# Derivations of Bloch (Majorana–Bloch) equation, von Neumann equation, and
Schrödinger–Pauli equation

Lihong V. Wang

###### Abstract

The transition from classical physics to quantum mechanics has been
mysterious. Here, we derive the space-independent von Neumann equation for
electron spin mathematically from the classical Bloch or Majorana–Bloch
equation, which is also derived. Subsequently, the space-independent
Schrödinger–Pauli equation is derived in both the quantum mechanical and
recently developed co-quantum dynamic frameworks.

###### Abstract

In the classic multi-stage Stern–Gerlach experiment conducted by Frisch and
Segrè, the Majorana (Landau–Zener) or Rabi formulae diverge afar from the
experimental observation while the physical mechanism for electron-spin
collapse remains unidentified. Here, introducing the physical co-quantum
concept provides a plausible physical mechanism and predicts the experimental
observation in absolute units without fitting (i.e., no parameters adjusted)
highly accurately. Further, the co-quantum concept is corroborated by
statistically reproducing exactly the wave function, density operator, and
uncertainty relation for electron spin.

## Keywords

Bloch equation; Majorana–Bloch equation; von Neumann equation; Liouville–von
Neumann equation; Schrödinger–Pauli equation; Schrödinger equation;
Landau–Lifshitz–Gilbert equation; Electron spin

##  1 Introduction

This manuscript is revised from Ref. [1] by adding Appendix B — Derivation of
Bloch or Majorana–Bloch equation.

The Schrödinger equation, as a postulate, is a cornerstone in quantum
mechanics. The transition from classical physics to quantum mechanics,
however, remains a mystery. Various approaches to obtaining the time-dependent
Schrödinger equation have been investigated [2, 3, 4, 5, 6, 7]. Recently,
Schleich et al. generalized the Hamilton–Jacobi equation to reach the
Schrödinger equation [8]. Most notably, Feynman used the path integral to
attain the same equation [9].

We investigate the transition from classical physics to the space-independent
Schrödinger–Pauli equation for electron spin. In classical electrodynamics,
the motion of the magnetic dipole moment of an electron is governed by the
Bloch (Majorana–Bloch) equation. Majorana stated that both the classical and
the quantum-mechanical treatments on spin flip of atoms moving in a magnetic
quadrupole field require integration of the same differential equations [10,
11]. It is known that the space-independent Schrödinger–Pauli equation or von
Neumann equation (also known as the Liouville–von Neumann equation) for a
unitary two-level system can be converted to the Bloch equation or its analog
[12, 13, 14]. However, the inverse conversion that would complete the two-way
transitions has not been found in the literature.

Here, the classical Bloch equation for electron spin is mathematically
converted to the space-independent von Neumann equation for a pure state of a
two-level spin system. Subsequently, the space-independent Schrödinger–Pauli
equation is derived in both frameworks of quantum mechanics and recently
developed co-quantum dynamics (CQD, see Appendix A). Therefore, the inverse
conversion is shown, and the two-way transitions for a pure state of electron
spin between the classical Bloch equation and the space-independent
Schrödinger–Pauli equation are established.

##  2 Derivation from Bloch equation to space-independent von Neumann equation

We start with the classical Bloch equation derived in Appendix B for an
electron in a magnetic field,

$\frac{d\vec{\mu}}{dt}=\gamma\vec{\mu}\times\vec{B},$ (1)

where $\vec{\mu}$ denotes the magnetic dipole moment of the electron, $\gamma$
the gyromagnetic ratio, $t$ time, and $\vec{B}$ the magnetic flux density.
Substitution of $\vec{\mu}=\frac{\hbar}{2}\gamma\hat{\mu}$ yields

$\frac{\hbar}{2}\frac{d\hat{\mu}}{dt}=\frac{\hbar}{2}\gamma\hat{\mu}\times\vec{B}.$
(2)

Here, $\hbar$ denotes the reduced Planck constant, $\frac{\hbar}{2}$ the spin
angular momentum of the electron, and $\hat{\mu}$ a unit vector. The unit
vector is expressed as

$\hat{\mu}=\begin{pmatrix}\sin\theta\cos\phi\\\ \sin\theta\sin\phi\\\
\cos\theta\end{pmatrix},$ (3)

where $\theta$ and $\phi$ denote the polar and azimuthal angles.

We now resort to the Pauli vector,

$\vec{\sigma}=\sigma_{x}\hat{x}+\sigma_{y}\hat{y}+\sigma_{z}\hat{z},$ (4)

as a mathematical tool, which transforms real-space ($x,y,z$) vectors and
operations using complex numbers. The Pauli matrices are given by

$\sigma_{x}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad\sigma_{y}=\begin{pmatrix}0&-i\\\
i&0\end{pmatrix},\quad\sigma_{z}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix},\quad\sigma_{0}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}.$ (5)

Note that the Pauli matrices, related to quaternions, are applied beyond
quantum mechanics.

