Question

Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio $\mathbf{\gamma }$ at rest in a static magnetic field${\mathbf{B}}_{\mathbf{0}}\widehat{\mathbf{k}}$ precesses at the Larmor frequency${\mathbf{\omega }}_{\mathbf{0}}\mathbf{=}{\mathbf{\gamma B}}_{\mathbf{0}}$ (Example 4.3). Now we turn on a small transverse radiofrequency (rf) field,${\mathit{B}}_{\mathbf{\text{rf}}}\mathbf{\left[}\mathbf{cos}\mathbf{\right(}\mathit{\omega }\mathit{t}\mathbf{)}\widehat{\mathbf{ı}}\mathbf{-}\mathbf{sin}\mathbf{(}\mathit{\omega }\mathit{t}\mathbf{\left)}\hat{\mathbf{j}}\mathbf{\right]}\mathit{\$}\mathbf{,}$ so that the total field is

role="math" localid="1659004119542" $\mathbf{B}\mathbf{=}{\mathit{B}}_{\mathbf{\text{rf}}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{\omega }\mathit{t}\mathbf{\right)}\widehat{\mathbf{ı}}\mathbf{-}{\mathit{B}}_{\mathbf{\text{rf}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{\omega }\mathit{t}\mathbf{\right)}\hat{\mathbf{j}}\mathbf{+}{\mathbf{B}}_{\mathbf{0}}\widehat{\mathbf{k}}$

(a) Construct the $\mathbf{2}\mathbf{\times }\mathbf{2}$Hamiltonian matrix (Equation 4.158) for this system.

(b) If $\mathit{\chi }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\left(\begin{array}{l}a\left(t\right)\\ b\left(t\right)\end{array}\right)$is the spin state at time $\mathbf{t}$, show that

$\dot{\mathbf{a}}\mathbf{=}\frac{\mathbf{i}}{\mathbf{2}}\mathbf{(}{\mathbf{\Omega e}}^{\mathbf{i\omega t}}\mathbf{b}\mathbf{+}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{a}\mathbf{)}\mathbf{:}\dot{\mathbf{b}}\mathbf{=}\frac{\mathbf{i}}{\mathbf{2}}\mathbf{(}{\mathbf{\Omega e}}^{\mathbf{-}\mathbf{i\omega t}}\mathbf{a}\mathbf{-}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{b}\mathbf{)}$

where $\mathbf{\Omega }\mathbf{\equiv }{\mathbf{\gamma B}}_{\mathbf{\text{rf}}}$is related to the strength of the rf field.

(c) Check that the general solution for$\mathbf{a}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ and$\mathbf{b}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ in terms of their initial values${\mathbf{a}}_{\mathbf{0}}$ and${\mathbf{b}}_{\mathbf{0}}$ is

role="math" localid="1659004637631" $\begin{array}{l}\mathbf{a}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{a}}_{\mathbf{0}}\mathbf{cos}\mathbf{(}{\mathbf{\omega }}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{+}\frac{\mathbf{i}}{{\mathbf{\omega }}^{\mathbf{\text{'}}}}\mathbf{[}{\mathbf{a}}_{\mathbf{0}}\mathbf{(}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{-}\mathbf{\omega }\mathbf{)}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{\Omega }\mathbf{]}\mathbf{sin}\mathbf{(}{\mathbf{\omega }}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{\}}{\mathbf{e}}^{\mathbf{i\omega t}\mathbf{/}\mathbf{2}}\\ \mathbf{b}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{b}}_{\mathbf{0}}\mathbf{cos}\mathbf{(}{\mathbf{\omega }}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{+}\frac{\mathbf{i}}{{\mathbf{\omega }}^{\mathbf{\text{'}}}}\mathbf{[}{\mathbf{b}}_{\mathbf{0}}\mathbf{(}\mathbf{\omega }\mathbf{-}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{)}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{\Omega }\mathbf{]}\mathbf{sin}\mathbf{(}{\mathbf{\omega }}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{\}}{\mathbf{e}}^{\mathbf{-}\mathbf{i\omega t}\mathbf{/}\mathbf{2}}\end{array}$

Where

${\mathbf{\omega }}^{\mathbf{\text{'}}}\mathbf{\equiv }\sqrt{{\mathbf{(}\mathbf{\omega }\mathbf{-}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\Omega }}^{\mathbf{2}}}$

