BrainFlux

Bloch sphere and NMR

Instrument

Drag the arrow's tip to tip the spin, or press a pulse button; drag anywhere else on the sphere to turn it

Spin ½ · the quantum state of the arrow

The arrow is the direction of the spin, ⟨S⟩, of the quantum state

$|\psi\rangle=\cos\frac{\theta}{2}\,\text{|↑⟩}+e^{i\varphi}\sin\frac{\theta}{2}\,\text{|↓⟩}$

P(↑) P(↓)

↑ is along B₀, the lower energy for a proton. P(↑) is the chance of finding the spin up if you measured it along B₀.

Drag the tank's peak along to retune it (the switches of SW1 follow); the red line is the protons' frequency

Drag a pulse along the timeline to move it; click one to choose it and set its angle and axis on the right

Experiments

Start here

The red arrow is the magnetization of a sample of protons, in units of what the magnet gives it once the protons have settled: pointing up, along the magnet's field, is that settled state. The arrow spins round the field as a top spins round the vertical, far too fast to watch (five million turns a second in the TeachSpin's magnet), so the page slows time down and says by how much.

Drag the arrow's tip to tip it, or press a pulse button: a short burst of radio waves that turns the arrow by the angle you choose. The plots under the sphere are what the instrument records: the voltage the spinning arrow makes in a coil. Switch to the rotating frame to turn with the arrow: then only the pulses and the slow changes are left to watch.

The tabs go on from one spin to many (why the signal dies away, how an echo brings it back, and the three times T₂*, T₂ and T₁ side by side), to the pulse sequences that measure a sample, and to the Earth's-field board, which tips the protons without any radio pulse. The buttons at the top load the numbers of the TeachSpin (5 MHz) or of the board (2 kHz).

1. The spin goes round the field at the Larmor frequency

One spin, tipped 45° from the field $B_0$, seen in the lab. It goes round $B_0$ at the Larmor frequency $f_0=\gamma B_0/2\pi$, with $\gamma/2\pi=42.577$ MHz/T for a proton: 5.000 MHz in the TeachSpin's 117.4 mT, slowed here so that one turn takes a second. Drag the tip to another angle: a turn takes the same time at any angle, because only the field sets the rate. Under More, set $B_0$ to 234.9 mT: twice as fast. The coil lies along $x$, and the voltage in it (the lower plot) follows $dM_x/dt$: an oscillation at $f_0$ as big as the part of the arrow across the field, $|M_{xy}|$.

2. A π/2 pulse tips the spin over, and the coil starts to ring

The arrow starts along $B_0$, with nothing across it and no signal. A pulse of RF at the Larmor frequency turns it down to the equator. In the lab it spirals down, twelve and a half turns round $B_0$ while it goes a quarter turn down, because the RF's field $B_1$ (2.35 mT) is 50 times weaker than $B_0$. Then it goes round freely, all of it across the field, and the coil rings at full size: the free induction. The pulse lasted 2.50 μs, as $\theta=\gamma B_1\tau$ says. Press π/2 about x′ again: a second quarter turn takes it to the bottom, and the ringing stops.

3. The rotating frame: the same pulse without the spinning

Turn with the RF, at the Larmor frequency, and $B_0$ is gone: in this frame nothing moves the spin but $B_1$, which stands still along $x'$. The spiral of experiment 2 becomes a plain quarter turn about $x'$, from $z$ to $y'$. The mixer's two outputs, $M_{x'}$ and $M_{y'}$, are the arrow's coordinates in this frame; the coil's oscillation is the same arrow seen from the lab. Press Back to equilibrium, then π/2 about y′: the arrow goes to $-x'$. The axis of a pulse is the phase of its RF.

4. Off resonance: the pulse misses, and the mixer beats

The RF is 50 kHz above the Larmor frequency, half the rate $\gamma B_1/2\pi=100$ kHz at which $B_1$ turns the spin. In the frame of the RF, $B_0$ is no longer quite cancelled: $\Delta B=-2\pi\delta f/\gamma$ is left along $z$, and during the pulse the spin turns about the tilted effective field $B_\mathrm{eff}$, so it misses the equator. The "π/2" leaves $M_z=0.053$; 100 kHz off it would leave 0.197. After the pulse the spin goes round $\Delta B$ at 50 kHz in this frame, so the mixer's outputs swing, a quarter period apart, while the envelope $|M_{xy}|$ holds still. That beat is how the TeachSpin is tuned: turn the RF's frequency until the mixer stops beating.

