# Quantum Mechanic Treatment

## Introduction

NMR employs strong magnetic field which interact with the intrinsic spin of the nucleus to form degenerate energy levels. Treating this phenomenon classically is mathematically tedious and/or impossible. Here we outline the fundamentals of NMR in quantum mechanical terms.

## Zeeman Interaction

If we begin with the historical Schrödinger equation

$H|\psi \rangle =E|\psi \rangle$

we can immediately realize that by defining the Hamiltonian of the system will result with an corresponding energy level of the wave function.

The Hamiltonian for a magnetic vector ($$\vec{\mu}$$ in J/T)in a magnetic field (also a vector), ($$\vec{B}$$ in T) is given by the expression

$H=- \vec{\mu} \cdot \vec{B}$

We then see that the Hamiltonian has units of energy (J). Therefore operating the Hamiltonian on a wavefinction will result in an energy eigenvalue! We can further refine the Hamiltonian by relating the magnetic moment of a nucleus to the particles angular momentum ari\sing from the intrinsic spin, I, of the nucleus thus

$\vec{\mu}=-\gamma \vec{I}$

where gamma is the ratio of the angular momentum to the magnetic moment. Then the Hamiltonian may be rewritten as

$H=-\gamma \vec{I} \cdot \vec{B}$

Assuming that the magnetic field lies along the Z-axis, then

$H_z =- \gamma (I_x\hat{i} +I_y \hat{j} + I_z \hat{k}) \cdot (B_0 \hat{k})=- \gamma B_0I_z=-\omega_0I_z$

where $$\omega_0$$ is the Larmor frequency. Therefore if the $$|\psi \rangle$$ is an eigenfunction of H then the $$|\pm \rangle$$ spin states are eigenfunctions of the total and z-component of angular momentum. We can describe this is three ways in a DC magnetic field:

Operator:

$H_z|+ \rangle=- \omega_0 I_z|+ \rangle=-\dfrac{\omega_0}{2}|+ \rangle$

$H_z|-\rangle=- \omega_0 I_z|-\rangle=-\dfrac{\omega_0}{2}|-\rangle$

Matrix:

$H_z=$

Energy Level

Insert picture

These representations rely on the fact that spin states are eigenfunctions of the angular momentum and therefore have the following properties, with I and m defining the state of the system.

$I^2|I,m \rangle=I(I+1)|I,m \rangle$

$I_z|I,m \rangle=m|I,m \rangle$

$I_{+1}|I,m \rangle=-\sqrt{\dfrac{1}{2}(I(I+1)-m(m+1))}|I,m_{+1} \rangle$

$I_{-1}|I,m \rangle=\sqrt{\dfrac{1}{2}(I(I+1)-m(m-1))} |I,m_{-1} \rangle$

where

$I_x=\dfrac{1}{\sqrt{2}}(I_{+1}-I_{-1})$

$I_y=\dfrac{i}{\sqrt{2}}(I_{+1}+I_{-1})$

Note that in contrast to the harmonic oscillator, the $$|I,m \rangle$$ spin basis is orthogonal and complete. This means for the spin $$I$$, there are only $$2I+1$$ spin states and the the overlap integral $$\langle I,m|I',m' \rangle=\delta_{mm'} \delta_{II'}$$ holds. This formalism holds equally well for $$I>1/2$$ particles as well.

While it is relatively straightforward to understand the Zeeman interaction, understanding the effects of radio frequency pulses on the Hamiltonian needs careful consideration. To examine this we will look at a modification to the Stern-Gerlach experiment made by Rabi in the 1930's.

\since no one has bothered to discuss the Stern Gerlach experiment in any detail on the chemwiki a brief glimpse into the initial experiment is necessary. Essentially a beam of silver atoms was shot through a magnetic field and instead of a \single line of atoms or \single spot, depending on the dispersion of the atoms, two spots were observed. This of course was due to the the fact that Zeeman splitting had occurred due to silvers spin=1/2 nature. A figure is provided below for reference.

