# Reformulating the NMR Quantum Mechanics Fundamental Aspects:
Spin$\tfrac{1}{2}$ Evolution in Magnetic Field Gradients

aTarek Khalil 111E-mail address<EMAIL_ADDRESS>
and
bDaniel Grucker 222 E-mail address<EMAIL_ADDRESS>
###### Abstract

Magnetization of a spin$\tfrac{1}{2}$ set is determined by means of their
individual wave function. The theoretical treatment based on the fundamental
axioms of quantum mechanics and solving explicitly Schrödinger equation gives
the evolution of spin$\tfrac{1}{2}$ system driven by magnetic fields. In this
work we consider the energy of the spin system interacting with magnetic
fields and all the other parts of the energy as a constant reservoir. Solving
this complete Hamiltonian can explain the measured sign change of nuclear spin
noise spectra compared to conventional NMR experiments and the possibility to
make NMR images without radio frequency magnetic fields. The possibility to
make NMR images without RF magnetic fields is an indication that entangled
spin$\tfrac{1}{2}$ states can be manipulated by magnetic field gradients,
opening a new way to perform quantum computation by NMR.

a Department of Physics, Faculty of Sciences(V), Lebanese University,
Nabatieh, Lebanon

b Laboratoire Icube, UMR 7357 CNRS/UdS, Université de Strasbourg, France

Keywords: Quantum spin$\tfrac{1}{2}$ systems, Nuclear-spin noise, Magnetic
resonance imaging.

## Introduction

The work of Norbert Müller and Alexej Jerschow [1] on nuclear spin noise
imaging has reactivated the fundamental idea given by Felix Bloch [2] in 1946
that N nuclei with magnetic moment $\mu$ can be measured, even in absence of
magnetic field, because of statistically incomplete cancellation. A review on
nuclear spin noise which corresponds to this idea can be found in reference
[3]. Generally, most of quantum systems are treated in contact with an
environment which may be of different types. These systems correspond to a so
called open quantum systems [4]. The aim of this work is to give the frame for
a quantum mechanical description of the evolution of a system consisting of an
ensemble of $N$ identical time dependent spin$\tfrac{1}{2}$ moving particles
which experience the presence of an external perturbation, as for instance
main static and gradient magnetic fields with or without radio frequency
fields. In other words we deal with spin$\frac{1}{2}$ system driven by
external time-dependent perturbations. Solving step by step the Schrödinger
equation and defining the complex susceptibility of one spin$\tfrac{1}{2}$ as
the product of the amplitude of one spin state time the complex conjugate of
the other state permits to retrieve the expression of the magnetization, the
absence of a significant induced current by an ensemble of spin$\tfrac{1}{2}$
at equilibrium, the existence of an induced current after a RF magnetic field
pulse in magnetic resonance experiments and the existence of the spin noise.
This derivation should be helpful for new magnetic resonance imaging sequences
or for developing quantum computing with spin $\tfrac{1}{2}$ using magnetic
field gradients, in order to create qubits not only by the different local
magnetic fields existing in a molecule. The theoretical description will be
tested with the main results of NMR experiments.

##  1 Theoretical description

###  1.1 One spin$\tfrac{1}{2}$ in a static magnetic field

For one spin$\tfrac{1}{2}$ in a $\boldsymbol{\vec{B}_{0}}$ magnetic field, it
is well known that the potential part of the Hamiltonian is:

$\displaystyle\begin{split}\mathscr{H}_{1}&=-\boldsymbol{\vec{\mu}}\cdot\boldsymbol{\vec{B}_{0}}=-\gamma\boldsymbol{\vec{I}}_{1}\cdot\boldsymbol{\vec{B}_{0}},\end{split}$
(1)

where $\gamma$ is the gyromagnetic ratio of the spin and
$\boldsymbol{\vec{I}}_{1}$ is the spin that verifies the following equation:

$\boldsymbol{\vec{I}}_{1}=\frac{1}{2}\hbar\boldsymbol{\vec{\sigma}},$ (2)

with $\boldsymbol{\vec{\sigma}}$ a vector defined by the following Pauli
matrices:

$\displaystyle\sigma^{X}=\begin{pmatrix}0&1\\\ 1&0\\\ \end{pmatrix},$
$\displaystyle\sigma^{Y}=\begin{pmatrix}0&-i\\\ i&\phantom{-}0\\\
\end{pmatrix},$ $\displaystyle\sigma^{Z}=\begin{pmatrix}1&\phantom{-}0\\\
0&-1\\\ \end{pmatrix}.$ (3)

The matrix notation of equation (1) is given by:

$\displaystyle\mathscr{H}_{1}$
$\displaystyle=-\tfrac{1}{2}\gamma\hbar\begin{pmatrix}B_{Z}&B_{X}-iB_{Y}\\\
B_{X}+iB_{Y}&-B_{Z}\\\ \end{pmatrix}.$ (4)

By convention, the vector
$\boldsymbol{\vec{B}_{0}}=\left(B_{X},B_{Y},B_{Z}\right)$ is placed in the
$Oz$-direction, that is $\boldsymbol{\vec{B}_{0}}=\left(0,0,B_{0}\right)$ and
the $xOy$-plane is called the transverse plane. Therefore the matrix form of
the Hamiltonian reduces to:

$\displaystyle\mathscr{H}_{1}$
$\displaystyle=-\tfrac{1}{2}\gamma\hbar\begin{pmatrix}B_{0}&\phantom{-}0\\\
0&-B_{0}\\\ \end{pmatrix}.$ (5)

The two eigenvalues of the Hamiltonian $\mathscr{H}_{1}$, which give the
energy of the quantum states, are: $E_{-}=-\frac{1}{2}\gamma\hbar B_{0}$ and
$E_{+}=\frac{1}{2}\gamma\hbar B_{0}$ which have the following two
corresponding eigenvectors:

$\displaystyle\left\lvert-\right\rangle=\begin{pmatrix}0\\\ 1\\\
\end{pmatrix},$ $\displaystyle\left\lvert+\right\rangle=\begin{pmatrix}1\\\
0\\\ \end{pmatrix}.$ (6)

We can describe the wave function as:

$\displaystyle\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}=r_{1}e^{i\phi_{1}}\left\lvert-\right\rangle+r_{2}e^{i\phi_{2}}\left\lvert+\right\rangle=\begin{pmatrix}r_{2}e^{i\phi_{2}}\\\
r_{1}e^{i\phi_{1}}\\\ \end{pmatrix}\text{with}$ $\displaystyle
r_{1}^{2}+r_{2}^{2}=1$ (7)

The 1 subscript corresponds to the low energy or ground level. The time
evolution of this quantum system (far from the speed of light) is given by the
Schrödinger equation:

$\displaystyle\frac{i\hbar\partial\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}}{\partial
t}$ $\displaystyle=\mathscr{H}\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}.$ (8)

If we write

$\displaystyle\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}$
$\displaystyle=\begin{pmatrix}x_{2}(t)\\\ x_{1}(t)\\\ \end{pmatrix},$ (9)

then we have the following differential equations:

$\displaystyle\begin{split}i\hbar\begin{pmatrix}\dot{x}_{2}(t)\\\
\dot{x}_{1}(t)\end{pmatrix}&=-\tfrac{1}{2}\gamma\hbar\begin{pmatrix}B_{0}&\phantom{-}0\\\
0&-B_{0}\\\ \end{pmatrix}\cdot\begin{pmatrix}x_{2}(t)\\\ x_{1}(t)\\\
\end{pmatrix}\\\
&=\tfrac{1}{2}\hbar\begin{pmatrix}\phantom{-}\omega_{0}x_{2}(t)\\\
-\omega_{0}x_{1}(t)\\\ \end{pmatrix},\end{split}$ (10) where
$\displaystyle\omega_{0}$ $\displaystyle=-\gamma B_{0}$ (11)

where $\omega_{0}$ is the resonance or Larmor frequency. The following time
dependent wave function is the obvious solution of this system of equations:

$\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}=\begin{pmatrix}r_{2}e^{-i\bigl{(}\tfrac{\omega_{0}t}{2}-\phi_{2}\bigr{)}}\\\
r_{1}e^{+i\bigl{(}\tfrac{\omega_{0}t}{2}+\phi_{1}\bigr{)}}\\\
\end{pmatrix}=\begin{pmatrix}e^{-i\tfrac{\omega_{0}t}{2}}&0\\\
0&e^{+i\tfrac{\omega_{0}t}{2}}\\\
\end{pmatrix}\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}=A(\omega_{0},t)\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}$
(12)

which defines the spin evolution time operator $A(\omega_{0},t)$ of
spin$\tfrac{1}{2}$ in a static magnetic field. We notice that $A$ is a $4\pi$
period function and that the $\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}$ is the
wave function given by equation (7).

