Chapter 11, Problem 11.29P

(a)

To determine

The $2\times 2$ Hamiltonian matrix for the given system.

Answer to Problem 11.29P

The $2\times 2$ Hamiltonian matrix for the given system is $H=-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_o& B_{rf}{e}^{i\omega t}\\ B_{rf}{e}^{-i\omega t}& -B_o\end{array}\right)$.

Explanation of Solution

Write the expression to find the Hamiltonian matric for a spinning charge particle in a magnetic field $B$, Equation 4.158

    $H=-\gamma B\cdot S$        (I)

Here, $S$ is the appropriate spin matrix, $B$ is the magnetic field, $\gamma$ is the proportionality constant and $H$ is the Hamiltonian matrix.

Write the expression for appropriate spin matrix

    $S=\frac{\hslash }{2}\sigma$        (II)

Where,

${\sigma }_x=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, ${\sigma }_y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$ and ${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

Given, the total magnetic field

$B=B_{rf}\cos \left(\omega t\right)\widehat{i}-B_{rf}\sin \left(\omega t\right)\widehat{j}+B_0\widehat{k}$        (III)

Where, $B_x=B_{rf}\cos \left(\omega t\right)$, $B_y=-B_{rf}\sin \left(\omega t\right)$ and $B_z=B_0$.

Solving equation (I),

$H=-\gamma \left(B_x{S}_x+B_y{S}_y+B_z{S}_z\right)$

Substitute equation (II) in the above equation,

$\begin{array}{c}H=-\gamma \frac{\hslash }{2}\left(B_x{\sigma }_x+B_y{\sigma }_y+B_z{\sigma }_z\right)\\ =-\gamma \frac{\hslash }{2}\left(B_x\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)+B_y\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)+B_z\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\right)\\ =-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_z& B_x-i{B}_y\\ B_x+i{B}_y& -B_z\end{array}\right)\\ =-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_o& B_{rf}\left(\cos \omega t+i\sin \omega t\right)\\ B_{rf}\left(\cos \omega t-i\sin \omega t\right)& -B_o\end{array}\right)\end{array}$

Solving further,

$H=-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_o& B_{rf}{e}^{i\omega t}\\ B_{rf}{e}^{-i\omega t}& -B_o\end{array}\right)$

Conclusion:

Thus, the $2\times 2$ Hamiltonian matrix for the given system is $H=-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_o& B_{rf}{e}^{i\omega t}\\ B_{rf}{e}^{-i\omega t}& -B_o\end{array}\right)$.

(b)

To determine

Show that $\dot{a}=\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)$ and $\dot{b}=\frac{i}{2}\left(\Omega e^{-i\omega t}a-{\omega }_{0}b\right)$, if $\chi \left(t\right)=\left(\begin{array}{c}a\left(t\right)\\ b\left(t\right)\end{array}\right)$ is the spin state at time $t$.

Answer to Problem 11.29P

It has been proved that $\dot{a}=\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)$ and $\dot{b}=\frac{i}{2}\left(\Omega e^{-i\omega t}a-{\omega }_{0}b\right)$, if $\chi \left(t\right)=\left(\begin{array}{c}a\left(t\right)\\ b\left(t\right)\end{array}\right)$ is the spin state at time $t$.

Explanation of Solution

Write the time-dependent Schrodinger equation

    $i\hslash \dot{\chi }=H\chi$        (IV)

Given,

$\chi \left(t\right)=\left(\begin{array}{c}a\left(t\right)\\ b\left(t\right)\end{array}\right)$

Thus, $\dot{\chi }\left(t\right)=\left(\begin{array}{c}\dot{a}\\ \dot{b}\end{array}\right)$

Substitute the above relations in equation (IV)

$i\hslash \left(\begin{array}{c}\dot{a}\\ \dot{b}\end{array}\right)=H\left(\begin{array}{c}a\\ b\end{array}\right)$

From part (a), $H=-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_o& B_{rf}{e}^{i\omega t}\\ B_{rf}{e}^{-i\omega t}& -B_o\end{array}\right)$. The above equation becomes,