5. Many spins: the signal dies, though no spin does

A real sample is many spins, and the magnet's field is not quite the same through it, so their frequencies spread, here over $\Delta f=500$ Hz (the full width of a Lorentzian). After a π/2 pulse they all start along $y'$ and fan out, the faster ones (orange) ahead and the slower (violet) behind. Their sum, which is what the coil sees, shrinks and dies away: the free induction decay, falling by $1/e$ in $T_2^*=1/(\pi\Delta f)=0.64$ ms. No spin has lost anything: each arrow is as long as ever. The spectrum is the line the spread makes, 500 Hz wide at half power: $1/(\pi T_2^*)$.

6. The spin echo: a π pulse brings the signal back

As in experiment 5, with a π pulse about $x'$ at $\tau=1.5$ ms, when the fan has spread all the way round. It flips the fan over: the faster spins are now behind by as much as they were ahead, and they catch up at $2\tau$. The fan closes and the signal comes back, an echo (Hahn, 1950), along $-y'$. It is smaller than the FID's start, 0.86 of it, because each spin has also lost some of itself with $T_2=20$ ms: $e^{-2\tau/T_2}$. The echo undoes the spread, not that loss. Slide $T_2$ down and the echo shrinks. $T_1$ (25 ms) pulls the sum's $M_z$ back up all the while: 0.08 by 5 ms.

7. Three clocks: T₂*, T₂ and T₁ on one plot

A made-up sample whose three times lie a decade apart: $T_1=100$ ms, $T_2=10$ ms, and a spread of the field (a Lorentzian, $\Delta f=286.5$ Hz wide) that makes $T_2^*=1.00$ ms, as $1/T_2^*=1/T_2+\pi\Delta f$. One π/2 pulse, then the clock speeds up as it goes, a decade every three seconds, and the plot's time axis is logarithmic, so each clock is a step of its own. The sum of the spins, what the coil sees, dies first, with $T_2^*$: the spins fan out and cancel. At 1 ms the sum is down to 0.37 while each spin's own part across the field (blue) is still 0.90 long: the fan shortens no spin. That part fades only with $T_2$, exactly as $e^{-t/T_2}$; on the sphere it is the radius of the blue ring the spins' tips lie on. Last, $M_z$ (green) climbs back as $1-e^{-t/T_1}$ and lifts the ring until it closes on the axis. Fitted to the curves: 1.00, 10.0 and 100 ms. The teeth are another run, a CPMG train with a π pulse about $y'$ every 5 ms: each echo closes the fan, and its top lands on the blue curve (the second, at 10 ms, on its $1/e$), never above it, as a sum can be no longer than the spins it adds. The echoes undo the fan, not $T_2$: their fit gives 10.0 ms. The bar over the plot is the rule $T_2^*\le T_2\le2T_1$. Slide $T_2$ and its mark moves along it; push it past $2T_1$ and $T_1$ follows. Narrow the spread and $T_2^*$ closes up on $T_2$; untick Many spins and it is $T_2$: one spin has no fan. Why $2T_1$: whatever turns a spin over to take its energy away, which is $T_1$, costs it its phase too, at half that rate.

8. T₂ from a train of echoes

A π/2 pulse, then sixteen π pulses about $y'$, at $\tau$, $3\tau$, $5\tau$… with $\tau=2$ ms. Each refocuses the fan, so an echo comes every 4 ms, out to 64 ms, and the echoes' heights fall as $e^{-t/T_2}$: one run measures $T_2$. The fit beside the sphere gives 20.2 ms for this oil's 20 ms. With one echo per run (Spin echo, then Scan τ) it takes twelve runs. This train is Carr and Purcell's (1954) with Meiboom and Gill's change (1958): the π pulses about $y'$, along the spins, not about $x'$.

9. Why Meiboom and Gill: pulses that are slightly wrong

The same train with the π pulses about $x'$ (Carr–Purcell), and every pulse 10% too long: a π that is really 198°, as a width set by hand can be. About $x'$ each pulse's error adds to the last, the echoes alternate in sign, and they die in half the time: the fit says $T_2=10.6$ ms. Press CPMG, keep the error at 10%: about $y'$, along the spins, an error in one pulse is undone by the next, and the fit gives 20.3 ms again. A short pulse, −10%, fails the same way about $x'$ (10.6 ms).