Multiplying both sides of the Bloch equation (Eq. 2) by $\vec{\sigma}$ from
the right yields

$\frac{\hbar}{2}\left(\frac{d}{dt}\hat{\mu}\right)\cdot\vec{\sigma}=\frac{\hbar}{2}\gamma(\hat{\mu}\times\vec{B})\cdot\vec{\sigma}.$
(6)

Merging the factors on the left-hand side and splitting the right-hand side
produces

$\frac{\hbar}{2}\frac{d}{dt}(\hat{\mu}\cdot\vec{\sigma})=\frac{\hbar}{2}\gamma\left(\frac{1}{2}(\hat{\mu}\times\vec{B})\cdot\vec{\sigma}-\frac{1}{2}(\vec{B}\times\hat{\mu})\cdot\vec{\sigma}\right).$
(7)

Applying the following mathematical identity—known as the Pauli vector
identity—for any two vectors $\vec{a}$ and $\vec{b}$,

$(\vec{a}\cdot\vec{\sigma})(\vec{b}\cdot\vec{\sigma})=(\vec{a}\cdot\vec{b})\sigma_{0}+i(\vec{a}\times\vec{b})\cdot\vec{\sigma},$
(8)

we obtain

$(\hat{\mu}\cdot\vec{\sigma})(\vec{B}\cdot\vec{\sigma})=(\hat{\mu}\cdot\vec{B})\sigma_{0}+i(\hat{\mu}\times\vec{B})\cdot\vec{\sigma}$
(9)

and

$(\vec{B}\cdot\vec{\sigma})(\hat{\mu}\cdot\vec{\sigma})=(\vec{B}\cdot\hat{\mu})\sigma_{0}+i(\vec{B}\times\hat{\mu})\cdot\vec{\sigma}.$
(10)

Subtraction of the above two equations produces

$(\hat{\mu}\cdot\vec{\sigma})(\vec{B}\cdot\vec{\sigma})-(\vec{B}\cdot\vec{\sigma})(\hat{\mu}\cdot\vec{\sigma})=i\left[(\hat{\mu}\times\vec{B})\cdot\vec{\sigma}-(\vec{B}\times\hat{\mu})\cdot\vec{\sigma}\right].$
(11)

This equation can be rewritten as
$\frac{1}{2}\left[(\hat{\mu}\cdot\vec{\sigma}),(\vec{B}\cdot\vec{\sigma})\right]=i(\hat{\mu}\times\vec{B})\cdot\vec{\sigma}$.
Substituting into Eq. 7 yields

$i\hbar\frac{d}{dt}\left(\frac{1}{2}\hat{\mu}\cdot\vec{\sigma}\right)=\frac{1}{2}\hbar\gamma\left((\frac{1}{2}\hat{\mu}\cdot\vec{\sigma})(\vec{B}\cdot\vec{\sigma})-(\vec{B}\cdot\vec{\sigma})(\frac{1}{2}\hat{\mu}\cdot\vec{\sigma})\right).$
(12)

We define

$\rho=\frac{1}{2}(\hat{\mu}\cdot\vec{\sigma}+\sigma_{0}).$ (13)

Adding $\sigma_{0}$ above ensures that $\rho$ has a unit trace.

Substituting Eq. 3–5 into Eq. 13 produces

$\rho=\begin{pmatrix}\cos^{2}\frac{\theta}{2}&\cos\frac{\theta}{2}\sin\frac{\theta}{2}\exp(-i\phi)\\\
\cos\frac{\theta}{2}\sin\frac{\theta}{2}\exp(i\phi)&\sin^{2}\frac{\theta}{2}\end{pmatrix},$
(14)

which reproduces the familiar quantum mechanical density matrix for a pure
state of electron spin. One may express the density matrix using the state
vector in outer-product form as follows:

$\displaystyle\rho=$ $\displaystyle\begin{pmatrix}\cos\frac{\theta}{2}\\\
\sin\frac{\theta}{2}\exp(i\phi)\end{pmatrix}\otimes\begin{pmatrix}\cos\frac{\theta}{2}\\\
\sin\frac{\theta}{2}\exp(i\phi)\end{pmatrix}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}\cos\frac{\theta}{2}\\\
\sin\frac{\theta}{2}\exp(i\phi)\end{pmatrix}\cdot\begin{pmatrix}\cos\frac{\theta}{2}&\sin\frac{\theta}{2}\exp(-i\phi)\end{pmatrix}.$
(15)

We next define

$H=-\frac{1}{2}\hbar\gamma\vec{B}\cdot\vec{\sigma},$ (16)

which reproduces the familiar quantum mechanical Hamiltonian for electron
spin. Note that $\frac{1}{2}\hbar\gamma$ is the magnitude of the magnetic
dipole moment.

