1, 1/2, 1/π. The three are the same 1/2 only in three dimensions.
To be candid, what has been shown is only that the condition for the phase to be single-valued is s×4π = 2π times an integer. Why particles of spin 1/2 exist in nature is not answered. The relation between spin and the statistics of particles is not touched.
Keywords: spin, geometric phase, solid angle, double cover, spinor, dimension, Planck constant.
Scope (what is not claimed)
No new theorem or law is claimed. The geometric phase acquired when a direction is carried around a loop (Berry 1984), the quantisation condition over a whole sphere (Dirac 1931), the change of sign of spin 1/2 under a rotation by 2π, and the fact that fractional spin is allowed in two dimensions (Leinaas and Myrheim 1977, Wilczek 1982) are all known. Why particles of spin 1/2 exist is not answered. What is counted is only that the allowed s are restricted to integer multiples of 1/2; which s appear in nature is not treated. The relation between spin and statistics is not treated. The sign under exchange of particles is not entered into. No theory in two dimensions is built. That s is undetermined in two dimensions is only seen through the multiplication rule of rotations. No claim is made that particles exist in higher dimensions. The matrices are only listed against the dimension. Relation to earlier papers: Paper 906 counted that carrying a polarisation around a closed path shifts its phase by half the enclosed solid angle, and Paper 68 showed that this is the same thing as the phase of a spin-1/2 qubit. Here that “half” is read as the size s of the spin, and it is counted that the phase is s times the solid angle for s equal to 1, 3/2 and 2 as well. Paper 147 showed that the period of the circle is 2π and the period of the sphere is 4π, and that whether quantisation occurs is decided by the coefficients. Here it is seen that the ratio of those two periods is the smallest spin, 1/2. Paper 104 treated the Clifford algebra in which spinors live. Here those matrices are actually built in three to six dimensions and it is counted that a rotation by 2π gives −1. Paper 129 showed that the 4π in the uncertainty bound ħ/2 is the product of 2 and 2π and not a solid angle. Here it is placed beside the fact that the 4π inside the 1/2 of spin is a genuine solid angle. Paper 1083 showed that setting the mass to 4π turns h/m into ħ/2 and that its 2 moves with the dimension. Here it is counted that the 1/2 of spin, which has the same form 2π÷4π, does not move when the dimension is raised. What is added: computing that the phase is s times the solid angle for general s, obtaining the condition s×4π = 2π times an integer from the ratio of the phases counted on the northern and southern sides, changing the dimension to count that the 1/2 of spin does not move, and separating the three “halves of ħ” by the dimension.
1. The question
Electrons, protons and neutrons have spin $1/2$. Spin is the amount of rotation a particle carries, and measured in units of Planck’s constant $\hbar$ it takes only values that are integer multiples of $1/2$, such as $1/2$, 1 and $3/2$. Why do $1/3$ and $0.3$ not appear? And where does the number $1/2$ come from?
The earlier papers of this corpus come very close to this question. Paper 906 counted that carrying a polarisation around a closed path shifts its phase by half the enclosed solid angle. Paper 68 showed that this is the same thing as the phase of a spin-$1/2$ qubit. This paper extends that “half” to a general size $s$ of the spin and counts the condition $s$ must satisfy.
2. Carrying the direction around a loop shifts the phase by $s$ times the solid angle
Among the states of a particle of spin $s$, consider the one in which the spin is best aligned with a direction $\mathbf{n}$. The direction $\mathbf{n}$ can be represented by a point on the sphere. Moving the direction along a closed path on the sphere and back gives the state a phase. The phase for going around three directions $A$, $B$, $C$ in turn can be measured as the argument of $\langle A|C\rangle\langle C|B\rangle\langle B|A\rangle$.
