Derivation of twovaluedness and angular momentum of spin1/2 from rotation of 3sphere
Home > Quantum mechanics > Spin of quantum mechanics
2022/05/25
Published 2013/05/19
K. Sugiyama^{[1]}
In this paper, we derive the twovaluedness and angular momentum of spin1/2 from a rotation of 3sphere.
Figure 38: Wave function of spin1/2 particle
The spin wave function changes its sign with a 360degree rotation. Therefore, if one assumes the existence of an angledependent spin1/2 wave function, the wave function takes on two different values at angles 0 and 360 degrees. This strange behavior is called "twovaluedness of spin1/2." Due to this "twovaluedness of spin1/2," it is believed that there is no angledependent spin1/2 wave function.
To solve this mystery of "twovaluedness of spin1/2" requires a revolution in the concept of the wave function. In the past, Kaluza and Klein interpreted the electromagnetic field function as a onedimensional circle. Similarly, this paper interprets the wave function as a threedimensional sphere.


This threedimensional sphere is decomposed as follows.




The circle in the WX plane in the above equation is shown in Figure 38. When the angle θ rotates 360 degrees, this circle is flipped over. This property is the true nature of "twovaluedness of spin1/2."
CONTENTS
1.3.1 History of research of spin
1.3.2 History of research of manifold
1.4 New construction method of this paper
2 Confirmation of the traditional research of the spin
2.1 Pauli matrices and quaternion
2.2 Confirmation of the experiment of twovaluedness and angular momentum of spin
2.2.1 Verification of twovaluedness of the spin by experiment
2.2.2 Verification of angular momentum of the spin by an experiment
3 Derivation of the twovaluedness and angular momentum of spin
3.1 Derivation of the twovaluedness of spin
3.1.1 Consideration of 3sphere
3.1.2 Taking a view of 2sphere by the simultaneous sections method
3.1.3 Taking a view of 3sphere by the simultaneous sections method
3.1.4 Taking a view of 3sphere by the simultaneous sections method (The other way)
3.1.5 Even torus and odd torus
3.2 Derivation of angular momentum of spin
3.2.1 Construction of 1dimensional helical space
3.2.2 Construction of 2dimensional helical space
3.2.3 Construction of 3dimensional helical space
3.2.4 Consideration of 3dimensional helical space
Kaluza and Klein introduced a onedimensional sphere in extradimensional space to unify gravitational and electromagnetic forces. In this paper, we introduce a threedimensional sphere in extradimensional space to derive the twovaluedness and angular momentum of spin1/2.
Many researchers have attempted to quantize gravity, but without success. This quantization of gravity has become an important issue in physics. One way to quantize gravity was to interpret a point particle as a string, a onedimensional manifold. Therefore, it can be inferred that interpreting the wave function of spin as a manifold is an effective way to quantize gravity.
George Eugene Uhlenbeck and Samuel Abraham Goudsmit discovered the spin of the electron in 1925. Wolfgang Pauli formulated the spin by the Pauli matrices in 1927. Paul Adrien Maurice Dirac derived the spin by the Dirac equation in 1928.
Albert Einstein constructed the general theory of relativity by the 4dimensional Riemann manifold in 1916. Theodor Kaluza^{[2]} and Oskar Klein^{[3]} constructed proposed in the KaluzaKlein theory by the 1dimensional circle in 1926.
In order to express the spin, Pauli defined the following Pauli matrices in 1927.

(2.1) 

(2.2) 

(2.3) 

(2.4) 
The products are shown below.

(2.5) 
We define the unit quaternions by this Pauli matrices.
(Unit quaternions)

(2.6) 

(2.7) 

(2.8) 

(2.9) 
The products are shown below.

(2.10) 
These are the matrix representations of the following unit quaternions.