Substituting Eq. 13 and 16 into Eq. 12 yields

$i\hbar\frac{d}{dt}(\rho-\frac{1}{2}\sigma_{0})=H(\rho-\frac{1}{2}\sigma_{0})-(\rho-\frac{1}{2}\sigma_{0})H.$
(17)

Eliminating the identity matrix $\sigma_{0}$ (Eq. 5) produces the von Neumann
equation,

$i\hbar\frac{d}{dt}\rho=H\rho-\rho H=[H,\rho].$ (18)

Therefore, the classical Bloch equation is mathematically converted to the
space-independent von Neumann equation for a pure state of electron spin.

##  3 Quantum mechanical derivation from von Neumann equation to space-
independent
Schrödinger–Pauli equation

While the Schrödinger equation naturally evolves to the von Neumann equation
[15], the inverse process holds for a pure state [16]. The quantum mechanical
density matrix for a pure state is given by

$\rho=|\hat{\mu}\rangle\langle\hat{\mu}|.$ (19)

The ket and bra vectors [15] are given by

$|\hat{\mu}\rangle=\begin{pmatrix}\cos\frac{\theta}{2}\\\
\sin\frac{\theta}{2}\exp(i\phi)\end{pmatrix}$ (20)

and

$\langle\hat{\mu}|=\begin{pmatrix}\cos\frac{\theta}{2}&\sin\frac{\theta}{2}\exp(-i\phi)\end{pmatrix}.$
(21)

Substituting Eq. 19 into Eq. 18 yields

$i\hbar\frac{d}{dt}\left(|\hat{\mu}\rangle\langle\hat{\mu}|\right)=\left[H,|\hat{\mu}\rangle\langle\hat{\mu}|\right].$
(22)

Expansion produces

$\left(i\hbar\frac{d}{dt}|\hat{\mu}\rangle\right)\langle\hat{\mu}|+|\hat{\mu}\rangle\left(i\hbar\frac{d}{dt}\langle\hat{\mu}|\right)=(H|\hat{\mu}\rangle)\langle\hat{\mu}|-|\hat{\mu}\rangle(\langle\hat{\mu}|H).$
(23)

Combining terms results in

$\left(i\hbar\frac{d}{dt}|\hat{\mu}\rangle-H|\hat{\mu}\rangle\right)\langle\hat{\mu}|=|\hat{\mu}\rangle\left(-i\hbar\frac{d}{dt}\langle\hat{\mu}|-\langle\hat{\mu}|H\right).$
(24)

Multiplying both sides by $|\hat{\mu}\rangle$ from the right and using
$\langle\hat{\mu}|\hat{\mu}\rangle=1$ yields

$\left(i\hbar\frac{d}{dt}|\hat{\mu}\rangle-H|\hat{\mu}\rangle\right)=|\hat{\mu}\rangle\left(-i\hbar\frac{d}{dt}\langle\hat{\mu}|-\langle\hat{\mu}|H\right)|\hat{\mu}\rangle.$
(25)

If the bra and ket vectors [15] originate from independent realizations, both
sides must vanish for the equation to hold. Therefore, we reach

$i\hbar\frac{d}{dt}|\hat{\mu}\rangle=H|\hat{\mu}\rangle,$ (26)

which is the space-independent Schrödinger–Pauli equation.

Conversely, we now assume that the bra and ket vectors originate from
identical realizations and examine the consequence. Differentiating
$\langle\hat{\mu}|\hat{\mu}\rangle=1$ yields

$\left(\frac{d}{dt}\langle\hat{\mu}|\right)|\hat{\mu}\rangle+\langle\hat{\mu}|\left(\frac{d}{dt}|\hat{\mu}\rangle\right)=0.$
(27)

Substituting into Eq. 25 and using Eq. 19 produces

$(1-\rho)\left(i\hbar\frac{d}{dt}|\hat{\mu}\rangle-H|\hat{\mu}\rangle\right)=0.$
(28)

If the matrix, $1-\rho$, is full rank, multiplying both sides from the left by
the inverse matrix uniquely yields the space-independent Schrödinger–Pauli
equation. However, from Eq. 14, we have the determinant, $|1-\rho|=0$; thus,
the matrix is singular. Consequently, the space-independent Schrödinger–Pauli
equation can be reached as only a sufficient but not necessary condition. To
reach the Schrödinger–Pauli equation as a sufficient and necessary condition,
we assume that the bra and ket vectors [15] originate from independent
realizations.

