NMR: RF-pulse response away from equilibrium

In summary, a 90° RF pulse applied to a system in a non-equilibrium state will rotate the magnetization vector by 90° around the x axis, resulting in a new magnetization vector given by the product of the rotation matrix and the initial magnetization vector. This is demonstrated in an example of a system with initial magnetization (Mx, My, Mz) = (a, b, c).
  • #1
Nogniks
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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
 
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  • #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.
 
  • #3
Nogniks said:
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?

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 .)
 
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Likes Nogniks and DrClaude
  • #4
Thank you very much for this perfect answer odietrich! :D
 

Related to NMR: RF-pulse response away from equilibrium

1. What is NMR and how does it work?

NMR stands for nuclear magnetic resonance, and it is a technique used to study the structure and behavior of molecules. It works by applying a strong magnetic field to a sample, which causes the nuclei of certain atoms to align with the field. Radiofrequency pulses are then used to excite the nuclei, and the resulting signals are analyzed to obtain information about the sample's molecular structure.

2. What is the purpose of using RF-pulses in NMR?

RF-pulses are used in NMR to manipulate the orientation of nuclei and induce transitions between energy levels. This allows for the analysis of various properties of the sample, such as chemical composition, molecular structure, and dynamics.

3. How does the RF-pulse response differ between equilibrium and non-equilibrium conditions?

In equilibrium conditions, the RF-pulse response is described by the well-known Bloch equations, which govern the behavior of nuclear spins in a steady magnetic field. In non-equilibrium conditions, such as when the sample is subjected to a sudden change in magnetic field, the RF-pulse response is more complex and can provide information about the sample's relaxation properties.

4. What factors can affect the RF-pulse response away from equilibrium?

The RF-pulse response away from equilibrium can be affected by various factors, such as the strength and duration of the RF-pulse, the strength of the magnetic field, and the properties of the sample, including its molecular structure and chemical environment.

5. How is the RF-pulse response away from equilibrium used in NMR studies?

The RF-pulse response away from equilibrium is often used to study the relaxation properties of molecules, which can provide valuable information about their dynamic behavior. This can be useful in a variety of applications, including drug discovery, materials science, and medical diagnostics.

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