Rabi modified this experiment by adding a coil of wire which generated a magnetic field perpendicular to the applied magnetic field from the stern gerlach experiment. By selecting only 1 of the nuclei either +/- 1/2 and running these particles through the radio frequency coil Rabi noted a characteristic nutation of the magnetization shown below.

insert figure

We can see then that the number of particles appearing on the screen is characterisitc of the frequency. This behavior is denoted as the Rabi or nutation frequency. This experiment serves as a good bridge into the NMR experiment. First, we must calculate the appropriate Hamiltonian for the sitation. We have two magnetic fields, one is the large external magnetic field which splits the degenerate ground states into energy levels, and the second is the osciallting magnetic field from the coil. As such the magnetic field experienced by the spins is

$\vec{B}=B_0 \hat{k} + 2B_1 \cos(\omega_0 t) \hat{i}$

Here we define the coil to be along the x direction and the external magnetic field about the z direction. \since the field is oscillating in time, a \co\sine function is the appropriate choice to model the behavior of this magnetic field.

We may now express the Hamiltonian in eith operator or matrix form as

$H=-\gamma \vec{I} \cdot \vec{B} = -\gamma B_0I_z - 2 \gamma B_1 \cos(\omega_o t)I_x=-\omega_0 I_z-2 \omega_1 \cos(\omega_0 t)I_x$

insert matrix here.

We now see that our Hamiltonian become time dependent. In order to solve the observation made by Rabi we must use the time depedent Schrodinger equation.

$\dfrac{d}{dt}|\psi (t) \rangle = -iH(t)| \psi (t) \rangle$

where H(t) is in frequency units and |$$\psi$$(t=0)>=|+>. Assuming that the Hamiltonian is time independent we can use a time dependent unity operator $$\hat{U}$$ defined as

$|\psi (t) \rangle=\hat{U}(t)| \psi (0)>$

$\hat{U}(t)=e^{-iHt}$

that transforms the initial wavefunction |$$\psi$$ (0)>to a wavefunction at a given time |$$\psi$$ (t) \rangle.

insert figure

This issue with the case above is that we assumed the Hamiltonian is time-independent, which is clearly not the our case. Therefore, we need to solve this problem by transforming the time-dependent Schrodinger equation into a reference frame that renders the Hamiltonian time independent.

Consider the development of the time evolution of |Q(t) \rangle under the following conditions

$|\psi (t) \rangle=R(t)|Q(t) \rangle$

$Q(t) \rangle=R^+(t)| \psi (t) \rangle$

$R(t)R^+(t)=R^+(t)R(t)=1$

As you can see R commutes and the adjoint give the unity operator. We can then rewrite the Schrodinger equation as

$\dfrac{d}{dt}| \psi (t) \rangle=\dfrac{d}{dt}R(t)|Q(t) \rangle=[\dfrac{d}{dt}R(t)]|Q(t) \rangle + R(t)\dfrac{d}{Dt}|Q(t) \rangle$

$=-iH(t)| \psi (t) \rangle=-iH(t)R(t)|Q(t) \rangle$

Mow if we multiple both sides of the equation by the adjoint then we can rearrange to obtain a new form of the Schrodinger equations

$[R^+(t) \dfrac{d}{dt}R(t)]|Q(t) \rangle + \dfrac{d}{dt}|Q(t) \rangle=-iR^+(t)H(t)R(t)|Q(t) \rangle$

$\dfrac{d}{dt}|Q(t) \rangle=-iH_{int}(t)|Q(t) \rangle$

where

$H_{int}=R^+(t)H(t)R(t)-iR^+(t)\dfrac{d}{dt}R(t)$

To apply this to our situation consider the case where

$R(t)=e^{-iH_z t}=e^{i \omega_0 I_zt}$

$R^+=e^{-i \omega_0 I_z t}$

Then we can develop the following Identities

$R(t)=\begin{bmatrix} e^{\dfrac{i \omega_0 t}{2}} &0\\0& e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix}$

$R^{\dagger}(t)=\begin{bmatrix} e^{\dfrac{-i \omega_0 t}{2}} &0\\0& e^{\dfrac{i \omega_0 t}{2}} \end{bmatrix}$

in which we can easily take derivitives of leading to

$\dfrac{d}{dt}R(t)=\begin{bmatrix}\dfrac{d}{dt}e^{\dfrac{i \omega_0 t}{2}} &0 \\ 0 & \dfrac{d}{dt}e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix}$

$\dfrac{d}{dt}R(t)=\begin{bmatrix}\dfrac{i \omega_0}{2}e^{\dfrac{i \omega_0 t}{2}} &0 \\ 0 & \dfrac{-i \omega_0}{2}e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix}$

$\dfrac{d}{dt}R(t)=i \omega_0 \begin{bmatrix} \dfrac{1}{2} &0\\0&\dfrac{1}{2}\end{bmatrix} \cdot \begin{bmatrix}e^{\dfrac{i \omega_0 t}{2}} &0\\0& e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix}$

