91影视

(a)

Consider a particle with spin of $1/2$and gyromagnetic ratio of $\mathrm{纬}$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 ${蝇}_0=纬B_0$Turn a transverse radio-frequency field, $B_{\text{rf}}\left[\cos \right(蝇t)\widehat{谋}鈭�\sin (蝇t\left)\widehat{j}\right]$, so the total magnetic field that the particle experiences is: $\mathbf{B}=\underset{{\mathrm{B}}_{\mathrm{z}}}{\underset{铃�}{{\mathrm{B}}_{\text{rf}}\cos \left(\mathrm{蝇t}\right)}}\underset{{\mathrm{B}}_{\mathrm{y}}}{\underset{铃�}{\widehat{\mathrm{i}}鈭�{\mathrm{B}}_{\text{rf}}\sin \left(\mathrm{蝇t}\right)}}\widehat{\mathrm{j}}+\underset{{\mathrm{B}}_{\mathrm{z}}}{\underset{铃�}{{\mathrm{B}}_0}}\widehat{\mathrm{k}}$

The spin Hamiltonian is given by:

$\mathrm{H}=鈭�\mathrm{纬}\mathbf{B}鈰�\mathbf{S}$

Where;

$\mathbf{S}=\frac{\mathrm{鈩�}}{2}蟽$and,

${蟽}_x=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right];\text{鈥夆赌夆赌�}{蟽}_y=\left[\begin{array}{cc}0& 鈭�i\\ i& 0\end{array}\right];\text{鈥夆赌夆赌�}{蟽}_z=\left[\begin{array}{cc}1& 0\\ 0& 鈭�1\end{array}\right]$

The Hamiltonian is therefore:

$\begin{array}{c}\mathrm{H}=鈭�\mathrm{纬}({\mathrm{B}}_{\mathrm{x}}{\mathrm{S}}_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}{\mathrm{S}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{z}}{\mathrm{S}}_{\mathrm{z}})\\ =鈭�\mathrm{纬}\frac{\mathrm{鈩�}}{2}({\mathrm{B}}_{\mathrm{x}}{\mathrm{蟽}}_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}{\mathrm{蟽}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{z}}{\mathrm{蟽}}_{\mathrm{z}})\\ =鈭�\frac{\mathrm{纬}\mathrm{鈩�}}{2}\left[{\mathrm{B}}_{\mathrm{x}}\left(\begin{array}{c}01\\ 1\end{array}\right)+{\mathrm{B}}_{\mathrm{y}}\left(\begin{array}{cc}0& 鈭�\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)+{\mathrm{B}}_{\mathrm{z}}\left(\begin{array}{cc}1& 0\\ 0& 鈭�1\end{array}\right)\right]\\ =鈭�\frac{\mathrm{纬}\mathrm{鈩�}}{2}\left(\begin{array}{cc}{\mathrm{B}}_{\mathrm{z}}& {\mathrm{B}}_{\mathrm{x}}鈭�{\mathrm{iB}}_{\mathrm{y}}\\ {\mathrm{B}}_{\mathrm{x}}+{\mathrm{iB}}_{\mathrm{y}}& 鈭�{\mathrm{B}}_{\mathrm{z}}\end{array}\right)\end{array}$

$\begin{array}{c}\mathrm{H}=鈭�\frac{\mathrm{纬}\mathrm{鈩�}}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rf}}(\mathrm{cos蝇t}+\mathrm{isin蝇t})\\ {\mathrm{B}}_{\text{rf}}(\mathrm{cos蝇t}鈭�\mathrm{isin蝇t})& 鈭�{\mathrm{B}}_0\end{array}\right)\\ =鈭�\frac{\mathrm{纬}\mathrm{鈩�}}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rr}}{\mathrm{e}}^{\mathrm{i蝇t}}\\ {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{鈭�\mathrm{i蝇t}}& 鈭�{\mathrm{B}}_0\end{array}\right)\\ =鈭�\frac{\mathrm{纬}\mathrm{鈩�}}{2}\left(\begin{array}{cc}{\mathrm{B}}_0& {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{\mathrm{i蝇t}}\\ {\mathrm{B}}_{\text{rf}}{\mathrm{e}}^{鈭�\mathrm{i蝇t}}& 鈭�{\mathrm{B}}_0\end{array}\right)\end{array}$