(2.11) 
William Rowan Hamilton discovered the unit quaternions in 1843.
H. Rauch^{[4]} and S. A. Werner^{[5]} verified the twovaluedness of the spin by the Neutron Interference Experiment in 1975. In this section, we confirm the twovaluedness of the spin.
We can express the wave function of a particle rotating about the zaxis by the Pauli matrices as follows.

(2.12) 
The rotating the angle of the rotation θ by 360 degrees does not bring it back to the same state, but to the state with the opposite phase. The rotating the angle of the rotation θ by 720 degrees brings it back to the original state.
Figure 21: Experiment for verification of twovaluedness of spin
We divide Neutron to path L and path R. Neutron of the path L goes through a domain without magnetic field. Neutron of the path R goes through a domain with magnetic field. As a result, the magnetic field changes the phase of the neutron of path R. Quantity of the change of the phase is as follows.

(2.13) 

(2.14) 
Here, the variable ω is the angular frequency of precession of the spin of the neutron. The variable T is the time neutron passes through the magnetic field. The variable g_{n} is a gfactor. Constant e is the elementary charge. Variable B is the strength of the magnetic field. Variable m is the mass of the neutron.
The neutron that passed along the path L and path R joins at the position I. We can observe it at position E or position F.
Since superposition of a wave function occurs when it joins at position E or position F, we can observe the phase shift. The phase shift was observed as the result of the experiment actually.
It has been clarified that a spin has twovaluedness by this experiment.
Albert Einstein and Wander Johannes de Haas^{[6]} verified the angular momentum of the spin by the following experiment in 1915.
Figure 22: The experiment to verify the angular momentum of the spin
The experiment was performed as follows.
We apply the magnetic field to the disk of the magnetic material. Then, we make the disk stationary state. After that, we stop the magnetic field. Then disk begins to turn around. This effect is called "Einsteinde Haas effect." It has been clarified that a spin has angular momentum by this experiment.
The point particle cannot rotate, because the point particle has the radius of rotation zero. We need infinite momentum to get finite angular momentum by the radius of rotation zero.
We can express angular momentum L by using the radius r and momentum p as follows. The operator × is outer product.

(3.1) 
If L is finite and radius r is zero, momentum p becomes infinite.
On the other hand, we cannot derive the twovaluedness of the spin by a rotation of 2dimensional surface of a sphere (2sphere). Therefore, we consider the rotation of 3dimensional surface of a sphere (3sphere).
We can express 3sphere S^{3} by combining two 3dimensional solid sphere B^{3}_{1} and B^{3}_{2} in the following figure.
Figure 31: 3sphere
3sphere has 6 kinds of spin R_{1}, R_{2}, R_{3}, R_{4}, R_{5}, and R_{6} like the following figure.
Figure 32: Rotation of 3sphere
It is not difficult to consider the spin R_{1}, R_{2}, R_{3}. However, it is difficult to consider spin R_{4}, R_{5}, R_{6}.
It is difficult to consider 3sphere because the 3sphere exists in the 4dimensional space. Then, we try to consider the 3sphere by taking a view of two sections of 3sphere simultaneously. We call the method to take a view of the two sections simultaneously like this the simultaneous sections method.
First, we try to apply the simultaneous sections method to 2sphere because it is easier to consider 2sphere than 3sphere.
We suppose that 2sphere S^{2} in extra 3dimensional space specified by the coordinates (X, Y, Z). If the radius of the 2sphere is 1, 2sphere satisfies the following equation.

(3.2) 
We can express this sphere by a sectional view of XY plane and the position on the Zaxis.

(3.3) 

(3.4) 
Here, angle θ satisfies the following equation.