$\dfrac{d}{dt}R(t)=i\omega_0 I_z \cdot e^{i \omega_0 tI_z}$

We can also calculate

$R^{\dagger} (t) I_x R(t)= \begin{bmatrix} e^{\dfrac{-i \omega_0 t}{2}} &0 \\ 0& e^{\dfrac{i \omega_0 t}{2}} \end{bmatrix} \cdot \begin{bmatrix} 0& \dfrac{1}{2} \\ \dfrac{1}{2} &0 \end{bmatrix} \cdot \begin{bmatrix} e^{\dfrac{i \omega_0 t}{2}} &0\\0& e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix}$

$R^{\dagger}(t)I_xR(t)=\begin{bmatrix} 0&\dfrac{1}{2}e^{\dfrac{-i \omega_0 t}{2}}\\ \dfrac{1}{2}e^{\dfrac{i \omega_0 t}{2}}&0 \end{bmatrix}$

$R^{\dagger}(t)I_xR(t)=\cos \omega t \begin {bmatrix} 0& \dfrac{1}{2} \\ \dfrac{1}{2} &0 \end{bmatrix} +\sin \omega t \begin {bmatrix} 0 & \dfrac{-i}{2} \\ \dfrac{i}{2} & 0\end{bmatrix} =\cos \omega_0 t I_x +\sin \omega_0 t I_y$

Similarly we can show (left as an exercise for the reader)

$R^{\dagger}(t)I_xR(t)=\cos \omega_0 t I_y -\sin \omega_0 t I_x$

We now have all the pieces we need to solve the Hamiltonian. We will show this in multiple ways

### Matrix Approach

$R^{\dagger}(t) \dfrac{d}{dt} R(t) =\begin{bmatrix} e^{\dfrac{-i \omega_0 t}{2}} &0\\0& e^{\dfrac{i \omega_0 t}{2}} \end{bmatrix} \cdot \begin{bmatrix}\dfrac{i \omega_0}{2}e^{\dfrac{i \omega_0 t}{2}} &0 \\ 0 & \dfrac{-i \omega_0}{2}e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix} =\begin{bmatrix} \dfrac{i \omega_0}{2} & 0 \\ 0 & \dfrac{-i \omega_0}{2} \end{bmatrix}$

$R^{\dagger}(t) H(t) R(t) =\begin{bmatrix} e^{\dfrac{-i \omega_0 t}{2}} &0\\0& e^{\dfrac{i \omega_0 t}{2}} \end{bmatrix} \cdot \begin{bmatrix} \dfrac{-\omega_0}{2} & - \omega_1 \cos \omega_0 t \\ -\omega_1 \cos \omega_0 t & \dfrac{\omega_0}{2} \end{bmatrix} \cdot \begin{bmatrix}e^{\dfrac{i \omega_0 t}{2}} &0 \\ 0 & e^{\dfrac{-i \omega_0 t}{2}} \end{bmatrix} = \begin{bmatrix} \dfrac{-\omega_0}{2} & - \omega_1 \cos \omega_0 t e^{-i \omega_0 t} \\ -\omega_1 \cos \omega_0 t e^{i \omega_0 t} & \dfrac{\omega_0}{2} \end{bmatrix}$

Then

$H_{int} =-\omega_1 \begin{bmatrix} 0& 1/2 \\ 1/2 &0 \end{bmatrix} - \omega_1 \cos2 \omega_0 t \begin{bmatrix} 0 &1/2\\1/2&0 \end{bmatrix} - \omega_1 \sin 2 \omega_0 t \begin{bmatrix} 0 &-i/2 \\i/2 &0 \end{bmatrix}$

$H_{int}=\omega_1 I_x - \omega_1 \cos 2 \omega_0 t I_x - \omega_1 \sin 2 \omega_0 t I_y$

### Operator Approach

We now wish to solve the Hamiltonian u\sing the I operators we have defined previously.