10. T₁ by inversion recovery

A π pulse turns $M$ to $-z$; it climbs back with $T_1$, as $1-2e^{-\tau/T_1}$; a π/2 at $\tau$ tips what it has reached over, and the FID's start reads it, with its sign on the mixer. The scan runs it for fifteen delays and plots the readings: below zero for short $\tau$, through zero at $\tau=T_1\ln2$, up to the full FID. The fit gives $T_1=25.0$ ms and the null at 17.3 ms. Click the plot there: the sequence runs at that $\tau$, and the sphere shows why the FID vanishes: $M$ has climbed back exactly to nothing, so the π/2 has nothing to tip. On the TeachSpin the null is the quick way: find the $\tau$ where the FID vanishes and divide by $\ln 2=0.693$. (The detector's envelope, which has no sign, folds the curve up at the null.)

11. Noise, and averaging it away

An FID with thermal noise: a signal-to-noise ratio of 5, so the noise is 0.20 of the full signal, r.m.s. Repeat and average: the signal is the same each time and adds up, the noise does not, and the average of $N$ runs has $1/\sqrt N$ of the noise: 0.050 after 16, 0.012 after 256. Watch the fit of $T_2^*$ beside the sphere: from one run it says 0.545 ms, 12% short of the true 0.617; from 16, 0.606; from 256, 0.618. It fits the mixer's output in phase with the FID, which noise leaves at zero on average. The envelope $|M_{xy}|$ does not go to zero: noise alone has a mean length $\sigma\sqrt{\pi/2}$, 0.25 here, which would make the FID look longer.

12. The Earth's-field board: polarize, switch off, listen

No magnet and no RF pulse. A litre of water sits in a coil whose axis lies across the Earth's field (east–west). For 3 s a current of 1.5 A makes 7.84 mT along the coil, 170 times the Earth's 46.0 μT, and the protons build a magnetization along it, as $1-e^{-t/T_1}$: 0.699 of all of it by 3 s for tap water ($T_1=2.5$ s). Then the current is switched off in 0.4 ms, before the protons can follow, so $M$ is left across the Earth's field: 98.9% of it. That is the π/2 pulse. It goes round $B_E$ at 1960 Hz and makes 0.78 μV in the same coil, dying away with $T_2^*=0.86$ s. The coil and its capacitors are a tank tuned to the protons, which multiplies it by 36.6: 28.6 μV. The receiver's gain and filters take that to 0.56 V at OUT (the plot under the board's cycle), 45 times the receiver's noise. Press Hear it: OUT, as the board's headphones give it.

13. Sudden or adiabatic: switch it off slowly and nothing rings

The same, with the current switched off in 300 ms instead of 0.4. As it falls, the total field turns from the coil's axis to the Earth's; slowly enough, the spins follow it round, and $M$ ends up along $B_E$, with nothing across it to ring: 4.5% is left, an FID of 0.034 μV, 25 mV at OUT, twice the receiver's noise. The field turns only at the very end of the ramp, through its last 1/170, so what matters is $A=\gamma B_E^2\tau/B_p$: 0.029 for the board's 0.4 ms (sudden), 1 at 13.8 ms (69% left across), 22 at 300 ms (adiabatic). The plainer $\gamma B_E\tau$ would say 4.9 at 0.4 ms, and be wrong. Watch the sphere through the switch-off at each speed.

14. Which way to lay the coil: the FID goes as sin² of its angle to the field

Here the coil lies level, pointing north–south. In Claremont the field dips 58.9° below the horizontal towards magnetic north, so the coil's axis is at $\alpha=58.9°$ to $B_E$. The current builds $M$ along the coil; switched off suddenly, only its part across the field, $M\sin\alpha$, goes round $B_E$: the tip sweeps a cone about the field, not its equator. And the coil picks up only the part of that turning $M$ along its own axis, $\sin\alpha$ of it again: the EMF goes as $\sin^2\alpha$, 0.733 here. The FID at OUT is 0.384 V against 0.56 V east–west: 0.68 of it, a little less than $\sin^2\alpha$, because during the 0.4 ms switch-off the spins follow the field's first turning towards $B_E$ by a few degrees, and at a slant that comes straight off the part across (78.5% of $M$ is left across, not $\sin\alpha$ = 85.6%). Press Upright (31.1°): 0.133 V. Drag the coil's end on the sphere along $B_E$: nothing rings. The handbook says lay it level and east–west (magnetic): square to the field anywhere on Earth. North–south is nearly as good only where the field dips steeply, upright only near the magnetic equator, where the field lies nearly level. East by the map instead of by the compass (11° off here, $\alpha$ = 84.2°) costs 2 to 3%: 1% by $\sin^2\alpha$, the rest for the switch-off's few degrees, which at any slant come off the part across. (With the current the other way round they would add to it instead: at a slant, which way round the coil is wired matters a little.)