$\begin{array}{c}i\hslash \left(\begin{array}{c}\dot{a}\\ \dot{b}\end{array}\right)=-\gamma \frac{\hslash }{2}\left(\begin{array}{cc}{B}_o& B_{rf}{e}^{i\omega t}\\ B_{rf}{e}^{-i\omega t}& -B_o\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)\\ \left(\begin{array}{c}\dot{a}\\ \dot{b}\end{array}\right)=i\frac{\gamma }{2}\left(\begin{array}{c}{B}_{o}a+B_{rf}{e}^{i\omega t}b\\ B_{rf}{e}^{-i\omega t}a-B_{o}b\end{array}\right)\end{array}$

Thus,

$\begin{array}{c}\dot{a}=i\frac{\gamma }{2}\left(B_{o}a+B_{rf}{e}^{i\omega t}b\right)\\ =\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)\end{array}$

$\begin{array}{c}\dot{b}=i\frac{\gamma }{2}\left(B_{rf}{e}^{-i\omega t}a-B_{o}b\right)\\ =\frac{i}{2}\left(\Omega e^{-i\omega t}a-{\omega }_{0}b\right)\end{array}$

Hence proved.

Conclusion:

It has been proved that $\dot{a}=\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)$ and $\dot{b}=\frac{i}{2}\left(\Omega e^{-i\omega t}a-{\omega }_{0}b\right)$, if $\chi \left(t\right)=\left(\begin{array}{c}a\left(t\right)\\ b\left(t\right)\end{array}\right)$ is the spin state at time $t$.

(c)

To determine

Whether the general solution for $a\left(t\right)$ and $b\left(t\right)$, in terms of their initial values $a_0$ and $b_0$ is $a\left(t\right)=\left\{a_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[a_0\left({\omega }_0-\omega \right)+b_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{i\omega t/2}$ and $b\left(t\right)=\left\{b_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[b_0\left({\omega }_0-\omega \right)+a_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{-i\omega t/2}$ where, $\omega =\sqrt{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$.

Answer to Problem 11.29P

It has been proved that the general solution for $a\left(t\right)$ and $b\left(t\right)$, in terms of their initial values $a_0$ and $b_0$ is $a\left(t\right)=\left\{a_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[a_0\left({\omega }_0-\omega \right)+b_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{i\omega t/2}$ and $b\left(t\right)=\left\{b_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[b_0\left({\omega }_0-\omega \right)+a_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{-i\omega t/2}$ where, $\omega =\sqrt{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$.

Explanation of Solution

From part (b), $\dot{a}=\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)$ and $\dot{b}=\frac{i}{2}\left(\Omega e^{-i\omega t}a-{\omega }_{0}b\right)$.

Using the above relations, $b=\frac{1}{\Omega }{e}^{-i\omega t}\left(\frac{2}{i}\dot{a}-{\omega }_{0}a\right)$

Differentiate $\dot{a}$,

$\begin{array}{c}\ddot{a}=\frac{i}{2}\left(\Omega i\omega e^{i\omega t}b+\Omega e^{i\omega t}\dot{b}+{\omega }_0\dot{a}\right)\\ =\frac{i}{2}\left(\Omega i\omega e^{i\omega t}\left[\frac{1}{\Omega }{e}^{-i\omega t}\left(\frac{2}{i}\dot{a}-{\omega }_{0}a\right)\right]+\Omega e^{i\omega t}\frac{i}{2}\left(\Omega e^{-i\omega t}a-{\omega }_{0}b\right)+{\omega }_0\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)\right)\\ =-\frac{\omega }{2}\left(\frac{2}{i}\dot{a}-{\omega }_{0}a\right)-\frac{1}{4}\Omega e^{i\omega t}\left(\Omega e^{-i\omega t}a-{\omega }_0\frac{1}{\Omega }{e}^{-i\omega t}\left(\frac{2}{i}\dot{a}-{\omega }_{0}a\right)\right)+\frac{i}{2}{\omega }_0\dot{a}\\ =i\omega \dot{a}-\frac{1}{4}\left({\Omega }^2+{\omega }_0^2-2\omega {\omega }_0\right)a\end{array}$

Rewrite the above equation in the form of second order differential equation,

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

Solving the terms inside the bracket,

$\begin{array}{c}{\omega }_0^2-2\omega {\omega }_0={\omega }_0^2-2\omega {\omega }_0+{\omega }^2-{\omega }^2\\ ={\left(\omega -{\omega }_0\right)}^2-{\omega }^2\end{array}$

Substitute the above relation in equation (V)

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

The characteristic equation is

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

Solving the above quadratic equation,

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

Where, $\omega \text{'}=\sqrt{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$.