##  4 CQD derivation from von Neumann
equation to space-independent
Schrödinger–Pauli equation

In Section 3, the independent realizations are implicit in the bra and ket
vectors (Eq. 20 and 21). On the basis of explicit independent realizations in
CQD (see Appendix), we repeat the derivation. Henceforth, subscripted $e$ and
$n$ denote the electron and nucleus, respectively, in the same atom.

The CQD pre-collapse state function is denoted by
$|\hat{\mu}_{e}\copyright\hat{\mu}_{n}\rangle$, where the co-quantum,
$\hat{\mu}_{n}$, is prefixed with $\copyright$ for clarity.
$|\hat{\mu}_{e}\copyright\hat{\mu}_{n}\rangle$ represents the principal
quantum, $\hat{\mu}_{e}$, accompanied with $\hat{\mu}_{n}$.

For a given $\hat{\mu}_{e}$, the CQD prediction expressions for two
independent realizations are written in dual spaces as follows:

$|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle=C_{1+}(\hat{\mu}_{e},\hat{\mu}_{n1})|+z\rangle+C_{1-}(\hat{\mu}_{e},\hat{\mu}_{n1})\exp(+i\phi_{e})|-z\rangle$
(29)

and

$\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|=C_{2+}(\hat{\mu}_{e},\hat{\mu}_{n2})\langle+z|+C_{2-}(\hat{\mu}_{e},\hat{\mu}_{n2})\exp(-i\phi_{e})\langle-z|.$
(30)

Numbers in subscripts denote independent realizations. Each binary coefficient
represents either one or zero according to the CQD branching condition.

As shown in Section 2, the von Neumann equation was derived without ensemble
averaging. Thus, we start with the following pre-averaging density operator:

$\rho_{0}=|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|.$
(31)

To illustrate the parallelism with Section 3, we keep the original wording as
much as possible below.

The von Neumann equation becomes

$i\hbar\frac{d}{dt}(|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|)=[H,|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|],$
(32)

where $H$ is assumed to be shared by the two realizations. Expansion yields

$\displaystyle\left(i\hbar\frac{d}{dt}|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\right)\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|+|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\left(i\hbar\frac{d}{dt}\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|\right)$
$\displaystyle=$
$\displaystyle(H|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle)\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|-|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle(\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|H).$
(33)

Rearranging terms produces

$\displaystyle\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|$
$\displaystyle=$
$\displaystyle|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\left(-i\hbar\frac{d}{dt}\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|-\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|H\right).$
(34)

Multiplying both sides by $|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle$ from
the right gives

$\displaystyle\left[\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\right]\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle$
$\displaystyle=$
$\displaystyle|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\left(-i\hbar\frac{d}{dt}\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|-\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle.$
(35)

From Eq. 30, we have

$\displaystyle\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle$
$\displaystyle=$ $\displaystyle
C_{2+}^{2}\langle+z|+z\rangle+C_{2-}^{2}\langle-z|-z\rangle=C_{2+}^{2}+C_{2-}^{2}=1.$
(36)

Therefore, we reach

$\displaystyle\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle$
$\displaystyle=$
$\displaystyle|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\left(-i\hbar\frac{d}{dt}\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|-\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n2}|H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle,$
(37)

which can be written alternatively as

$\displaystyle\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle$
$\displaystyle=$
$\displaystyle|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\left(\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle\right)^{\dagger}|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle.$
(38)

Because this equation holds for any two independent realizations, both sides
must vanish, yielding

$\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle=0$
(39)

and

$\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n2}\rangle=0.$
(40)

Ensemble averaging either equation over all co-quantum realizations produces
the space-independent Schrödinger–Pauli equation,

$\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\rangle=0.$ (41)

Conversely, we now assume that the bra and ket vectors originate from
identical realizations and examine the consequence. Replacing each subscripted
2 with 1 in Eq. 37 yields

$\displaystyle\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle$
$\displaystyle=$
$\displaystyle|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\left(-i\hbar\frac{d}{dt}\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n1}|-\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n1}|H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle.$
(42)

Replacing each subscripted 2 with 1 in Eq. 36 gives

$\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n1}|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle=1.$
(43)

Differentiation yields

$\left(\frac{d}{dt}\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n1}|\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle+\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n1}|\left(\frac{d}{dt}|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\right)=0.$
(44)

Substitution into Eq. 42 gives

$\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle=|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle\langle\hat{\mu}_{e}\copyright\hat{\mu}_{n1}|\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle.$
(45)

Replacing each subscripted 2 with 1 in Eq. 31 and substituting it into the
above equation produces

$(1-\rho_{0})\left(i\hbar\frac{d}{dt}-H\right)|\hat{\mu}_{e}\copyright\hat{\mu}_{n1}\rangle=0.$
(46)

If the matrix, $1-\rho_{0}$, is full rank, multiplying both sides from the
left by the inverse matrix uniquely yields the space-independent
Schrödinger–Pauli equation. However, from Eq. 14, which holds for $\rho_{0}$
as well, we have the determinant, $|1-\rho_{0}|=0$; thus, the matrix is
singular. Therefore, the space-independent Schrödinger–Pauli equation can be
reached as only a sufficient but not necessary condition. To reach the
Schrödinger–Pauli equation as a sufficient and necessary condition, we assume
that the bra and ket vectors [15] originate from independent realizations,
which is explicit here.