15. Tune the tank to the protons

The coil and the capacitors that the eight switches of SW1 put across it make a resonant circuit, a tank: it multiplies the FID by $|H|$, up to $Q=\omega L/R=37$ at its resonance $f_T=1/2\pi\sqrt{LC}$, in a band 52.6 Hz wide. Here every switch is OFF, as a new board comes: only the clamp's 0.39 nF is across the coil, the tank rings at 49 kHz, far above the band, and the FID passes at ×1.00: 17 mV at OUT, 1.3 times the receiver's noise. The handbook's recipe (§4): the protons here are at 42.577 Hz/μT × 46.03 μT = 1959.8 Hz; the coil's 26.8 mH resonates there with $C=1/4\pi^2f^2L$ = 246.2 nF; of the switches' 3.3, 6.8, 13.3, 26.8, 53.8, 106.8, 210 and 420 nF, 210 + 26.8 + 6.8 + 3.3 come nearest, switches 1 2 4 7, and with the clamp's that makes 247.29 nF: the tank at 1955.6 Hz, 4.3 Hz below the protons. Click those four switches, or drag the tank's peak, or press Tune to the protons: ×36.6, 0.56 V at OUT. One step either side is still ×35.6 (switches 2 4 7) or ×30.9 (3 4 7): each step of 3.3 nF moves the tank 13 Hz.

16. Another place: the protons slide off the tank

The board, tuned in Claremont, taken where the Earth's field is 50.0 μT, as further north. The protons are now at 2129 Hz, 173 Hz above the tank, more than three of its widths: on the tank's plot the red line stands out on the tail. The gain there is ×5.34, not ×36.6, and the FID at OUT is 0.090 V instead of 0.56, though the EMF in the coil is a little larger, 0.86 μV, as it grows with the frequency. Press Tune to the protons: the recipe asks for 208.7 nF, switches 2 3 4 5 6 give 207.89 nF with the clamp's, the tank is at 2132.8 Hz and the gain ×40.2: 0.67 V. Another day needs nothing: the field drifts through a day by a few tens of nanotesla, about a hertz, and the tank is 52.6 Hz, 1240 nT, wide. It is a new place that needs the switches reset.

17. The tank narrows the noise band

Here the tank is tuned 55 Hz below the protons (switches 1 2 3 4 7). The coil's 8.9 Ω has thermal noise, $\sqrt{4kTR}$ = 0.38 nV/√Hz, in series with the EMF, so the tank multiplies it just as it multiplies the FID: read on the right axis, the tank's curve is that noise, 14 nV/√Hz at its peak. In all it is $\sqrt{kT/C}$, 0.125 μV for these 260.6 nF, whatever $R$ and $Q$ are, as equipartition says ($\tfrac12C\langle V^2\rangle=\tfrac12kT$), and nearly all of it in the tank's band. The preamplifier adds its own 9.8 nV/√Hz (dashed), flat over the receiver's whole band, 0.7 to 4.2 kHz: most of the 12.6 mV r.m.s. at OUT. Off the tank's top the protons get ×15.4: 0.24 V at OUT, 19 times the noise. Press Tune to the protons: the FID and the coil's own noise there are both ×36.6, and the FID is 45 times the noise at OUT, which has hardly changed. The tank does not lower the noise: it lifts the FID, and the coil's noise with it, above the preamplifier's, in its band, so that the receiver adds only 1.7 dB to the coil's own. Press All off and Hear it: ×1, and the whistle is lost in the hiss. (The spectrum still finds the line: in its own 0.5 Hz there is little noise.)