(3.5) 
We show the 2sphere that is applied the simultaneous sections method to in the following figure.
Figure 33: The simultaneous sections of 2sphere
We express the radius of the circle in the XY plane and position Z at the angle θ.
Table 31: The radius of the circle in the XY plane and position Z at the angle θ
Angle θ 
Radius of the circle in the XY plane 
Position Z 
0° 
0 
1 
90° 
1 
0 
180° 
0 
1 
We can consider the structure of the 2sphere by taking a view of the radius of the circle in the XY plane and position Z simultaneously, like this.
Then, we apply the simultaneous sections method to 3sphere.
We suppose that 3sphere S^{3} in 4dimensional space specified by the coordinates (W,X,Y,Z). If the radius of the 3sphere is 1, 3sphere satisfies the following equation.

(3.6) 
We can express this sphere by a sectional view of XYZ space and position on the Waxis.

(3.7) 

(3.8) 
We show the 3sphere that is applied the simultaneous sections method to in the following figure.
Figure 34: The simultaneous sections of 3sphere
We express the radius of the sphere in XYZ space plane and position Z at the angle θ.
Table 32: The radius of the sphere in XYZ space and position Z at the angle θ
Angle θ 
Radius of the sphere in XYZ space 
Position Z 
0° 
0 
1 
90° 
1 
0 
180° 
0 
1 
In this section, we divided 3sphere to 2sphere and position on an axis. However, we can divide 3sphere by the other way, too. We consider the way in the next section.
We suppose that 3sphere S^{3} in extra 4dimensional space specified by the coordinates (W,X,Y,Z). If the radius of the 3sphere is 1, 3sphere satisfies the following equation.

(3.9) 
We can express this sphere by a sectional view of WX plane and YZ plane.

(3.10) 

(3.11) 
This is the Hopf fibration which Heinz Hopf found in 1931.
We show the 3sphere that is applied the simultaneous sections method to in the following figure.
Figure 35: The simultaneous sections of 3sphere (The other way)
We express the radius of the circle in WX plane and circle in the YZ plane at the angle θ.
Table 33: The radius of the circle in WX plane and circle in the YZ plane at the angle θ
Angle θ 
Radius of the circle in WX plane 
Radius of the circle in the YZ plane 
0° 
0 
1 
90° 
1 
0 
180° 
0 
1 
Here we can connect the circle in the WX plane at the angle θ = 0° and the circle in the WX plane at the angle θ = 180° because they have the same radius 0. In addition, we can also connect the circle in the YZ plane at the angle θ = 0° and the circle in the YZ plane at the angle θ = 180° because they have the same radius 1.
Therefore, we can interpret the angle θ as an angle of rotation of the manifold.
This rotation turns the circle inside out. For example, the circle in the YZ plane is turned inside out at the angle of rotation θ = 180°. Therefore, this rotation is strange spin that is different from the normal spin.
We call the strange spin "toric spin." In addition, we call normal spin "spheric spin."
Here we express 3sphere as follows.

(3.12) 

(3.13) 
We express the 3sphere by the simultaneous sections method in the following figure.
Figure 36: Wave function of spin1 particle(WXθ)
Figure 37: Wave function of spin1 particle(YZθ)
We can interpret the above torus as a wave function of spin1 particle. We can express it by the complex function as follows.

(3.14) 
Next, we express 3sphere as follows.

(3.15) 

(3.16) 
We can express the 3sphere by the simultaneous sections method in the following figure.
Figure 38: Wave function of spin1/2 particle (WXθ)
Figure 39: Wave function of spin1/2 particle (YZθ)
We can interpret the above torus as a wave function of spin1/2 particle. We can express it by the complex function as follows.

(3.17) 
Here we express 3sphere as follows.

(3.18) 

(3.19) 
Variable n is an integer. We call the torus that has even n even torus. We call the torus that has odd n odd torus.
In this paper, we interpret spin as a rotation of 3sphere. Why does the rotation of 3sphere have the same angular momentum as the angular momentum in the 3dimensional normal space.
In this section, we consider the possibility that the 3sphere connects to the 3dimensional normal space.
We can construct 1dimensional helical space as follows.
Figure 310: Construction of 1dimensional helical space
We explain the transformation of each step in the following table.
Table 34: Construction of 1dimensional helical space
Step 
Method of construction 

1 
If we remove one point from a circle, we can get an arc. On the other hand, if we remove one point from a segment of a line, we can get two boundaries. 

