Nuclear Magnetic Resonance & Imaging

4 hours ~23 min read

Nuclear Magnetic Resonance & Imaging

A nucleus is 101510^{-15} m across and buried inside an atom, and yet a hospital down the road turns billions of them over on command, twice a second, and reads back a picture of your brain. The trick is entirely in this course's toolkit: precession, resonance, and a magnetic field that tells the truth about position.

Learning Objectives

After this lesson you will be able to:

  1. Explain why flipping a nuclear spin requires a transverse field that rotates, and state the resonance condition ω=γB0\omega = \gamma B_0.
  2. Construct the rotating frame, derive the effective field Beff\mathbf{B}_{\text{eff}}, and solve for Rabi flopping on and off resonance.
  3. Compute pulse durations: show that the flip angle is γB1t\gamma B_1 t, and design π/2\pi/2 and π\pi pulses from realistic hardware numbers.
  4. Explain the chemical shift, read an ethanol spectrum, and state the field homogeneity a chemical-shift measurement demands.
  5. Describe the spin echo and distinguish T1T_1, T2T_2, and T2T_2^{*}.
  6. Show how a linear field gradient converts frequency into position, explain what an MRI machine actually measures, and describe how blood flow is imaged.

Intuition

A note on where you are. Freericks' §2.4 is independent of §2.3 — nothing here uses entanglement, Bell inequalities, or the analyzer loop. What it does use is the very first thing this course established: a magnetic moment tied to an angular momentum precesses about a magnetic field at the Larmor rate. If you skipped straight to this lesson, R.1.1 is the only prerequisite, and this is the lesson where that precession stops being an obstacle to be averaged away and becomes the instrument.

Everything in R.1 treated precession as a nuisance we had to understand in order to trust the Stern-Gerlach apparatus. Here it becomes the whole point. We want to reach into a sample, take hold of the spin of a hydrogen nucleus, and turn it over — without touching it, because it is a proton, and without heating the sample, because in the medical case the sample is a person.

The only handle we have is a magnetic field, and the only thing a field does to a spin is make it precess. So the plan writes itself: a spin sitting along z^\hat z will tip over if we make it precess about a horizontal axis. Apply a horizontal field and wait a half-turn.

That plan fails immediately, and the reason it fails is the interesting part. There is already a large vertical field B0B_0 present — we need it, for reasons that will become clear — and the spin is already whipping around it at the Larmor rate, typically tens of megahertz. From the spin's point of view a static horizontal field is not horizontal at all: it sweeps past at 6363 million times a second, pushing the spin one way and then the other, so the net effect over any appreciable time is nothing. The spin just wobbles.

The fix is the whole subject. Make the horizontal field rotate, at exactly the frequency at which the spin is already precessing. Then, in a frame turning along with the spin, the transverse field stands still, and the spin can precess about it steadily and tip over. That matching condition is resonance, and it is why the technique is called nuclear magnetic resonance [Fre, §2.4].

Get resonance right and everything else follows: the frequency you must supply tells you about the local field, which tells you about chemistry (the chemical shift) and, if you deliberately make the field vary across the sample, about position. That last step is magnetic resonance imaging, and it is one line of algebra away.


Theory

Precession, for a nucleus

From R.1.1: a magnetic moment tied to angular momentum by μ=γS\boldsymbol\mu = \gamma\mathbf{S} obeys dS/dt=γS×Bd\mathbf{S}/dt = \gamma\,\mathbf{S}\times\mathbf{B}, which preserves S|\mathbf{S}| and the component along B\mathbf{B}, leaving precession about the field at the Larmor frequency ω0=γB0\omega_0 = \gamma B_0. Nothing changes for a nucleus except the numerical value of γ\gamma, which is smaller than the electron's by roughly the mass ratio. For the proton — a spin-12\tfrac12 nucleus, and the one that matters, since the human body is mostly water and fat —

γp2π=42.58 MHz/T. \frac{\gamma_p}{2\pi} = 42.58\ \text{MHz/T} .

So a 1.51.5 T clinical magnet puts protons at 63.963.9 MHz and a 33 T magnet at 127.7127.7 MHz: the FM band and just above it. This is why the transverse field is supplied by radio waves. A circularly polarized radio wave is precisely a magnetic field vector rotating in a plane, and tuning the transmitter tunes the rotation rate.

One number is worth having before we go on, because it explains the entire engineering of the field. The two spin states differ in energy by ω0\hbar\omega_0, and at body temperature the thermal population difference is

ω02kBT=(1.055×1034)(2π×63.9×106)2(1.381×1023)(310)5×106, \frac{\hbar\omega_0}{2k_BT} = \frac{(1.055\times10^{-34})(2\pi\times 63.9\times10^6)}{2(1.381\times10^{-23})(310)} \approx 5\times10^{-6} ,

five parts per million. Only that tiny excess contributes any signal at all; the rest cancels. NMR works only because a cubic centimetre of water holds about 7×10227\times10^{22} protons, so five parts per million is still 3×10173\times10^{17} spins. It also explains why the magnet is enormous and expensive: signal scales with B0B_0 through the polarization and through the induced voltage, so roughly as B02B_0^2, and there is no cheap substitute.

