How does magentic field change spin values of a proton or electron? [closed]








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I am asking this in context of NMR. Firstly I wanna say that I thought that spin values of a proton or electrons were intrinsic. I didn't know that you can change spin states from one value to another. But now I read that this is not the case. So my question is that I am reading that prior to application of magnetic field, spin values of proton are equal. However when magnetic field is applied, the of the two spin values change? How does this happen? I am not advanced in physics so if anyone can explain this in detail and in layman terms it would be great, Thanks. 




The value of the spin is not changed. A proton is a spin-1/2 particle, whether it is in a magnetic field or not.


However, all spin-1/2 particles have so-called "spin up" and "spin down" states. The thing that is changed is the *energies* of these up and down spin states. You may wish to think of the hydrogen nucleus as a bar magnet in a magnetic field. The case where the magnet is aligned with the field is more stable; when the magnet is opposed to the field it is less stable. For example, a compass pointing North is more stable. If you reach inside and turn the compass needle around 180 degrees so that it points South, you need to input energy to hold the needle in that orientation; that's because that orientation is less stable.


This analogy should be described very thoroughly in various places on the Internet. For example, Chemguide has an [entire page on it](http://www.chemguide.co.uk/analysis/nmr/background.html#top), replete with diagrams.


(Disclaimer: quantum mechanically a proton does *not* necessarily have to be either in the spin-up or spin-down state. For more details, consult an NMR textbook. The difference in energy also can be explained by QM and is due to the presence of a magnetic term in the Hamiltonian: $\hat{H} = -\hat{\mu}\_z B\_0 = -\gamma \hat{I}\_{\!z} B\_0$ where $\gamma$ is the gyromagnetic ratio, $\hat{I}\_{\!z}$ the projection of the nuclear spin along the $z$-axis, and $B\_0$ the strength of the external magnetic field. Therefore, the "up" state which has $\hat{I}\_{\!z}|\alpha\rangle = +\hbar/2$ has a magnetic energy of $E = \langle\alpha|\hat{H}|\alpha\rangle = -\gamma B\_0 \hbar/2$.)


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