##  5 Discussion and summary

Quantum mechanics, celebrated for its countless triumphs, still poses open
questions as discussed insightfully in recent monographs [17, 18, 19, 20].
Various thought experiments have been proposed [21, 22, 23]. The transition
from classical physics to quantum mechanics remains an open question. In the
Copenhagen interpretation, an electron spin is considered to be simultaneously
in both eigenstates, and its wavefunction collapses statistically upon
measurement to either eigenstate [13]. The collapse of wavefunction is stated
separately as a measurement postulate because it cannot be modeled by the
original Schrödinger equation [18].

We now extend the Bloch equation to the Landau–Lifshitz–Gilbert equation [24],

$\frac{d\hat{\mu}}{dt}=\gamma\hat{\mu}\times\vec{B}-k_{i}\hat{\mu}\times\frac{d\hat{\mu}}{dt}.$
(47)

Here, the dimensionless $k_{i}$ is called the induction factor, which is used
in CQD to explain the collapse of wavefunction (see Appendix). Although this
equation was originally intended for condensed matter, the underlying physical
mechanism for the added term is compatible with CQD. In fact, the author had
developed CQD before realizing its connection with the Landau–Lifshitz–Gilbert
equation. Setting $k_{i}=0$ recovers the Bloch equation.

Following the same procedure shown in Section 2, the Landau–Lifshitz– Gilbert
equation (Eq. 47) can be converted to the following nonlinear variant of the
von Neumann equation:

$i\hbar\frac{d}{dt}\rho-\hbar
k_{i}\left[\frac{d\rho}{dt},\rho\right]=[H,\rho].$ (48)

One may compare CQD with the existing quantum mechanical theories for
collapse, e.g., the Ghirardi–Rimini–Weber model [25], continuous spontaneous
localization model [26, 27], and the “Wavefunction Is the System Entity”
(WISE) interpretation [28].

Attempted conversion to a variant of the Schrödinger–Pauli equation has
yielded only

$i\hbar\frac{d}{dt}|\hat{\mu}\rangle-\hbar
k_{i}(1-\rho)\frac{d}{dt}|\hat{\mu}\rangle=H|\hat{\mu}\rangle,$ (49)

where $\rho$ cannot be eliminated yet. This situation may be related to the
fact that the collapse of wavefunction cannot be modeled by the original
Schrödinger equation [18].

In summary, the classical Bloch equation has been shown to lead to the space-
independent von Neumann equation and the space-independent Schrödinger–Pauli
equation, both for a pure state of electron spin. While it is known that the
space-independent Schrödinger–Pauli equation or von Neumann equation for a
unitary two-level system can be converted to the Bloch equation or its analog
[12, 13, 14], the inverse conversion for electron spin is shown here. It is
first shown that the Bloch equation and the space-independent von Neumann
equation are equivalent for a pure state of electron spin. Further conversion
from the space-independent von Neumann equation to the space-independent
Schrödinger–Pauli equation as a both sufficient and necessary condition is
proven under the assumption of independent realizations of the bra and ket
vectors, which is implicit in quantum mechanics but explicit in CQD. Without
such an assumption, the space-independent Schrödinger–Pauli equation is only a
sufficient but not necessary condition to either the classical Bloch equation
or the space-independent von Neumann equation. The presented transition from
classical physics to quantum mechanics can potentially lead to new insight
into some of the open questions.

## Acknowledgments

The author thanks Dr. Zhe He for discussing the manuscript and verifying the
derivations, Prof. JT Shen for discussing the work, and Dr. Kelvin Titimbo and
Suleyman Kahraman for discussing Appendix B.

## Declarations

Funding: No funding was received for conducting this study.
Conflicts of interest/Competing interests: The authors have no relevant
financial or non-financial interests to disclose.
Availability of data and material: All data used in this study are available
from the author upon reasonable request.
Code availability: All custom codes used in this study are available from the
author upon reasonable request.

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##  Appendix A Co-quantum dynamics

An essential excerpt on co-quantum dynamics (CQD)[29, 30, 31, 32] is provided
below for completeness.