###  1.2 The effect of a RF magnetic field

In magnetic resonance experiments, a transition between the two states
$\left\lvert-\right\rangle$ and $\left\lvert+\right\rangle$ is obtained by a
$B_{1}$ magnetic field rotating in the transverse plane, the $(xOy)$-plane.
This field is generated by an electromagnetic RF wave obtained by an
oscillating current in a solenoid surrounding the spin system. In this case
the Hamiltonian has non-diagonal elements due to the RF magnetic field
$\boldsymbol{\vec{B}_{1}}=(B_{1}cos(\omega t)),B_{1}sin(\omega t),0)$ rotating
around $\boldsymbol{\vec{B}_{0}}$ with an angular velocity $\omega$:

$\displaystyle\mathscr{H}_{\mathrm{RF}}(t)$
$\displaystyle=-\frac{1}{2}\gamma\hbar\begin{pmatrix}B_{0}&B_{1}e^{-i\omega
t}\\\ B_{1}e^{i\omega t}&-B_{0}\\\ \end{pmatrix}.$ (13)

In this case the time evolution is no longer trivial. We have the following
differential equations:

$\displaystyle\begin{split}\frac{i\hbar\partial\bigl{\lvert}\psi_{\mathrm{RF}}(t)\bigr{\rangle}}{\partial
t}&=\mathscr{H}_{\mathrm{RF}}(t)\begin{pmatrix}x_{2}(t)\\\ x_{1}(t)\\\
\end{pmatrix}\\\
&=\tfrac{1}{2}\hbar\begin{pmatrix}\omega_{0}&\omega_{1}e^{-i\omega t}\\\
\omega_{1}e^{i\omega t}&-\omega_{0}\\\
\end{pmatrix}\cdot\begin{pmatrix}x_{2}(t)\\\ x_{1}(t)\\\
\end{pmatrix},\end{split}$ (14) where $\displaystyle\omega_{1}$
$\displaystyle=-\gamma B_{1}.$ (15)

At resonance, without RF ($B_{1}=\omega_{1}=0$) these equations obviously
reduce to equation (10) and with RF we need to solve the following
differential equations:

$\displaystyle\begin{pmatrix}\dot{x}_{2}(t)\\\ \dot{x}_{1}(t)\\\
\end{pmatrix}$
$\displaystyle=-\frac{i}{2}\begin{pmatrix}\omega_{0}x_{2}(t)+\omega_{1}e^{-i\omega
t}x_{1}(t)\\\ \omega_{1}e^{i\omega t}x_{2}(t)-\omega_{0}x_{1}(t)\\\
\end{pmatrix}.$ (16)

To solve these two equations we make the following substitutions:

$\displaystyle p(t)$ $\displaystyle=x_{2}(t)e^{i\tfrac{\omega_{0}}{2}t},$
(17a) $\displaystyle q(t)$
$\displaystyle=x_{1}(t)e^{-i\tfrac{\omega_{0}}{2}t}.$ (17b)

Therefore we have:

$\displaystyle\dot{p}(t)$
$\displaystyle=\left(-\frac{i}{2}\left(\omega_{0}x_{2}(t)+\omega_{1}e^{-i\omega
t}x_{1}(t)\right)+\frac{i\omega_{0}}{2}x_{2}(t)\right)e^{i\tfrac{\omega_{0}}{2}t}$
$\displaystyle=-\frac{i\omega_{1}}{2}e^{-i\omega
t}x_{1}(t)e^{i\tfrac{\omega_{0}}{2}t}$ (18)
$\displaystyle=-\frac{i\omega_{1}}{2}q(t)e^{i(\omega_{0}-\omega)t}$ and
$\displaystyle\dot{q}(t)$
$\displaystyle=\left(-\frac{i}{2}\left(\omega_{1}e^{i\omega
t}x_{2}(t)-\omega_{0}x_{1}(t)\right)-\frac{i\omega_{0}}{2}x_{1}(t)\right)e^{-i\tfrac{\omega_{0}}{2}t}$
$\displaystyle=-\frac{i\omega_{1}}{2}e^{i\omega
t}x_{2}(t)e^{-i\tfrac{\omega_{0}}{2}t}$ (19)
$\displaystyle=-\frac{i\omega_{1}}{2}p(t)e^{-i(\omega_{0}-\omega)t}.$

When we take the second derivative of $p(t)$ we obtain the following second
order differential equation:

$\displaystyle\begin{split}\ddot{p}(t)&=-\frac{i\omega_{1}}{2}\dot{q}(t)e^{i(\omega_{0}-\omega)t}+i(\omega_{0}-\omega)\dot{p}(t)\\\
&=-\frac{i\omega_{1}}{2}\left(-\frac{i\omega_{1}}{2}p(t)e^{-i\bigl{(}(\omega_{0}-\omega)t\bigr{)}}e^{i(\omega_{0}-\omega)t}\right)+i(\omega_{0}-\omega)\dot{p}(t)\\\
&=i(\omega_{0}-\omega)\dot{p}(t)-\frac{\omega_{1}^{2}}{4}p(t).\end{split}$
(20)

Let $\lambda_{\pm}$ be the solutions of the equation

$\lambda^{2}-i(\omega_{0}-\omega)\lambda+\frac{\omega_{1}^{2}}{4}=0.$ (21)

We have

$\displaystyle\lambda_{\pm}=i\frac{\Omega\pm\Delta}{2}$ with
$\displaystyle\Omega=\omega_{0}-\omega$
$\displaystyle\Delta=\sqrt{\Omega^{2}+\omega_{1}^{2}}$ (22)

and

$\displaystyle p(t)$
$\displaystyle=C_{1}e^{\lambda_{+}t}+C_{2}e^{\lambda_{-}t},$ (23)

therefore:

$\displaystyle\begin{split}x_{2}(t)&=p(t)e^{-\tfrac{i\omega_{0}t}{2}}\\\
&=\left(C_{1}e^{\lambda_{+}t}+C_{2}e^{\lambda_{-}t}\right)e^{-\tfrac{i\omega_{0}t}{2}}\\\
&=e^{-i\tfrac{\omega}{2}t}\left(C_{1}e^{i\tfrac{\Delta}{2}t}+C_{2}e^{-i\tfrac{\Delta}{2}t}\right),\end{split}$
(24a) with initial value $\displaystyle x_{2}(0)$
$\displaystyle=r_{2}e^{i\phi_{2}}.$ (24b)

A similar computation for $x_{1}(t)$ leads to:

$\displaystyle x_{1}(t)$ $\displaystyle=e^{\tfrac{i\omega
t}{2}}\left(C_{3}e^{i\tfrac{\Delta}{2}t}+C_{4}e^{-i\tfrac{\Delta}{2}t}\right),$
(25a) $\displaystyle x_{1}(0)$ $\displaystyle=r_{1}e^{i\phi_{1}}.$ (25b)

As $x_{1}(t),x_{2}(t)$ verify the initial value and the differential equations
(16)

$\displaystyle C_{1}+C_{2}=r_{2}e^{i\phi_{2}},$ $\displaystyle
C_{3}+C_{4}=r_{1}e^{i\phi_{1}},$ (26a)
$\displaystyle(\Delta+\Omega)C_{1}+\omega_{1}C_{3}=0,$
$\displaystyle(\Delta-\Omega)C_{2}-\omega_{1}C_{4}=0,$ (26b)

The computation of the constants $C_{i}$ is now straightforward. We have

$\displaystyle C_{1}$
$\displaystyle=\tfrac{1}{2}\left((1-\tfrac{\Omega}{\Delta})r_{2}e^{i\phi_{2}}-\tfrac{\omega_{1}}{\Delta}r_{1}e^{i\phi_{1}}\right),$
(27a) $\displaystyle C_{2}$
$\displaystyle=\tfrac{1}{2}\left((1+\tfrac{\Omega}{\Delta})r_{2}e^{i\phi_{2}}+\tfrac{\omega_{1}}{\Delta}r_{1}e^{i\phi_{1}}\right),$
(27b) $\displaystyle C_{3}$
$\displaystyle=\tfrac{1}{2}\left(-\tfrac{\omega_{1}}{\Delta}r_{2}e^{i\phi_{2}}+(1+\tfrac{\Omega}{\Delta})r_{1}e^{i\phi_{1}}\right),$
(27c) $\displaystyle C_{4}$
$\displaystyle=\tfrac{1}{2}\left(\tfrac{\omega_{1}}{\Delta}r_{2}e^{i\phi_{2}}+(1-\tfrac{\Omega}{\Delta})r_{1}e^{i\phi_{1}}\right).$
(27d)

We can write the evolution of the spin wave function in matrix notation,
therefore equation (12) becomes:

$\displaystyle\bigl{\lvert}\psi_{\mathrm{RF}}(t)\bigr{\rangle}$
$\displaystyle=\begin{pmatrix}x_{2}(t)\\\ x_{1}(t)\\\
\end{pmatrix}=E_{RF}(\omega,t)\cdot
R_{RF}(\omega,\omega_{0},\omega_{1},t)\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}$
(28)

The evolution matrix is now dependent on $\omega$ the RF frequency and
$\omega_{1}$ a frequency proportional to the amplitude of RF magnetic field in
addition to the time and the magnetic field $B_{0}$.
As shown by equations (24a),(25a) and (27), E and R matrices are:

$\displaystyle E_{RF}$ $\displaystyle=\begin{pmatrix}e^{-\tfrac{i\omega
t}{2}}&0\\\ 0&e^{\tfrac{i\omega t}{2}}\\\ \end{pmatrix};$ $\displaystyle
R_{RF}=\begin{pmatrix}\cos\tfrac{\Delta
t}{2}-\tfrac{i\Omega}{\Delta}\sin\tfrac{\Delta
t}{2}&-i\tfrac{\omega_{1}}{\Delta}\sin\tfrac{\Delta t}{2}\\\
-i\tfrac{\omega_{1}}{\Delta}\sin\tfrac{\Delta t}{2}&\cos\tfrac{\Delta
t}{2}+\tfrac{i\Omega}{\Delta}\sin\tfrac{\Delta t}{2}\\\ \end{pmatrix}$ (29)

These expressions are modified if the angular velocity $\omega$ of the RF
magnetic field verifies the resonance condition $\omega=\omega_{0}$. In that
case where $\Omega=0$ and $\Delta=\omega_{1}$, the evolution matrix becomes
$A_{RF}$:

$\displaystyle A_{RF}(\omega_{0},\omega_{1},t)$
$\displaystyle=\begin{pmatrix}e^{-\tfrac{i\omega_{0}t}{2}}&0\\\
0&e^{\tfrac{i\omega_{0}t}{2}}\end{pmatrix}\cdot\begin{pmatrix}\cos\tfrac{\omega_{1}t}{2}&-i\sin\tfrac{\omega_{1}t}{2}\\\\[5.0pt]
-i\sin\tfrac{\omega_{1}t}{2}&\cos\tfrac{\omega_{1}t}{2}\end{pmatrix}.$ (30)

Therefore the frequency reference is the same for $\omega_{0}$ and $\omega$
which gives an experimental way to define a precise time scale with its origin
given by the electrical current used in the RF coil.
In magnetic resonance experiments the effect of a RF pulse is usually
described [5] as a rotation with an angle of $\theta_{1}=\omega_{1}t$ in the
spin space around the vector
$\boldsymbol{\vec{u}}=\left(u_{x},u_{y},u_{z}\right)$ as:

$\displaystyle\boldsymbol{R_{u,\theta_{1}}^{\tfrac{1}{2}}}$
$\displaystyle=\begin{pmatrix}\cos\tfrac{\theta_{1}}{2}-iu_{z}\sin\tfrac{\theta_{1}}{2}&\left(-iu_{x}-u_{y}\right)\sin\tfrac{\theta_{1}}{2}\\\\[5.0pt]
\left(-iu_{x}+u_{y}\right)\sin\tfrac{\theta_{1}}{2}&\cos\tfrac{\theta_{1}}{2}+iu_{z}\sin\tfrac{\theta_{1}}{2}\end{pmatrix}.$
(31)

In NMR $\boldsymbol{\vec{u}}$ lies in the transverse plane:
$\boldsymbol{\vec{u}}=\left(1,0,0\right)$ where the $x$-direction is given by
the phase of the RF pulse (in other words by the beginning of the RF pulse)
and therefore rotations induced in NMR are described by:

$\displaystyle\boldsymbol{R_{(1,0,0),\theta_{1}}^{\tfrac{1}{2}}}$
$\displaystyle=\begin{pmatrix}\cos\tfrac{\theta_{1}}{2}&-i\sin\tfrac{\theta_{1}}{2}\\\\[5.0pt]
-i\sin\tfrac{\theta_{1}}{2}&\cos\tfrac{\theta_{1}}{2}\end{pmatrix}.$ (32)

We notice that this is exactly the transformation matrix that we have computed
at resonance in equation (30), except that the time evolution
$E(\omega_{0},t)$ is missing in this equation. Therefore in the general case,
equation (28) gives the evolution of the wave function and equation (29) shows
that the evolution of the spin system is a time evolution given by matrix
$E(\omega,t)$ which depends on the RF frequency followed by a rotation of
angle $\Delta t=\sqrt{(\omega_{0}-\omega)^{2}+\omega_{1}^{2}}\ t$ around the
vector:

$\boldsymbol{\vec{u}_{\omega}}=\left(u_{x}=\dfrac{\omega_{1}}{\sqrt{(\omega_{0}-\omega)^{2}+\omega_{1}^{2}}},\
u_{y}=0,\
u_{z}=\dfrac{\omega_{0}-\omega}{\sqrt{(\omega_{0}-\omega)^{2}+\omega_{1}^{2}}}\right).$
(33)

It is important to note that matrix product is non-commutative. In other
words, the spin evolution in presence of a RF field is, according to equations
(28) and (29), a matrix product of a time evolution operator and a rotation
operator. The wave function after the pulse is given by:

$\displaystyle\bigl{\lvert}\psi_{\mathrm{RF}}(t)\bigr{\rangle}$
$\displaystyle=A_{RF}\cdot\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}=E_{RF}\cdot
R\cdot\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}$ (34)

which is the same as the equation derived in Abragam textbook [7] but here we
have calculated explicitly the $A_{RF}$ matrix. Therefore at resonance the
wave function at the end of the RF pulse is equal to the wave function
multiplied by the rotation matrix of angle $\theta_{1}=\omega_{1}t$ which is
then multiplied by the evolution matrix but there is no time sequence between
these two operations. This elementary description of the spin dynamics
retrieve the main characteristics of NMR experiments, but it is not complete.

###  1.3 The other part of the Hamiltonian.

Usually the Hamiltonian is developed only for the part of the system under
investigation. For example the hydrogen atom in a magnetic field is studied
for the energy levels of the electron without any reference to the proton or
the quarks. The effect of the magnetic field on the proton is usually seen as
a second order perturbation of the electron spin system. In H nuclear magnetic
resonance, the studied system is the proton but the magnetic field has an
effect on the electron and on the proton. For several hydrogen atoms the
kinetic part of each proton is dependent on the structure of the spin system
and the kinetic energy of the proton ($p^{2}/2m$) has to be considered. More
precisely the kinetic energy is
$\frac{-\hbar^{2}}{2m}\nabla^{2}(F(\mathbf{r},t))$ where F is a function of
space and time dependent on the spin system. For our purpose we consider the
energy of the hydrogen atom due to the proton kinetic energy and to the
nuclear forces to be very great compared to the magnetic-spin interaction.
This energy will be called $K^{\prime}$. The time fluctuations of $K^{\prime}$
are very small, it is why the fluctuations are usually neglected in the
evolution of the Hamiltonian $\mathscr{H}_{T}(t)$ but it can not be canceled
because it has important effects as it will be seen through this work. Hence,
for a spin system that experiences an external $\boldsymbol{\vec{B}_{0}}$ the
complete Hamiltonian for one spin$\tfrac{1}{2}$ is defined by:

$\displaystyle\mathscr{H}_{T}(t)$
$\displaystyle=K^{\prime}-\vec{\mu}\cdot\vec{B}_{0}$ (35)
$\displaystyle=\tfrac{1}{2}\hbar\begin{pmatrix}\omega_{0}+\frac{2K^{\prime}}{\hbar}&\frac{2K^{\prime}}{\hbar}\\\
\frac{2K^{\prime}}{\hbar}&-\omega_{0}+\frac{2K^{\prime}}{\hbar}\\\
\end{pmatrix}$

and the Schrödinger equation becomes:

$\displaystyle\begin{pmatrix}\dot{x}_{2T}(t)\\\ \dot{x}_{1T}(t)\\\
\end{pmatrix}$
$\displaystyle=-\frac{i}{2}\begin{pmatrix}(\omega_{0}+K)x_{2T}(t)+Kx_{1T}(t)\\\
Kx_{2T}(t)-(\omega_{0}-K)x_{1T}(t)\\\ \end{pmatrix}$ (36)

with $K=2K^{\prime}/\hbar$. We call the $K$ parameter measured in hertz the
rest constant.
Here to solve these equations we make the following substitutions:

$\displaystyle p(t)$ $\displaystyle=x_{2}(t)e^{i\tfrac{\omega_{0}+K}{2}t},$
$\displaystyle q(t)=x_{1}(t)e^{-i\tfrac{\omega_{0}-K}{2}t}.$ (37)

which gives if $\omega_{0}$ and $K$ are time independent:

$\displaystyle\ddot{p}(t)-i\omega_{0}\dot{p}(t)+\frac{K^{2}}{4}p(t)=0$
$\displaystyle\ddot{q}(t)+i\omega_{0}\dot{q}(t)+\frac{K^{2}}{4}q(t)=0$ (38)

which are the same equation as that obtained in equation (20) with
$\Omega=\omega_{0}$ and $\omega_{1}=K$. The solution of these equations is:

$\displaystyle\begin{pmatrix}C_{1}e^{i\tfrac{\Delta-K}{2}t}+C_{2}e^{-i\tfrac{\Delta+K}{2}t}\\\\[8.00003pt]
C_{3}e^{i\tfrac{\Delta-K}{2}t}+C_{4}e^{-i\tfrac{\Delta+K}{2}t}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}e^{-i\tfrac{K}{2}t}&0\\\
0&e^{-i\tfrac{K}{2}t}\\\
\end{pmatrix}\cdot\begin{pmatrix}C_{1}e^{i\tfrac{\Delta}{2}t}+C_{2}e^{-i\tfrac{\Delta}{2}t}\\\\[8.00003pt]
C_{3}e^{i\tfrac{\Delta}{2}t}+C_{4}e^{-i\tfrac{\Delta}{2}t}\end{pmatrix}$ (39)

where $C_{1},C_{2},C_{3},C_{4}$ are identical to the coefficients obtained in
the previous case given by equations (27) where $\Delta$ is
$\sqrt{\omega_{0}^{2}+K^{2}}$ and exchanging $\Omega$ with $\omega_{0}$ and
$\omega_{1}$ with $K$. This solution is again the product of an evolution
matrix with a rotation matrix:

$\displaystyle\bigl{\lvert}\psi_{\mathrm{T}}(t)\bigr{\rangle}$
$\displaystyle=E_{T}\cdot R_{T}\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}$ (40)

with

$\displaystyle E_{T}$ $\displaystyle=\begin{pmatrix}e^{-\tfrac{iKt}{2}}&0\\\
0&e^{-\tfrac{iKt}{2}}\\\ \end{pmatrix};$ $\displaystyle
R_{T}=\begin{pmatrix}\cos\tfrac{\Delta
t}{2}-\tfrac{i\omega_{0}}{\Delta}\sin\tfrac{\Delta
t}{2}&-i\tfrac{K}{\Delta}\sin\tfrac{\Delta t}{2}\\\
-i\tfrac{K}{\Delta}\sin\tfrac{\Delta t}{2}&\cos\tfrac{\Delta
t}{2}+\tfrac{i\omega_{0}}{\Delta}\sin\tfrac{\Delta t}{2}\\\ \end{pmatrix}$

Therefore the evolution of the wave function of one spin is given by the
product of the spin evolution operator and a spin rotation of angle
$\theta=\sqrt{\omega_{0}^{2}+K^{2}}\>t$ around
$\vec{u}(K/\Delta,0,\omega_{0}/\Delta)$. The evolution operator induces no
oscillation between the two states therefore it seems that no magnetic
resonance signal can be detected but we will see that if the magnetic field is
not homogeneous the spin motion introduces oscillations between the states
even without RF magnetic fields.
It is important to notice that the rest constant $K$ of the Hamiltonian, has a
tremendous effect on the spin motion in a homogeneous magnetic field because
there is no more $\omega_{0}$ in the $E_{T}$ evolution part of the wave
function given by equation (40).

###  1.4 General description of one spin$\tfrac{1}{2}$

In the general case the Hamiltonian of a spin$\tfrac{1}{2}$ is dependent on
the time evolution of the magnetic field $B(t)$, therefore it can be written
as:

$\displaystyle\mathscr{H}_{1}$
$\displaystyle=-\tfrac{1}{2}\hbar\begin{pmatrix}\gamma B_{Z}-K&\gamma
B_{X}-i\gamma B_{Y}-K\\\ \gamma B_{X}+i\gamma B_{Y}-K&-\gamma
B_{Z}-K\end{pmatrix}$ (41)

and the Schrödinger equation becomes:

$\displaystyle\begin{pmatrix}\dot{x}_{2}\\\ \dot{x}_{1}\\\ \end{pmatrix}$
$\displaystyle=-\frac{i}{2}\begin{pmatrix}(\omega_{Z}+K)x_{2}+(\omega_{X}-i\omega_{Y}+K)x_{1}\\\
(\omega_{X}+i\omega_{Y}+K)x_{2}-(\omega_{Z}-K)x_{1}\end{pmatrix}$ (42)

with $K=2K^{\prime}/\hbar$, $\omega_{X}=-\gamma B_{X}$, $\omega_{Y}=-\gamma
B_{Y}$ , $\omega_{Z}=-\gamma B_{Z}$ . This equation gives the spin motion
whatever the magnetic fields not only with an important magnetic field in the
Z direction direction as stated in conventional NMR description.
We look for a constant solution by an adequate exponential substitution
$p(t)=x_{2}(t)e^{A(t)}$ and $q(t)=x_{1}(t)e^{B(t)}$ which their derivatives
give:

$\displaystyle\dot{p}$
$\displaystyle=p\left[\dot{A}-\tfrac{i}{2}(\omega_{Z}+K)\right]-\tfrac{i}{2}(\omega_{X}-i\omega_{Y}+K)qe^{(A-B)}$
(43a) $\displaystyle\dot{q}$
$\displaystyle=q\left[\dot{B}+\tfrac{i}{2}(\omega_{Z}-K)\right]-\tfrac{i}{2}(\omega_{X}+i\omega_{Y}+K)pe^{(B-A)}$
(43b)

If we take the parameters at one time $A=\tfrac{i}{2}(\omega_{Z}+K)t$,
$B=-\tfrac{i}{2}(\omega_{Z}-K)t$, one obtains:

$\displaystyle\ddot{p}$
$\displaystyle-i\omega_{Z}\dot{p}+\tfrac{1}{4}(\omega_{X}^{2}+\omega_{Y}^{2}+K^{2}+2K\omega_{X})p=0$
(44a) $\displaystyle\ddot{q}$
$\displaystyle+i\omega_{Z}\dot{q}+\tfrac{1}{4}(\omega_{X}^{2}+\omega_{Y}^{2}+K^{2}+2K\omega_{X})q=0$
(44b)

In case of constant magnetic field $B(0,0,B_{0})$ we retrieve equations (38)
but the comparison with these equations show that the general derivation
introduces a coupling of the rest constant and the X component of the magnetic
field in case of non constant magnetic fields. The solutions of equations (44)
is, as usually, obtained by the constant solution then varying the constants.
The solution of these equations, equivalent to equations (24-25), is:

$\displaystyle x_{2}(t)$
$\displaystyle=p(t)e^{-A(t)}=e^{-\tfrac{i}{2}(\omega_{Z}+K)t}\left(C_{1}e^{\lambda_{+}t}+C_{2}e^{\lambda_{-}t}\right)$
(45a) $\displaystyle x_{1}(t)$
$\displaystyle=q(t)e^{-B(t)}=e^{\tfrac{i}{2}(\omega_{Z}-K)t}\left(C_{3}e^{-\lambda_{-}t}+C_{4}e^{-\lambda_{+}t}\right)$
(45b)

where

$\displaystyle\lambda_{\pm}=\frac{i\left(\omega_{Z}\pm\Delta\right)}{2},$
$\displaystyle\Delta=\sqrt{\omega_{Z}^{2}+\omega_{1}^{2}},$
$\displaystyle\omega_{1}^{2}=\omega_{X}^{2}+\omega_{Y}^{2}+K^{2}+2K\omega_{X}$
(46) $\displaystyle\omega_{Z}=\omega_{0}-K$ (47)

The $C_{i}$ coefficients are given by equations (27) where $\Omega$ is
substituted by $\omega_{Z}$ ($\Omega=\omega_{0}-\omega=\omega_{Z}$).
This substitution is not only a mathematical tool but it has the physical
significance that the off resonance of the RF is equivalent to a magnetic
field oscillation in the magnetic field direction which is the Z direction.
The general evolution of the spin wave function is:

$\displaystyle\bigl{\lvert}\psi(t)\bigr{\rangle}$
$\displaystyle=\begin{pmatrix}x_{2}(t)\\\ x_{1}(t)\\\
\end{pmatrix}=E(\omega_{0},K,t)\cdot
R(\omega,\omega_{0},\omega_{1},t)\cdot\bigl{\lvert}\psi_{1}(0)\bigr{\rangle},$
(48)

The solution with the parameters used for the differential equation resolution
is:

$\displaystyle\bigl{\lvert}\psi(t)\bigr{\rangle}$
$\displaystyle=\begin{pmatrix}x_{2}(t)\\\
x_{1}(t)\end{pmatrix}=\begin{pmatrix}e^{-\frac{i}{2}(\Omega+K)t}&0\\\
0&e^{\frac{i}{2}(\Omega-K)t}\end{pmatrix}\cdot\begin{pmatrix}a&b\\\
b&\bar{a}\end{pmatrix}\cdot\begin{pmatrix}r_{2}e^{i\phi_{2}}\\\
r_{1}e^{i\phi_{1}}\end{pmatrix}$ (49)

with

$\displaystyle a=\cos\tfrac{\Delta
t}{2}-\tfrac{i\Omega}{\Delta}\sin\tfrac{\Delta t}{2}$ $\displaystyle
b=-i\tfrac{\omega_{1}}{\Delta}\sin\tfrac{\Delta t}{2}$

This equation shows that evolution of the spin states between the two states
exists if it exists a magnetic field of intensity about K ($\Omega=\omega_{Z}$
in the general case). If $\Omega=0$ the two states have the same time
evolution $e^{-\frac{i}{2}Kt}$. We will see that the presence of another spin
with its magnetic moment
$\boldsymbol{\vec{\mu}}=\frac{1}{2}\gamma\hbar\boldsymbol{\vec{\sigma}}$ will
always introduce a local magnetic field.

###  1.5 Two spins $\tfrac{1}{2}$

For two spins we consider a four dimensional Hilbert space which is a tensor
product of the two dimensional Hilbert spaces of the one spin system. We will
distinguish two cases: one for an homogeneous magnetic field
$\boldsymbol{\vec{B}_{0}}$ and one where the two spins are in different
magnetic fields.