18. T₁ from the polarizing time

The FID's start is the magnetization built while the current flowed, $1-e^{-t_p/T_1}$. Measure T₁ runs the cycle polarizing for 0.2 to 4.7 s, the board's range, and fits the starts: $T_1=2.49$ s for tap water's 2.5. Each cycle you run adds its own point. Try Copper sulphate: its ions take the protons' energy far faster, and $T_1$ falls to 0.3 s, so the FID reaches full size within a second of polarizing.

19. The Earth's field to a nanotesla

Listening for 4.8 s, the spectrum's line is 0.37 Hz wide, $1/(\pi T_2^*)$, and its peak falls at 1959.82 Hz: $B=f/42.577$ Hz/μT = 46 030.0 nT, the field that was set. One nanotesla moves the line 0.043 Hz, a tenth of its width, and an FID this clean locates the peak far better than that, so the board reads the field to a few nanotesla: enough to watch it drift through the day. Set the Earth's field to 50 μT, as further north: the peak gives 49 999.8 nT, a little less cleanly, as the tank, still tuned for Claremont, gives the FID only ×5.3 there (experiment 16).

20. A steel tool nearby

Steel near the coil bends the field, so it differs across the bottle: 100 nT here, not 5. The protons' frequencies spread over 4.26 Hz, $T_2'=1/(\pi\Delta f)$ falls from 1.5 s to 75 ms, and the FID dies within a tenth of a second ($T_2^*=72$ ms), its line 4.4 Hz wide. The protons have lost nothing: an echo would bring them back, but the board has no π pulse. The cure is to take the steel away: a steel desk, a car or a radiator nearby kills the signal.

How it works

The Bloch equations. A magnetization $\mathbf M$ in a field $\mathbf B$ precesses and relaxes, $$\frac{d\mathbf M}{dt}=\gamma\,\mathbf M\times\mathbf B-\frac{M_x\hat x+M_y\hat y}{T_2}-\frac{M_z-M_0}{T_1}\,\hat z ,$$ with $\gamma/2\pi=42.577$ MHz/T for protons. The first term turns $\mathbf M$ about $\mathbf B$ at $\omega_0=\gamma B$, clockwise seen from the tip of $\mathbf B$; $T_1$ brings $M_z$ back to its equilibrium $M_0$, and $T_2$ takes away the part across the field. The page integrates them exactly, as rotations about the field of the moment and exponentials. $M$ is in units of $M_0$ (of the magnetization the polarizing field gives, for the board).

The rotating frame. Seen from a frame turning at $\omega$ about $z$, the same equation holds with $\mathbf B$ replaced by $\mathbf B+\boldsymbol\omega/\gamma$: at $\omega=-\gamma B_0$ (the precession's own sense) $B_0$ vanishes. An RF field $B_1\cos\omega t$ along the coil is two fields turning opposite ways; the one turning with the spins stands still in this frame, along $x'$ (or where its phase says), and turns $\mathbf M$ by $\theta=\gamma B_1\tau$ in a pulse of length $\tau$. The other turns at $2\omega$ and averages away (the page leaves it out).

The effective field. With the RF at $f_0+\delta f$, what is left of $B_0$ in its frame is $\Delta B=-2\pi\delta f/\gamma$ along $z$, and $\mathbf M$ turns about $\mathbf B_\mathrm{eff}=B_1\hat x'+\Delta B\hat z$, tilted from $z$ by $\beta$ with $\tan\beta=\gamma B_1/2\pi\delta f$, at $\sqrt{(\gamma B_1)^2+(2\pi\delta f)^2}$. From $z$, $M_z=\cos^2\beta+\sin^2\beta\cos\big(\sqrt{1+(2\pi\delta f/\gamma B_1)^2}\,\theta\big)$.

The signal. A coil along $x$ links a flux proportional to $M_x$, and its EMF is $-d\Phi/dt\propto -dM_x/dt$. With $M_x=M_{x'}\cos\omega t+M_{y'}\sin\omega t$ the fast factor dominates, so the EMF is $\omega_0$ times the transverse magnetization: the signal grows as $\omega_0 M_0\propto B_0^2$ ($M_0\propto B_0$ too). Multiplying it by the reference, $\cos\omega t$ and $\sin\omega t$, and filtering gives $M_{x'}$ and $M_{y'}$: the mixer's outputs. Their length $|M_{xy}|$ is the detector's envelope. Thermal noise is added to the two outputs, independently at each point; the signal-to-noise ratio is the full signal over its r.m.s. Averaging $N$ runs leaves the signal and $1/\sqrt N$ of the noise.