$R^{\dagger}(d)\dfrac{d}{dt}R(t)=e^{-i\omega_0I_z} \cdot i \omega_0 I_z \cdot e^{i \omega_0 I_z}=i \omega_0 I_z$

$R^{\dagger}(t)H(t)R(t)=e^{-i\omega_0I_z} (-\omega_0 I_z - 2 \omega_1 I_x) e^{i \omega_0 I_z}$

$= -\omega_0 I_z -2 \omega_1 \cos \omega_0 t I_x E^{i \omega_0 t I_z} I_xe^{-i \omega_otI_z}$

$=-\omega_0 I_z - 2 \omega_1 \cos^2 \omega_0 t I_x - 2\omega_1 \cos \omega_0 t \sin \omega_0 t I_y$

$=-\omega_0 I_z -\omega_1 I_x -\omega_1 \cos2 \omega_0 tI_x -\omega_1 \sin2 \omega_0 t I_y$

$H_{int}(t)=-\omega_1 I_x -\omega_1 \cos2 \omega_0 t I_x - \omega_1 \sin 2\omega_0 t I_y$

This is the exact expression we obtained u\sing the matrix approach!

## Rotating Wave Approximation

It can be shown that the time dependent part of the Hamiltonian modifies the first term in the above equation by $$\frac{\omega_1^2}{\omega_0}$$ (Need to prove this). Standard NMR spectrometers typically output $$\omega_1$$ on the order of kHz, whereas the larmor frequency is on the order of MHz and we can neglect the time department part of the Hamiltonian. This is known as the Rotating Wave Approximation. The Hamiltonian becomes

$H_{int} = -\omega_1 I_x =-\omega_1 \begin{bmatrix}0&\frac{1}{2} \\ \frac{1}{2}&0 \end{bmatrix}$

We now need to find the describe the initial state of the system

$|Q(0)>=R^{\dagger}(t=0)>=\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} =\begin{pmatrix} 0 \\1 \end{pmatrix}$

Now we if we diagonalize the matrix then we can find the state of the system at any time t. The diagonalized Hamiltonian is then

$U(t)=\begin{bmatrix} \cos \frac{\omega_1 t}{2} & i\sin \frac{\omega_1 t}{2} \\ i\sin \frac{\omega_1 t}{2} & \cos \frac{\omega_1 t}{2}\end{bmatrix}$

We can now evaluate |Q(t) \rangle as \Q(t) \rangle=U(t)|Q(0)

$|Q(t) \rangle=\begin{bmatrix} \cos \frac{\omega_1 t}{2} & i\sin \frac{\omega_1 t}{2} \\ i \sin \frac{\omega_1 t}{2} & \cos \frac{\omega_1 t}{2}\end{bmatrix} \begin{pmatrix} 0 \\1 \end{pmatrix} = \begin{bmatrix} i\sin \frac{\omega_1 t}{2} \\ \cos \frac{ \omega_1 t}{2}\end{bmatrix}$

Finally recognizing that $$|\Psi(t) \rangle$$=R(t)|Q(t). There for

$|\Psi(t) \rangle=\begin{bmatrix} e^{\frac{i \omega_0 t}{2}} & 0 \\ 0& e^{\frac{-i \omega_0 t}{2}}\end{bmatrix} \begin{bmatrix} i\sin \frac{\omega_1 t}{2}/ \\cos \frac{ \omega_1 t}{2}\end{bmatrix} = \begin{bmatrix} i\sin \frac{\omega_1 t}{2} e^{\frac{i \omega_0 t}{2}}\\ \cos \frac{\omega_1 t}{2} e^{\frac{-i \omega_0 t}{2}}\end{bmatrix}$

$= i\sin \frac{\omega_1 t}{2} e^{\frac{i \omega_0 t}{2}} |-\rangle + \cos \frac{\omega_1 t}{2} e^{\frac{-i \omega_0 t}{2}} |+>$

$=c_-(t)|-\rangle+c_+(t)|+>$

The probability that Rabi obsereved is given by multiplying the complex congugates which are calculated as

$P_{|-\rangle}=\frac{1}{2}-\frac{1}{2}\cos \omega_1t$

$P_{|+>}=\frac{1}{2}+\frac{1}{2}\cos \omega_1t$

The Hamiltonian we found to govern Rabi's experiment is the exact Hamiltonian used in the NMR experiment, with two slight modifications. First, Rabi was able to select an initial state, while in NMR we have a thermal equilibrium population. Secondly we measure a transverse magnetization instead of |+> and |-\rangle.

## Liouville Von Neumann Equation

\since we are dealing with a system that can be described by both a statistical and quantum mechanical aspects, we need a way to unify these descriptions. The solution is the Liouville-von Neumann equation, which describes how the density operator evolves in time. More specifically, it describes the time evolution of a mixed state, which is directly applicable to NMR, as we want to know how the off-diagonal elements of the density matrix evolve after an applied RF. The density matrix is evolving in time, thus we must derive a differential equation to describe this behavior then integrate it with each step of the NMR experiment. As usual we begin with the time-dependent Schroedinger Equation.