####  1.5.1 Two spins in an homogeneous magnetic field

In the case of an homogeneous magnetic field, each spin experiences the same
magnetic field and the Hamiltonian of the system transforms a 2-dimensional
Hilbert space in 4-dimensional Hilbert space. In order to preserve the scalar
product, the Kronecker sum is used. Kronecker sum is defined by:

$\displaystyle A\oplus_{K}B=A\otimes I_{n}+I_{m}\otimes B,$ (50)

with $\otimes$ the tensor product with identity matrices $I$ of dimensions $n$
the first dimension of matrix $B$ and $m$ the second dimension of matrix $A$,
(the important property of Kronecker sum is that
$e^{A\oplus_{K}B}=e^{A}\otimes e^{B}$).
For matrices 2x2 one obtains:

$\displaystyle A=\begin{pmatrix}a&b\\\ c&d\\\ \end{pmatrix},$ $\displaystyle
B=\begin{pmatrix}\alpha&\beta\\\ \gamma&\delta\\\ \end{pmatrix},$ (51)

$\displaystyle A\oplus_{K}B=\begin{pmatrix}a+\alpha&b&\beta&0\\\
c&d+\alpha&0&\beta\\\ \gamma&0&a+\delta&b\\\ 0&\gamma&c&d+\delta\\\
\end{pmatrix}$ (52)

The Hamiltonian of two spin$\tfrac{1}{2}$ in an homogeneous magnetic field is
therefore:

$\displaystyle\mathscr{H}_{2,\mathrm{Hom}}=-\tfrac{1}{2}\gamma\hbar\bigl{[}(\sigma_{1}^{x}\oplus_{K}\sigma_{2}^{x}\bigr{)}B_{X}+\bigl{(}\sigma_{1}^{y}\oplus_{K}\sigma_{2}^{y}\bigr{)}B_{Y}+\bigl{(}\sigma_{1}^{z}\oplus_{K}\sigma_{2}^{z}\bigr{)}B_{Z}\bigr{]},$
(53)

with

$\displaystyle\sigma_{p}^{x}=\begin{pmatrix}K_{p}&K_{p}+1\\\ K_{p}+1&K_{p}\\\
\end{pmatrix},$ $\displaystyle\sigma_{p}^{y}=\begin{pmatrix}K_{p}&K_{p}-i\\\
K_{p}+i&K_{p}\\\ \end{pmatrix},$ (54)

and

$\displaystyle\sigma_{p}^{z}=\begin{pmatrix}K_{p}+1&K_{p}\\\ K_{p}&K_{p}-1\\\
\end{pmatrix}.$ $\displaystyle K_{p}=-2K^{\prime}_{p}/\hbar$ (55)

$K_{p}=-K/\omega_{p}$ is dimensionless as shown in
$\mathscr{H}_{2,\mathrm{Hom}}$ through the matrices
$(\sigma_{p}^{x},\sigma_{p}^{y},\sigma_{p}^{z})$, the minus sign indicates
that energy reference is given by the spin system, $\omega_{p}$ is the Larmor
frequency for each $p$ nucleus measured in the range of tens to hundreds of
megahertz in frequency space [6] for high resolution NMR. $K^{\prime}_{p}$ has
the same dimension as $\hbar$. Note that the matrices are no more Pauli
matrices but are still the components of a vector
$\boldsymbol{\vec{\sigma}_{p}}$.
In very homogeneous magnetic field where $\boldsymbol{\vec{B}}=(0,0,B)$ and
$B_{X}=B_{Y}=0$ the Hamiltonian (53) can be written:

$\displaystyle\mathscr{H}_{2,\mathrm{B}}=-\frac{1}{2}\gamma\hbar$ (56)
$\displaystyle\begin{pmatrix}(K_{1}+K_{2}+2)B&K_{1}B&K_{2}B&0\\\
K_{1}B&(K_{1}+K_{2})B&0&K_{2}B\\\ K_{2}B&0&(K_{1}+K_{2})B&K_{1}B\\\
0&K_{2}B&K_{1}B&(K_{1}+K_{2}-2)B\end{pmatrix}$

This equation shows that for these two spins the rest constant of the
Hamiltonian becomes important by introducing a coupling between the spins even
without any component in the orthogonal plane to the magnetic field direction.
The wave function can be obtained by the eigenvalues of the Hamiltonian which
is no more a tensor product of the individual wave functions because the
kinetic part is dependent on the position of the spin. The important thing is
that the two spins are no more a linear combination of the two spin states but
a four spin state which can be considered as entangled spin states.

####  1.5.2 Two spins in different magnetic fields

If the magnetic fields experienced by spins $A$ and $B$ are respectively
$\boldsymbol{\vec{B}_{A}}$ and $\boldsymbol{\vec{B}_{B}}$, then the
Hamiltonian of the system is:

$\displaystyle\mathscr{H}_{2}$
$\displaystyle=-\left((\boldsymbol{\vec{\mu}_{A}}\cdot\boldsymbol{\vec{B}_{A}})\oplus_{K}(\boldsymbol{\vec{\mu}_{B}}\cdot\boldsymbol{\vec{B}_{B}})\right).$
(57)

The two spins are now distinguishable by their magnetic environment which is
always greater than the magnetic field heterogeneity and $K_{A}=K_{B}=0$
because the kinetic part can be included in the local magnetic field. If we
write this Hamiltonian in matrix form, we obtain the following:

$\displaystyle\mathscr{H}_{2}$
$\displaystyle=\\!-\frac{\gamma\hbar}{2}\\!\begin{pmatrix}B_{AZ}+B_{BZ}&B_{BX}-iB_{BY}&B_{AX}-iB_{AY}&0\\\
B_{BX}+iB_{BY}&B_{AZ}-B_{BZ}&0&B_{AX}-iB_{AY}\\\
B_{AX}+iB_{AY}&0&B_{BZ}-B_{AZ}&B_{BX}-iB_{BY}\\\
0&B_{AX}+iB_{AY}&B_{BX}+iB_{BY}&-B_{AZ}-B_{BZ}\end{pmatrix}\\!\\!,$ (58)

If we suppose that both magnetic fields are in the $Oz$-direction, this matrix
reduces to:

$\displaystyle\mathscr{H}_{2}$
$\displaystyle=-\frac{\gamma\hbar}{2}\begin{pmatrix}B_{A}+B_{B}&0&0&0\\\
0&B_{A}-B_{B}&0&0\\\ 0&0&B_{B}-B_{A}&0\\\ 0&0&0&-(B_{A}+B_{B})\end{pmatrix}.$
(59)

We directly obtain the eigenvalues of the Hamiltonian (59):

$\displaystyle E_{++}$ $\displaystyle=\frac{1}{2}\gamma\hbar(B_{A}+B_{B}),$
(60a) $\displaystyle E_{+-}$
$\displaystyle=\frac{1}{2}\gamma\hbar(B_{A}-B_{B}),$ (60b) $\displaystyle
E_{-+}$ $\displaystyle=-\phantom{-}\frac{1}{2}\gamma\hbar(B_{A}-B_{B}),$ (60c)
$\displaystyle E_{--}$
$\displaystyle=-\phantom{-}\frac{1}{2}\gamma\hbar(B_{A}+B_{B}).$ (60d)

This leads to the following time-dependent wave function:

$\displaystyle\bigl{\lvert}\psi_{2}(t)\bigr{\rangle}$
$\displaystyle=\begin{pmatrix}r_{4}e^{-\tfrac{i}{2}\bigl{(}(\omega_{A}+\omega_{B})t-\phi_{4}\bigr{)}}\\\
r_{3}e^{-\tfrac{i}{2}\bigl{(}(\omega_{A}-\omega_{B})t-\phi_{3}\bigr{)}}\\\
r_{2}e^{\tfrac{i}{2}\bigl{(}(\omega_{A}-\omega_{B})t-\phi_{2}\bigr{)}}\\\
r_{1}e^{\tfrac{i}{2}\bigl{(}(\omega_{A}+\omega_{B})t-\phi_{1}\bigr{)}}\end{pmatrix},$
(61)

where:

$\displaystyle\omega_{A}$ $\displaystyle=-\gamma B_{A},$ (62a)
$\displaystyle\omega_{B}$ $\displaystyle=-\gamma B_{B}.$ (62b)

The wave function for two independent spins can be written as a tensor product
of the wave function of each spin:

$\bigl{\lvert}\psi_{2}(t)\bigr{\rangle}=\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}_{\omega_{A}}\\!\\!\otimes\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}_{\omega_{B}},$
(63)

This is simply done by taking equation (63) as the proper definition for the
case of two spins $\tfrac{1}{2}$ and using equation (12) for each single wave
function to obtain the following wave function for two spins:

$\displaystyle\begin{split}\bigl{\lvert}\psi_{2}(t)\bigr{\rangle}&=\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}_{\omega_{A}}\\!\\!\otimes\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}_{\omega_{B}}\\\
&=\begin{pmatrix}r_{A2}e^{-i\bigl{(}\tfrac{\omega_{A}t}{2}+\phi_{A1}\bigr{)}}\\\
r_{A1}e^{i\bigl{(}\tfrac{\omega_{A}t}{2}+\phi_{A1}\bigr{)}}\end{pmatrix}\otimes\begin{pmatrix}r_{B2}e^{-i\bigl{(}\tfrac{\omega_{B}t}{2}+\phi_{B2}\bigr{)}}\\\
r_{B1}e^{i\bigl{(}\tfrac{\omega_{B}t}{2}+\phi_{B1}\bigr{)}}\end{pmatrix}\\\
&=\begin{pmatrix}r_{4}e^{-i\bigl{(}\tfrac{(\omega_{A}+\omega_{B})t}{2}+\phi_{4}\bigr{)}}\\\
r_{3}e^{-i\bigl{(}\tfrac{(\omega_{A}-\omega_{B})t}{2}+\phi_{3}\bigr{)}}\\\
r_{2}e^{i\bigl{(}\tfrac{(\omega_{A}-\omega_{B})t}{2}+\phi_{2}\bigr{)}}\\\
r_{1}e^{i\bigl{(}\tfrac{(\omega_{A}+\omega_{B})t}{2}+\phi_{1}\bigr{)}}\end{pmatrix}.\end{split}$
(64)

with

$\displaystyle r_{1}$ $\displaystyle=r_{A1}r_{B1}$
$\displaystyle\phi_{1}=\phi_{A1}+\phi_{B1}$ (65) $\displaystyle r_{2}$
$\displaystyle=r_{A2}r_{B1}$ $\displaystyle\phi_{2}=\phi_{A2}+\phi_{B1}$
$\displaystyle r_{3}$ $\displaystyle=r_{A1}r_{B2}$
$\displaystyle\phi_{3}=\phi_{A1}+\phi_{B2}$ $\displaystyle r_{4}$
$\displaystyle=r_{A2}r_{B2}$ $\displaystyle\phi_{4}=\phi_{A2}+\phi_{B2}$

We see that we retrieve the fundamental rule:

$\displaystyle\sum r_{i}^{2}=1$ (66)

only if the phase of each spin is distinguishable $\phi_{A}$ or $\phi_{B}$. In
this case the two spins are not in an entangled state.

####  1.5.3 Two spins in different magnetic fields with a RF field

Using the tensor product notation we can directly compute the wave function
for two spins in an inhomogeneous magnetic field with an RF magnetic field
$\boldsymbol{\vec{B}_{1}}$ rotating around $\boldsymbol{\vec{B}_{0}}$ with an
angular velocity $\omega$:

$\displaystyle\bigl{\lvert}\psi_{2}(t)\bigr{\rangle}$
$\displaystyle=\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}_{\omega_{A}}\\!\\!\otimes\bigl{\lvert}\psi_{1}(t)\bigr{\rangle}_{\omega_{B}}$
$\displaystyle=\Bigl{(}E(\omega)\cdot
R(\omega_{A})\cdot\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}_{\omega_{A}}\Bigr{)}\otimes\Bigl{(}E(\omega)\cdot
R(\omega_{B})\cdot\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}_{\omega_{B}}\Bigr{)}$
(67) $\displaystyle=\bigl{(}E(\omega)\otimes
E(\omega)\bigr{)}\cdot\bigl{(}R(\omega_{A})\otimes
R(\omega_{B})\bigr{)}\cdot\Bigl{(}\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}_{\omega_{A}}\\!\\!\otimes\bigl{\lvert}\psi_{1}(0)\bigr{\rangle}_{\omega_{B}}\Bigr{)}.$

####  1.5.4 Two spins$\tfrac{1}{2}$ in the general case

If the rest constant of the Hamiltonian is not neglected, the eigenvalues of
the Hamiltonian are no more reduced to no off diagonal components but are
given by the Kronecker sum of the Hamiltonian (41) and the wave function is
given by solving an extension to matrices (4X4) of the differential equations
(49).

###  1.6 N spins$\tfrac{1}{2}$

The description we have obtained for the Hamiltonian of two spins can be
generalized for a $N$ spin system by decomposing it in q parts of homogeneous
magnetic field $\boldsymbol{\vec{B_{q}}}(B_{qX},B_{qY},B_{qZ})$ with $n_{q}$
spins in each homogeneous part. The Hamiltonian of the spins in each spatial
domain q is given by the Kronecker sum of the $\sigma$ three components:

$\displaystyle\mathscr{H}_{n_{q},\mathrm{Hom}}=-\tfrac{1}{2}\gamma\hbar\left((\bigoplus_{p=1}^{n_{q}}\sigma_{p}^{x}\bigr{)}B_{qX}+(\bigoplus_{p=1}^{n_{q}}\sigma_{p}^{y}\bigr{)}B_{qY}+(\bigoplus_{p=1}^{n_{q}}\sigma_{p}^{z}\bigr{)}B_{qZ}\right),$
(68)

The $\sigma$ components are still defined by the (54) equations. Therefore
this Hamiltonian can be understood as it has the dot product of a new magnetic
moment $\boldsymbol{\vec{\mu}_{q}}$ and the local magnetic field
$\boldsymbol{\vec{B}_{q}}$ with $n_{q}$ spins in each domain.
The Hamiltonian of the N=$\sum_{1}^{q}n_{q}$ spins is given by the
generalization of equation (57):

$\displaystyle\mathscr{H}_{N}=-\left(\bigoplus_{q=1}^{q}(\boldsymbol{\vec{\mu}_{q}}\cdot\boldsymbol{\vec{B}_{q}})\right).$
(69)

In each homogeneous magnetic field $B_{q}$ where there is only one wave
function corresponding to $n_{q}$ entangled spins obey the Fermi-Dirac
statistics for spin$\tfrac{1}{2}$. The q domains which correspond to
$\boldsymbol{\vec{\mu}_{q}}$ distinguishable magnetic moments, obey the
Boltzmann statistics currently used for NMR experiments explanations.

##  2 NMR experiments

A way to test this theoretical description of spin evolution is to look at
proton NMR which is a very good example for studying isolated spins.

###  2.1 The nuclear spin Hamiltonian

If we consider one Hydrogen nucleus its total energy is $m^{{}^{\prime}}c^{2}$
where $m^{{}^{\prime}}$ is the mass of the moving nucleus and its magnetic
energy $-\mu B$ is tremendously low compared to the total energy. In an
homogeneous magnetic field the Hamiltonian is given by Equations (53) for two
spins and equation (68) for N spins.
In NMR the applied magnetic field is far greater than local magnetic
fluctuations arising from electron spin and space becomes oriented by an
homogeneous magnetic field which by convention is the z-axis. There is another
more implicit convention which is the change of the reference frame which
introduces the gyromagnetic ratio and a change of the origin of the energy
scale. The origin of the energy scale becomes by the Larmor frequency of the
spin under investigation no more an independent physical constant. $K_{p}$
defined by equation (55) and used in equations (56) and (68) can be understood
as the ratio that contributes in the part of magnetic energy from the total
energy. The Hamiltonian for $N$ nuclear spins is given by an extended matrix
based on the equation (56) where for $K_{p}=0$ only the energy of the first
and the last of the $2^{N}$ states are non zero because the intermediary
states cancel two by two. Therefore only these two states have a time
evolution: $r_{1}e^{i(\omega_{0}t+\phi_{1})}$ and
$r_{2^{N}}e^{-i(\omega_{0}t-\phi_{2^{N}})}$ but there is no transition between
these two states because only $r_{1}^{2}+r_{2}^{2}=1$ and
$r_{2^{N}}^{2}+r_{2^{N}-1}^{2}=1$ are states of the same spin. Therefore there
is no evolution of the spin states and we retrieve the important experimental
fact that there is no NMR signal at equilibrium.

###  2.2 Complex susceptibility of one spin$\tfrac{1}{2}$

In NMR, the measurement of the spin system is obtained by the current induced
in a tuned $LCR$ circuit. The influence of the sample is determined from its
complex susceptibility $\chi=\chi^{\prime}(\omega)+i\chi"(\omega)$ which
expresses the linear relationship between magnetization and magnetic field
[8]. It is well known that after an adequate $RF$ pulse an electrical current
is induced in the $LCR$ circuit.
For one spin we propose to define the complex susceptibility by:

$\displaystyle\chi=\gamma\hbar\left(\bar{x}_{1}(t)x_{2}(t)\right),$ (70)

where $\bar{x}_{1}(t)$ is the complex conjugate of the amplitude of state
$\left\lvert-\right\rangle$ and $x_{2}(t)$ the amplitude of state
$\left\lvert+\right\rangle$ of the wave function of the spin. As the amplitude
of the states are related by $\lvert x_{1}(t)\rvert^{2}+\lvert
x_{2}(t)\rvert^{2}=1$ the relative phase information is given by the complex
conjugate of one of the two amplitudes. In a static magnetic field $B_{0}$
very homogeneous and without the rest constant $K$, the wave function is given
by the conventional equation (12) where
$x_{1}(t)=r_{1}e^{i(\tfrac{\omega_{0}t}{2}+\phi_{1})}$,
$x_{2}(t)=r_{2}e^{-i(\tfrac{\omega_{0}t}{2}-\phi_{2})}$and the definition
(70), equal to

$\displaystyle\chi_{0}(t)=\gamma\hbar
r_{1}r_{2}e^{-i(\omega_{0}t+\phi_{1}-\phi_{2})}$ (71)