For each of $s=1/2$, 1, $3/2$ and 2 the spin matrices were built, the state was rotated with the matrix exponential, and this phase was counted. Four triangles were chosen at random.
| Triangle | Solid angle $\Omega$ | $s=1/2$ | $s=1$ | $s=3/2$ | $s=2$ |
|---|---|---|---|---|---|
| 1 | $-0.112725$ | $+0.056363$ | $+0.112725$ | $+0.169088$ | $+0.225451$ |
| 2 | $-0.251788$ | $+0.125894$ | $+0.251788$ | $+0.377683$ | $+0.503577$ |
| 3 | $+5.636735$ | $-2.818368$ | $+0.646450$ | $-2.171918$ | $+1.292900$ |
| 4 | $-5.031613$ | $+2.515806$ | $-1.251573$ | $+1.264233$ | $-2.503146$ |
All sixteen phases agreed with $-s\Omega$ folded into $2\pi$, with a largest departure of $5.151\times10^{-14}$. For the small triangles (1 and 2) the phase is proportional to $s$. For the large triangles (3 and 4) $s\Omega$ exceeds $\pi$ and appears folded back by $2\pi$.
The same holds for a smooth path. The parallel at 60 degrees from the pole encloses a solid angle of $\pi$ on its northern side. Going around this parallel in 768 steps gave a phase of $-1.570790$ for $s=1/2$, $-3.141580$ for $s=1$, $+1.570816$ for $s=3/2$ and $+0.000026$ for $s=2$. The values of $-s\pi$ folded are, in order, $-1.570796$, $-3.141593$, $+1.570796$ and $+0.000000$, and the largest departure is $2.628\times10^{-5}$. The “half the solid angle” counted in Paper 906 is this formula for $s=1/2$.
3. The condition for north and south to give the same answer
There is a catch here. A closed path on the sphere divides the sphere into two caps. The “enclosed solid angle” may be said to be that of the northern cap, or equally that of the southern cap. If the northern solid angle is $\Omega$, the southern one is $4\pi-\Omega$, and the path appears to be traversed in the opposite sense. The phase factor is $e^{-is\Omega}$ counted on the north and $e^{+is(4\pi-\Omega)}$ counted on the south.
One loop cannot have two phases. The ratio of the two factors is $e^{i4\pi s}$, and this must be 1.
$$s\times4\pi = 2\pi\times(\text{integer})\quad\Longleftrightarrow\quad s=\frac{\text{integer}}{2}$$| $s$ | $s\times4\pi\div2\pi$ | Mismatch of the two counts $|e^{i4\pi s}-1|$ |
|---|---|---|
| $1/2$ | $1.000000$ | $0.000000$ |
| $1$ | $2.000000$ | $0.000000$ |
| $3/2$ | $3.000000$ | $0.000000$ |
| $2$ | $4.000000$ | $0.000000$ |
| $1/3$ | $0.666667$ | $1.732051$ |
| $1/4$ | $0.500000$ | $2.000000$ |
| $0.3$ | $0.600000$ | $1.902113$ |
When $s$ is an integer multiple of $1/2$ the mismatch is 0. For $s=1/3$ or $0.3$ the phase counted on the north and the phase counted on the south do not agree. The smallest positive $s$ satisfying the condition is $2\pi\div4\pi=0.500000$.
This is where the $1/2$ of spin comes from. The $2\pi$ in the numerator is the angle through which a phase turns once and returns. The $4\pi$ in the denominator is the solid angle of the whole sphere. Paper 147 counted that the period of the circle is $2\pi$ and the period of the sphere is $4\pi$. The $1/2$ of spin is the ratio of those two periods. It has the same form as the condition Dirac obtained for a magnetic monopole (Dirac 1931).
4. A rotation by 2π changes the sign
The same can be seen by rotating the state directly. The operation of rotating by an angle $\theta$ about an axis was applied to the states with $s=1/2$ to 2.
| $s$ | After a rotation by $2\pi$ (multiple of the identity) | After a rotation by $4\pi$ |
|---|---|---|
| $1/2$ | $-1.000000$ | restored |
| $1$ | $+1.000000$ | restored |
| $3/2$ | $-1.000000$ | restored |
| $2$ | $+1.000000$ | restored |
When $s$ is a half-odd number ($1/2$, $3/2$), a rotation by $2\pi$ does not restore the state but changes its sign. It takes $4\pi$ to restore it. The sign is $(-1)^{2s}$, which corresponds to whether the “$s\times4\pi\div2\pi$” of the previous section is odd or even. That a particle of spin $1/2$ changes sign under a rotation by $2\pi$ has actually been measured, in an experiment that splits neutrons into two paths and rotates only one of them with a magnetic field (Rauch et al. 1975).