$i \bar{h} \frac{d}{dt} \Psi (t) = H(t)\Psi (t)$

Lets also look at the thermal equilibrium of the spins. U\sing Boltzmann statistics the initial condition of our spins is described by

$c_+(0)c^*_+(0)=\frac{Ne^{-\frac{E_+}{kT}}}{Z}$

$c_-(0)c^*_-(0)=\frac{Ne^{-\frac{E_-}{kT}}}{Z}$

where

$Z=[e^{-\frac{E_-}{kT}}+e^{-\frac{E_+}{kT}}$

and

$c_+(0)c^*_-(0)=0$

$c_-(0)c^*_+(0)=0$

Defining

$| \Psi(t) \rangle=c_+(t)\+> +c_-(t)|-\rangle$

$\langle \Psi(t)|=c^*_+(t)\langle+| +c^*_-(t)<-|\ we immediately realize that we have a density operator! \[\rho(t)=|\Psi (t) \rangle<\Psi (t) |=c_+(t)c^*_+(t)\+>\langle+| + c_-(t)c^*_-(t)|-\rangle<-|+c_+(t)c^*_-(t)|+><-|+c_-(t)c^*_+(t)|-\rangle\langle+|$

Then the time evolution is

$\frac{d}{dt}\rho(t)=\frac{d}{dt}|\Psi (t) \rangle<\Psi (t) | +|\Psi (t) \rangle\frac{d}{dt}< \Psi (t)|=-iH(t)|\Psi (t) \rangle<\Psi (t)|+i|\Psi (t) \rangle<\Psi (t)|H(t)$

which gives the Liouville-von Neumann Equation as

$\frac{d}{dt}\rho(t)=i[H(t),\rho (t)]$

Lets now look at how this evolves in a simples NMR experiment.

Initially we can define the density matrix at thermal equilibrium as

$\rho (0)=\begin{bmatrix} c_+(0)c^*_+(0)&c_+(0)c^*_-(0)\\c_-(0)c^*_+(0)&c_-(0)c^*_-(0) \end{bmatrix} = \frac{1}{Z} \begin{bmatrix}e^{-\frac{E_+}{kT}}&0 \\ 0&e^{-\frac{E_-}{kT}} \end{bmatrix}$

### High Temperature Approximation

We can further refine the inital density operator by substituting in the equations for the energy levels E+ and E-.

$E_(\pm)=\mp \frac{\bar{h} \omega_0}{2}$

but \since we know the kT>>$$\omega_0$$ then we can use a high temperature approximation of

$e^{\pm\frac{\bar{h} \omega_0}{2kT}}=1 \pm \frac{\bar{h} \omega_0}{2kT}$

ans

$Z=2$

resulting in a new density operator as

$\rho (0)\begin{bmatrix} \frac{1}{2} +\frac{\bar{h} \omega_0}{2kT} & 0 \\ 0 & \frac{1}{2}- \frac{\bar{h} \omega_0}{2kT} \end{bmatrix}=\frac{N}{2}1 + \frac{N \bar{h} \omega_0}{2kT}I_z$

Normally the magnetization constant is taken to be a scaling factor and the unity operator is invariant to applied fields which firther reduces the density matrix to

$\rho(0)=I_z$

## NMR Observables

In a simple NMR experiment, the magnetization along the externally applied magnetix field is knocked into the x-y plane u\sing a rf pulse. As the spins recover along the external field direction (z direction) they precess near their Larmor frequencies through a coil which in turn produces an EMF. In order for an EMF to be generated the preces\sing magnetization must be perpendicular to the coil. According the Kirchoff

$EMF=-\mu_0 \eta A \frac{d}{dt} M_x$

as is immediatley evident the magnetizaion is time dependent and our problem is essentially reduced to

$M_x(t)=<T_x>=<\Psi (t)| I_x| \Psi (t) \rangle$

which is the expectation value of Ix. If instead we consider a generic operator <O> then

$| \Psi (t) \rangle= \sum_n c_n(t)|n>$

$< \Psi (t)|= \sum_n c^*_n(t)<n|$

$<O>=<\Psi (t)| O| \Psi (t) \rangle=\sum_{n,m}c_n(t)c^*_m(t)<m|O|n>$

and recalling that

$<n|\rho (t)|m>=c_n(t)c^*_m(t)$

results in

$<O> \sum_{n,m} <n|\rho (t)|m><m|O|n>=Tr[\rho(t)O]$

Now for Mx

$M_x=<I_x>=TR[I_x \rho (t)$

$\rho (t)=I_z$

$H(t)=-\omega_0I_z-2\omega_1 \cos\omega_0 t I_x$

Then all we need to do is calculate $$\rho (t)$$ and <Ix>!

