# Spin precession vs alignment in magnetic field

1. Aug 3, 2014

### 43arcsec

The spin of an arbitrarily oriented electron precesses in the presence of a magnetic field (see Feyman lectures 10-7). The energy states also split due to magnetic fields (see Feynman lectures 12-4). I am trying to understand how to think about spin precession, spin flipping, and spin alignment.

Let's say we are near absolute zero with 0 magnetic field and a proton is spinning in an arbitrary direction. Then a magnetic field is turned on making a -45 degree angle with the proton's spin. This proton should begin to precess around its axis. The magnetic field however, has created 2 energy levels, one lower energy level where the spin is oriented parallel to the magnetic field, and a higher energy level antiparallel. How do these energy levels relate to the proton spinning and precessing at the -45 degrees?

If some energy is withdrawn from the system by lowering the temperature, then the proton should want to flip to the lower energy level. So, does the proton flip to +45 degrees and continue to precess or does it align directly with the magnetic field or something else? If it does flip, or anything other than precess, than what is the quantum mechanical description of the spin of the proton changing its orientation?

2. Aug 4, 2014

### king vitamin

What is the level of your understanding of quantum mechanics? I'll write the following assuming that you've taken a semester of undergrad QM, but if you're at a lower level, it might be hard to understand.

Let's just work at zero temperature and talk about adding/subtracting energy since thermal effects are complicated. If a proton is prepared with its spin along a certain axis, and a magnetic field is turned on along a 45 degree angle from this axis, the resulting state is not in a stationary energy state. The two stationary energy states (which split from being degenerate, as you say), are the state with the spin pointed along the magnetic field, and the state with the spin anti-aligned with the magnetic field. Assuming a Zeeman coupling, H = B.S, the proton will initially be in a superposition of these two energy states, and this superposition will change in time as the spin "precesses." The average energy will be conserved unless you take energy out. The precession is also a quantum effect, and it's really the average (or expectation value) of spin which is precessing.

3. Aug 4, 2014

### Jilang

43arcsec, hello and welcome to the forum. I had this same question, is the force enough to make the spin flip? I never really go the bottom of it but was left with the impression that the energy levels were not that large in the scheme of things quantum.

4. Aug 4, 2014

### 43arcsec

Hi King Vitamin, and thanks for your reply. I am certainly no quantum expert, but I can get through most of the math. I probably didn't state my question very well, I'm finding a well asked question is harder than it would seem. Let me see if I can parse this out a little better, probably more for me to crystallize my own misunderstanding than it is for you to understand it.

Let's say we have prepared a proton with its spin aligned with a magnetic field, and let's say we have chosen two base states, |1> aligned with the field, and |0> anti-aligned. It is in a stationary state which is precisely base state |1> and left unperturbed, it will stay like that forever. Well, for fun, we apply an electric field which is tuned to exactly kick it over 90 degrees so now its spin is exactly perpendicular to the magnetic field. At this point, the proton's spin begins to precess in the plan perpendicular to the magnetic field, and its state can be described as a superposition a|0>+b|1>. So, let me stop for a question:

Base state |1> is a lower energy than base state |0>, so given that the proton is precessing exactly perpendicular to the magnetic field, are the coefficients a and b the same or because the energy of base state |1> is lower, is b greater than a?

Ok, so let's have some more fun: we double the magnetic field. We know this increases the energy, E=mu B, but there should be no flipping or alignment, instead from what I can tell, all this does is increase the precession frequency: 2 mu Bz / hbar (Feynam lectures section 10-7). So let me stop for another question:

After turning up the magnetic field, can the proton still be described with the same state, a|0>+b|1>?

What I am getting at here is, I don't understand how a magnetic field aligns proton spins. Thanks again for looking.

5. Aug 4, 2014

### 43arcsec

Hi Jilang, thanks for your reply. I am glad someone else had the same question. As you can see from my post to King Vitamin, I am still at a loss to answer your question. As best I can tell, magnetic fields do not flip or align proton spins. Hopefully King Vitamin or someone on Physics Forum will be able to straighten us out.

6. Aug 8, 2014

### king vitamin

Sorry for the late reply 43arcsec, hopefully you still see this.

Let's begin with the proton aligned with the B field along the z-axis, what you call |1>. The -z-axis is what you call |0>.

Then, you kick the proton into an eigenstate of S_x, so let's just take the new state to be (|0>+|1>)/sqrt(2). In this simplified problem, this is just a coordinate choice. Now, the time dependence of the wavefunction (assuming the hamiltonian H = μBS_z) is given by

$$\frac{1}{\sqrt{2}}\left( | 0 \rangle e^{i \omega t} + | 1 \rangle e^{- i \omega t} \right)$$

where ω = μBS/hbar (S is the spin magnitude). You can now verify that for all times, <S_z> = 0, <S_x> = Scos(2ωt), and <S_y> = Ssin(2ωt), so that the expectation value of the spin rotates around the S_z axis with frequency 2ω. Increasing the magnetic field just increases the frequency here (if the initial state weren't perpendicular to the field, this wouldn't be true). <H> = 0 before and after increasing the magnetic field since the spin is perpendicular to it.