In typical Stern–Gerlach experiments, the dominant motion of $\hat{\mu}_{e}$
is precession about the main field, and the secondary motion is collapse due
to induction—with the following trend:

$\tan\frac{\theta_{e}(t)}{2}=\tan\frac{\theta_{e}(0)}{2}\exp\left[-\text{sgn}(\theta_{n}-\theta_{e})k_{i}|\Delta\phi_{e}(t)|\right].$
(50)

Here, $\Delta\phi_{e}$ denotes the traversed azimuthal angle (i.e., the
phase). As time evolves, $\theta_{e}$ approaches either 0 or $\pi$ according
to the following branching condition:

$\text{sgn}(\theta_{n}-\theta_{e})=\begin{cases}1&\text{if
}\theta_{n}>\theta_{e},\\\ 0&\text{if }\theta_{n}=\theta_{e},\\\
-1&\text{else}.\end{cases}$ (51)

Therefore, $\hat{\mu}_{e}$ collapses to either $+z$ or $-z$.

The CQD pre-collapse state function is denoted by
$|\hat{\mu}_{e}\copyright\hat{\mu}_{n}\rangle$, where the co-quantum,
$\hat{\mu}_{n}$, is prefixed with $\copyright$ for clarity.
$|\hat{\mu}_{e}\copyright\hat{\mu}_{n}\rangle$ represents $\hat{\mu}_{e}$
accompanied with $\hat{\mu}_{n}$.

The CQD prediction expression for Stern–Gerlach experiments is written as

$|\hat{\mu}_{e}\copyright\hat{\mu}_{n}\rangle=C_{+}(\hat{\mu}_{e},\hat{\mu}_{n})|+z\rangle+C_{-}(\hat{\mu}_{e},\hat{\mu}_{n})\exp(i\phi_{e})|-z\rangle.$
(52)

The equal sign functions as a right arrow ($\rightarrow$) because the right-
hand side predicts the measurement outcome. A given $\hat{\mu}_{e}$ collapses
to either $+\hat{z}$ or $-\hat{z}$ according to the branching condition (Eq.
51). The two real and positive $C$ coefficients take on mutually exclusive
binary values while $\exp(i\phi_{e})$ captures the phase information. If
$\theta_{n}>\theta_{e}$, then $C_{+}=1$ and $C_{-}=0$; if
$\theta_{n}<\theta_{e}$, $C_{+}=0$ and $C_{-}=1$. In either case, $C_{+}\cdot
C_{-}=0$ and $C_{+}+C_{-}=1$.

##  Appendix B Derivation of Bloch or Majorana–Bloch
equation

Majorana formulated the equation commonly known as the “Bloch equation” [10,
11] fourteen years before Bloch published his version in 1946 [33]. Whereas
Majorana applied the equation to an atom, Bloch asserted that it is valid only
for macroscopic magnetization. The author proposes that the Bloch equation be
renamed the Majorana–Bloch equation. Since the derivation of the Bloch
equation is absent in both Majorana’s and Bloch’s works, as well as in the
literature, to the best of our knowledge, we present our derivation for
completeness using two models below.

###  B.1 Current-loop model

We first use a current loop to model the angular momentum $\vec{S}$ and the
magnetic moment $\vec{\mu}$. The Newton’s second law in a rotational system
states

$\frac{d}{dt}\vec{S}=\vec{\tau},$ (53)

where $\vec{\tau}$ denotes the torque.

In the presence of an external magnetic field $\vec{B}$, the differential
torque due to the Lorentz force can be written as

$\displaystyle d\vec{\tau}$
$\displaystyle=\vec{r}\times\left(Id\vec{r}\times\vec{B}\right)$
$\displaystyle=I\left(\vec{r}\cdot\vec{B}\right)d\vec{r}-I\left(\vec{r}\cdot
d\vec{r}\right)\vec{B}$
$\displaystyle=I\left(\vec{r}\cdot\vec{B}\right)d\vec{r},$ (54)

where $\vec{r}$ denotes the position vector relative to the center of the
loop, $I$ the current, and $d\vec{r}$ a differential segment of the loop.

Without losing generality, we center-align the loop with the $xy$ plane and
further align the $x$-axis such that $B_{y}=0$. Using the azimuthal angle
$\phi$, we express

$\vec{r}=R\cos(\phi)\hat{x}+R\sin(\phi)\hat{y},$ (55)

where $R$ denotes the radius of the current loop. Consequently, we have

$\vec{r}\cdot\vec{B}=R\cos(\phi)B_{x}$ (56)

and

$d\vec{r}=-R\sin(\phi)d\phi\hat{x}+R\cos(\phi)d\phi\hat{y}.$ (57)

Substituting the preceding two equations into Eq. 54, we complete the integral
$\vec{\tau}=\int d\vec{\tau}$ over the loop, yielding

$\vec{\tau}=\pi IR^{2}\hat{z}\times\vec{B}.$ (58)

