13.6 Nu­clear Mag­netic Res­o­nance

Nu­clear mag­netic res­o­nance, or NMR, is a valu­able tool for ex­am­in­ing nu­clei, for prob­ing the struc­ture of mol­e­cules, in par­tic­u­lar or­ganic ones, and for med­ical di­ag­no­sis, as MRI. This sec­tion will give a ba­sic quan­tum de­scrip­tion of the idea. Lin­ear al­ge­bra will be used.

13.6.1 De­scrip­tion of the method

First demon­strated in­de­pen­dently by Bloch and Pur­cell in 1946, NMR probes nu­clei with net spin, in par­tic­u­lar hy­dro­gen nu­clei or other nu­clei with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. Var­i­ous com­mon nu­clei, like car­bon and oxy­gen do not have net spin; this can be a bless­ing since they can­not mess up the sig­nals from the hy­dro­gen nu­clei, or a lim­i­ta­tion, de­pend­ing on how you want to look at it. In any case, if nec­es­sary iso­topes such as car­bon 13 can be used which do have net spin.

It is not ac­tu­ally the spin, but the as­so­ci­ated mag­netic di­pole mo­ment of the nu­cleus that is rel­e­vant, for that al­lows the nu­clei to be ma­nip­u­lated by mag­netic fields. First the sam­ple is placed in an ex­tremely strong steady mag­netic field. Typ­i­cal fields are in terms of Tesla. (A Tesla is about 20 000 times the strength of the mag­netic field of the earth.) In the field, the nu­cleus has two pos­si­ble en­ergy states; a ground state in which the spin com­po­nent in the di­rec­tion of the mag­netic field is aligned with it, and an el­e­vated en­ergy state in which the spin is op­po­site {N.33}. (De­spite the large field strength, the en­ergy dif­fer­ence be­tween the two states is ex­tremely small com­pared to the ther­mal ki­netic en­ergy at room tem­per­a­ture. The num­ber of nu­clei in the ground state may only ex­ceed those in the el­e­vated en­ergy state by say one in 100 000, but that is still a large ab­solute num­ber of nu­clei in a sam­ple.)

Now per­turb the nu­clei with a sec­ond, much smaller and ra­dio fre­quency, mag­netic field. If the ra­dio fre­quency is just right, the ex­cess ground state nu­clei can be lifted out of the low­est en­ergy state, ab­sorb­ing en­ergy that can be ob­served. The res­o­nance fre­quency at which this hap­pens then gives in­for­ma­tion about the nu­clei. In or­der to ob­serve the res­o­nance fre­quency very ac­cu­rately, the per­turb­ing rf field must be very weak com­pared to the pri­mary steady mag­netic field.

In Con­tin­u­ous Wave NMR, the per­turb­ing fre­quency is var­ied and the ab­sorp­tion ex­am­ined to find the res­o­nance. (Al­ter­na­tively, the strength of the pri­mary mag­netic field can be var­ied, that works out to the same thing us­ing the ap­pro­pri­ate for­mula.)

In Fourier Trans­form NMR, the per­tur­ba­tion is ap­plied in a brief pulse just long enough to fully lift the ex­cess nu­clei out of the ground state. Then the de­cay back to­wards the orig­i­nal state is ob­served. An ex­pe­ri­enced op­er­a­tor can then learn a great deal about the en­vi­ron­ment of the nu­clei. For ex­am­ple, a nu­cleus in a mol­e­cule will be shielded a bit from the pri­mary mag­netic field by the rest of the mol­e­cule, and that leads to an ob­serv­able fre­quency shift. The amount of the shift gives a clue about the mol­e­c­u­lar struc­ture at the nu­cleus, so in­for­ma­tion about the mol­e­cule. Ad­di­tion­ally, neigh­bor­ing nu­clei can cause res­o­nance fre­quen­cies to split into sev­eral through their mag­netic fields. For ex­am­ple, a sin­gle neigh­bor­ing per­turb­ing nu­cleus will cause a res­o­nance fre­quency to split into two, one for spin up of the neigh­bor­ing nu­cleus and one for spin down. It is an­other clue about the mol­e­c­u­lar struc­ture. The time for the de­cay back to the orig­i­nal state to oc­cur is an­other im­por­tant clue about the lo­cal con­di­tions the nu­clei are in, es­pe­cially in MRI. The de­tails are be­yond this au­thor's knowl­edge; the pur­pose here is only to look at the ba­sic quan­tum me­chan­ics be­hind NMR.