Trying to flip the initial spin to, say, 45 degrees might clarify some issues with how increasing B changes the dynamics.

7. Aug 8, 2014

### Jilang

King Vitamin, thanks for your nice explanation. Does changing the temperature make any difference at all?

8. Aug 8, 2014

### WannabeNewton

It changes the statistical distribution of spin orientations in the usual way.

9. Aug 8, 2014

### 43arcsec

Hi King Vitamin, and no worry at all regarding the delay. I'm just delighted there are other people out there who understand this stuff and are willing to share what they know. Thanks.

Forgive me if I'm being dense, but getting back to my last question about the proton precessing in the plane perpendicular to the magnetic field. You chose the state to be:

[tex] \frac{1/\sqrt(2)} (|0>+|1>)[\tex]

This implies that the 2 states |0> and |1> are equally likely, which makes some sense because the spin is exactly half way between the two. What is troubling me is that because |0> is a higher energy than |1>, is it still correct to say their supposition state has equal contributions from the two? I'd think that state |1> should have a greater weight in the supposition state.

As for alignment, maybe you can just help me qualitatively before I get lost in the details. Almost every MRI page you look at says you start by turning on a magnetic field and the protons line up. Well, from your answer and what I've read in Feyman, there is no "lining up", but instead there is precession. Are the people who describe MRI missing something or is it I?

Thanks again for your insights, I really appreciate it.

10. Aug 8, 2014

### WannabeNewton

I hope you don't mind me answering for now.

The state $|\psi, t_0 \rangle \equiv \frac{1}{\sqrt{2}}(|0 \rangle + |1\rangle)$ is just an initial condition in the spin precession from the Schrodinger equation i.e. it's just an initial state. We have the spin aligned with the $z$ axis initially, that is, it's in an eigenstate of $S_z$, and an electric field is applied which is meant to kick the spin into the plane perpendicular to the $z$ axis. We can always arrange for the electric field to be such that it kicks the spin into the eigenstate $|\psi,t_0 \rangle$ of $S_x$. This is an instantaneous perturbation of the system, which you can imagine as quickly turning on and off the electric field. As such the spin doesn't notice the magnetic field until after it has been kicked. Then $|\psi,t_0 \rangle$ precesses under $i\hbar \partial_t |\psi \rangle = -\vec{\mu} \cdot \vec{B} |\psi \rangle$.

There are two different things you seem to be conflating. In the case of MRI we have an ensemble of spins which are all in eigenstates of $S_z$ so they are randomly oriented either parallel to $z$ or antiparallel to it. We then apply a magnetic field along $z$ and this causes certain spins in the ensemble to flip sign, with temperature playing a role. This is not precession. There cannot be precession when the initial state of the spin is parallel to the applied magnetic field. Some of the spins flip because the system wants to minimize its Helmholtz free energy $F = E - TS$ which corresponds to maximizing entropy $S$ at high temperatures and minimizing average internal energy $E$ at low temperatures.

11. Aug 9, 2014

### king vitamin

WannabeNewton is correct that we assume an instantaneous "kick," so there no previous memory of the initial state. This isn't a bad approximation for modeling MRI experiments.

The difference between what you read in Feynman and famous MRI experiments (and the way an NMR machine functions) has to do with the difference between a simple idealized model and more complex models which accurately capture experiments. If you work with spins at zero temperature, and apply a constant homogeneous magnetic field at an angle to their magnetic moments, they will ideally spin forever around the axis. However, treating physical systems always leads to decay of the spins back into their lowest energy state. These are due to temperature, as WanabeNewton says, and come from considerations from equilibrium stat mech (although I would add that standard MRI/NMR experiments and procedures do involve spin precession - usually driven by a perpendicular oscillating field in addition to the time-independent field). The average decay times are called T1 and T2 for the z-component and perpendicular components respectively, and these are called spin-lattice relaxation and spin-spin relaxation for reasons that I don't understand (historical?). You might be interested in the Wiki article on NMR relaxation for more info: http://en.wikipedia.org/wiki/Relaxation_(NMR [Broken])

There's another decay rate called T2* due to magnetic field inhomogeneities across several spins, but I'll refrain from saying much more since this is already a lot of info.

Last edited by a moderator: May 6, 2017
12. Aug 9, 2014

### 43arcsec

King and Wannabe, thanks so much for your replies, I believe I am finally getting somewhere. Please tell me if this understanding is correct:

While spin-lattice relaxation (as measured with the T1 time) usually applies to the time it takes after the radio frequency bump for the spin to recover to its thermodynamically preferred z-aligned state (measured at 1-1/e the way there), isn't it also true that the exact same mechanism is responsible for the alignment of the spins after the magnetic field is first turned on?

So instead of beginning the MRI discussion as, we turn on a magnetic field and the proton spins line up with the magnetic field; we should say, to be more complete, we turn on a magnetic field and any proton whose spin is not already aligned with the magnetic field undergoes spin-lattice relaxation that aligns the proton's spin with the magnetic field.

However, Wannable wrote,

which still gives me some doubt. Why would the ensemble all be in one eigenstate or another before the magnetic field is applied.