5. Changing the dimension
Does this $1/2$ depend on space having three dimensions? Matrices that rotate a spinor (the quantity corresponding to spin $1/2$) were built in $d$ dimensions, and the spinor was rotated within one plane. The matrices are assembled from $d$ matrices that change sign when their order is exchanged and square to 1 (the Clifford algebra treated in Paper 104).
| $d$ | Components of the spinor | Spinor rotated by $2\pi$ | Spinor rotated by $4\pi$ | Vector rotated by $2\pi$ |
|---|---|---|---|---|
| 3 | 2 | $-1.000000$ | restored | restored |
| 4 | 4 | $-1.000000$ | restored | restored |
| 5 | 4 | $-1.000000$ | restored | restored |
| 6 | 8 | $-1.000000$ | restored | restored |
In each of three to six dimensions the spinor is multiplied by $-1$ after $2\pi$ and restored after $4\pi$. That the matrices satisfy their rules, and that rotating the spinor turns a vector through the right angle, was checked in every dimension with departures smaller than $10^{-12}$. The $1/2$ of spin stays $1/2$ when the dimension is raised.
Only two dimensions are different. A rotation of the plane is fixed by a single angle $\theta$, and the operation of multiplying the state by $e^{is\theta}$ satisfies the multiplication rule of rotations for any real $s$ (checked for $s=1/2$, $1/3$ and $0.3$). In a plane there is no sphere of directions. The condition of the previous section, recounting on the north and the south, does not arise at all. It is known that in a two-dimensional world spins that are not integer multiples of $1/2$ are allowed (Leinaas and Myrheim 1977, Wilczek 1982).
6. Three “halves of $\hbar$”
The form $\hbar/2$ appears three times in quantum mechanics. As the lower bound on the product of the uncertainties of position and momentum, as the smallest value of spin, and as the $h/m$ counted in Paper 1083 when the mass is set to $4\pi$. The three are listed against the dimension.
| $d$ | Uncertainty bound | Smallest spin | $h/m$ (mass taken as the surface area of the unit sphere) |
|---|---|---|---|
| 2 | $0.500000$ | undetermined | $1.000000$ |
| 3 | $0.500000$ | $0.500000$ | $0.500000$ |
| 4 | $0.500000$ | $0.500000$ | $0.318310$ |
| 5 | $0.500000$ | $0.500000$ | $0.238732$ |
| 6 | $0.500000$ | $0.500000$ | $0.202642$ |
The three are $1/2$ together only in the row for three dimensions. The $1/2$ of uncertainty comes from the shape of a wave along one axis and does not know the dimension (Paper 129). The $1/2$ of spin is $1/2$ as long as there is a sphere of directions, and is undetermined in two dimensions where there is none. The $1/2$ of $h/m$ moves, because the placed mass changes with the dimension.
The $1/2$ of spin and the $1/2$ of $h/m$ can both be written as “$2\pi\div4\pi$”. The $2\pi$ in the numerator is, in both, one turn of a phase. What differs is the denominator. The $4\pi$ of spin is the solid angle of the whole sphere and cannot be chosen. The $4\pi$ of $h/m$ is a number placed as the unit of mass. As Paper 129 showed, the $4\pi$ in the $\hbar/2=h/(4\pi)$ of uncertainty is neither: it is the product of 2 and $2\pi$. The same characters $4\pi$ stand for three different things in the three formulas.
7. Relation to earlier papers
Paper 68 took a census of the twos that appear in the corpus and counted the 2 of spinors (double cover, 720 degrees) as one that shares its origin with none of the others. This paper has recounted that 2 as “the period $4\pi$ of the sphere $\div$ the period $2\pi$ of the circle”. What the name double cover says is that a rotation by $2\pi$ does not restore the state and a rotation by $4\pi$ does; the $-1.000000$ in the tables is its content.