13.6.2 The Hamil­ton­ian

The mag­netic fields will be as­sumed to be of the form

$$\skew2\vec{\cal B}= {\cal B}_0 {\hat k} + {\cal B}_1 \left( {\hat\imath}\cos\omega t - {\hat\jmath}\sin\omega t\right) %$$ (13.44)

where ${\cal B}_0$ is the Tesla-strength pri­mary mag­netic field, ${\cal B}_1$ the very weak per­turb­ing field strength, and $\omega$ is the fre­quency of the per­tur­ba­tion.

The com­po­nent of the mag­netic field in the $xy$-​plane, ${\cal B}_1$, ro­tates around the $z$-​axis at an­gu­lar ve­loc­ity $\omega$. Such a ro­tat­ing mag­netic field can be achieved us­ing a pair of prop­erly phased coils placed along the $x$ and $y$ axes. (In Fourier Trans­form NMR, a sin­gle per­tur­ba­tion pulse ac­tu­ally con­tains a range of dif­fer­ent fre­quen­cies $\omega$, and Fourier trans­forms are used to take them apart.) Since the ap­pa­ra­tus and the wave length of a ra­dio fre­quency field is very large on the scale of a nu­cleus, spa­tial vari­a­tions in the mag­netic field can be ig­nored.

Now sup­pose you place a spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ nu­cleus in the cen­ter of this mag­netic field. As dis­cussed in sec­tion 13.4, a par­ti­cle with spin will act as a lit­tle com­pass nee­dle, and its en­ergy will be low­est if it is aligned with the di­rec­tion of the am­bi­ent mag­netic field. In par­tic­u­lar, the en­ergy is given by

$$H = - \vec \mu \cdot \skew2\vec{\cal B}$$

where $\vec\mu$ is called the mag­netic di­pole strength of the nu­cleus. This di­pole strength is pro­por­tional to its spin an­gu­lar mo­men­tum ${\skew 6\widehat{\vec S}}$:

$$\vec \mu = \gamma {\skew 6\widehat{\vec S}}$$

where the con­stant of pro­por­tion­al­ity $\gamma$ is called the gy­ro­mag­netic ra­tio. The nu­mer­i­cal value of the gy­ro­mag­netic ra­tio can be found as

$$\gamma = \frac{g q}{2 m}$$

In case of a hy­dro­gen nu­cleus, a pro­ton, the mass $m_{\rm p}$ and charge $q_p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e$ can be found in the no­ta­tions sec­tion, and the pro­ton's ex­per­i­men­tally found $g$-​fac­tor is $g_p$ $\vphantom0\raisebox{1.5pt}{$=$}$ 5.59.

The bot­tom line is that you can write the Hamil­ton­ian of the in­ter­ac­tion of the nu­cleus with the mag­netic field in terms of a nu­mer­i­cal gy­ro­mag­netic ra­tio value, spin, and the mag­netic field:

$$H = - \gamma {\skew 6\widehat{\vec S}}\cdot \skew2\vec{\cal B} %$$ (13.45)

Now turn­ing to the wave func­tion of the nu­cleus, it can be writ­ten as a com­bi­na­tion of the spin-up and spin-down states,

$$\Psi= a{\uparrow}+b{\downarrow},$$

where ${\uparrow}$ has spin $\frac12\hbar$ in the $z$-​di­rec­tion, along the pri­mary mag­netic field, and ${\downarrow}$ has $-\frac12\hbar$. Nor­mally, $a$ and $b$ would de­scribe the spa­tial vari­a­tions, but spa­tial vari­a­tions are not rel­e­vant to the analy­sis, and $a$ and $b$ can be con­sid­ered to be sim­ple num­bers.