In the general case we have to use matrix (49) describing the spin wave
evolution and the complex susceptibility becomes:

$\displaystyle\chi(t)=\gamma\hbar e^{-i\Omega
t}\left\\{(a\>\bar{b}\>r_{2}^{2}+a\>b\>r_{1}^{2})+r_{1}r_{2}(b\>\bar{b}\>e^{i(\phi_{1}-\phi_{2})}+a^{2}e^{i(\phi_{2}-\phi_{1})})\right\\}$
(72)

with

$\displaystyle(a\>\bar{b}\>r_{2}^{2}+a\>b\>r_{1}^{2})=\tfrac{\omega_{1}}{\Delta}(\tfrac{\Omega}{\Delta}sin^{2}\tfrac{\Delta
t}{2}+i\cos\tfrac{\Delta t}{2}\sin\tfrac{\Delta
t}{2})(r_{2}^{2}-r_{1}^{2})=D_{1}(r_{2}^{2}-r_{1}^{2})$ (73)

and

$\displaystyle
r_{1}r_{2}\left(\tfrac{\omega_{1}^{2}}{\Delta^{2}}sin^{2}\tfrac{\Delta
t}{2}e^{i(\phi_{1}-\phi_{2})}+(cos^{2}\tfrac{\Delta
t}{2}-\tfrac{\Omega^{2}}{\Delta^{2}}sin^{2}\tfrac{\Delta
t}{2}-2i\tfrac{\Omega}{\Delta}\cos\tfrac{\Delta t}{2}\sin\tfrac{\Delta
t}{2})e^{i(\phi_{2}-\phi_{1}}\right)$ (74)

and

$\displaystyle\Omega=\omega_{Z};$
$\displaystyle\omega_{1}^{2}=\omega_{X}^{2}+\omega_{Y}^{2}+K^{2}+2K\omega_{X};$
$\displaystyle\Delta=\sqrt{\omega_{Z}^{2}+\omega_{1}^{2}}$ (75)

These equations show that K of the Hamiltonian is included in the $\omega_{1}$
parameter. The susceptibility of one spin can be written as:

$\displaystyle\chi(t)=\gamma\hbar\>e^{-i\>\Omega\>t}\left\\{D_{1}(r_{2}^{2}-r_{1}^{2})+r_{1}r_{2}(D_{2}\>e^{i(\phi_{1}-\phi_{2})}+D_{3}\>e^{i(\phi_{2}-\phi_{1})}\right\\}$
(76)

For N independent spins $\tfrac{1}{2}$, we can consider two cases the first
one where the rest constant K is not taken into account (K=0) and the second
case where K is present in the Hamiltonian of the spin set.

###  2.3 Low field NMR experiments

Magnetic resonance experiments was, at the beginning of the NMR story,
performed with homogeneous magnetic fields of about 0.1T with an homogeneity
better than $\Delta B_{0}/B_{0}=10^{-5}$. By convention the Z direction of the
reference frame is given by the direction of the magnetic field. As the gyro-
magnetic ratio of the proton is $\gamma=2.7\times
10^{8}\>rad\>s^{-1}\>T^{-1}$. According to $\omega_{Z}=-\gamma B$, for a
magnetic field of 1T and an homogeneity of $10^{-5}$, the angular velocities
are typically $\omega_{Z}=-2.7\times 10^{8}rad\>s^{-1}$ and $2.7\times
10^{3}\>rad\>s^{-1}$ for $\omega_{X}$ and $\omega_{Y}$. For a 3 $cm^{3}$ water
sample there is 1/3 mole of water which corresponds to $N=2\times 10^{24}$
protons.
As there is no NMR signal at equilibrium, the way to induce it is to use a RF
magnetic field. In this case the complex susceptibility of N spins is the sum
of the susceptibility of one spin given by equation (76) :

$\displaystyle\chi_{N}(t)$
$\displaystyle=\sum_{q=1}^{q=N}\int_{-\pi}^{\pi}\chi_{q}(t)d\phi_{q}$ (77)

The difference $\phi_{q}=\phi_{2q}-\phi_{1q}$ is random but is defined for
each spin therefore the integral over $\phi_{q}$ is null and the total
susceptibility of N independent spins is:

$\displaystyle\chi_{N}(t)$
$\displaystyle=\sum_{q=1}^{q=N}\int_{-\pi}^{\pi}\chi_{q}(t)d\phi_{q}=\gamma\hbar
Ne^{-i\>\Omega\>t}D_{1}\sum_{q=1}^{q=N}(r_{2q}^{2}-r_{1q}^{2})$ (78)

with

$\displaystyle
D_{1}=\tfrac{\omega_{1}}{\Delta}(\tfrac{\Omega}{\Delta}\sin^{2}{\tfrac{\Delta
t}{2}}+i\cos{\tfrac{\Delta t}{2}}\sin{\tfrac{\Delta t}{2}})$

The parameters are:

$\displaystyle\Omega=\omega_{Z}$
$\displaystyle\Delta=\sqrt{\omega_{Z}^{2}+\omega_{1}^{2}},$
$\displaystyle\omega_{1}^{2}=\omega_{X}^{2}+\omega_{Y}^{2}+K^{2}+2K\omega_{X}$
(79)

In NMR experiments, the RF magnetic field $\omega_{Y}=\omega_{X}$ used
corresponds to a 1 mT oscillating magnetic field with an amplitude much more
than the proton K fluctuations and about 1000 times less than the main
magnetic field. The $\omega$ RF frequency used is precisely at the resonance
frequency: $\omega=\omega_{0}=\omega_{Z}$ and with K equal to 0 because all
the magnetic fields present are far greater than the K parameter fluctuations,
therefore $(\omega_{1}=\sqrt{2}\omega_{X}$ and $\Delta=\omega_{0}$). With
these parameters the complex susceptibility of the N spins becomes:

$\displaystyle\chi_{N}(t)$
$\displaystyle=\gamma\hbar\>e^{-i\omega_{0}t}\tfrac{\sqrt{2}\omega_{X}}{\omega_{0}}\left[\sin^{2}{\tfrac{\omega_{0}t}{2}}+i\tfrac{1}{2}\sin\omega_{0}t\right]N\sum_{q=1}^{q=N}(r_{2q}^{2}-r_{1q}^{2})$
(80)
$\displaystyle=\gamma\hbar\>\tfrac{\sqrt{2}\omega_{X}}{2\omega_{0}}\left[e^{-i\omega_{0}t}-\>e^{-i\>2\omega_{0}t}\right]N\sum_{q=1}^{q=N}(r_{2q}^{2}-r_{1q}^{2})$

where $N\sum(r_{2q}^{2}-r_{1q}^{2})$ corresponds to the Boltzmann distribution
of the N spins in the N first $2^{N}$ spin states. This equation retrieves the
less known experimental fact that there are two resonances which can be
detected one at $\omega_{0}$ and another at $2\omega_{0}$ [7]. The NMR signal
is maximum when the pulse has a duration $t_{p}$ such as
$\omega_{X}t_{p}=\pi/2$ corresponding to the so called $\pi/2$ pulse.

###  2.4 High resolution NMR

High resolution NMR is characterized by very homogeneous high magnetic fields
typically $\Delta B_{0}/B_{0}=10^{-8}$. In this case the magnetic field
experienced by one spin becomes dependent on the orientation of neighborhood
spins and the Hamiltonian is given by equation (57). If all the magnetic
fields are in the same direction the Hamiltonian is given by equation (59).
For example in ethanol ($CH_{3}-CH_{2}-OH$) solution there are three groups of
equivalent protons: the 3 protons of the $CH_{3}$ methyl group, the two
protons of the $CH_{2}$ group and the proton of the OH group. Schematically,
the protons of the methyl group experience a magnetic local field
$\omega_{Zl}$ dependent on the orientation of the neighborhood two spins. The
local magnetic fields are $\omega_{Zl}=\omega_{Z}+\epsilon$ if the two spins
are in (Z) direction, $\omega_{Zl}=\omega_{Z}$ if one spin is in (Z) direction
and the other in (-Z) direction this combination is twice more probable than
the preceding one and $\omega_{Zl}=\omega_{Z}-\epsilon$ if both spins are in
(-Z) direction. Therefore the NMR signal characterized by the amplitude of
each $\omega_{Zl}$ frequencies is composed of 2 neighborhood protons +1.
Therefore for the protons of the methyl group of ethanol molecules there are 3
frequencies with relative amplitude (1,2,1). The wave function with a RF field
is according to equation (1.5.3) the tensor product of the evolution matrix of
each spin and the tensor product of the rotation matrices. Therefore the NMR
spectrum of the methyl group is still three resonances with the same relative
amplitudes (1,2,1). Again the amplitude of the NMR signal is maximum after a
$\pi/2$ pulse.