The rotating frame

Now the central calculation. Apply a static field along z^\hat z and a transverse field of magnitude B1B_1 rotating in the xxyy plane at angular frequency ω\omega, in the same sense as the precession:

B(t)=B0z^+B1(cosωtx^sinωty^). \mathbf{B}(t) = B_0\hat z + B_1\bigl(\cos\omega t\,\hat x - \sin\omega t\,\hat y\bigr) .

This is a time-dependent problem and looks unpleasant. Move to a frame rotating about z^\hat z at the same rate ω\omega, so that the transverse field is stationary in it. A vector's rate of change in the two frames differs by the frame's rotation, and applying that to S\mathbf{S} turns the equation of motion into a static one:

(dSdt)rot=γS×Beff,Beff=(B0ωγ)z^+B1x^, \left(\frac{d\mathbf{S}}{dt}\right)_{\text{rot}} = \gamma\,\mathbf{S}\times\mathbf{B}_{\text{eff}}, \qquad \mathbf{B}_{\text{eff}} = \left(B_0 - \frac{\omega}{\gamma}\right)\hat z + B_1\hat x' ,

with x^\hat x' the (now stationary) direction of the transverse field. The rotation has subtracted ω/γ\omega/\gamma from the vertical field — the frame's own turning does part of the precession for us, and only the remainder is left to act.

The physics is now visible by inspection. Define the detuning Δ=ω0ω\Delta = \omega_0 - \omega and the Rabi frequency ω1=γB1\omega_1 = \gamma B_1. Then γBeff=Δz^+ω1x^\gamma\mathbf{B}_{\text{eff}} = \Delta\hat z + \omega_1\hat x', and in the rotating frame the spin simply precesses about that fixed axis at rate

Ω=Δ2+ω12. \Omega = \sqrt{\Delta^2 + \omega_1^2} .

On resonance, ω=ω0\omega = \omega_0 so Δ=0\Delta = 0: the vertical part cancels exactly, the effective field is purely transverse, and the spin precesses about x^\hat x' at ω1\omega_1. Starting from +z\lvert{+z}\rangle it tips smoothly down through the xx'zz plane,

Sz(t)=2cos(ω1t),P(z,t)=sin2 ⁣(ω1t2), \langle S_z\rangle(t) = \frac{\hbar}{2}\cos\left(\omega_1 t\right) , \qquad P(-z, t) = \sin^2\!\left(\frac{\omega_1 t}{2}\right) ,

reaching /2-\hbar/2 at ω1t=π\omega_1 t = \pi and returning at 2π2\pi. This periodic turning over and back is Rabi flopping. Note that even in the rotating frame nothing has been "pushed"; the spin is doing the only thing it can do, precessing, about an axis we manufactured.

Off resonance, the effective field is tilted out of the plane by an angle β\beta with sinβ=ω1/Ω\sin\beta = \omega_1/\Omega, so the spin precesses on a cone that no longer reaches the south pole:

P(z,t)=ω12ω12+Δ2sin2 ⁣(Ωt2). P(-z, t) = \frac{\omega_1^2}{\omega_1^2 + \Delta^2}\,\sin^2\!\left(\frac{\Omega t}{2}\right) .

The prefactor is a Lorentzian in the detuning with half-width Δ=ω1\Delta = \omega_1. That single formula carries the whole story: the flip is complete only exactly on resonance, and off resonance the spin merely nods — Freericks' "dips down and up a little bit." It also tells us the linewidth: a stronger drive flips faster but responds over a wider range of frequencies, so sharp spectroscopy needs weak, long pulses and fast control needs strong, short ones. That trade-off runs through every qubit control problem in Term 4.

When to stop

The flip angle after a resonant pulse of duration tt is

α=ω1t=γB1t. \alpha = \omega_1 t = \gamma B_1 t .

A π/2\pi/2 pulse (α=90°\alpha = 90°) tips a spin from the axis into the transverse plane, where it precesses and induces a voltage in a pickup coil — this is how the signal is created. A π\pi pulse (α=180°\alpha = 180°) inverts it completely.

Freericks makes a question of this [Fre, §2.4.4], and it is worth answering sharply because the wrong answer is very tempting. The duration depends on B1B_1, not on B0B_0. The big field sets the frequency you must transmit at; the little field sets how long you must transmit for. They are independent knobs, and conflating them is the commonest confusion in the subject.