(d) If the particle starts out with spin up (i.e. ${\mathbf{a}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{b}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$,), find the probability of a transition to spin down, as a function of time. Answer:$\mathit{P}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{\Omega }}^{\mathbf{2}}\mathbf{/}\mathbf{[}{\mathbf{(}\mathbf{\omega }\mathbf{-}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\Omega }}^{\mathbf{2}}\mathbf{]}\mathbf{\}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{(}{\mathbf{\omega }}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}$

(e) Sketch the resonance curve,

role="math" localid="1659004767993" $\mathbf{P}\mathbf{\left(}\mathbf{\omega }\mathbf{\right)}\mathbf{=}\frac{{\mathbf{\Omega }}^{\mathbf{2}}}{{\mathbf{(}\mathbf{\omega }\mathbf{-}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\Omega }}^{\mathbf{2}}}\mathbf{,}$

as a function of the driving frequency$\mathbf{\omega }$ (for fixed ${\mathbf{\omega }}_{\mathbf{0}}$and$\mathbf{\Omega }$ ). Note that the maximum occurs at$\mathbf{\omega }\mathbf{=}{\mathbf{\omega }}_{\mathbf{0}}$ Find the "full width at half maximum,"$\mathbf{\Delta \omega }$

(f) Since ${\mathbf{\omega }}_{\mathbf{0}}\mathbf{=}{\mathbf{\gamma B}}_{\mathbf{0}}$we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the factor of the proton is to be measured, using a static field of 10,000 gauss and an rf field of amplitude gauss. What will the resonant frequency be? (See Section for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.)

Short answer

a) Hamiltonian matrix $\mathrm{H}=-\frac{\mathrm{\gamma }\mathrm{\hslash }}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{\mathrm{i\omega t}}\\ {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{-\mathrm{i\omega t}}& -{\mathrm{B}}_0\end{array}\right)$

b) The strength of the field $\begin{array}{l}\dot{\mathrm{a}}=\frac{\mathrm{i}}{2}({\mathrm{\Omega e}}^{\mathrm{i\omega t}}\mathrm{b}+{\mathrm{\omega }}_0\mathrm{a})\\ \dot{\mathrm{b}}=\frac{\mathrm{i}}{2}({\mathrm{\Omega e}}^{-\mathrm{i\omega t}}\mathrm{a}-{\mathrm{\omega }}_0\mathrm{b})\end{array}$

c) The general solution $\begin{array}{l}\mathrm{a}\left(\mathrm{t}\right)=\left[{\mathrm{a}}_0\cos \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)+\frac{\mathrm{i}}{{\mathrm{\omega }}^{\text{'}}}({\mathrm{b}}_0\mathrm{\Omega }+{\mathrm{a}}_0({\mathrm{\omega }}_0-\mathrm{\omega }))\sin \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\right]{\mathrm{e}}^{\mathrm{i\omega t}/2}\\ \mathrm{b}\left(\mathrm{t}\right)=\left[{\mathrm{b}}_0\cos \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)+\frac{\mathrm{i}}{{\mathrm{\omega }}^{\text{'}}}({\mathrm{a}}_0\mathrm{\Omega }-{\mathrm{b}}_0({\mathrm{\omega }}_0-\mathrm{\omega }))\sin \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\right]{\mathrm{e}}^{-\mathrm{i\omega t}/2}\end{array}$

d) The probability of a transition$\mathrm{P}\left(\mathrm{t}\right)=\frac{{\mathrm{\Omega }}^2}{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2}{\sin }^2\left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)$

e) The resonance curve $\mathrm{\Delta \omega }=2\mathrm{\Omega }$

f) The width of the resonance curve$\begin{array}{l}{\mathrm{v}}_{\text{res}}=4.26\times {10}^7\text{Hz\hspace{0.17em}}\\ \mathrm{\Delta v}=85.2\text{Hz}\end{array}$

Step-by-step solution

Hamiltonian matrix.