You can use the con­cise no­ta­tions of lin­ear al­ge­bra by com­bin­ing $a$ and $b$ in a two-com­po­nent col­umn vec­tor (more pre­cisely, a spinor),

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

In those terms, the spin op­er­a­tors be­come ma­tri­ces, the so-called Pauli spin ma­tri­ces of sec­tion 12.10,

$${\widehat S}_x = \frac{\hbar}{2} \left(\begin{array}{rr} 0... ...}{2} \left(\begin{array}{rr} 1 & 0\ 0 & -1\end{array}\right)$$ (13.46)

Sub­sti­tu­tion of these ex­pres­sions for the spin, and (13.44) for the mag­netic field into (13.45) gives af­ter clean­ing up the fi­nal Hamil­ton­ian:

$$H = - \frac{\hbar}2 \left( \begin{array}{cc} \omega_0 &... ...ga_0 = \gamma {\cal B}_0 \quad \omega_1 = \gamma {\cal B}_1 %$$ (13.47)

The con­stants $\omega_0$ and $\omega_1$ have the di­men­sions of a fre­quency; $\omega_0$ is called the “Lar­mor fre­quency.” As far as $\omega_1$ is con­cerned, the im­por­tant thing to re­mem­ber is that it is much smaller than the Lar­mor fre­quency $\omega_0$ be­cause the per­tur­ba­tion mag­netic field is small com­pared to the pri­mary one.

13.6.3 The un­per­turbed sys­tem

Be­fore look­ing at the per­turbed case, it helps to first look at the un­per­turbed so­lu­tion. If there is just the pri­mary mag­netic field af­fect­ing the nu­cleus, with no ra­dio-fre­quency per­tur­ba­tion $\omega_1$, the Hamil­ton­ian de­rived in the pre­vi­ous sub­sec­tion sim­pli­fies to

$$H = -\frac{\hbar}2 \left( \begin{array}{cc} \omega_0 & 0 \\ 0 & -\omega_0 \end{array} \right)$$

The en­ergy eigen­states are the spin-up state, with en­ergy $-\frac12\hbar\omega_0$, and the spin-down state, with en­ergy $\frac12\hbar\omega_0$.

The dif­fer­ence in en­ergy is in rel­a­tivis­tic terms ex­actly equal to a pho­ton with the Lar­mor fre­quency $\omega_0$. While the treat­ment of the elec­tro­mag­netic field in this dis­cus­sion will be clas­si­cal, rather than rel­a­tivis­tic, it seems clear that the Lar­mor fre­quency must play more than a su­per­fi­cial role.

The un­steady Schrö­din­ger equa­tion tells you that the wave func­tion evolves in time like ${\rm i}\hbar\dot\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H\Psi$, so if $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a{\uparrow}+b{\downarrow}$,

$${\rm i}\hbar\left(\begin{array}{c}\dot a\ \dot b\end{array... ...{array} \right) \left(\begin{array}{c}a\ b\end{array}\right)$$

The so­lu­tion for the co­ef­fi­cients $a$ and $b$ of the spin-up and -down states is:

$$a = a_0 e^{{\rm i}\omega_0 t/2} \qquad b = b_0 e^{-{\rm i}\omega_0 t/2}$$

if $a_0$ and $b_0$ are the val­ues of these co­ef­fi­cients at time zero.

Since $\vert a\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert a_0\vert^2$ and $\vert b\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert b_0\vert^2$ at all times, the prob­a­bil­i­ties of mea­sur­ing spin-up or spin-down do not change with time. This was to be ex­pected, since spin-up and spin-down are en­ergy states for the steady sys­tem. To get more in­ter­est­ing physics, you re­ally need the un­steady per­tur­ba­tion.