Paper 906 did not enter into the origin of the phase and gave only the numbers for light. This paper, too, has not derived why the phase is proportional to the solid angle. What has been counted is that the coefficient of proportionality is the size $s$ of the spin, and that unless $s$ is an integer multiple of $1/2$ the bookkeeping on the sphere does not close.
Closing
Closing: the 1/2 of spin is one turn of phase, 2π, divided by the solid angle of the whole sphere, 4π. Carrying the direction of a state of spin s around a loop on the sphere shifts the phase by s times the enclosed solid angle; for s equal to 1/2, 1, 3/2 and 2 the departure over four triangles was at most 5.151×10⁻¹⁴. The same loop gives the same phase counted on the northern and on the southern cap only when s×4π is an integer multiple of 2π, and the smallest s is $\mathbf{0.500000}$. A rotation by 2π multiplies the states with s equal to 1/2 and 3/2 by $\mathbf{-1.000000}$, and a rotation by 4π restores them. Spinors in three to six dimensions are also multiplied by −1 by a rotation of 2π, so the 1/2 of spin does not move when the dimension is raised. In two dimensions it is undetermined. The 1/2 of uncertainty, the 1/2 of spin and the 1/2 of h divided by the mass separate in three ways when the dimension is changed. To be candid, why particles of spin 1/2 exist has not been answered. What has been answered is why spin cannot be 1/3 or 0.3. The separator is whether the 4π in the denominator is the solid angle of the sphere or a placed mass.
References
- Berry, M. V. (1984). Quantal Phase Factors Accompanying Adiabatic Changes. Proceedings of the Royal Society of London A.
- Dirac, P. A. M. (1931). Quantised Singularities in the Electromagnetic Field. Proceedings of the Royal Society of London A.
- Leinaas, J. M., & Myrheim, J. (1977). On the Theory of Identical Particles. Il Nuovo Cimento B.
- Rauch, H., Zeilinger, A., Badurek, G., Wilfing, A., Bauspiess, W., & Bonse, U. (1975). Verification of Coherent Spinor Rotation of Fermions. Physics Letters A.
- Sakurai, J. J. (1994). Modern Quantum Mechanics. Addison-Wesley.
- Wilczek, F. (1982). Quantum Mechanics of Fractional-Spin Particles. Physical Review Letters.
- Yamagishi, Y. (2026). The Qubit as a Hopf Fibration [Paper 68]. YOI: https://www.yamagishiyuuki.com/yoi/68; DOI: 10.5281/zenodo.21642238
- Yamagishi, Y. (2026). The Algebra That Holds the Metric [Paper 104]. YOI: https://www.yamagishiyuuki.com/yoi/104; DOI: 10.5281/zenodo.21836876
- Yamagishi, Y. (2026). $\hbar$ Is Not on the Mathematical Side of the Inequality [Paper 129]. YOI: https://www.yamagishiyuuki.com/yoi/129; DOI: 10.5281/zenodo.22079681
- Yamagishi, Y. (2026). The Same $H^2$, and with Different Coefficients Nothing Is Quantised [Paper 147]. YOI: https://www.yamagishiyuuki.com/yoi/147; DOI: 10.5281/zenodo.22083144
- Yamagishi, Y. (2026). Carrying a Polarisation Around a Closed Path Shifts Its Phase by Half the Enclosed Solid Angle [Paper 906]. YOI: https://www.yamagishiyuuki.com/yoi/906; DOI: 10.5281/zenodo.22803812
- Yamagishi, Y. (2026). Setting $m=4\pi$ Turns $h/m$ into $\hbar/2$, and Its 2 Has a Different Root from the 2 of the Uncertainty Bound [Paper 1083]. YOI: https://www.yamagishiyuuki.com/yoi/1083
Reference
Yamagishi, Y. (2026). The 1/2 of Spin Is One Turn of Phase, 2π, Divided by the Full Solid Angle, 4π [Paper 1087]. YOI: https://www.yamagishiyuuki.com/yoi/1087
ORCID: 0009-0006-8587-3450
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The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 1087. This paper is a preprint and has not been peer reviewed.