ectron. A general equation that include possible intrinsic dynamics associated with an elementary particle can be written as

2 2 μ 2 Ψ ( r , r 1 , , r n ) + V ( r ) Ψ ( r , r 1 , , r n ) + s = 1 N ( 2 2 μ s s 2 Ψ ( r , r 1 , , r n ) + V s ( r s ) Ψ ( r , r 1 , , r n ) ) = E Ψ ( r , r 1 , , r n ) (49)

where each potential V s ( r s ) is needed to be determined for a particular dynamics associated with the quantum particle under investigation. Even though the quantities m s have the dimension of mass they should be considered as parameters of the equation because they are related to the intrinsic dynamics that must be determined based on the characteristics of the motion under consideration. If all intrinsic dynamics are independent then Equation (49) can be separated into a system of equations as follows

2 2 μ 2 ψ ( r ) + V ( r ) ψ ( r ) = E 0 ψ ( r ) (50)

2 2 μ 1 1 2 χ ( r 1 ) + V 1 ( r 1 ) χ ( r 1 ) = E 1 χ ( r 1 ) (51)

2 2 μ N N 2 χ ( r N ) + V N ( r N ) χ ( r N ) = E N χ ( r N ) (52)

where E 1 + E 2 + + E N = E . For example, if we assume that there are N 1 two-dimensional and N 2 three-dimensional intrinsic dynamics so that N 1 + N 2 = N , and all intrinsic dynamics have the intrinsic potentials of the form V s ( r s ) = A s / r s then using Equations (8) and (42) we would obtain an expression for the total energy spectrum as

E ( n , n s , m s ) = μ 2 2 ( Z q 2 4 π ε 0 ) 2 1 n 2 s = 1 N 1 A s 2 μ s 2 2 ( n s + m s + 1 2 ) 2 s = 1 N 2 A s 2 μ s 2 2 n s 2 (53)

As an example for the case of a three-dimensional intrinsic dynamics, let us consider an intrinsic dynamics that can be described as a spin dynamics of a photon when it is absorbed and then emitted from a hydrogen atom. If the photon exhibits a three-dimensional intrinsic dynamics then we would obtain not only the normal three-dimensional Schrödinger wave equation for the hydrogen atom but also an intrinsic three-dimensional Schrödinger wave equation for the photon, similar to the system of equations given in Equations (32) and (33). In this case the total energy spectrum can be found as

E ( n , n s ) = μ 2 2 ( Z q 2 4 π ε 0 ) 2 1 n 2 A s 2 μ s 2 2 n s 2 (54)

When the electron of the hydrogen atom at the energy level absorbs a photon and moves to a higher energy level n , we may suggest that the photon also changes its energy levels from the level n s to the level n s . We then obtain the new total energy level

E ( n , n s ) = μ 2 2 ( Z q 2 4 π ε 0 ) 2 1 ( n ) 2 A s 2 μ s 2 2 ( n s ) 2 (55)

If we also assume that the energy difference E ( n , n s ) E ( n , n s ) equals the Planck energy h ν then we obtain

h ν = μ 2 2 ( Z q 2 4 π ε 0 ) 2 ( 1 n 2 1 ( n ) 2 ) + A s 2 μ s 2 2 ( 1 n s 2 1 ( n s ) 2 ) (56)

The quantity μ s may be identified with the mass of a photon. It is seen that, unless the photon is massive, i.e. μ s 0 , Equation (56) reduces to the familiar energy spectrum of the hydrogen atom as shown in quantum mechanics.

6. Conclusion

We have shown in this work the possibility to formulate the spin dynamics associated with a quantum particle using Schrödinger equation in quantum mechanics. Contrary to the general assumption that spin dynamics belongs to the domain of relativistic quantum mechanics that cannot be represented by a wavefunction, we have shown that spin dynamics can be formulated by a non-relativistic Schrödinger wave equation by considering possible intrinsic dynamics conferred on quantum particles. Similar to the normal dynamics, intrinsic dynamics can also be expressed in terms of Schrödinger wave equation by using intrinsic coordinates. Since intrinsic coordinates are independent to external coordinates, the total Schrödinger wave equation can be separated into a system of Schrödinger wave equations each of which can be solved separately to obtain exact solutions and their corresponding eigenvalues for the energy. To illustrate, we have applied the formulations to the spin angular momentum for the electron of a hydrogen atom and shown that the quantum numbers associated with the spin angular momentum can take half-integral values, and these results can be used to explain the Stern-Gerlach experiment and other experiments that involve the electron spin resonance. Furthermore, we have also applied the formulation to a possible spin dynamics associated with the radiation of a photon from a hydrogen atom.

Acknowledgements

We would like to thank the reviewers for their constructive comments and we would also like to thank Jane Gao of the administration of JMP for her editorial advice during the preparation of this work.