Numbers, for a typical B1=10 μB_1 = 10\ \muT:

ω12π=42.58×106×105=426 Hz,tπ=πγB1=12×426=1.17 ms, \frac{\omega_1}{2\pi} = 42.58\times10^{6}\times10^{-5} = 426\ \text{Hz}, \qquad t_{\pi} = \frac{\pi}{\gamma B_1} = \frac{1}{2\times426} = 1.17\ \text{ms},

with tπ/2=587 μt_{\pi/2} = 587\ \mus. So the carrier oscillates at 6464 MHz while the envelope lasts about a millisecond — some 75,00075{,}000 carrier cycles inside one pulse. The two timescales are utterly separated, which is exactly what makes the rotating-frame picture so accurate.

The chemical shift

If every proton in the sample felt exactly B0B_0, NMR would be a one-line experiment. They do not. The electrons around a nucleus circulate in the applied field and generate a small opposing field of their own, so the nucleus feels

Blocal=B0(1σ),ω0=γB0(1σ), B_{\text{local}} = B_0\,(1 - \sigma) , \qquad \omega_0 = \gamma B_0 (1 - \sigma) ,

where the shielding constant σ\sigma is set by the chemical environment. The resulting shift in resonance frequency is the chemical shift, quoted in parts per million of the carrier so that it is the same number on any spectrometer. It is typically 001010 ppm for protons: at a 400400 MHz instrument, 11 ppm is 400400 Hz.

The classic demonstration is ethanol, C2H5OH\mathrm{C_2H_5OH}, whose six hydrogens fall into three chemically distinct groups — the three on the methyl carbon, the two on the carbon next to the oxygen, and the one bonded to the oxygen itself. A low-resolution spectrum shows three peaks with areas in the ratio 3:2:13:2:1, which counts the hydrogens in each group directly. That alone is a structural fingerprint.

At higher resolution each peak splits further, because each group of protons feels the small magnetic fields of its neighbours' nuclear spins. The methyl peak becomes a triplet (three sublines) and the neighbouring pair becomes a quartet (four). The pattern follows the "n+1n+1 rule": a group with nn equivalent neighbouring spin-12\tfrac12 nuclei splits into n+1n+1 lines, because nn spin-12\tfrac12 neighbours have n+1n+1 distinguishable total projections. The methyl group has two neighbours, so 2+1=32 + 1 = 3; the pair has three neighbours, so 3+1=43 + 1 = 4. The hydroxyl proton exchanges rapidly between molecules and typically shows as a single, solvent-dependent line whose position wanders.

This is why chemists care. The peak positions identify the functional groups, the areas count the nuclei, and the splitting patterns reveal which groups are bonded to which — a molecular structure read straight off a frequency axis.

It also sets a punishing hardware requirement, which is the substance of two of Freericks' problems [Fre, §§2.4.3, 2.4.5]. Resolving 0.10.1 ppm demands that B0B_0 be uniform across the sample to one part in 10710^7, or about a microtesla in a 9.49.4 T magnet. If the field is less uniform than the shift you are trying to see, every peak broadens into a smear and the chemical shift is simply lost — the sample is reporting the magnet's imperfections rather than its own chemistry. Hence shim coils, spinning sample tubes, and a great deal of money.

The spin echo

Field inhomogeneity is not only a spectroscopy problem; it also destroys the signal in the time domain, and Hahn's 1950 fix for it is one of the prettiest tricks in physics.

Start with the spins along z^\hat z and apply a π/2\pi/2 pulse to lay them into the transverse plane, where they precess together and induce a signal. They do not stay together. Each sits in a slightly different local field, so each precesses at a slightly different rate, and the fan of spin vectors dephases — splays out around the circle — until they point in all directions and the net signal is gone. The characteristic time for this is called T2T_2^{*}, and most of it is the magnet's fault.

Now wait a time τ\tau and apply a π\pi pulse about the same transverse axis. This reflects the whole fan: the spins that had run ahead are now behind by exactly as much as they had gained. But each spin is still in its own local field, so each keeps its own rate — the fast ones are still fast. After a second interval τ\tau, every spin has made up exactly the deficit it was given, and the whole fan converges: the signal reappears as a spin echo at time 2τ2\tau.

flowchart LR
    A["spins along z"] -->|"pi/2 pulse"| B["all in the<br/>transverse plane"]
    B -->|"evolve tau"| C["dephased fan<br/>fast ahead of slow"]
    C -->|"pi pulse"| D["fan reflected<br/>fast now behind"]
    D -->|"evolve tau"| E["refocused<br/>ECHO at 2 tau"]

The beauty of it is that we never needed to know what the local fields were. Any static spread of precession rates is undone, whatever its cause. What is not undone is anything that changes between the two intervals — a spin that diffuses into a different field, or that exchanges energy with its surroundings. So the echo amplitude decays as τ\tau is lengthened, and the time constant of that decay is the genuine, irreducible transverse relaxation time T2T_2:

1T2=1T2+γΔB02, \frac{1}{T_2^{*}} = \frac{1}{T_2} + \gamma\,\frac{\Delta B_0}{2} ,

the second term being the recoverable part. Separately, the spins also drift back toward z^\hat z, because that is the low-energy direction and energy leaks into the surroundings; the time constant for the longitudinal magnetization to recover is T1T_1. The two are physically different — T1T_1 requires energy exchange, T2T_2 only phase scrambling — so T22T1T_2 \le 2T_1 always. In a mobile liquid the molecular tumbling averages the local fields so effectively that the two nearly coincide; in solids and in tissue T2T_2 is far shorter. Brain white matter at 1.51.5 T has T1700T_1 \approx 700 ms and T280T_2 \approx 80 ms, while cerebrospinal fluid has T14T_1 \approx 4 s and T22T_2 \approx 2 s. Those differences are the contrast in an MRI image — the machine is not photographing anatomy, it is photographing relaxation times.

From spectroscopy to imaging

Everything so far demanded a uniform field. Imaging is what you get by deliberately ruining that, in a controlled way [Fre, §2.4.6].

Superimpose on B0B_0 a field gradient GG, a small field that grows linearly with position:

B(z)=B0+Gzω(z)=γ(B0+Gz). B(z) = B_0 + G z \quad\Longrightarrow\quad \omega(z) = \gamma\left(B_0 + G z\right) .

Precession frequency is now a linear function of position. Since the received signal's amplitude at each frequency counts the spins precessing at that frequency, the measured spectrum is the projection of the proton density along zz. Frequency has become a ruler. Rotate the gradient direction, take another projection, repeat, and reconstruct the density from its projections exactly as a CT scanner does — which is what Lauterbur did in 1973, and what Mansfield turned into a fast, practical slice-by-slice method.

A real scanner uses three gradients and gets the job done in one pass per line:

  • Slice selection. Apply GzG_z during a shaped RF pulse of bandwidth Δf\Delta f. Only spins in the slab where ω(z)\omega(z) falls inside the pulse's band are excited, a slab of thickness Δz=2πΔf/(γGz)\Delta z = 2\pi\Delta f/(\gamma G_z). Everything outside is untouched.
  • Frequency encoding. Apply GxG_x while the signal is read out, so position along xx maps to frequency and one Fourier transform of the recorded signal gives a whole line.
  • Phase encoding. Apply GyG_y briefly before readout so that spins acquire a position-dependent phase, and step its strength from one repetition to the next. The set of readouts then samples a two-dimensional grid, and a 2D inverse FFT gives the image.

Formally the machine measures the Fourier transform of the spin density, S(k)=ρ(r)eikrd3rS(\mathbf{k}) = \int \rho(\mathbf{r})\,e^{-i\mathbf{k}\cdot\mathbf{r}}\,d^3r with k=γGdt\mathbf{k} = \gamma\int\mathbf{G}\,dt, and the gradients steer a trajectory through that k\mathbf{k}-space. Resolution follows from arithmetic: with G=10G = 10 mT/m the frequency changes by γG/2π=426\gamma G/2\pi = 426 Hz per millimetre, so resolving 11 mm requires resolving 426426 Hz, which requires reading out for at least 1/4262.31/426 \approx 2.3 ms. Stronger gradients buy resolution or speed; they also demand fearsome amplifiers and are the source of the famous banging noise, as the coils flex against B0B_0.

Why linear? Any monotonic field profile assigns a unique frequency to each position, so an image could in principle be recovered. Linearity is what makes the recovery a plain Fourier transform: the k\mathbf{k}-space samples are then uniformly spaced and the reconstruction is an FFT rather than a warped, non-uniform inversion that costs resolution and signal-to-noise. A non-monotonic profile is fatal — two different positions share a frequency and no algorithm can separate them, which is exactly what a badly shimmed magnet does to the edges of its bore.

Blood flow, and fMRI

Freericks asks whether motion spoils the imaging [Fre, §2.4.7]. It changes it, and the changes are more useful than the damage.

The complication is real: a proton that moves between excitation and readout is encoded at one position and detected at another, producing blurring and, with pulsatile flow, ghost images repeated across the picture. Cardiac gating and flow-compensating gradient shapes exist precisely to suppress this.

But the same sensitivity is a measurement. In time-of-flight angiography, repeated rapid pulses saturate the stationary spins in a slice — they never fully recover between pulses — while blood flowing into the slice arrives fully relaxed and gives a much larger signal. Vessels light up against a dark background with no contrast agent at all. In phase-contrast imaging, a bipolar gradient pulse (equal and opposite lobes) returns static spins to zero net phase while a spin moving along the gradient accumulates a phase proportional to its velocity, so the phase map is a velocity map, signed and quantitative.