Many spins. The ensemble is 4096 spins whose offsets from $f_0$ follow a Lorentzian of full width $\Delta f$, taken at its quantiles, not at random, so that their sum decays as $e^{-t/T_2'}$ with $T_2'=1/(\pi\Delta f)$, and with $T_2$ as $e^{-t/T_2^*}$, $1/T_2^*=1/T_2+1/T_2'$. The sphere draws a few of them, evenly spread through the line. A finite ensemble never quite cancels: a floor of about 1.5% is left after the FID has gone. The power spectrum $|S(f)|^2$ of $A\,e^{-t/T_2^*}$ is a Lorentzian $1/(\pi T_2^*)$ wide at half height. The pulses act on the spins far out in the Lorentzian's tails less well than on the rest: with 2.5 μs pulses an echo is 0.4% lower than $e^{-2\tau/T_2}$.

The three clocks. After a π/2 pulse the field in the rotating frame is along $z$ for every spin ($\delta\omega_i/\gamma$ of it), so each spin's own part across the field only turns and fades: $|\mathbf M_{\perp i}(t)|=|\mathbf M_{\perp i}(0)|\,e^{-t/T_2}$ whatever $T_1$ and the spread are, and $M_z(t)=1-(1-M_z(0))\,e^{-t/T_1}$. The plot draws three things: the mean of the first over the 4096 spins; $M_z$; and the length of the sum's part, $|\sum_i\mathbf M_{\perp i}|/N$, which for a Lorentzian spread is $e^{-t/T_2^*}$ (to the 0.4% residue of a finite ensemble). The 2.5 μs pulse tips the line's far tails a little less: just after it the spins' mean part across is 0.9994, and $M_z$ is 0.0009. By the triangle inequality the sum's part is never longer than the mean of the spins' own; an echo can bring it back up to that and no further, so $T_2^*\le T_2$. Each time is the $\tau$ of $A\,e^{-t/\tau}$ fitted by least squares to its curve from 0.1 ms ($1-M_z$ for $T_1$; for $T_2^*$ down to 3% of the start, past which only the residue is left), shown once the curve has fallen to $1/e$. Why $T_2\le2T_1$: for a spin ½ the relaxation that moves it between ↑ and ↓, at the rates that make $1/T_1$, destroys the coherence between them at half that rate, and anything else that scrambles the phase only adds to it: $1/T_2=1/(2T_1)+1/T_\varphi$. On the page's sphere, whose top is the equilibrium: a state tipped a small $\theta$ from it has $d|\mathbf M|^2/dt=\theta^2(1/T_1-2/T_2)$, which a $T_2$ longer than $2T_1$ would make positive, carrying the state out of the sphere. The page keeps $T_2\le2T_1$ as either is slid. The experiment's clock runs at $v_0\max(1,t/t_0)$, with $t_0=0.1$ ms: a decade of $t$ every 3 s on screen.

The board. While the current flows, the protons relax towards the polarizing field, $B_p\approx170\,B_E$ along the coil; during the switch-off they relax towards the field of the moment. A linear ramp of $B_p$ to zero in $\tau$ turns the total field from the coil's axis to $B_E$, fastest at the end, through its last radian in $\tau B_E/B_p$. The spins follow while the field turns slower than they precess, $A=\gamma B_E^2\tau/B_p\gg1$ (adiabatic); when $A\ll1$ (sudden) they follow only the field's first, slow turning and are left almost where they were: tilted 8.7° towards $B_E$ at the board's 0.4 ms, 98.9% across, and 2.5° at a tenth of that time. The FID's EMF is the board's design value, 1.199 μV for a litre of water fully polarized at 1.5 A and precessing at 1960 Hz, scaled by the current and the field.