(a)

Consider a particle with spin of $1/2$and gyromagnetic ratio of $\mathrm{\gamma }$the particle is at rest in a static magnetic field $B_0{\widehat{k}}_{\text{t}}$in this case it processes at the Larmor frequency ${\omega }_0=\gamma B_0$Turn a transverse radio-frequency field, $B_{\text{rf}}\left[\cos \right(\omega t)\widehat{ı}-\sin (\omega t\left)\widehat{j}\right]$, so the total magnetic field that the particle experiences is: $\mathbf{B}=\underset{{\mathrm{B}}_{\mathrm{z}}}{\underbrace{{\mathrm{B}}_{\text{rf}}\cos \left(\mathrm{\omega t}\right)}}\underset{{\mathrm{B}}_{\mathrm{y}}}{\underbrace{\widehat{\mathrm{i}}-{\mathrm{B}}_{\text{rf}}\sin \left(\mathrm{\omega t}\right)}}\widehat{\mathrm{j}}+\underset{{\mathrm{B}}_{\mathrm{z}}}{\underbrace{{\mathrm{B}}_0}}\widehat{\mathrm{k}}$

The spin Hamiltonian is given by:

$\mathrm{H}=-\mathrm{\gamma }\mathbf{B}\cdot \mathbf{S}$

Where;

$\mathbf{S}=\frac{\hslash }{2}\sigma$and,

${\sigma }_x=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right];{\sigma }_y=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right];{\sigma }_z=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$

The Hamiltonian is therefore:

$\begin{array}{c}\mathrm{H}=-\mathrm{\gamma }({\mathrm{B}}_{\mathrm{x}}{\mathrm{S}}_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}{\mathrm{S}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{z}}{\mathrm{S}}_{\mathrm{z}})\\ =-\mathrm{\gamma }\frac{\mathrm{\hslash }}{2}({\mathrm{B}}_{\mathrm{x}}{\mathrm{\sigma }}_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}{\mathrm{\sigma }}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{z}}{\mathrm{\sigma }}_{\mathrm{z}})\\ =-\frac{\mathrm{\gamma }\mathrm{\hslash }}{2}\left[{\mathrm{B}}_{\mathrm{x}}\left(\begin{array}{c}01\\ 1\end{array}\right)+{\mathrm{B}}_{\mathrm{y}}\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)+{\mathrm{B}}_{\mathrm{z}}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\right]\\ =-\frac{\mathrm{\gamma }\mathrm{\hslash }}{2}\left(\begin{array}{cc}{\mathrm{B}}_{\mathrm{z}}& {\mathrm{B}}_{\mathrm{x}}-{\mathrm{iB}}_{\mathrm{y}}\\ {\mathrm{B}}_{\mathrm{x}}+{\mathrm{iB}}_{\mathrm{y}}& -{\mathrm{B}}_{\mathrm{z}}\end{array}\right)\end{array}$

$\begin{array}{c}\mathrm{H}=-\frac{\mathrm{\gamma }\mathrm{\hslash }}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rf}}(\mathrm{cos\omega t}+\mathrm{isin\omega t})\\ {\mathrm{B}}_{\text{rf}}(\mathrm{cos\omega t}-\mathrm{isin\omega t})& -{\mathrm{B}}_0\end{array}\right)\\ =-\frac{\mathrm{\gamma }\mathrm{\hslash }}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rr}}{\mathrm{e}}^{\mathrm{i\omega t}}\\ {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{-\mathrm{i\omega t}}& -{\mathrm{B}}_0\end{array}\right)\\ =-\frac{\mathrm{\gamma }\mathrm{\hslash }}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{\mathrm{i\omega t}}\\ {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{-\mathrm{i\omega t}}& -{\mathrm{B}}_0\end{array}\right)\end{array}$

(1)

The spin state at time.

(b)

Let$\mathrm{\chi }\left(\mathrm{t}\right)$be the spin state at time$t$which is given by:

role="math" localid="1659005644415" $\chi \left(t\right)=\left(\begin{array}{l}a\left(t\right)\\ b\left(t\right)\end{array}\right)$

To find$\dot{a}\left(t\right)$ and$\dot{b}\left(t\right)$in order to do this we need to apply Schrodinger equation on this state, the time dependent Schrodinger is given by:

$\mathrm{ih}\dot{\mathrm{\chi }}=\text{H}\mathrm{\chi }$

Using the result of part (a) we get:

Thus:

$\begin{array}{c}\dot{\mathrm{a}}=\mathrm{i}\frac{\mathrm{\gamma }}{2}({\mathrm{B}}_0\mathrm{a}+{\mathrm{B}}_{\text{rf}}{\mathrm{c}}^{\mathrm{i\omega t}}\mathrm{b})\\ \dot{\mathrm{b}}=-\mathrm{i}\frac{\mathrm{\gamma }}{2}({\mathrm{B}}_{00}\mathrm{b}-{{\mathrm{B}}_{\text{rf}\mathrm{e}}}^{\mathrm{i\omega t}}\mathrm{a})\end{array}$