But first, to un­der­stand the quan­tum processes bet­ter in terms of the ideas of non­quan­tum physics, it will be help­ful to write the un­steady quan­tum evo­lu­tion in terms of the ex­pec­ta­tion val­ues of the an­gu­lar mo­men­tum com­po­nents. The ex­pec­ta­tion value of the $z$-​com­po­nent of an­gu­lar mo­men­tum is

$$\langle S_z\rangle = \vert a\vert^2 \frac{\hbar}{2} - \vert b\vert^2 \frac{\hbar}{2}$$

To more clearly in­di­cate that the value must be in be­tween $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$$\raisebox{.5pt}{$/$}$​2 and $\hbar$$\raisebox{.5pt}{$/$}$​2, you can write the mag­ni­tude of the co­ef­fi­cients in terms of an an­gle $\alpha$, the “pre­ces­sion an­gle”,

$$\vert a\vert = \vert a_0\vert \equiv \cos(\alpha/2) \qquad \vert b\vert = \vert b_0\vert \equiv \sin(\alpha/2)$$

In terms of the so-de­fined $\alpha$, you sim­ply have, us­ing the half-an­gle trig for­mu­lae,

$$\langle S_z\rangle = \frac{\hbar}{2} \cos\alpha$$

The ex­pec­ta­tion val­ues of the an­gu­lar mo­menta in the $x$ and $y$ di­rec­tions can by found as the in­ner prod­ucts $\langle\Psi\vert{\widehat S}_x\Psi\rangle$ and $\langle\Psi\vert{\widehat S}_y\Psi\rangle$, chap­ter 4.4.3. Sub­sti­tut­ing the rep­re­sen­ta­tion in terms of spin­ors and Pauli spin ma­tri­ces, and clean­ing up us­ing the Euler for­mula (2.5), you get

$$\langle S_x\rangle = \frac{\hbar}{2} \sin\alpha \cos(\omega... ...\rangle = - \frac{\hbar}{2} \sin\alpha \sin(\omega_0 t+\alpha)$$

where $\alpha$ is some con­stant phase an­gle that is fur­ther unim­por­tant.

The first thing that can be seen from these re­sults is that the length of the ex­pec­ta­tion an­gu­lar mo­men­tum vec­tor is $\hbar$$\raisebox{.5pt}{$/$}$​2. Next, the com­po­nent with the $z$-​axis, the di­rec­tion of the pri­mary mag­netic field, is at all times $\frac12\hbar\cos\alpha$. That im­plies that the ex­pec­ta­tion an­gu­lar mo­men­tum vec­tor is un­der a con­stant an­gle $\alpha$ with the pri­mary mag­netic field.

Fig­ure 13.18: Lar­mor pre­ces­sion of the ex­pec­ta­tion spin (or mag­netic mo­ment) vec­tor around the mag­netic field.

The com­po­nent in the $xy$-​plane is $\frac12\hbar\sin\alpha$, and this com­po­nent ro­tates around the $z$-​axis, as shown in fig­ure 13.18, caus­ing the end point of the ex­pec­ta­tion an­gu­lar mo­men­tum vec­tor to sweep out a cir­cu­lar path around the mag­netic field $\skew2\vec{\cal B}$. This ro­ta­tion around the $z$-​axis is called “Lar­mor pre­ces­sion.” Since the mag­netic di­pole mo­ment is pro­por­tional to the spin, it traces out the same con­i­cal path.