### Matrix Approach

Like we did in the above treatment for the roatating wave, we need to move $$\rho (t)$$ into a different representation.

$\sigma(t)=R^{\dagger}(t) \rho(t)R(t)$

$\rho(t)=R(t) \sigma(t)R^{\dagger}(t)$

$\frac{d}{dt} \sigma(t)=-iH[H_{int}, \sigma(t)]$

and remembering that

$R(t)=e^{i \omega_0 t I_z}$

$\sigma(0)=\rho(0)$

We get

$U(t)=\begin{bmatrix} \cos \frac{\omega_1 t}{2}&i\sin \frac{\omega_1 t}{2}\\ i\sin \frac{\omega_1 t}{2}&\cos \frac{\omega_1 t}{2}\end{bmatrix}$

$U^{\dagger}(t)=\begin{bmatrix} \cos \frac{\omega_1 t}{2}&-i\sin \frac{\omega_1 t}{2}\\ -i\sin \frac{\omega_1 t}{2}&\cos \frac{\omega_1 t}{2}\end{bmatrix}$

which gives

$\sigma(t)= U(t) \sigma(0) U^{\dagger}(t)$

$=\begin{bmatrix} \cos \frac{\omega_1 t}{2}&i\sin \frac{\omega_1 t}{2}\\ i\sin \frac{\omega_1 t}{2}&\cos \frac{\omega_1 t}{2}\end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} \cos \frac{\omega_1 t}{2}&-i\sin \frac{\omega_1 t}{2}\\ -i\sin \frac{\omega_1 t}{2}&\cos \frac{\omega_1 t}{2}\end{bmatrix}=\frac{1}{2}\begin{bmatrix} \cos \omega_1 t & -i\sin \omega_1 t\\-i\sin \omega_1 t&\cos \omega_1 t\end{bmatrix}$

Now we can use $$\sigma(t)$$ to find $$\rho (t)$$.

$\rho(t)=R(t) \sigma(t) R^{\dagger}(t)=\frac{1}{2} \begin{bmatrix} e^{\frac{i \omega_0 t}{2}} & 0 \\ 0 &e^{\frac{-i \omega_0 t}{2}} \end{bmatrix} \begin{bmatrix} \cos \omega_1 t & -i\sin \omega_1 t\\-i\sin \omega_1 t&\cos \omega_1 t\end{bmatrix} \begin{bmatrix} e^{\frac{-i \omega_0 t}{2}} & 0 \\ 0 &e^{\frac{i \omega_0 t}{2}} \end{bmatrix} =\begin{bmatrix} \cos \omega_1 t & -i\sin \omega_1 t e^{i \omega_0 t}\\-i\sin \omega_1 t e^{-i \omega_0 t}&\cos \omega_0 t \end{bmatrix}$

Now lets look at the observables Ix and Iy.

$<I_x> = Tr{I_x \rho(t)}=Tr{ \begin{bmatrix} 0 & 1/2 \\ 1/2 & 0 \end {bmatrix} \begin{bmatrix} \cos \omega_1 t & -i\sin \omega_1 t e^{i \omega_0 t}\\-i\sin \omega_1 t e^{-i \omega_0 t}&\cos \omega_1 t \end{bmatrix}}$

$=\frac{sin \omega_1t sin \omega_0 t}{2}$

Then <Iy> is

$<I_y>=\frac{sin \omega_1 t cos \omega_0 t}{2}$

And <Iz> is

$<I_z>=\frac{cos \omega_1 t}{2}$

### Operator Approach

Alternatively we can use the operator approach, which is much less mathemtically intensive.

$\sigma(0)=\rho (0) =I_z$

$H_{int}=-\omega_1 I_x$

$\sigma (t)=e^{-i H_{int} t} \sigma(0) e^{i H_{int} t}=cos \omega_1 t I_z +sin \omega_1 t I_y$

$\rho (t) = R(t) \sigma(t) R^{\dagger} (t) = cos \omega_1 t I_z + sin \omega_1 t cos\omega_0t I_y + sin\omega_1 t sin \omega_0 t I_x$

Using

$Tr [I_n I_m] =\frac{1}{2} \delta_{nm}$

we obtain the same expectation values as before.