Phase-Locking Proton Precession with 90° RF at Lamour Frequency

[amylase]

As title. Thank you.

In the context of MRI, why/how does applying 90 degree radiofrequency at Lamour frequency put proton precessions into phase?

I understand 90 degree RF minimises net magnetic moment, such that there is least difference between number of protons in parallel (with external field) and number of protons in antiparallel alignment (ie. those at higher energy states, and are against external field).

The 90 degree RF also puts proton spins into phase. Why / how does the 90 degree RF wave manage to put protons into phase please?

Thank you.

 

[K^2]

Let me use the classical picture to illustrate. So think of each spin as a magnetized gyro. Quantum picture ends up working almost exactly the same, but the algebra is a bit more complex.

You have a B field from the main magnet. Most of the proton spins align with that external magnetic field. Now you apply a transverse field that rotates with Lamour frequency. Let's go into a coordinate system that rotates with the applied transverse field. What does the proton experience in that frame of reference? Why, it experiences just the transverse field, which now has a fixed direction perpendicular to the spin. The spin, naturally, begins to precess around that transverse field. The length of the 90° pulse is such that the spins complete exactly 1/4 of revolution around that transverse field.

But now, in the rotating frame, every proton was precessing around the same transverse field with exactly the same phase. So they all ended up at the exactly the same angle, which is perpendicular to both the transverse field and the magnet's constant field at the moment 90° pulse cut out. So all of the spins ended up being in the exact same phase.

Of course, not quite all of the spins started out parallel to the B field, due to finite temperature, so not quite all of the spins will be in phase after 90° pulse either. This is one of many factors responsible for finite peak widths.