Functional MRI adds one more layer. Haemoglobin carrying oxygen is diamagnetic; without it, deoxyhaemoglobin is paramagnetic and distorts the local field around each vessel, shortening T2T_2^{*} and darkening the image. When a region of the brain becomes active, blood flow to it increases by more than the extra oxygen consumed, so the fraction of deoxyhaemoglobin falls, T2T_2^{*} lengthens, and the signal rises — by a few percent at 33 T. This is the BOLD contrast (blood-oxygen-level dependent), and it is worth being clear about what it is: an indirect, haemodynamic proxy for neural activity, lagging it by four to six seconds, not a picture of thought. Other nuclei give other windows; phosphorus-31 spectroscopy, for instance, tracks ATP and phosphocreatine and so reports on metabolism directly.

Why this belongs in a quantum computing curriculum

NMR is not a detour. It is the oldest and most refined technology for controlling individual quantum two-level systems, and essentially every technique used to drive a superconducting or trapped-ion qubit was invented by NMR spectroscopists first: resonant π\pi and π/2\pi/2 pulses, composite pulses robust against amplitude errors, spin echo and its multi-pulse descendants (now called dynamical decoupling), Ramsey interferometry, and the T1/T2T_1/T_2 vocabulary in which every qubit datasheet is written.

It was also, briefly, the leading quantum computing platform. In 2001 Vandersypen and collaborators ran Shor's algorithm on a seven-spin molecule and factored 15, using exactly the pulse sequences described above. Liquid-state NMR quantum computing then hit a wall, and the reason is the five-parts-per-million number from the start of this lesson: the usable signal from a thermal ensemble falls off exponentially with the number of qubits, so the approach does not scale past about a dozen. The control physics survived the platform and is now everywhere — see 1.6.2 Two-Level Dynamics & Rabi Oscillations for the same Hamiltonian done as quantum dynamics, and Term 4's hardware course for its descendants.


Hands-on (Python)

import numpy as np

gamma_2pi = 42.58e6          # proton gyromagnetic ratio / 2pi, Hz/T

# --- 1. Larmor frequencies and pulse durations ---------------------------
print(" B0 (T)   f0 (MHz)")
for B0 in (0.5, 1.5, 3.0, 9.4):
    print(f"  {B0:4.1f}    {gamma_2pi * B0 / 1e6:8.2f}")

print("\n B1 (uT)   f1 (Hz)   t_pi/2 (us)   t_pi (us)")
for B1 in (1e-6, 1e-5, 1e-4):
    f1 = gamma_2pi * B1                       # Rabi frequency in Hz
    print(f"  {B1*1e6:5.0f}    {f1:8.1f}   {1/(4*f1)*1e6:9.1f}   {1/(2*f1)*1e6:8.1f}")
# The frequency comes from B0; the DURATION comes from B1 alone.

# --- 2. Rabi flopping, on and off resonance ------------------------------
def p_flip(t, w1, delta):
    Om = np.hypot(w1, delta)
    return (w1**2 / Om**2) * np.sin(Om * t / 2) ** 2

w1 = 2 * np.pi * 426.0                        # rad/s, for B1 = 10 uT
t_pi = np.pi / w1
print(f"\n detuning/w1   P(flip) at t = t_pi   max possible flip")
for r in (0.0, 0.5, 1.0, 2.0, 5.0):
    d = r * w1
    grid = np.linspace(0, 20 * t_pi, 20001)
    print(f"   {r:5.1f}         {p_flip(t_pi, w1, d):.4f}              "
          f"{p_flip(grid, w1, d).max():.4f}")
# On resonance the flip is complete; at a detuning of one Rabi frequency the
# best achievable flip is already only 1/2 -- a Lorentzian of half-width w1.

# --- 3. Spin echo: refocusing a static spread of rates -------------------
rng = np.random.default_rng(0)
N, tau = 20000, 5e-3
dw = 2 * np.pi * rng.normal(0, 40.0, N)       # +/- 40 Hz static inhomogeneity

def coherence(phases):
    return abs(np.mean(np.exp(1j * phases)))

free = dw * 2 * tau                            # no pi pulse: dephase for 2 tau
echo = dw * tau - dw * tau                     # pi pulse at tau flips the sign
print(f"\nsignal after 2 tau, free induction : {coherence(free):.4f}")
print(f"signal after 2 tau, with pi pulse  : {coherence(echo):.4f}")
# ~0.03 vs exactly 1: the static spread is undone without ever measuring it.