(1)

(b)

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

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

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

$\mathrm{ih}\stackrel{藱}{\mathrm{蠂}}=\text{H}\mathrm{蠂}$

Using the result of part (a) we get:

Thus:

$\begin{array}{c}\stackrel{藱}{\mathrm{a}}=\mathrm{i}\frac{\mathrm{纬}}{2}({\mathrm{B}}_0\mathrm{a}+{\mathrm{B}}_{\text{rf}}{\mathrm{c}}^{\mathrm{i蝇t}}\mathrm{b})\\ \stackrel{藱}{\mathrm{b}}=鈭�\mathrm{i}\frac{\mathrm{纬}}{2}({\mathrm{B}}_{00}\mathrm{b}鈭�{{\mathrm{B}}_{\text{rf}\mathrm{e}}}^{\mathrm{i蝇t}}\mathrm{a})\end{array}$

Let$\mathrm{惟}鈮�{\mathrm{纬叠}}_{\text{rr}}\mathrm{\$},$ thus:

$\begin{array}{l}\stackrel{藱}{\mathrm{a}}=\frac{\mathrm{i}}{2}({\mathrm{惟别}}^{\mathrm{i蝇t}}\mathrm{b}+{\mathrm{蝇}}_0\mathrm{a})\\ \stackrel{藱}{\mathrm{b}}=\frac{\mathrm{i}}{2}({\mathrm{惟别}}^{鈭�\mathrm{i蝇t}}\mathrm{a}鈭�{\mathrm{蝇}}_0\mathrm{b})\end{array}$ 鈥︹�� (2)

$\mathrm{b}=\frac{1}{\mathrm{惟}}{\mathrm{e}}^{鈭�\mathrm{i蝇t}}\left(\frac{2}{\mathrm{i}}\stackrel{藱}{\mathrm{a}}鈭�{\mathrm{蝇}}_0\mathrm{a}\right)$ 鈥︹�� (3)

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

role="math" localid="1659006175845" $\begin{array}{c}\stackrel{篓}{\mathrm{a}}=\frac{\mathrm{i}}{2}({\mathrm{惟i蝇e}}^{\mathrm{i蝇t}}\mathrm{b}+\mathrm{惟}{\mathrm{e}}^{\mathrm{i蝇t}}\stackrel{藱}{\mathrm{b}}+{\mathrm{蝇}}_0\stackrel{藱}{\mathrm{a}})\\ =\frac{\mathrm{i}}{2}\left({\mathrm{惟i蝇e}}^{\mathrm{i蝇t}}\frac{1}{\mathrm{惟}}{\mathrm{e}}^{鈭�\mathrm{i蝇t}}\left(\frac{2}{\mathrm{i}}\stackrel{藱}{\mathrm{a}}鈭�{\mathrm{蝇}}_0\mathrm{a}\right)+{\mathrm{惟别}}^{\mathrm{i蝇t}}\frac{\mathrm{i}}{2}({\mathrm{惟别}}^{鈭�\mathrm{i蝇t}}\mathrm{a}鈭�{\mathrm{蝇}}_0\mathrm{b})+{\mathrm{蝇}}_0\stackrel{藱}{\mathrm{a}}\right)\\ =鈭�\frac{\mathrm{蝇}}{2}\left(\frac{2}{\mathrm{i}}\stackrel{藱}{\mathrm{a}}鈭�{\mathrm{蝇}}_0\mathrm{a}\right)鈭�\frac{1}{4}{\mathrm{惟别}}^{\mathrm{i蝇t}}({\mathrm{惟别}}^{鈭�\mathrm{i蝇t}}\mathrm{a}鈭�{\mathrm{蝇}}_0\frac{1}{\mathrm{惟}}{\mathrm{e}}^{鈭�\mathrm{i蝇t}}\left(\frac{2}{\mathrm{i}}\stackrel{藱}{\mathrm{a}}鈭�{\mathrm{蝇}}_0\mathrm{a}\right)+\frac{\mathrm{i}}{2}{\mathrm{蝇}}_0\stackrel{藱}{\mathrm{a}}\\ =\mathrm{i蝇}\stackrel{藱}{\mathrm{a}}鈭�\frac{1}{4}({\mathrm{惟}}^2+{\mathrm{蝇}}_0^2鈭�2{\mathrm{蝇蝇}}_0)\mathrm{a}\end{array}$