Cau­tion should be used against at­tach­ing too much im­por­tance to this clas­si­cal pic­ture of a pre­cess­ing mag­net. The ex­pec­ta­tion an­gu­lar mo­men­tum vec­tor is not a phys­i­cally mea­sur­able quan­tity. One glar­ing in­con­sis­tency in the ex­pec­ta­tion an­gu­lar mo­men­tum vec­tor ver­sus the true an­gu­lar mo­men­tum is that the square mag­ni­tude of the ex­pec­ta­tion an­gu­lar mo­men­tum vec­tor is $\hbar^2$$\raisebox{.5pt}{$/$}$​4, three times smaller than the true square mag­ni­tude of an­gu­lar mo­men­tum.

13.6.4 Ef­fect of the per­tur­ba­tion

In the pres­ence of the per­turb­ing mag­netic field, the un­steady Schrö­din­ger equa­tion ${\rm i}\hbar\dot\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H\Psi$ be­comes

$${\rm i}\hbar\left(\begin{array}{c}\dot a\ \dot b\end{array... ...rray} \right) \left(\begin{array}{c}a\ b\end{array}\right) %$$ (13.48)

where $\omega_0$ is the Lar­mor fre­quency, $\omega$ is the fre­quency of the per­tur­ba­tion, and $\omega_1$ is a mea­sure of the strength of the per­tur­ba­tion and small com­pared to $\omega_0$.

The above equa­tions can be solved ex­actly us­ing stan­dard lin­ear al­ge­bra pro­ce­dures, though the the al­ge­bra is fairly sti­fling {D.75}. The analy­sis brings in an ad­di­tional quan­tity that will be called the “res­o­nance fac­tor”

$$f = \sqrt{\frac{\omega_1^2}{(\omega - \omega_0)^2 + \omega_1^2}}$$ (13.49)

Note that $f$ has its max­i­mum value, one, at res­o­nance, i.e. when the per­tur­ba­tion fre­quency $\omega$ equals the Lar­mor fre­quency $\omega_0$.

The analy­sis finds the co­ef­fi­cients of the spin-up and spin-down states to be:

$$a = \left[ a_0 \left( \cos\bigg(\frac{\omega_1 t}{2f}\bigg) - {\rm i}... ...f \sin\bigg(\frac{\omega_1 t}{2f}\bigg) \right] e^{{\rm i}\omega t/2}\quad\quad$$ (13.50)

$$b = \left[ b_0 \left( \cos\bigg(\frac{\omega_1 t}{2f}\bigg) + {\rm i}... ... \sin\bigg(\frac{\omega_1 t}{2f}\bigg) \right] e^{-{\rm i}\omega t/2}\quad\quad$$ (13.51)

where $a_0$ and $b_0$ are the ini­tial co­ef­fi­cients of the spin-up and spin-down states.

This so­lu­tion looks pretty for­bid­ding, but it is not that bad in ap­pli­ca­tion. The pri­mary in­ter­est is in nu­clei that start out in the spin-up ground state, so you can set $\vert a_0\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $b_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Also, the pri­mary in­ter­est is in the prob­a­bil­ity that the nu­clei may be found at the el­e­vated en­ergy level, which is

$$\vert b\vert^2 = f^2 \sin^2\bigg(\frac{\omega_1 t}{2f}\bigg)$$ (13.52)

That is a pretty sim­ple re­sult. When you start out, the nu­clei you look at are in the ground state, so $\vert b\vert^2$ is zero, but with time the rf per­tur­ba­tion field in­creases the prob­a­bil­ity of find­ing the nu­clei in the el­e­vated en­ergy state even­tu­ally to a max­i­mum of $f^2$ when the sine be­comes one.

Fig­ure 13.19: Prob­a­bil­ity of be­ing able to find the nu­clei at el­e­vated en­ergy ver­sus time for a given per­tur­ba­tion fre­quency $\omega$.

