# I NMR: RF-pulse response away from equilibrium

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1. Jan 19, 2017

### Nogniks

Hello everybody,

I've got a gap in my understanding regarding NMR systems, and I hope you can help: if the magnetisation of a proton is M0 in the Mz direction initially, a 90ox-RF-pulse will change the magnetisation to M0 in the My direction immediately after the pulse. But what happens when a 90ox-RF-pulse is applied to a system which, at the beginning of the pulse, is not in the equilibrium state (for example in a state (Mx,My,Mz) = (a,b,c)) - what will happen? As far as my understanding reaches, Mx=a won't change, but what happens to Mz and My?

I hope you can help, and thanks in advance!

Regards,
Thijs

2. Jan 19, 2017

### Jilang

Just a guess....M_z would change form as a multiple of a X b and M _y as a multiple of c X a. There might be a factor of a function of a,b and c in the denominator.

3. Jan 20, 2017

### odietrich

A $90°_x$ RF pulse acts on the magnetization $\mathbf{M}=(M_x,M_y,M_z)^T$ as a $90°$ rotation matrix $\mathsf{R}$ (around the $x$ axis) with $$\mathsf{R}=\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix}$$ i. e. the magnetization vector is rotated by $90°$ around the $x$ axis. The resulting magnetization after the RF pulse is the product $\mathsf{R} \mathbf{M}$. In your example, this is $\mathbf{M}=(a,-c,b)$.

(Cf. the article "Extended phase graphs: Dephasing, RF pulses, and echoes - pure and simple" by M. Weigel, http://dx.doi.org/10.1002/jmri.24619 [Broken].)

Last edited by a moderator: May 8, 2017
4. Jan 20, 2017

### Nogniks

Thank you very much for this perfect answer odietrich! :D