# --- 4. Gradient encoding: the spectrum IS the projection ----------------
G = 10e-3                                      # T/m
z = np.linspace(-0.10, 0.10, 400)              # 20 cm field of view
rho = (np.abs(z) < 0.06) * 1.0                 # a slab ...
rho += (np.abs(z - 0.02) < 0.01) * 1.5         # ... with a bright inclusion
f = gamma_2pi * G * z                          # frequency assigned to position
print(f"\nfrequency ruler        : {gamma_2pi * G / 1e3:.1f} kHz per metre"
      f"  =  {gamma_2pi * G / 1e3 * 1e-3:.3f} kHz per mm")
print(f"readout time for 1 mm  : {1 / (gamma_2pi * G * 1e-3) * 1e3:.2f} ms")
print(f"bandwidth over 20 cm   : {(f.max() - f.min()) / 1e3:.1f} kHz")
# Reading the spectrum back out recovers rho(z) exactly, because the map
# z -> f is linear and therefore invertible with a plain Fourier transform.

Exercises

E1 (easy). A scanner runs at B0=3B_0 = 3 T. (a) What radio frequency flips protons? (b) The same scanner is used for sodium-23, with γ/2π=11.26\gamma/2\pi = 11.26 MHz/T. What frequency does that need? (c) If the transmitter is left tuned to the proton frequency, what happens to the sodium nuclei?

Solution

(a) f0=42.58×3=127.7f_0 = 42.58 \times 3 = 127.7 MHz.

(b) f0=11.26×3=33.8f_0 = 11.26 \times 3 = 33.8 MHz — a completely different band. Different nuclei are addressed by different frequencies in the same magnet, which is what makes multinuclear spectroscopy possible without changing hardware.

(c) Essentially nothing. The detuning is Δ/2π94\Delta/2\pi \approx 94 MHz while a typical Rabi frequency is a few hundred hertz, so the maximum achievable flip is ω12/(ω12+Δ2)(426/9.4×107)22×1011\omega_1^2/(\omega_1^2 + \Delta^2) \sim (426/9.4\times10^7)^2 \approx 2\times10^{-11}. The sodium spins wobble by an unmeasurable amount and stay put. Resonance is extraordinarily selective.

E2 (easy). A pulse of B1=20 μB_1 = 20\ \muT is applied on resonance. How long must it last to be a π/2\pi/2 pulse? A π\pi pulse? If the amplifier delivers only half the intended B1B_1 but the timing is unchanged, what flip angle results, and what fraction of the intended signal does a "π/2\pi/2" pulse produce?

Solution

ω1/2π=42.58×106×2×105=852\omega_1/2\pi = 42.58\times10^6\times 2\times10^{-5} = 852 Hz. A π/2\pi/2 pulse needs ω1t=π/2\omega_1 t = \pi/2, i.e. t=1/(4×852)=293 μt = 1/(4\times852) = 293\ \mus; a π\pi pulse needs 587 μ587\ \mus.

At half amplitude the flip angle halves: the "π/2\pi/2" pulse delivers 45°45°. The transverse magnetization it creates is sinα\sin\alpha of the full value, so sin45°=0.707\sin 45° = 0.707 — about 71% of the intended signal, a 29% loss from a 50% amplitude error. Small errors cost far less than that, because sinα\sin\alpha is stationary at α=90°\alpha = 90°: a 5%5\% amplitude error loses only 1sin(85.5°)=0.3%1 - \sin(85.5°) = 0.3\% of the signal. A π\pi pulse enjoys no such protection — its error is first order — which is why composite pulses were invented to protect inversions rather than excitations.

E3 (medium). A spectrometer operates at 500500 MHz for protons. (a) What is B0B_0? (b) Two peaks are separated by 2.12.1 ppm; how many hertz apart are they? (c) To resolve peaks 0.050.05 ppm apart, how uniform must B0B_0 be across the sample, in tesla? (d) What happens to the ethanol spectrum if the field is ten times less uniform than that?

Solution

(a) B0=500/42.58=11.74B_0 = 500/42.58 = 11.74 T.

(b) 2.1 ppm×500 MHz=2.1×106×5×108=10502.1\ \text{ppm}\times 500\ \text{MHz} = 2.1\times10^{-6}\times5\times10^{8} = 1050 Hz.

(c) Resolving 0.050.05 ppm means the unwanted spread must be well below 5×1085\times10^{-8} of B0B_0: ΔB5×108×11.74=5.9×107\Delta B \lesssim 5\times10^{-8}\times 11.74 = 5.9\times10^{-7} T, roughly half a microtesla — about one hundredth of the Earth's field, across the whole sample volume. This is what shim coils and sample spinning are for.

(d) Each line broadens to about 0.50.5 ppm, which is comparable to the spacing of the fine structure. The triplet and quartet merge into single humps, so the JJ-coupling information — and with it the bonding topology — is lost. The three main group peaks, separated by more than 1 ppm, survive as broad bumps, so you would still count 3:2:13:2:1 and identify the functional groups, but you could no longer say which group is attached to which. Resolution buys structure, not just prettiness.