Con­tin­u­ing the per­tur­ba­tion be­yond that time is bad news; it de­creases the prob­a­bil­ity of el­e­vated states again. As fig­ure 13.19 shows, over ex­tended times, there is a flip-flop be­tween the nu­clei be­ing with cer­tainty in the ground state, and hav­ing a prob­a­bil­ity of be­ing in the el­e­vated state. The fre­quency at which the prob­a­bil­ity os­cil­lates is called the “Rabi flop­ping fre­quency”. The au­thor’s sources dif­fer about the pre­cise de­f­i­n­i­tion of this fre­quency, but the one that seems to be most log­i­cal is $\omega_1$$\raisebox{.5pt}{$/$}$​$f$.

Fig­ure 13.20: Max­i­mum prob­a­bil­ity of find­ing the nu­clei at el­e­vated en­ergy.

Any­way, by keep­ing up the per­tur­ba­tion for the right time you can raise the prob­a­bil­ity of el­e­vated en­ergy to a max­i­mum of $f^2$. A plot of $f^2$ against the per­turb­ing fre­quency $\omega$ is called the “res­o­nance curve,“ shown in fig­ure 13.20. For the per­tur­ba­tion to have max­i­mum ef­fect, its fre­quency $\omega$ must equal the nu­clei's Lar­mor fre­quency $\omega_0$. Also, for this fre­quency to be very ac­cu­rately ob­serv­able, the spike in fig­ure 13.20 must be nar­row, and since its width is pro­por­tional to $\omega_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\gamma{\cal B}_1$, that means the per­turb­ing mag­netic field must be very weak com­pared to the pri­mary mag­netic field.

Fig­ure 13.21: A per­turb­ing mag­netic field, ro­tat­ing at pre­cisely the Lar­mor fre­quency, causes the ex­pec­ta­tion spin vec­tor to come cas­cad­ing down out of the ground state.

There are two qual­i­ta­tive ways to un­der­stand the need for the fre­quency of the per­tur­ba­tion to equal the Lar­mor fre­quency. One is geo­met­ri­cal and clas­si­cal: as noted in the pre­vi­ous sub­sec­tion, the ex­pec­ta­tion mag­netic mo­ment pre­cesses around the pri­mary mag­netic field with the Lar­mor fre­quency. In or­der for the small per­tur­ba­tion field to ex­ert a long-term down­ward torque on this pre­cess­ing mag­netic mo­ment as in fig­ure 13.21, it must ro­tate along with it. If it ro­tates at any other fre­quency, the torque will quickly re­verse di­rec­tion com­pared to the mag­netic mo­ment, and the vec­tor will start go­ing up again. The other way to look at it is from a rel­a­tivis­tic quan­tum per­spec­tive: if the mag­netic field fre­quency equals the Lar­mor fre­quency, its pho­tons have ex­actly the en­ergy re­quired to lift the nu­clei from the ground state to the ex­cited state.

At the Lar­mor fre­quency, it would naively seem that the op­ti­mum time to main­tain the per­tur­ba­tion is un­til the ex­pec­ta­tion spin vec­tor is ver­ti­cally down; then the nu­cleus is in the ex­ited en­ergy state with cer­tainty. If you then al­low na­ture the time to probe its state, every nu­cleus will be found to be in the ex­cited state, and will emit a pho­ton. (If not messed up by some col­li­sion or what­ever, lit­tle in life is ideal, is it?) How­ever, ac­cord­ing to ac­tual de­scrip­tions of NMR de­vices, it is bet­ter to stop the per­tur­ba­tion ear­lier, when the ex­pec­ta­tion spin vec­tor has be­come hor­i­zon­tal, rather than fully down. In that case, na­ture will only find half the nu­clei in the ex­cited en­ergy state af­ter the per­tur­ba­tion, pre­sum­ably de­creas­ing the ra­di­a­tion yield by a fac­tor 2. The clas­si­cal ex­pla­na­tion that is given is that when the (ex­pec­ta­tion) spin vec­tor is pre­cess­ing at the Lar­mor fre­quency in the hor­i­zon­tal plane, the ra­di­a­tion is most eas­ily de­tected by the coils lo­cated in that same plane. And that closes this dis­cus­sion.