The coil's angle. With the coil's axis $\hat n$ at the angle $\alpha$ to $B_E$ (along $\hat z$), $\hat n=\hat z\cos\alpha+\hat e\sin\alpha$, the polarizing field and so $\mathbf M$ lie along $\hat n$. Switched off suddenly, the part along the field, $M\cos\alpha$, stays and the part across, $M\sin\alpha$, goes round $B_E$: $\mathbf M(t)=M\cos\alpha\,\hat z+M\sin\alpha\,(\hat e\cos\omega_0t-\hat f\sin\omega_0t)$, with $\hat f$ square to both. By reciprocity the coil's flux is proportional to $\mathbf M\cdot\hat n=M\cos^2\alpha+M\sin^2\alpha\cos\omega_0t$, so its EMF, $-d\Phi/dt$, has the amplitude $\omega_0M\sin^2\alpha$: one $\sin\alpha$ for what precesses, one for what the coil sees of it. For a coil lying level at the compass bearing $\varphi$ from magnetic north, where the field dips by $I$, $\cos\alpha=\cos I\cos\varphi$: east–west is square to the field for any $I$; north–south gives $\sin^2I$ (0.733 in Claremont, $I$ = 58.9°), upright $\cos^2I$. With the angle, the field during the switch-off turns from $\hat n$ to $\hat z$, its last radian in $\tau B_E/(B_p\sin\alpha)$.

The tank. The EMF is induced in the coil itself, in series with its inductance $L$ (26.8 mH) and resistance $R$ (8.86 Ω); the bank's capacitance $C$ is across the coil, and the preamplifier reads the voltage on it. So the tank multiplies the EMF by $$H(f)=\frac{1}{1-\omega^2LC+i\omega RC},$$ 1 at low frequencies, $Q=\omega_TL/R$ with the phase −90° at its resonance $f_T=1/2\pi\sqrt{LC}$, and nothing, at −180°, far above; $|H|^2$ is $f_T/Q=R/2\pi L$ = 52.6 Hz wide at half height, whatever $C$ is. It answers in $2Q/\omega_T$ = 6 ms, far quicker than the FID changes, so the FID on the tank is the EMF times $H(f_0)$. The bank is the clamp's 0.39 nF and eight switches of 3.3 to 420 nF in binary steps of 3.3 nF; the handbook tunes it by $C=1/4\pi^2f_0^2L$ and the nearest sum. The receiver then multiplies by 101 (the instrumentation amplifier), filters with a fourth-order high-pass at 0.7 kHz and a second-order low-pass at 4.1 kHz (0.97 at 1960 Hz), and multiplies by 201 (GAIN at HIGH). These are the board's design numbers, and the page's tests check them against its design arithmetic.

The noise. The coil's resistance has thermal noise $\sqrt{4kTR}$ per √Hz in series with the EMF, so on the tank it is $\sqrt{4kTR}\,|H(f)|$: the tank's own shape, 14 nV/√Hz at the top. Integrated over all frequencies, $\int_0^\infty4kTR|H|^2df=kT/C$: the equipartition theorem's $\tfrac12C\langle V^2\rangle=\tfrac12kT$, 0.128 μV for 247 nF, whatever $R$ is (the bias resistors' current noise through the tank adds 0.6%). The preamplifier's own 9.8 nV/√Hz is flat, so over the receiver's band it gives 0.61 μV on the tank's scale, 12.3 of the 12.6 mV r.m.s. at OUT. Near $f_T$ the coil's noise is the larger, and the receiver adds only 1.7 dB to it: the signal-to-noise ratio is then the coil's own, which no better amplifier could improve. The page draws OUT at 48 kHz for the 10 ms plot and for Hear it: the FID times the tank's and the filters' gain and phase at $f_0$, plus white noise through a digital copy of the tank (the coil's) and alone (the preamplifier's), both through the filters, times the gain. For what is read in a narrow band at $f_0$ (the spectrum, the FID's start, the $T_1$ scan) it adds noise of the density at $f_0$ to the record.

Spin ½. For one proton the arrow is the expectation value $\langle\mathbf S\rangle$ (times $2/\hbar$) of the state $\cos\frac\theta2\,\text{|↑⟩}+e^{i\varphi}\sin\frac\theta2\,\text{|↓⟩}$: every pure state of a spin ½ is a point on the sphere. Its Hamiltonian $H=-\gamma\mathbf S\cdot\mathbf B$ is linear in $\mathbf S$, and the components of $\mathbf S$ close under commutation, so Ehrenfest's theorem gives $d\langle\mathbf S\rangle/dt=\gamma\langle\mathbf S\rangle\times\mathbf B$ exactly: the quantum spin's arrow moves just as a classical magnetization does, and so does any sum of them. A sum shorter than one spin is a mixed state.

New in 2026 for BrainFlux. The Earth's-field board is Jason Gallicchio's (github.com/gallicchio/earth-nmr-pcb). Source: github.com/gallicchio/brainflux, MIT licence.