So, the second order ODE with constant coefficient,

$\stackrel{篓}{\mathrm{a}}鈭�\mathrm{i蝇}\stackrel{藱}{\mathrm{a}}+\frac{1}{4}({\mathrm{惟}}^2+{\mathrm{蝇}}_0^2鈭�2{\mathrm{蝇蝇}}_0)\mathrm{a}=0$

Write the last two terms in the bracket as:

${\mathrm{蝇}}_0^2鈭�2{\mathrm{蝇蝇}}_0={\mathrm{蝇}}_0^2鈭�2{\mathrm{蝇蝇}}_0+{\mathrm{蝇}}^2鈭�{\mathrm{蝇}}^2={(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2鈭�{\mathrm{蝇}}^2$

Thus:

$\stackrel{篓}{\mathrm{a}}鈭�\mathrm{i蝇}\stackrel{藱}{\mathrm{a}}+\frac{1}{4}({\mathrm{惟}}^2+{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2鈭�{\mathrm{蝇}}^2)\mathrm{a}=0$ 鈥︹赌�..(4)

The characteristic equation is therefore:

${\mathrm{位}}^2鈭�\mathrm{i蝇位}+\frac{1}{4}({\mathrm{惟}}^2+{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2鈭�{\mathrm{蝇}}^2)=0$

The roots of this equation,

$\begin{array}{c}\mathrm{位}=\frac{\mathrm{i蝇}卤\sqrt{鈭�\mathrm{蝇}鈭�({\mathrm{惟}}^2+{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2鈭�{\mathrm{蝇}}^2)}}{2}\\ =\frac{\mathrm{i}}{2}(\mathrm{蝇}卤\sqrt{{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2+{\mathrm{惟}}^2})\\ =\frac{\mathrm{i}}{2}(\mathrm{蝇}卤{\mathrm{蝇}}^{\text{'}})\end{array}$

Where;

${\mathrm{蝇}}^{\text{'}}鈮�\sqrt{{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2+{\mathrm{惟}}^2}$

The general solution is:

$\begin{array}{c}\mathrm{a}\left(\mathrm{t}\right)={\mathrm{Ae}}^{{\mathrm{位}}_1}+{\mathrm{Be}}^{{\mathrm{位}}_2}\\ ={\mathrm{Ae}}^{\mathrm{i}(\mathrm{蝇}+{\mathrm{蝇}}^{\text{'}})\mathrm{t}/2}+{\mathrm{Be}}^{\mathrm{i}(\mathrm{蝇}鈭�{\mathrm{蝇}}^{\text{'}})\mathrm{t}/2}\end{array}$鈥︹赌︹赌︹赌︹赌�(5)

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

$\mathrm{b}\left(\mathrm{t}\right)={\mathrm{Ce}}^{\mathrm{i}(鈭�\mathrm{蝇}+{\mathrm{蝇}}^{\text{'}})\mathrm{l}/2}+{\mathrm{De}}^{\mathrm{i}(鈭�\mathrm{蝇}鈭�{\mathrm{蝇}}^{\text{'}})\mathrm{t}/2}$ 鈥︹赌︹赌︹赌︹赌�(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

$\begin{array}{l}{\mathrm{a}}_0=\mathrm{A}+\mathrm{B}\text{鈥夆赌夆赌�}\\ {\mathrm{b}}_0=\mathrm{C}+\mathrm{D}\end{array}$鈥︹赌︹赌︹赌�(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$t=0$

鈥︹�� (8)