###  2.5 Nuclear spin noise experiments

Felix Bloch [2] has predicted the theoretical possibility of an emission from
nuclear spins without any magnetic field. The first observation of random
noise emission from nuclear spins was performed by Sleator, Hahn, Hilbert and
Clarke [9, 10] using a superconducting SQUID. At very low temperature without
applied magnetic field the other part of the Hamiltonian becomes important and
can explain an emission from nuclear spins via the spin motion in the magnetic
fields generated by the neighborhood spins and can be measured.
In very homogeneous magnetic fields where $\Delta B_{Y}/B_{0}=10^{-8}$ the
rest constant of the Hamiltonian has also to be taken into account and can
explain nuclear spin emission without any RF magnetic field even at room
temperature. MCCoy and Ernst [12] have observed nuclear spin noise emission at
room temperature using a conventional (300 MHz) NMR spectrometer. They
compared the proton resonance of the methyl group in ethanol without any RF
pulse just by co-adding 8000 measurements with the proton resonance obtained
with a conventional one $\pi/2$ RF pulse. The most surprising feature of the
spectrum without any RF is the negative deviation compared to the spectrum
obtained by a $\pi/2$ excitation pulse and secondly an amplitude of the signal
which is about $10^{8}$ time smaller. Our derivation retrieves these main
characteristics.
The wave function without RF is given by an extension to N spins of equation
(49) where the spin system for the proton of the methyl protons is composed of
three local magnetic fields. The parameters of equation (78) which gives the
complex susceptibility of N spin is $\Omega=\omega_{0}$, $\omega_{1}=K$ and
$\Delta=\omega_{0}$ therefore the complex susceptibility is:

$\displaystyle\chi_{noise}(t)=\gamma\hbar\>\tfrac{K}{2\omega_{0}}\left[e^{-i\omega_{0}t}-\>e^{-i\>2\omega_{0}t}\right]N\sum_{q=1}^{q=N}(r_{2q}^{2}-r_{1q}^{2})$
(81)

which is the same as equation (80) for experiment with RF except where K is
replaced by $\sqrt{2}\omega_{X}$. The rest constant can be expressed in
frequency unit according to the substitution given by equation (37).
In NMR, $\omega_{X}$ is always positive because all the frequency measurements
are locked to the RF signal by the use of lock-in amplifiers, without such a
locking tool the phase of the electrical current becomes random and no
accumulation can be performed. As the kinetic part of the rest constant is
proportional to $-\hbar/m_{p}$ the complex susceptibility with no RF is always
negative. The amplitude of the complex susceptibility for a defined number of
spin is given by $\sqrt{2}\omega_{X}/\omega_{0}$ with RF and $K/\omega_{0}$
without RF. As K is about $6\times 10^{-8}rad/s$ for one proton it seems not
measurable but K is a scalar and in homogeneous magnetic field of $\Delta
B/B=10^{-8}$ a part of the N spins have the same rest constant and the complex
susceptibility becomes measurable. The possibility to measure a NMR signal is
always dependent on the homogeneity of the magnetic field even in low magnetic
field NMR experiments.

###  2.6 NMR imaging without RF experiments

####  2.6.1 External lock experiments

In an imaging experiment, where magnetic gradient fields are applied in all
space directions Müller and Jerschow [1] have obtained an image without RF
field, showing that entangled spins can be modified by magnetic field
gradients. They used a 500 MHz chemical spectrometer with magnetic field
gradients of 6 mT/m strength oriented in direction orthogonal to $B_{0}$ for
$X,Y=0^{o}$ to $X,Y=180^{o}$ by step of $6^{o}$. Each step was measured 512
times. To prevent any leakage or accidentally delivery RF power to the probe,
the RF power transmission cable was disconnected. The obtained image is a way
to visualize the spatial distribution of the rest constant. Again the
synchronization of NMR signals was obtained by the external lock of high
resolution NMR spectrometers which gives the time origin and $\omega_{X}$.

####  2.6.2  Experiments with no external lock

Usually NMR imaging is obtained by a first gradient applied in one direction
with a RF pulse in order to select a slice characterized by its position on
this first direction and its thickness given by resonance condition. After
that a gradient is applied in an orthogonal direction in order to give a phase
coherence and thereafter a gradient is applied in the third orthogonal
direction during the signal acquisition. In a previous work [11], NMR images
were obtained using such method with a very small RF power. In fact a very
small RF power is automatically applied in the used spectrometer with no
external lock. In the experiment described by [11] the RF power was 1 $\mu$W
which for a 10 cm diameter coil, is less than the electrical noise level in
such a coil. Therefore image in Figure 2B of [11] can be considered as an
image obtained without RF showing that conventional NMR imaging sequence can
be used for spin noise imaging if the signal acquisition is properly
synchronized. Another experimental fact showing the importance of
synchronization is that the MCCoy and Ernst experiment [12] is not
reproducible on a Bruker 7T BS70/30 system without any external lock for
synchronization.

###  2.7 Summary

In summary we analyzed different experimental effects that show the following
physical implications of the solutions of Schrödinger’s equation with the
complete Hamiltonian:

  * •

We retrieve for the low field NMR experiments the less known experimental fact
that there are two detectable resonances one at $\omega_{0}$ and another at
$2\omega_{0}$. Furthermore, the NMR signal is maximum when the pulse has a
duration corresponding to the so called $\pi/2$ pulse which shows that time
unit is dependent on the studied NMR frequency.

  * •

In the case of high resolution NMR for the protons of the methyl group of
ethanol molecules there are 3 frequencies with the relative amplitudes
(1,2,1), and the maximum amplitude of the NMR signal is still obtained after a
$\pi/2$ pulse.

  * •

Our derivation retrieves the main characteristics of the spectrum without any
RF, i.e. the nuclear spin noise experiments for the phase change and the NMR
signal amplitudes.

  * •

NMR imaging without RF experiments has been discussed in both cases with and
without external locks showing the experimental importance to define a zero
time of acquired signals in order to sum them adequately.

##  3  Conclusion

This work has shown that an elementary approach of the quantum description of
an ensemble of independent spin $\tfrac{1}{2}$ can retrieve the magnetic
resonance basic experimental facts by an adequate definition of the complex
susceptibility. This derivation completes the density matrix formalism which
is well adapted for describing NMR experiments where spin coupling by magnetic
fields induced by neighborhood spins have measurable effects. Experiments
using nuclear spin noise are difficult to be explained by the matrix density
formalism without introducing an extra probability factor and a stochastic
operator as in the paper of Field and Bain [13].
The present work has proposed that in very homogeneous magnetic field the non
external magnetic part of the Hamiltonian becomes important by introducing a
coupling between the spins even without any magnetic component in the plane
orthogonal to the magnetic field direction. In this case the
spins$\tfrac{1}{2}$ into the homogeneous magnetic field are in an entangled
state which can be used for quantum computing. NMR is still a good example of
an experimental quantum computer and the use of magnetic gradients can be a
way of increasing the number of qubits. The main result of our approach is to
predict that the quantum interference effects are best detected in a very
homogeneous magnetic field where many spins are in entangled spin states.
Magnetic field gradients can be used to initiate the system and the RF pulses
can be used for creating logical gates and for measuring the result of quantum
entanglements computation [14]. NMR images without RF is the proof that
encoding gradients can select a voxel where the magnetic field is sufficiently
homogeneous that it can be considered as an entangled NMR quantum bit.
Previous NMR experiments on quantum computing have shown that NMR qubits can
be manipulated in order to create quantum gates [15, 16]. Using magnetic field
gradients can tremendously increase the number of qubits compared to molecule
based qubits and can give a new way to do quantum computing by NMR especially
at low temperature where the loss of coherence effects are reduced.

## References

  * [1] Müller N and Jerschow A. (2006) PNAS 103, 6790-6792
  * [2] Bloch F. (1946) Phys. Rev. 70, 460-474
  * [3] Müller N, Jerschow A, Schlanitweit J (2013) eMagRes 2, 237-244 DOI 10.1002/9780470034590.emrstm1314
  * [4] Ángel Rivas, Susana F. Huelga and Martin B. Plenio, Rep.Prog. Phys.77 (2014) 094001
  * [5] Cohen-Tannoudji C, Diu B, LaloëF. Quantum Mechanics. New York: Wiley; (1977). p 972-975
  * [6] Ryan C. A., Negrevergne C., Laforest M., Knill E., and Laflamme R. (2008) Phys. Rev. A 78, 012328
  * [7] Abragam A. The principles of nuclear magnetism. Oxford: Clarendon; (1961). p 22-26
  * [8] Slichter CP. Principles of Magnetic Resonance. Berlin: Springer-Verlag; (1989). p 37
  * [9] Sleator T., Hahn E.L., Hilbert C., Clarke J. (1985) Phys. Rev. Lett. 55, 1742-1745
  * [10] Sleator T., Hahn E.L., Hilbert C., Clarke J. (1987) Phys. Rev. B 36, 1969-1980
  * [11] Van Schenk Brill K., Lahfadi J., Khalil T. and Grucker D. (2015) arXiv:1504.04882V1 [quant-phys], 1-17
  * [12] McCoy M.A. and Ernst R.R. (1989) Chem. Phys. Lett. 159, 587-593
  * [13] Field T.R. and Bain A.D.(2010) Appl. Magn. Reson. 38, 167-178
  * [14] Nielsen M. A. and Chuang I.L., Quantum Computation and Quantum Information. Cambridge University Press; (2002)
  * [15] Long G.L. et al. (2000) arXiv:0007077V1 [quant-phys], 1-3
  * [16] Vandersypen L.M.K. et al (2001) Nature 414, 883-887