My understanding is that there's an initial precession frequency ($\omega_0$) about the direction of magnetization ($z$). Then when you excite the nucleus into the spin state with the radio frequency pulse, the direction of Nucleus's Magnetic Moment rotates orthogonality ($x$) to the pulse direction ($y$) changing the bulk magnetization.

This bulk magnetization is still precessing about the direction of magnetization ($z$), so is the rotating frame just accounting for that precession?

I'm pretty sure the rotation ($\omega_1$) about the direction of pulse magnetization (that is about $x$) doesn't have anything to do with the frequency of the rotating frame. My understanding is that precession only relates to the pulse length.

I'm just not sure what the rotating frame is or how it relates to the measurement of induced current from free induction decay.

Edit: I posted this question in several locations, so here's a summary of the answers I got.

  • The Initial Precession Frequency (ω0) is the Rotating Frame.
  • The Pulse Rotation (ω1) is in the Rotating Frame.
  • I'm still not clear on how Bulk Magnetization develops after a pulse sequence.

1 Answer 1


The rotating frame is, as you say, just to account for the rotation about the $z$ axis. More precisely, it's a choice of frame that "undoes" the rotation. This also allows us to think of the RF magnetic field as simply a pulsed static magnetic field along a particular direction (say $x$). From this, we can easily see how the RF magnetic field simply rotates the magnetization in the $xz$ plane.

This also answers the question regarding $\omega_1$, since it is clear that the precession amount only depends on the pulse length and the strength of the pulsed RF field. $\omega_0$ was already taken into account in transforming to the rotating frame, and in a sense is no longer relevant.

  • $\begingroup$ The precession frequency doesn't change as you rotate the magnetization with an RF pulse. Hence, you can effectively think of every spin rotating in unison at all times, or that there is no effective difference between individual and bulk magnetization. This breaks down when you talk about $T_2$ due to inhomogeneities in the magnet, as individual spins will rotate at slightly different rates because of a slightly different magnetic field in $z$. $\endgroup$
    – Aaron
    Commented Oct 6, 2017 at 2:40
  • $\begingroup$ To be precise, the precession frequency is primarily determined by the specific particle you're looking at. For example, if you're doing proton NMR the precession frequency is determined by the proton's gyromagnetic ratio. This alone does not give any decay or frequency shift; you need additional processes for that. However, the point is that these additional processes make at most miniscule shifts in the precession frequency. Chemical shifts are measurements of frequency shifts, and they're measured in parts per million. $\endgroup$
    – Aaron
    Commented Oct 6, 2017 at 4:11
  • $\begingroup$ "bulk magnetization just decay without any precession in the free induction decay graph" -- if you have a pure water sample, with a good shim, and set the base frequency of the spectrometer to the H20 offset, this is exactly what you'll see -- a simple exponential decay. The FID, however, is typically measured at the resonance of TMS at the magnetic field of the experiment ("0 ppm") ... which is offset from water, so the FID of H20 will look like it precesses, which it does, relative to TMS. $\endgroup$
    – tesch1
    Commented Oct 6, 2017 at 8:12
  • $\begingroup$ $\gamma_n$ is the gyromagnetic ratio of the particle/compound you're studying, and $B$ is the applied magnetic field. Hence, $\gamma_n B$ is the resonance frequency of the sample. $\omega_s$ should be the frequency of your RF field. Hence, $\omega'$ measures the how far off resonance you are. In the example in your text, you calibrate $B$ and $\omega_s$ so that $\omega' = 0$ for TMS. Because of chemical shifts, if you measure water afterwards without adjusting $B$ or $\omega_s$ you will see a precession because of small changes in $\omega_n$. $\endgroup$
    – Aaron
    Commented Oct 6, 2017 at 15:12

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