$\begin{array}{c}A\frac{i}{2}(蝇+{蝇}^{\text{'}})+B\frac{i}{2}(蝇鈭�{蝇}^{\text{'}})=\frac{i惟}{2}(C+D)+{蝇}_0{a}_0蝇a_0+(A鈭�B){蝇}^{\text{'}}\\ \frac{i惟}{2}{b}_0+{蝇}_0{a}_{0}A鈭�B\\ \frac{1}{{蝇}^{\text{'}}}(b_0惟+a_0({蝇}_0鈭�蝇))\end{array}$

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:

$\mathrm{a}\left(\mathrm{t}\right)=\left[{\mathrm{a}}_0\cos \left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)+\frac{\mathrm{i}}{{\mathrm{蝇}}^{\text{'}}}({\mathrm{b}}_0\mathrm{惟}+{\mathrm{a}}_0({\mathrm{蝇}}_0鈭�\mathrm{蝇}))\sin \left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)\right]{\mathrm{e}}^{\mathrm{i蝇t}/2}$

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

$\mathrm{b}\left(\mathrm{t}\right)=\left[{\mathrm{b}}_0\cos \left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)+\frac{\mathrm{i}}{{\mathrm{蝇}}^{\text{'}}}({\mathrm{a}}_0\mathrm{惟}鈭�{\mathrm{b}}_0({\mathrm{蝇}}_0鈭�\mathrm{蝇}))\sin \left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)\right]{\mathrm{e}}^{鈭�\mathrm{i蝇t}/2}$

(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:

$\begin{array}{c}\mathrm{P}\left(\mathrm{t}\right)=\frac{{\mathrm{惟}}^2}{{\left({\mathrm{蝇}}^{\text{'}}\right)}^2}{\sin }^2\left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)\\ =\frac{{\mathrm{惟}}^2}{{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2+{\mathrm{惟}}^2}{\sin }^2\left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)\mathrm{P}\left(\mathrm{t}\right)\\ =\frac{{\mathrm{惟}}^2}{{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2+{\mathrm{惟}}^2}{\sin }^2\left(\frac{{\mathrm{蝇}}^{\text{'}}\mathrm{t}}{2}\right)\end{array}$

(e)

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

$\mathrm{P}\left(\mathrm{蝇}\right)=\frac{{\mathrm{惟}}^2}{{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2+{\mathrm{惟}}^2}$

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

$\begin{array}{c}\frac{{\mathrm{惟}}^2}{{(\mathrm{蝇}鈭�{\mathrm{蝇}}_0)}^2+{\mathrm{惟}}^2}=0.5\\ \mathrm{蝇}鈭�{\mathrm{蝇}}_0=卤\mathrm{惟}\end{array}$

The full width at half maximum is therefore:

$\mathrm{螖蝇}=2\mathrm{惟}$

(f)

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

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

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

To find the resonant frequency, which is:

${\mathrm{谓}}_{\text{res}}=\frac{{\mathrm{蝇}}_0}{2\mathrm{蟺}}$

But ${\mathrm{蝇}}_0={\mathrm{纬叠}}_0$

so:

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

$\begin{array}{c}{\mathrm{谓}}_{\text{res}}=\frac{(5.59)(1.6脳{10}^{鈭�19}\text{C})}{4\mathrm{蟺}(1.67脳{10}^{鈭�27}\text{kg})}(1\text{T})\\ =4.26脳{10}^7\text{Hz}\\ {\mathrm{谓}}_{\text{res}}=4.26脳{10}^7\text{Hz}\end{array}$

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

$\mathrm{螖谓}=\frac{\mathrm{螖蝇}}{2\mathrm{蟺}}=\frac{\mathrm{惟}}{\mathrm{蟺}}$

$\begin{array}{c}\mathrm{螖谓}=\frac{\mathrm{纬}}{\mathrm{蟺}}{\mathrm{B}}_{\text{rf}}\\ =\frac{\mathrm{纬}}{2\mathrm{蟺}}2{\mathrm{B}}_{\text{rf}}\\ ={\mathrm{谓}}_{\text{res}}\frac{2{\mathrm{B}}_{\text{rf}}}{{\mathrm{B}}_0}\end{array}$

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

$\begin{array}{c}\mathrm{螖谓}=(4.26脳{10}^7)(2脳{10}^{鈭�6})\\ =85.2\text{Hz}\end{array}$