The general solution $a\left(t\right)$ is

$\begin{array}{c}a\left(t\right)=A{e}^{{\lambda }_1}+B{e}^{{\lambda }_2}\\ =A{e}^{i\left(\omega +\omega \text{'}\right)t/2}+B{e}^{i\left(\omega -\omega \text{'}\right)t/2}\end{array}$        (VII)

The general solution $b\left(t\right)$ is

$b\left(t\right)=C{e}^{i\left(-\omega +\omega \text{'}\right)t/2}+D{e}^{i\left(-\omega -\omega \text{'}\right)t/2}$        (VIII)

For the initial condition, $t=0$

$a_0=A+B$ and $b_0=C+D$        (IX)

From part (b), $\dot{a}=\frac{i}{2}\left(\Omega e^{i\omega t}b+{\omega }_{0}a\right)$. Substitute equation (VIII) and (IX) in the relation,

$\begin{array}{c}\dot{a}=A\frac{i}{2}\left(\omega +\omega \text{'}\right)e^{i\left(\omega +\omega \text{'}\right)t/2}+B\frac{i}{2}\left(\omega -\omega \text{'}\right)e^{i\left(\omega -\omega \text{'}\right)t/2}\\ =\frac{i\Omega }{2}{e}^{i\omega t}\left[C{e}^{i\left(-\omega +\omega \text{'}\right)t/2}+D{e}^{i\left(-\omega -\omega \text{'}\right)t/2}\right]+{\omega }_{0}a\end{array}$

Substitute $t=0$ in the above equation,

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

Substitute equation (IX) and (X) in (VII)

$\begin{array}{c}a\left(t\right)=\left[A{e}^{i\omega \text{'}t/2}+B{e}^{-i\omega \text{'}t/2}\right]e^{i\omega t/2}\\ =\left[\left(A+B\right)\cos \left(\frac{\omega \text{'}t}{2}\right)+i\left(A-B\right)\sin \left(\frac{\omega \text{'}t}{2}\right)\right]e^{i\omega t/2}\\ =\left[a_0\cos \left(\frac{\omega \text{'}t}{2}\right)+\frac{i}{\omega \text{'}}\left(b_0\Omega +a_0\left({\omega }_0-\omega \right)\right)\sin \left(\frac{\omega \text{'}t}{2}\right)\right]e^{i\omega t/2}\end{array}$

And

$B\left(t\right)=\left[b_0\cos \left(\frac{\omega \text{'}t}{2}\right)+\frac{i}{\omega \text{'}}\left(a_0\Omega -b_0\left({\omega }_0-\omega \right)\right)\sin \left(\frac{\omega \text{'}t}{2}\right)\right]e^{-i\omega t/2}$

Hence proved.

Conclusion:

It has been proved that the general solution for $a\left(t\right)$ and $b\left(t\right)$, in terms of their initial values $a_0$ and $b_0$ is $a\left(t\right)=\left\{a_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[a_0\left({\omega }_0-\omega \right)+b_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{i\omega t/2}$ and $b\left(t\right)=\left\{b_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[b_0\left({\omega }_0-\omega \right)+a_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{-i\omega t/2}$ where, $\omega =\sqrt{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$.

(d)

To determine

The probability of a transition to spin down, as a function of time.

Answer to Problem 11.29P

The probability of a transition to spin down, as a function of time is $P\left(t\right)={\left(\frac{\Omega }{\omega \text{'}}\right)}^2{\sin }^2\left(\omega \text{'}t/2\right)$.