E4 (medium). Explain the spin echo carefully. (a) Why does a π\pi pulse refocus a static spread of precession rates? (b) Why does it fail to refocus diffusion through an inhomogeneous field? (c) A sample has T2=90T_2 = 90 ms and, in the scanner's field, T2=12T_2^{*} = 12 ms. What echo amplitude survives at 2τ=602\tau = 60 ms, and what would a free-induction signal give at the same time?

Solution

(a) After the first interval, a spin whose offset is δω\delta\omega has acquired phase δωτ\delta\omega\,\tau. The π\pi pulse reflects every phase about the pulse axis, sending ϕϕ\phi \to -\phi. During the second interval each spin accumulates +δωτ+\delta\omega\,\tau again, because it is still sitting in the same local field, so its total phase is δωτ+δωτ=0-\delta\omega\tau + \delta\omega\tau = 0 — independently of δω\delta\omega. Every spin returns to the same phase at once and the signal reappears. Crucially, we never had to know any δω\delta\omega: the refocusing is automatic.

(b) The cancellation relies on each spin experiencing the same offset in both intervals. A spin that diffuses to a different location samples a different field after the pulse, so its two accumulated phases no longer cancel and it contributes a residual random phase. This is why echo attenuation in a gradient is used to measure diffusion coefficients — the failure of the trick is itself an instrument.

(c) The echo decays with the true T2T_2: e60/90=e0.667=0.513e^{-60/90} = e^{-0.667} = 0.513, so about 51% survives. A free-induction signal decays with T2T_2^{*}: e60/12=e5=0.0067e^{-60/12} = e^{-5} = 0.0067, under 1%. Refocusing recovers a factor of about 76 in signal here — which is why virtually every clinical sequence is built on echoes rather than raw free induction.

E5 (hard). Freericks ends the chapter by asking you to explain an MRI machine to a non-specialist [Fre, §2.4.8]. Do it properly, as a technical brief a competent non-physicist could follow, covering: (a) what the large static field does to a nuclear spin; (b) how fast the rotating field must turn and why; (c) what happens when it does; (d) why the static field is deliberately made to vary across the patient, and what would happen if it did not; (e) what the machine actually measures when it makes an image. Include at least three quantitative statements.

Solution

(a) The static field. Every hydrogen nucleus in the body is a tiny magnet carrying angular momentum, so in a magnetic field it does not line up — it precesses, sweeping around the field direction like a tilted spinning top, at a rate strictly proportional to the field: 42.5842.58 MHz per tesla. In a 1.51.5 T scanner that is 63.963.9 MHz. The field also very slightly biases which way the spins point: about five nuclei in a million more along the field than against it. That minute imbalance is the entire signal, and it works only because a cubic centimetre of tissue contains around 7×10227\times10^{22} hydrogen nuclei.

(b) How fast the radio wave must rotate. To tip a spin over you must make it precess about a horizontal axis, so you apply a horizontal magnetic field — but the spin is already whirling around the vertical field 64 million times a second, and a horizontal field that stands still averages to nothing from the spin's point of view. The horizontal field must therefore rotate at exactly the precession rate, 63.963.9 MHz, so that the spin always sees it in the same direction. That matching is resonance, and a circularly polarized radio wave at the right frequency supplies it.

(c) What happens. In a frame rotating with the wave, the vertical field cancels out and the spin sees only the small horizontal field — about 10 μ10\ \muT, a ten-thousandth of the main field. It precesses about that, tipping steadily over. Turn the wave off after a quarter turn and the spins lie in the horizontal plane, where they precess together and induce a measurable voltage in a coil; that is the signal. The timing follows from the small field alone: at 10 μ10\ \muT, a half-turn takes about 1.21.2 ms. The big field says what frequency; the little field says how long.

(d) Why the field is made non-uniform. With a perfectly uniform field, every hydrogen nucleus in the body precesses at the same frequency, so the signal is a single tone and tells you only how much hydrogen is present — a single number for the whole patient, with no way to tell head from foot. Adding a field that increases steadily along the body makes the precession frequency increase steadily too, so that each slice precesses at its own frequency. Frequency has become a ruler: separating the received signal into its frequency components separates the body into its slices. With a typical gradient the frequency changes by about 426426 Hz per millimetre. The gradient must be linear so the map from position to frequency can be inverted by a straightforward Fourier transform; a field that rose and then fell would give two different places the same frequency, and no amount of computing could tell them apart.

(e) What is actually measured. Not anatomy, and not "hydrogen" alone. The machine records how much signal comes back from each frequency — hence each location — and how fast that signal fades. The fading times, T1T_1 and T2T_2, differ strongly between tissues: brain white matter has T280T_2 \approx 80 ms while cerebrospinal fluid has T22T_2 \approx 2 s, a factor of 25. By choosing the pulse timings the operator decides which of those differences dominates the picture, which is why the same patient in the same machine yields images that look nothing alike. An MRI image is a map of hydrogen density weighted by relaxation times — and because relaxation is exquisitely sensitive to the molecular environment, that turns out to be a superb map of soft tissue.


