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In descriptions of the spin echo measurement sequence, it says

  • first a $90°$ pulse turns the magnetization sum vector into the transversal plane where it rotates with the Larmor frequency

  • spins of different protons are in phase until the pulse is switched of begin to dephase

  • after a short waiting time (TE/2) a $180°$ pulse is sent.

  • the $180°$ pulse turns around the orientation of the precession

  • after waiting for TE/2 time we observe a rephasing of all protons (echo)

I don't understand the previous last point. How does the RF pulse turn around the direction of the precession?

Also: why does the pulse itself not rephase all rotating spins (like the $90°$ pulse did)?

(I assume the $180°$ pulse is simply twice as long as the $90°$ pulse.)

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How does the RF pulse turn around the direction of the precession?

It doesn’t. The precession direction is determined by the B0 field, which is assumed constant before and after the RF pulse. So after the refocusing pulse the spin continues precessing in the same direction and at the same rate as it did before.

The effect of the refocusing pulse is to flip the phase of the precessing spins. So if a spin had precessed in the positive direction to a phase of 34 degrees, then after the refocusing pulse its phase would be -34 degrees. From that point it would continue precessing in the positive direction and at the echo time would reach 0 phase.

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  • $\begingroup$ Ah ok, but I still don't get it completely I'm afraid. 1) Isn't the angle of the mangnetization (which does the precession) the same as the phase angle? 2) I assume right after the 90° pulse there is no phase shift between spins whatsoever (the pulse synchronizes the phases of two neighbouring spins), therefore I would assume that a pulse of twice the length (180° pulse) would also synchronize the phases of two neighbouring spins completely. $\endgroup$
    – mcExchange
    Nov 23, 2021 at 13:11
  • $\begingroup$ So a 180° pulse causes phase shift between two neighbouring spins but 90° pulse does not??? $\endgroup$
    – mcExchange
    Nov 23, 2021 at 13:18
  • $\begingroup$ The phase is the angle in the transverse plane. The x axis is 0 and the y axis is 90 deg phase. The magnetization starts on the z axis. The excitation pulse rotates the magnetization 90 deg about the y axis. That puts all of the magnetization at 0 deg phase in the transverse plane. Then it dephases to e.g. 17 deg phase. The refocusing pulse rotates it 180 deg about the x axis, so the magnetization that was at 17 deg phase is now at -17 deg. $\endgroup$
    – Dale
    Nov 23, 2021 at 14:15
  • $\begingroup$ On an atomic level: a spin has a direction (up or down with respect to Bo). Say two neighbouring spins are pointing up at first. Also they have random phases. Then a 90° HF pulse equalizes the phase of both of them and flips one of them down. Then they they start dephasing. Now comes the 180° pulse: It flips either of the two spins up or down and should also syncronize their phases. How can that operation turn a 17 deg into a - 17 deg phase? (When I get time, I'll draw and post a picture here) $\endgroup$
    – mcExchange
    Nov 29, 2021 at 17:51
  • $\begingroup$ @mcExchange you should stick with the classical view. The quantum view is more complicated and gains you nothing here. I am not sure how you are getting this idea that RF pulses synchronize or equalize pre-existing phase. It isn't correct. Probably you are mixing some classical and some quantum concepts. You need to go one way or the other, and purely classical is better for now. If you want to go pure quantum then all you have is the wavefunction, usually in the spin up-down basis. There are no little arrows pointing in a specific direction with a definite but unmeasured phase etc. $\endgroup$
    – Dale
    Nov 30, 2021 at 15:00

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