Explanation of Solution

Write the expression to find the probability of transition to spin down

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

Given, $a_0=1$ and $b_0=0$. From part (c), $b\left(t\right)=\left\{b_0\cos \left(\omega \text{'}t/2\right)+\frac{i}{\omega \text{'}}\left[b_0\left({\omega }_0-\omega \right)+a_0\Omega \right]\sin \left(\omega \text{'}t/2\right)\right\}{e}^{-i\omega t/2}$

Substitute $a_0=1$ and $b_0=0$ in the above equation,

$b\left(t\right)=\left\{\frac{i\Omega }{\omega \text{'}}\sin \left(\omega \text{'}t/2\right)\right\}{e}^{-i\omega t/2}$

Now, the probability of a transition to spin down is

$P\left(t\right)={\left(\frac{\Omega }{\omega \text{'}}\right)}^2{\sin }^2\left(\omega \text{'}t/2\right)$

Conclusion:

Thus, the probability of a transition to spin down, as a function of time is $P\left(t\right)={\left(\frac{\Omega }{\omega \text{'}}\right)}^2{\sin }^2\left(\omega \text{'}t/2\right)$.

(e)

To determine

Sketch the resonance curve, $P\left(\omega \right)=\frac{{\Omega }^2}{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$ as a function of driving frequency $\omega$ and determine the full width at half maximum, $\Delta \omega$.

Answer to Problem 11.29P

The resonance curve, $P\left(\omega \right)=\frac{{\Omega }^2}{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$ as a function of driving frequency $\omega$

the full width at half maximum is $\Delta \omega =2\Omega$.

Explanation of Solution

Given,

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

$\begin{array}{c}P\left(\omega \right)=\frac{1}{2}\\ \frac{1}{2}=\frac{{\Omega }^2}{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}\\ {\left(\omega -{\omega }_0\right)}^2+{\Omega }^2=2{\Omega }^2\\ {\left(\omega -{\omega }_0\right)}^2={\Omega }^2\end{array}$

Taking square root on both sides,

$\left(\omega -{\omega }_0\right)=\pm \Omega$

Thus, the full width at half maximum, $\Delta \omega$ is

$\Delta \omega =2\Omega$

Conclusion:

The resonance curve, $P\left(\omega \right)=\frac{{\Omega }^2}{{\left(\omega -{\omega }_0\right)}^2+{\Omega }^2}$ as a function of driving frequency $\omega$

the full width at half maximum is $\Delta \omega =2\Omega$.

(f)

To determine

The resonant frequency in a nuclear magnetic resonance experiment and the width of the resonance curve.

Answer to Problem 11.29P

The resonant frequency in a nuclear magnetic resonance experiment is $4.26\times {10}^7\text{ Hz}$ and the width of the resonance curve is $85.2\text{ Hz}$.

Explanation of Solution

Compare Equation 4.156 and 7.89, the gyro-magnetic ratio for a proton is

$\gamma =\frac{g_{p}e}{2m_p}$        (XI)

Where, $g_p=5.59$

Write the expression to find the resonant frequency

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

Where, ${\omega }_0=\gamma B_0$.

Substitute equation (XI) in the above equation,

${\nu }_{res}=\frac{g_{p}e}{4\pi m_p}{B}_0$

Substitute $5.59$ for $g_p$, $1.6\times {10}^{-19}\text{ C}$ for $e$, $1\text{ T}$ for $B_0$ and $1.67\times {10}^{-27}\text{ kg}$ for $m_p$ in the above equation to find the value of ${\nu }_{res}$

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

Write the expression to find the width of the resonance curve

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

Where, $\Delta \omega =\Omega =2\gamma B_{rf}$.

Thus, the above equation becomes,

$\begin{array}{c}\Delta \nu =\frac{\gamma }{2\pi }2B_{rf}\\ ={\nu }_{res}\frac{2B_{rf}}{B_0}\end{array}$

Substitute $4.26\times {10}^7\text{ Hz}$ for ${\nu }_{res}$, $1\times {10}^{-6}$ for $\frac{B_{rf}}{B_0}$ in the above equation

$\begin{array}{c}\Delta \nu =\left(4.26\times {10}^7\text{ Hz}\right)\left(2\right)\left(1\times {10}^{-6}\right)\\ =85.2\text{ Hz}\end{array}$

Conclusion:

Thus, the resonant frequency in a nuclear magnetic resonance experiment is $4.26\times {10}^7\text{ Hz}$ and the width of the resonance curve is $85.2\text{ Hz}$.