Checkpoint

  1. Why must the transverse field rotate rather than stand still, and what is the resonance condition?
  2. Write Beff\mathbf{B}_{\text{eff}} in the rotating frame and describe the motion on and off resonance. What sets the width of the resonance?
  3. A pulse must flip a spin from +z+z to z-z. What sets its frequency, and what sets its duration?
  4. What does a π\pi pulse refocus in a spin-echo sequence, and what does it fail to refocus? Distinguish T1T_1, T2T_2, and T2T_2^{*}.
  5. How does a linear field gradient turn frequency into position? What would go wrong with a non-monotonic field?
Answers
  1. The spin is already precessing about the large static field at ω0=γB0\omega_0 = \gamma B_0, so a static transverse field sweeps past it at that rate and its effect averages to nothing. A field rotating at ω0\omega_0 stands still in the spin's own rotating frame and can act coherently. The resonance condition is ω=γB0\omega = \gamma B_0.
  2. Beff=(B0ω/γ)z^+B1x^\mathbf{B}_{\text{eff}} = (B_0 - \omega/\gamma)\hat z + B_1\hat x', so γBeff=Δz^+ω1x^\gamma\mathbf{B}_{\text{eff}} = \Delta\hat z + \omega_1\hat x'. On resonance the vertical part cancels and the spin precesses about x^\hat x' at ω1=γB1\omega_1 = \gamma B_1, flipping completely. Off resonance it precesses about a tilted axis at Ω=Δ2+ω12\Omega = \sqrt{\Delta^2+\omega_1^2} and only reaches ω12/(ω12+Δ2)\omega_1^2/(\omega_1^2+\Delta^2) of a flip — a Lorentzian of half-width Δ=ω1\Delta = \omega_1, so the drive strength sets the linewidth.
  3. The frequency is set by the static field, ω=γB0\omega = \gamma B_0; the duration by the rotating field, tπ=π/(γB1)t_\pi = \pi/(\gamma B_1). They are independent: at B0=1.5B_0 = 1.5 T and B1=10 μB_1 = 10\ \muT the carrier is 63.963.9 MHz and the pulse lasts 1.171.17 ms.
  4. It refocuses any static spread of precession rates — magnet inhomogeneity, susceptibility variation — because each spin's phase error is reversed and then re-accumulated identically. It fails on anything that changes between the two intervals: diffusion into a different field, or energy exchange with the surroundings. T1T_1 is the recovery of magnetization along the field (needs energy exchange); T2T_2 is the irreducible loss of transverse phase coherence; T2T_2^{*} is the faster apparent decay including the recoverable inhomogeneous part, 1/T2=1/T2+γΔB0/21/T_2^{*} = 1/T_2 + \gamma\Delta B_0/2.
  5. With B(z)=B0+GzB(z) = B_0 + Gz the precession frequency ω(z)=γ(B0+Gz)\omega(z) = \gamma(B_0 + Gz) is a linear function of position, so the amplitude at each frequency counts the spins at the corresponding place and the spectrum is the projection of the spin density. Linearity makes the inversion a plain Fourier transform. A non-monotonic field maps two different positions to the same frequency, so their signals are irreversibly superimposed and no reconstruction can separate them.

Further Reading

  • [Fre] J. K. Freericks, Quantum Mechanics Done Right, §2.4 — the NMR/MRI story this lesson follows, with its Nobel-Prize timeline and the "explain the MRI machine" essay problem.
  • [Sak] Sakurai & Napolitano, §§2.1 and 5.5 — spin precession, the rotating frame, and magnetic resonance as time-dependent perturbation theory.
  • [Gri] Griffiths & Schroeter, §4.4.3 and Problem 4.50 — Larmor precession and the driven two-level system in wave-mechanics language.
  • [NC] Nielsen & Chuang, §7.7 — NMR quantum computation: pulse sequences, refocusing, pseudo-pure states, and why the ensemble approach stops scaling.
  • [ER] Eisberg & Resnick, Ch. 12 — nuclear moments and magnetic resonance in the historical/experimental route.
  • E. L. Hahn, "Spin Echoes", Phys. Rev. 80, 580 (1950); L. M. K. Vandersypen et al., Nature 414, 883 (2001) — the echo, and Shor's algorithm on a seven-spin molecule.
  • Classical Stern-Gerlach Experiment — where μ=γS\boldsymbol\mu = \gamma\mathbf{S} and the Larmor frequency come from.
  • 1.6.2 Two-Level Dynamics & Rabi Oscillations — the same driven two-level problem as quantum dynamics, with the rotating-wave approximation done properly.
  • 1.2.2 The Bloch Sphere — the geometry in which every pulse in this lesson is a rotation.

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