2 
We connect their boundaries. 

3 
If we repeat this process, we can connect many circles. 

4 
If we change the orientation of the circle, we can construct 1dimensional helical space. 

We can express 1dimensional helical space of the matrix representations of complex numbers as follows.

(3.20) 
The variables T and X are the coordinates of an extra space. The variable R is a radius of the extra space. The variable Θ is the angle in the extra space.
The symbols {E, I} are matrix representations of complex numbers.

(3.21) 

(3.22) 

(3.23) 
We express the coordinate x of a normal space by a wavelength λ as follows.

(3.24) 
Figure 311: 1dimensional helical space
If we combine the both ends, we can get 1dimensional helical circle.

(3.25) 

(3.26) 
Here, n is an integer. The variable Θ is the angle of rotation of the major radius of helical circle. The variable r is the major radius of helical circle. The variable R is the minor radius of helical circle. The variables {W, X} are coordinates of an extra space. The variables x are coordinates of a normal space.
We express the 1dimensional helical circle in the following figure.
Figure 312: 1dimensional helical circle
Is it possible to do the same thing in 2dimensional space? We consider it in the next section.
We can construct 2dimensional helical space as follows.
Figure 313: Construction of 2dimensional helical space
We explain the transformation of each step in the following table.
Table 35: Construction of 2dimensional helical space
Step 
Method of construction 

1 
If we remove one point from a 2sphere, we can get 2dimensional disk. On the other hand, if we remove one point from a plane, we can get a boundary like a circle. 

2 
We connect their boundaries. 

3 
If we repeat this process, we can connect many 2sphere. 

4 
If we change the orientation of the 2sphere, we can construct 2dimensional helical space. 

We cannot express 2dimensional helical space by the trigonometric functions. We cannot express 2dimensional helical space by the complex function, too. Therefore, I do not guess 2dimensional helical space exists. However, 3 dimensional helical space might exist. We consider it in the next section.
We can construct 3dimensional helical space as follows.
Figure 314: Construction of 3dimensional helical space
We explain the transformation of each step in the following table.
Table 36: Construction of 3dimensional helical space
Step 
Method of construction 

1 
If we remove one point from a 3sphere, we can get 3dimensional solid sphere. On the other hand, if we remove one point from 3dimensional space, we can get a boundary like 2sphere. 

2 
We connect their boundaries. 

3 
If we repeat this process, we can connect many 3sphere. 

4 
If we change the orientation of the 3sphere, we can construct 3dimensional helical space. 

We can express 3dimensional helical space of the matrix representations {E, I, J, K} of unit quaternions as follows.

(3.27) 
The variables {W, X, Y, Z} are coordinates of an extra space. {Θ_{1}, Θ_{2}, Θ_{3}}_{ }are the angle in the extra space. R is a radius of extra space.
The matrix representations {E, I, J, K} of unit quaternions are shown below.

(3.28) 

(3.29) 

(3.30) 

(3.31) 

(3.32) 
We express the coordinate (x, y, z) of the normal space by the wavelength {λ_{1}, λ_{2}, λ_{3}} as follows.

(3.33) 

(3.34) 

(3.35) 
Figure 315: 3dimensional helical space
If we combine the both ends, we can obtain 3dimensional helical sphere.

