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I NMR: RF-pulse response away from equilibrium

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  1. Jan 19, 2017 #1
    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. jcsd
  3. Jan 19, 2017 #2
    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.
     
  4. Jan 20, 2017 #3
    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
  5. Jan 20, 2017 #4
    Thank you very much for this perfect answer odietrich! :D
     
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