Let$\mathrm{\Omega }\equiv {\mathrm{\gamma B}}_{\text{rr}}\mathrm{\$},$ thus:

$\begin{array}{l}\dot{\mathrm{a}}=\frac{\mathrm{i}}{2}({\mathrm{\Omega e}}^{\mathrm{i\omega t}}\mathrm{b}+{\mathrm{\omega }}_0\mathrm{a})\\ \dot{\mathrm{b}}=\frac{\mathrm{i}}{2}({\mathrm{\Omega e}}^{-\mathrm{i\omega t}}\mathrm{a}-{\mathrm{\omega }}_0\mathrm{b})\end{array}$ …… (2)

Checking the general equation.

a. We need to solve the equations in part (b), first we need to write the first equation in (2) for $b$as

$\mathrm{b}=\frac{1}{\mathrm{\Omega }}{\mathrm{e}}^{-\mathrm{i\omega t}}\left(\frac{2}{\mathrm{i}}\dot{\mathrm{a}}-{\mathrm{\omega }}_0\mathrm{a}\right)$ …… (3)

The derivative of the first equation in (2), so the result will be in terms $\dot{a},\dot{b}$and $b$we substitute with $b$from (3) and with$\dot{b}$for (2)

role="math" localid="1659006175845" $\begin{array}{c}\ddot{\mathrm{a}}=\frac{\mathrm{i}}{2}({\mathrm{\Omega i\omega e}}^{\mathrm{i\omega t}}\mathrm{b}+\mathrm{\Omega }{\mathrm{e}}^{\mathrm{i\omega t}}\dot{\mathrm{b}}+{\mathrm{\omega }}_0\dot{\mathrm{a}})\\ =\frac{\mathrm{i}}{2}\left({\mathrm{\Omega i\omega e}}^{\mathrm{i\omega t}}\frac{1}{\mathrm{\Omega }}{\mathrm{e}}^{-\mathrm{i\omega t}}\left(\frac{2}{\mathrm{i}}\dot{\mathrm{a}}-{\mathrm{\omega }}_0\mathrm{a}\right)+{\mathrm{\Omega e}}^{\mathrm{i\omega t}}\frac{\mathrm{i}}{2}({\mathrm{\Omega e}}^{-\mathrm{i\omega t}}\mathrm{a}-{\mathrm{\omega }}_0\mathrm{b})+{\mathrm{\omega }}_0\dot{\mathrm{a}}\right)\\ =-\frac{\mathrm{\omega }}{2}\left(\frac{2}{\mathrm{i}}\dot{\mathrm{a}}-{\mathrm{\omega }}_0\mathrm{a}\right)-\frac{1}{4}{\mathrm{\Omega e}}^{\mathrm{i\omega t}}({\mathrm{\Omega e}}^{-\mathrm{i\omega t}}\mathrm{a}-{\mathrm{\omega }}_0\frac{1}{\mathrm{\Omega }}{\mathrm{e}}^{-\mathrm{i\omega t}}\left(\frac{2}{\mathrm{i}}\dot{\mathrm{a}}-{\mathrm{\omega }}_0\mathrm{a}\right)+\frac{\mathrm{i}}{2}{\mathrm{\omega }}_0\dot{\mathrm{a}}\\ =\mathrm{i\omega }\dot{\mathrm{a}}-\frac{1}{4}({\mathrm{\Omega }}^2+{\mathrm{\omega }}_0^2-2{\mathrm{\omega \omega }}_0)\mathrm{a}\end{array}$

So, the second order ODE with constant coefficient,

$\ddot{\mathrm{a}}-\mathrm{i\omega }\dot{\mathrm{a}}+\frac{1}{4}({\mathrm{\Omega }}^2+{\mathrm{\omega }}_0^2-2{\mathrm{\omega \omega }}_0)\mathrm{a}=0$

Write the last two terms in the bracket as:

${\mathrm{\omega }}_0^2-2{\mathrm{\omega \omega }}_0={\mathrm{\omega }}_0^2-2{\mathrm{\omega \omega }}_0+{\mathrm{\omega }}^2-{\mathrm{\omega }}^2={(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2-{\mathrm{\omega }}^2$

Thus:

$\ddot{\mathrm{a}}-\mathrm{i\omega }\dot{\mathrm{a}}+\frac{1}{4}({\mathrm{\Omega }}^2+{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2-{\mathrm{\omega }}^2)\mathrm{a}=0$ ……..(4)