(3.36) 

(3.37) 

(3.38) 

(3.39) 
Here, {n_{1, }n_{2,} n_{3}} are integers. {Θ_{1}, Θ_{2}, Θ_{3}} are the angles of rotation of the major radius of helical circle. The variable r is the major radius of helical circle. The variable R is the minor radius of helical circle. The variables {x, y, z} are the coordinates of normal space.
We can express the 3dimensional helical sphere symbolically in the following figure.
Figure 316: 3dimensional helical sphere
1dimensional helical space corresponded to complex. On the other hand, 3dimensional helical space corresponded to quaternions. I guess 2dimensional helical space does not exist because triples of numbers do not exist.
We can interpret a position in 3dimensional helical sphere as the position in normal 3dimensional space. Therefore, we can interpret an angular momentum in 3dimensional helical sphere as an angular momentum in normal 3dimensional space. In other words, we can interpret the spin of the quantum mechanics as the rotation of a particle.
In this paper, we derived the following property of the spin.
(1) Twovaluedness of a spin
(2) Angular momentum of a spin
Future issues are shown as follows.
 Derivation of Dirac equation
We introduced the 3sphere as the wave function in 3space in this paper.
We express the 3sphere by the quaternionic functions as follows.

(6.1) 

(6.2) 

(6.3) 

(6.4) 

(6.5) 

(6.6) 
This is the Hopf fibration.
We express the coordinate (W, X) for the rotational angle θ in the following figure.
Figure 61: Wave function of the particle of spin 1
The value of the quaternionic function f of the rotational angle 180 degrees becomes the (1) times of the value of the quaternionic function f of the rotational angle 0 degrees.

(6.7) 
The value of the quaternionic function f of the rotational angle 360 degrees becomes the same value of the quaternionic function f of the rotational angle 0 degrees.

(6.8) 
We interpret the manifold as the wave function of a particle of spin 1. We interpret the rotational angle of the manifold as the phase of the wave function. We interpret the surface area of the manifold as the absolute value of the wave function.
Here we change the angle θ to the half angle.

(6.9) 
Then, we express the quaternionic function as follows.

(6.10) 

(6.11) 

(6.12) 

(6.13) 

(6.14) 
We express the coordinate (W, X ) for the rotational angle θ in the following figure.
Figure 62: Wave function of a particle of spin 1/2
The value of the quaternionic function f of the rotational angle 360 degrees becomes the (1) times of the value of the quaternionic function f of the rotational angle 0 degrees.

(6.15) 
The value of the quaternionic function f of the rotational angle 720 degrees becomes the same value of the quaternionic function f of the rotational angle 0 degrees.

(6.16) 
We interpret the manifold as the wave function of a particle of spin 1/2. We interpret the rotational angle of the manifold as the phase of the wave function. We interpret the surface area of the manifold as the absolute value of the wave function.
Table 71: Spin and so on
# 
Term 
Explanation 
1 
Spin 
Rotation of the object that contains the axis of rotation. 
2 
Spheric spin 
Rotation that does not include the inside out circle. 
3 
Toric spin 
Rotation, including the inside out circle. 
Table 72: Helical space and so on
# 
Category 
Term 
1 
Space 
Helical space 
2 
Circle 
Helical circle 
3 
Sphere 
Helical sphere 
In writing this paper, I thank from my heart to NS who gave valuable advice to me.
[2] Kaluza T., On the Unification Problem of Physics (Zum Unitätsproblem in der Physik), Sitzungsber. Preuss. Akad. Wiss. Berlin. (Math. Phys.), 96, 6972, (1921).
[3] Klein O., Quantum theory and five dimensional theory of relativity (Quantentheorie und fünfdimensionale Relativitätstheorie), Zeitschrift für Physik, A37(12), 895906, (1926).
[4] Rauch H. et al., Verification of coherent spinor rotation of fermions, Phys. Lett., Volume A54, Pages 425427, (1975).
[5] Werner S. A. et al., Observation of the phase shift of a neutron due to precession in a magnetic field, Phys. Rev. Lett., Volume 35, Pages 10531055 (1975).
[6] Einstein A. & de Haas W. J., Experimental Proof of Ampère's Molecular Currents (Experimenteller Nachweis der Ampereschen Molekularströme), Deutsche Physikalische Gesellschaft, Verhandlungen 17, 152170 (1915).