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

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, 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 $E = \langle\alpha|\hat{H}|\alpha\rangle = -\gamma B_0 \hbar/2$.)

• Okay so you are saying that spin values are intrinsic. So a particular proton that is -1/2 will stay -1/2 forever except that within it there is a spin up and spin down. I always thought spin up was +1/2 and spin down was -1/2. I didn't know that within each there is spin up and spin down – TLo Feb 12 '17 at 0:55
• A proton has a spin of $1/2$. Positive $1/2$, not $\pm 1/2$! However, this spin can "point" in different directions: so-called up and so-called down. The proper terminology is that the projection of the spin can assume the values $+1/2$ ("up") and $-1/2$ ("down"). However, no matter which way the spin is pointing, its magnitude (i.e. its spin) is always $1/2$. This magnitude is invariant. – orthocresol Feb 12 '17 at 0:57
• Oh okay. That's so weird. I always assumed by -1/2 and +1/2 protons and electrons come in two types. So I just wanted to ask, I read often that spin states are quantized and that u can only have up and down states and not take any values between those up and down. But then other time I read that without magnetic field protons and electrons are spinning at randon and I see arrows depicted not only as up and down but also sideways, making it seem that u can have range of values from up and down. What are u supposed of make of this? – TLo Feb 12 '17 at 1:13
• It may perhaps not be a very good idea to read too much into it, because it can lead to misconceptions. It is a fundamental aspect of QM: before you measure the state of a particle, you cannot know whether it is spin up or spin down. It is best described by a certain percentage of up and down: for example, a proton might be 45% up and 55% down at a certain point of time. However, if you try to measure it, then you can only measure either up and down. You'd have a 45% chance of measuring up, and a 55% chance of measuring down. That's the real meaning of quantisation. – orthocresol Feb 12 '17 at 1:25
• That first sentence is found in typical introductory NMR explanations, but does not make any sense from a QM perspective. There is no difference between the states that a proton nucleus is allowed to adopt, with or without a magnetic field. All linear combinations of $|\alpha\rangle$ and $|\beta\rangle$, i.e. $|\psi\rangle=c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, are allowed states. The only difference is that under a magnetic field, the ensemble average of $|c_\beta|^2$ is slightly different from that of $|c_\alpha|^2$. Whether it is bigger or smaller depends on the sign of $\gamma$. – orthocresol Feb 12 '17 at 3:25