The characteristic equation is therefore:

${\mathrm{\lambda }}^2-\mathrm{i\omega \lambda }+\frac{1}{4}({\mathrm{\Omega }}^2+{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2-{\mathrm{\omega }}^2)=0$

The roots of this equation,

$\begin{array}{c}\mathrm{\lambda }=\frac{\mathrm{i\omega }\pm \sqrt{-\mathrm{\omega }-({\mathrm{\Omega }}^2+{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2-{\mathrm{\omega }}^2)}}{2}\\ =\frac{\mathrm{i}}{2}(\mathrm{\omega }\pm \sqrt{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2})\\ =\frac{\mathrm{i}}{2}(\mathrm{\omega }\pm {\mathrm{\omega }}^{\text{'}})\end{array}$

Where;

${\mathrm{\omega }}^{\text{'}}\equiv \sqrt{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2}$

The general solution is:

$\begin{array}{c}\mathrm{a}\left(\mathrm{t}\right)={\mathrm{Ae}}^{{\mathrm{\lambda }}_1}+{\mathrm{Be}}^{{\mathrm{\lambda }}_2}\\ ={\mathrm{Ae}}^{\mathrm{i}(\mathrm{\omega }+{\mathrm{\omega }}^{\text{'}})\mathrm{t}/2}+{\mathrm{Be}}^{\mathrm{i}(\mathrm{\omega }-{\mathrm{\omega }}^{\text{'}})\mathrm{t}/2}\end{array}$……………(5)

The solution of$b\left(t\right)$ is similar to$a\left(t\right)$ but with signs reversed on$\omega$and${\omega }_0$so the sign of ${\omega }^{\text{'}}$stays the same. So,

$\mathrm{b}\left(\mathrm{t}\right)={\mathrm{Ce}}^{\mathrm{i}(-\mathrm{\omega }+{\mathrm{\omega }}^{\text{'}})\mathrm{l}/2}+{\mathrm{De}}^{\mathrm{i}(-\mathrm{\omega }-{\mathrm{\omega }}^{\text{'}})\mathrm{t}/2}$ ……………(6)

To apply the initial conditions, which are$\mathrm{a}\left(0\right)={\mathrm{a}}_0$and $b\left(0\right)=b_0$From (6) and (7) we get

$\begin{array}{l}{\mathrm{a}}_0=\mathrm{A}+\mathrm{B}\\ {\mathrm{b}}_0=\mathrm{C}+\mathrm{D}\end{array}$…………(7)

To find the constant, substitute with the solutions (6) and (7) into the first equation of so we get:

role="math" localid="1659006658486"

At$t=0$

…… (8)

$\begin{array}{c}A\frac{i}{2}(\omega +{\omega }^{\text{'}})+B\frac{i}{2}(\omega -{\omega }^{\text{'}})=\frac{i\Omega }{2}(C+D)+{\omega }_0{a}_0\omega a_0+(A-B){\omega }^{\text{'}}\\ \frac{i\Omega }{2}{b}_0+{\omega }_0{a}_{0}A-B\\ \frac{1}{{\omega }^{\text{'}}}(b_0\Omega +a_0({\omega }_0-\omega ))\end{array}$

The three equations with four constants, we can't solve for them, so what we will do is to write the solution in equation (5) in terms of the cosine and sine then let collect the common terms, will notice that the constants appear as a sum of subtract between them, and we have this relation in (7) and (8).

Substitute from (7) and (8), we get:

$\mathrm{a}\left(\mathrm{t}\right)=\left[{\mathrm{a}}_0\cos \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)+\frac{\mathrm{i}}{{\mathrm{\omega }}^{\text{'}}}({\mathrm{b}}_0\mathrm{\Omega }+{\mathrm{a}}_0({\mathrm{\omega }}_0-\mathrm{\omega }))\sin \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\right]{\mathrm{e}}^{\mathrm{i\omega t}/2}$

The solution of $b\left(t\right)$is similar to $a\left(t\right)$, but with signs reversed on $\omega$and ${\omega }_0$so the sign of ${\omega }^{\text{'}}$stays the same, thus:

$\mathrm{b}\left(\mathrm{t}\right)=\left[{\mathrm{b}}_0\cos \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)+\frac{\mathrm{i}}{{\mathrm{\omega }}^{\text{'}}}({\mathrm{a}}_0\mathrm{\Omega }-{\mathrm{b}}_0({\mathrm{\omega }}_0-\mathrm{\omega }))\sin \left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\right]{\mathrm{e}}^{-\mathrm{i\omega t}/2}$

