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In NMR we apply a 90 degree radio frequency pulse to the sample. This causes magnetization vector to fall into transverse plane. Then the relaxation of initial phase coherence occurs. Why are the spins initially coherent?

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The spins are originally all aligned because they are all aligned with the magnetic field. So when you rotate all the spins 90º they are still all aligned but now at 90º to the magnetic field.

The spins lose coherence because they precess in the transverse plane at different rates so while they start aligned they become effectively randomly distributed in the transverse plane.

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I know this is kinda old but I find the answer of John is actually contradicting what I learned. My intuition was as well, that the loss of phase coherence is due to spins precessing around $B_0$ at different speeds. However, why would it then be, that large molecules like proteins or polymers show quicker $T_2$-relaxation, which is observable in broad lines? The common explanation for that is that big molecules have a slower tumbling motion, which in turn causes a broader range of frequencies emitted that are utilized by the dephasing relaxation mechanism (see fig. 8-1).

Now, this is only the why is coherence lost part and doesn't answer the original question of why phase coherence occurs in the first place. First and foremost, I can't tell you much more than 'because of quantum mechanics'. I can tell you however of why to my knowledge, John's answer is not only oversimplified, but wrong: Spin-$\frac{1}{2}$ particles (e.g. protons, the most common nucleus used for NMR) have two, and only two options of orienting themselves in a magnetic field. You do change the field slightly by your pulse, but your $B_0$ is way stronger and not turned off during the pulse. The only way I know to turn off $B_0$ is quenching, and for technical reasons it seems impossible to do otherwise. So your spin, being forced into the two states, can only ever change between those, when EM radiation of suitable frequency comes along. Dictated by quantum mechanics, you cannot just rotate a spin by an arbitrary angle. There is flip the spin or nothing at all. The one and only way that John is right under these circumstances (and then still misleading for calling it a rotation), is when the angle between the spins and $B_0$ is $45^\circ$. I cannot find anything on this angle right now, but I am pretty confident that it's not $45^\circ$.

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