Appendix

In this appendix we show in details the formulation of Maxwell field equations from the system of linear first order partial differential equations given in Equation (10) of Section 3. The system of equations given in Equation (10) can be written the following matrix form

( A 0 t + A 1 x + A 2 y + A 3 z ) ψ = A 4 J (1)

where ψ = ( ψ 1 , ψ 2 , ψ 3 , ψ 4 , ψ 5 , ψ 6 ) T , J = ( j 1 , j 2 , j 3 , 0 , 0 , 0 ) T and the matrices A i are given as follows

A 0 = ( 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 ) , A 1 = ( 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 ) , A 2 = ( 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 ) , A 3 = ( 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 ) , A 4 = ( μ 0 0 0 0 0 0 μ 0 0 0 0 0 0 μ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) (2)

The system of equations given in Equation (1) becomes

ψ 1 t + ψ 6 y ψ 5 z = μ j 1 (3)

ψ 2 t + ψ 4 z ψ 6 x = μ j 2 (4)

ψ 3 t + ψ 5 x ψ 4 y = μ j 3 (5)

ψ 4 t + ψ 3 y ψ 2 z = 0 (6)

ψ 5 t + ψ 1 z ψ 3 x = 0 (7)

ψ 6 t + ψ 2 x ψ 1 y = 0 (8)

Using the identification E = ( ψ 1 , ψ 2 , ψ 3 ) and B = ( ψ 4 , ψ 5 , ψ 6 ) , the above system of equations can be rewritten in the familiar form given in classical electrodynamics as

E = ρ e ϵ (9)

B = 0 (10)

× E + B t = 0 (11)

× B ϵ μ E t = μ j e (12)

where the charge density ρ e and the current density j e satisfy the conservation law

j e + ρ e t = 0 (13)

From the matrices A i given in Equation (2) we obtain

A 0 2 = ( 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 ) , A 1 2 = ( 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 ) ,

A 2 2 = ( 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ) , A 3 2 = ( 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ) ,

A 4 2 = ( μ 2 0 0 0 0 0 0 μ 2 0 0 0 0 0 0 μ 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) , A 1 A 2 + A 2 A 1 = ( 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 )

A 1 A 3 + A 3 A 1 = ( 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 ) , A 2 A 3 + A 3 A 2 = ( 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 )

A 0 A i + A i A 0 = 0 for i = 1 , 2 , 3 (14)

Now, if we apply the differential operator ( A 0 / t + A 1 / x + A 2 / y + A 3 / z ) to Equation (1) then we arrive at

(15)

From Equation (15), we obtain the following system of equations for the electric field E = ( E x , E y , E z ) = ( ψ 1 , ψ 2 , ψ 3 )

2 ψ 1 t 2 2 ψ 1 y 2 2 ψ 1 z 2 + x ( ψ 2 y + ψ 3 z ) = μ j 1 t (16)

2 ψ 2 t 2 2 ψ 2 x 2 2 ψ 2 z 2 + y ( ψ 1 x + ψ 3 z ) = μ j 2 t (17)

2 ψ 3 t 2 2 ψ 3 x 2 2 ψ 3 y 2 + z ( ψ 1 x + ψ 2 y ) = μ j 3 t (18)

If the electric field also satisfies Gauss’s law

E = ψ 1 x + ψ 2 y + ψ 3 z = ρ e ϵ (19)

then we obtain the following relations

x ( ψ 2 y + ψ 3 z ) = x ( ρ e ϵ ψ 1 x ) = 2 ψ 1 x 2 + x ( ρ e ϵ ) (20)

y ( ψ 1 x + ψ 3 z ) = y ( ρ e ϵ ψ 2 y ) = 2 ψ 2 y 2 + y ( ρ e ϵ ) (21)

z ( ψ 1 x + ψ 2 y ) = z ( ρ e ϵ ψ 3 z ) = 2 ψ 3 z 2 + z ( ρ e ϵ ) (22)

From Equations (16-18) together with relations given in Equations (20-22), we obtain, in vector form, the wave equation for the electric field as

2 E t 2 2 E = ( ρ e ϵ ) μ J e t (23)

where J e = ( j 1 , j 2 , j 3 ) . Similarly for the magnetic field B = ( B x , B y , B z ) = ( ψ 4 , ψ 5 , ψ 6 ) we obtain the following equations and relations

2 ψ 4 t 2 2 ψ 4 y 2 2 ψ 4 z 2 + x ( ψ 5 y + ψ 6 z ) = 0 (24)

2 ψ 5 t 2 2 ψ 5 x 2 2 ψ 5 z 2 + y ( ψ 4 x + ψ 6 z ) = 0 (25)

2 ψ 6 t 2 2 ψ 6 x 2 2 ψ 6 y 2 + z ( ψ 4 x + ψ 5 y ) = 0 (26)

B = ψ 4 x + ψ 5 y + ψ 6 z = 0 (27)

x ( ψ 5 y + ψ 6 z ) = 2 ψ 4 x 2 (28)

y ( ψ 4 x + ψ 6 z ) = 2 ψ 5 y 2 (29)

z ( ψ 4 x + ψ 5 y ) = 2 ψ 6 z 2 (30)

2 B t 2 2 B = 0 (31)

Cite this paper
Ho, V. (2019) A Formulation of Spin Dynamics Using Schrödinger Equation. Journal of Modern Physics, 10, 1374-1393. doi: 10.4236/jmp.2019.1011091.
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