The probability of a transition  

(d)

Now consider a particle that starts out with a spin up,$a_0=1$ and$b_0=0$ the probability of a transition to the spin down is:

$P\left(t\right)=\left|b\right(t){|}^2$

The result of part $\left(c\right)$we get:

$\begin{array}{c}\mathrm{P}\left(\mathrm{t}\right)=\frac{{\mathrm{\Omega }}^2}{{\left({\mathrm{\omega }}^{\text{'}}\right)}^2}{\sin }^2\left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\\ =\frac{{\mathrm{\Omega }}^2}{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2}{\sin }^2\left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\mathrm{P}\left(\mathrm{t}\right)\\ =\frac{{\mathrm{\Omega }}^2}{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2}{\sin }^2\left(\frac{{\mathrm{\omega }}^{\text{'}}\mathrm{t}}{2}\right)\end{array}$

The resonance curve.

(e)

Now we need to sketch the resonance curve, which is given by:

$\mathrm{P}\left(\mathrm{\omega }\right)=\frac{{\mathrm{\Omega }}^2}{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2}$

Let ${\mathrm{\omega }}_0=1$and $\mathrm{\Omega }=2$ Use any values, the graph is shown in the following figure. That the maximum occurs at $\mathrm{\omega }={\mathrm{\omega }}_0$to find at which point half of the maximum lie let set$\mathrm{P}\left(\mathrm{\omega }\right)=0.5$ and then solve for$\mathrm{\omega }$ so:

$\begin{array}{c}\frac{{\mathrm{\Omega }}^2}{{(\mathrm{\omega }-{\mathrm{\omega }}_0)}^2+{\mathrm{\Omega }}^2}=0.5\\ \mathrm{\omega }-{\mathrm{\omega }}_0=\pm \mathrm{\Omega }\end{array}$

The full width at half maximum is therefore:

$\mathrm{\Delta \omega }=2\mathrm{\Omega }$

The gyro-magnetic ration

(f)

The gyro-magnetic ratio for a proton is given by:

$\mathrm{\gamma }=\frac{{\mathrm{g}}_{\mathrm{p}}\mathrm{e}}{2{\mathrm{m}}_{\mathrm{p}}}$

Where${\mathrm{g}}_{\mathrm{p}}=5.59$

To find the resonant frequency, which is:

${\mathrm{\nu }}_{\text{res}}=\frac{{\mathrm{\omega }}_0}{2\mathrm{\pi }}$

But ${\mathrm{\omega }}_0={\mathrm{\gamma B}}_0$

so:

${\nu }_{\text{res}}=\frac{g_{p}e}{4\pi m_p}{B}_0$for ${\mathrm{B}}_0=10,000$gauss $=1\text{T}$, we get:

$\begin{array}{c}{\mathrm{\nu }}_{\text{res}}=\frac{(5.59)(1.6\times {10}^{-19}\text{C})}{4\mathrm{\pi }(1.67\times {10}^{-27}\text{kg})}(1\text{T})\\ =4.26\times {10}^7\text{Hz}\\ {\mathrm{\nu }}_{\text{res}}=4.26\times {10}^7\text{Hz}\end{array}$

To find the width of the resonance curve, which is given by:

$\mathrm{\Delta \nu }=\frac{\mathrm{\Delta \omega }}{2\mathrm{\pi }}=\frac{\mathrm{\Omega }}{\mathrm{\pi }}$

$\begin{array}{c}\mathrm{\Delta \nu }=\frac{\mathrm{\gamma }}{\mathrm{\pi }}{\mathrm{B}}_{\text{rf}}\\ =\frac{\mathrm{\gamma }}{2\mathrm{\pi }}2{\mathrm{B}}_{\text{rf}}\\ ={\mathrm{\nu }}_{\text{res}}\frac{2{\mathrm{B}}_{\text{rf}}}{{\mathrm{B}}_0}\end{array}$

For ${\mathrm{B}}_{\text{rf}}=0.01$ gauss $=1\times {10}^{-6}\text{T}$,we get:

$\begin{array}{c}\mathrm{\Delta \nu }=(4.26\times {10}^7)(2\times {10}^{-6})\\ =85.2\text{Hz}\end{array}$