Note that the magnetic moment of a current loop is defined as the product of
the current and the area enclosed by the loop, with the direction given by the
right-hand rule (perpendicular to the plane of the loop):

$\mu=I\cdot{\pi R^{2}}.$ (59)

Therefore, we obtain

$\vec{\tau}=\vec{\mu}\times\vec{B}.$ (60)

Substitution of this equation into Eq. 53 yields

$\frac{d}{dt}\vec{S}=\vec{\mu}\times\vec{B}.$ (61)

Using the gyromagnetic ratio,

$\gamma=\frac{\mu}{S},$ (62)

we obtain the Bloch or Majorana–Bloch equation:

$\frac{d}{dt}\vec{\mu}=\gamma\vec{\mu}\times\vec{B}.$ (63)

Given the charge $q$ and mass $m$ rotating at the angular velocity $\omega$,
we compute the gyromagnetic ratio classically as

$\gamma=\frac{\mu}{S}=\frac{I\cdot\pi
R^{2}}{mR^{2}\cdot\omega}=\frac{\frac{1}{2}q\omega R^{2}}{m\omega
R^{2}}=\frac{q}{2m}.$ (64)

###  B.2 Point-particle model

We next consider a point-particle model. For comparison, we keep some of the
above descriptions verbatim. The point mass $m$ circles with velocity
$\vec{v}$, subject to an intrinsic force $\vec{F_{i}}$ that governs the
circular motion, in addition to an extrinsic force $\vec{F_{e}}$. Newton’s
second law states

$m\frac{d}{dt}\vec{v}=\vec{F_{i}}+\vec{F_{e}}.$ (65)

Cross-multiplying the equation with the position vector of the point particle
$\vec{r}$ from the left-hand side yields

$\vec{r}\times m\frac{d}{dt}\vec{v}=\vec{r}\times\vec{F_{e}},$ (66)

where $\vec{r}\times\vec{F_{i}}\approx 0$ because
$\vec{F_{i}}\parallel\vec{r}$ holds over one cycle $T$ if the circular motion
is much faster than precession.

When $\vec{F_{e}}$ is provided by an external magnetic field $\vec{B}$ on the
charge $q$ of the point mass, substituting the Lorentz force formula yields

$\vec{r}\times
m\frac{d}{dt}\vec{v}=\vec{r}\times\left(q\vec{v}\times\vec{B}\right).$ (67)

Adding the above equation with

$\left(\frac{d}{dt}\vec{r}\right)\times m\vec{v}=\vec{v}\times m\vec{v}=0,$
(68)

we update the left-hand side and reach

$\frac{d}{dt}\left(\vec{r}\times
m\vec{v}\right)=\vec{r}\times\left(q\vec{v}\times\vec{B}\right).$ (69)

The left-hand side is related to the instantaneous angular momentum

$\vec{S_{i}}=\vec{r}\times m\vec{v},$ (70)

and the right-hand side is the instantaneous torque

$\vec{\tau_{i}}=\vec{r}\times\left(q\vec{v}\times\vec{B}\right).$ (71)

Consequently, we obtain the rewritten form:

$\frac{d}{dt}\vec{S_{i}}=\vec{\tau_{i}}.$ (72)

We now average both sides of the equation over one cycle $T$:

$\left\langle\frac{d}{dt}\vec{S_{i}}\right\rangle=\left\langle\vec{\tau_{i}}\right\rangle.$
(73)

The left-hand side can be expressed as

$\displaystyle\left\langle\frac{d}{dt}\vec{S_{i}}(t)\right\rangle$
$\displaystyle=\frac{1}{T}\int_{t-T/2}^{t+T/2}\frac{d\vec{S_{i}}(\tau)}{\partial\tau}\,d\tau$
$\displaystyle=\frac{1}{T}\left(\vec{S_{i}}(t+T/2)-\vec{S_{i}}(t-T/2)\right).$
(74)

A commuted form can be expressed below using the Leibniz integral rule:

$\displaystyle\frac{d}{dt}\langle\vec{S_{i}}(t)\rangle$
$\displaystyle=\frac{d}{dt}\left(\frac{1}{T}\int_{t-T/2}^{t+T/2}\vec{S_{i}}(\tau)\,d\tau\right)$
$\displaystyle=\frac{1}{T}\left(\vec{S_{i}}(t+T/2)-\vec{S_{i}}(t-T/2)\right).$
(75)

Comparing the above two equations leads to

$\left\langle\frac{d}{dt}\vec{S_{i}}(t)\right\rangle=\frac{d}{dt}\langle\vec{S_{i}}(t)\rangle.$
(76)

We use $\vec{S}$ to denote the average angular momentum
$\langle\vec{S_{i}}(t)\rangle$. One can also derive the above commutation as
follows:

$\displaystyle\left\langle\frac{d\vec{S}_{i}(t)}{dt}\right\rangle$
$\displaystyle=\left\langle\lim_{\Delta t\to 0}\frac{\vec{S}_{i}(t+\Delta
t)-\vec{S}_{i}(t)}{\Delta t}\right\rangle$ $\displaystyle=\lim_{\Delta t\to
0}\frac{\langle\vec{S}_{i}(t+\Delta
t)\rangle-\langle\vec{S}_{i}(t)\rangle}{\Delta t}$ $\displaystyle=\lim_{\Delta
t\to 0}\frac{\vec{S}(t+\Delta t)-\vec{S}(t)}{\Delta t}$
$\displaystyle=\frac{d\vec{S}(t)}{dt}.$ (77)

Substituting into Eq. 73 yields

$\frac{d}{dt}\vec{S}(t)=\left\langle\vec{\tau_{i}}\right\rangle.$ (78)

We now evaluate the right-hand side of the above equation for the average
torque, denoted by $\vec{\tau}$:

$\displaystyle\vec{\tau}$
$\displaystyle=\left\langle\vec{r}\times\left(q\vec{v}\times\vec{B}\right)\right\rangle$
$\displaystyle=q\left\langle\left(\vec{r}\cdot\vec{B}\right)\vec{v}-\left(\vec{r}\cdot\vec{v}\right)\vec{B}\right\rangle$
$\displaystyle=q\left\langle\left(\vec{r}\cdot\vec{B}\right)\vec{v}\right\rangle.$
(79)

Without losing generality, we center-align the circular trajectory with the
$xy$ plane and further align the $x$-axis such that $B_{y}=0$. The position
vector $\vec{r}$ versus time $t$ can be expressed as

$\vec{r}=R\cos(\omega t)\hat{x}+R\sin(\omega t)\hat{y},$ (80)

where $R$ denotes the radius of the circular motion and $\omega$ the angular
velocity. As the average is over one cycle, we choose $t$ such that the
initial phase is $0$. Consequently, we have

$\vec{r}\cdot\vec{B}=R\cos(\omega t)B_{x}$ (81)

and

$\vec{v}=\frac{d\vec{r}}{dt}=-R\omega\sin(\omega t)\hat{x}+R\omega\cos(\omega
t)\hat{y}.$ (82)

Substituting the preceding two equations into Eq. 79 and completing the
averaging over one cycle using

$\langle\cdot\rangle=\frac{1}{T}\int_{t-T/2}^{t+T/2}\cdot\;d\tau$ (83)

yield

$\vec{\tau}=\frac{1}{2}qvR\hat{z}\times\vec{B}.$ (84)

Note that the magnetic moment of the equivalent current loop is defined as the
product of the current and the area enclosed by the loop:

$\mu=\frac{q}{\frac{2\pi R}{v}}{\pi R^{2}}=\frac{1}{2}qvR.$ (85)

Therefore, we reach

$\vec{\tau}=\vec{\mu}\times\vec{B}.$ (86)

Substitution of this equation into Eq. 78 yields

$\frac{d}{dt}\vec{S}=\vec{\mu}\times\vec{B}.$ (87)

Using the gyromagnetic ratio,

$\gamma=\frac{\mu}{S},$ (88)

we obtain the Bloch or Majorana–Bloch equation:

$\frac{d}{dt}\vec{\mu}=\gamma\vec{\mu}\times\vec{B}.$ (89)

Given the charge $q$ and mass $m$, we compute the gyromagnetic ratio
classically as

$\gamma=\frac{\mu}{S}=\frac{\frac{1}{2}qvR}{mvR}=\frac{q}{2m}.$ (90)

Here, we use circular motion to emulate spin. In contrast, in relativistic
quantum mechanics, the momentum density in the Dirac wave field includes a
term related to the circulation of energy in the rest frame of the electron
[34].

One can extend the above equation from a point particle to an object of other
shapes, such as a loop or spheroid, rotating at $\omega$. The time evolution
of a differential element is given by

$\frac{1}{\gamma}\frac{d(d\vec{\mu})}{dt}=d\vec{\mu}\times\vec{B}.$ (91)

If $\gamma$ is uniform, integrating both sides over the entire object yields

$\frac{1}{\gamma}\frac{d}{dt}\left(\int{d\vec{\mu}}\right)=\left(\int{d\vec{\mu}}\right)\times\vec{B}.$
(92)

Equating each integral to the total magnetic moment $\vec{\mu}$ yields the
Bloch or Majorana–Bloch equation:

$\frac{1}{\gamma}\frac{d\vec{\mu}}{dt}=\vec{\mu}\times\vec{B}.$ (93)

If $\gamma$ is nonuniform, one may define an effective value using a $g$
factor.