3
$\begingroup$

enter image description here

According to Dr James Keeler of Cambridge University in one of his lectures at Queensland University, the angle theta can be any angle, but I struggle to see how this is true. There are only a certain number of I values that nuclei take and as illustrated in the figure above, each magnetic quantum number corresponds to a fixed theta. Is my logic flawed?

$\endgroup$
3
  • 1
    $\begingroup$ It depends whether you are talking about a single spin or a collection. $\endgroup$ – Buck Thorn Jan 11 '20 at 12:18
  • $\begingroup$ a single spin. I'd be interested to know about the collection case too. $\endgroup$ – TSA Jan 11 '20 at 12:32
  • $\begingroup$ As an aside, the coordinate system you drew is left-handed. There's nothing intrinsically wrong with that, but conventionally people work with right-handed systems. $\endgroup$ – orthocresol Jan 11 '20 at 14:57
5
$\begingroup$

You are correct that a given value of $m_I$ and $I$ does correspond to one single value of the angle $\theta$. For the most commonly used nuclei in NMR spectroscopy, the "allowed" values are $I = 1/2$ and $m_I = \pm 1/2$. The two states are commonly labelled $\alpha$ and $\beta$; these are the eigenstates of the angular momentum projection operator $\hat{I}_z$, such that

$$\begin{align} \hat{I}_z|\alpha\rangle &= \frac{1}{2}|\alpha\rangle \\ \hat{I}_z|\beta\rangle &= -\frac{1}{2}|\beta\rangle \end{align}$$

(treating $\hbar \equiv 1$). The problem is that you are assuming that the wavefunction (of a spin) can only be in an eigenstate. This is not true. It is perfectly valid, and indeed necessary (see below), to have states that are superpositions of eigenstates. In general, a spin can have any state

$$|\psi\rangle = c_\alpha |\alpha\rangle + c_\beta |\beta\rangle$$

where $|c_\alpha|^2 + |c_\beta|^2 = 1$ (normalisation), and in general the coefficients are time-dependent (see below). This is no longer an eigenstate of $\hat{I}_z$, and consequently is no longer well-described by one value of $m_I$, or for that matter, $\theta$. But that is not intrinsically a problem. The only circumstance in which a wavefunction must be described by an eigenstate is immediately following an observation, where the wavefunction $|\psi\rangle$ collapses to the eigenstate $|i\rangle$ with probability $|c_i|^2$. Otherwise, superposition states are entirely consistent with quantum mechanics.


A more generalised example is to think of the time-independent Schrödinger equation:

$$\hat{H}|\psi\rangle = E|\psi\rangle$$

which is an eigenvalue equation for the Hamiltonian, $\hat{H}$. It is true that only eigenstates of the Hamiltonian are admissible solutions to this equation. But that does not mean that the only physically permissible states for the system are eigenstates of the Hamiltonian. The time-independent SE only governs systems that are not evolving in time, or stationary states. We need something else to describe systems that do evolve with time, which is the time-dependent Schrödinger equation:

$$\hat{H}|\Psi\rangle = i\hbar\frac{\partial |\Psi\rangle}{\partial t}$$

The most general solution to the TDSE is, indeed, a superposition of stationary states (the Internet has lots of more thorough explanations, or read Griffiths' Introduction to Quantum Mechanics):

$$|\Psi\rangle = \sum_i c_i \exp\left(-\frac{iE_it}{\hbar}\right) |\psi_i\rangle $$

where the $c_i$'s are time-independent. Indeed, it is necessary for us to have superposition states, because this allows the coefficients of each eigenstate to evolve in a time-dependent fashion (because of the exponential term, which evolves at different frequencies for different eigenstates). If a system is prepared in an eigenstate of the Hamiltonian, then it can never evolve into another eigenstate (the exponential evolves, but if there is only one eigenstate, it is no more than a simple phase factor).

This might not seem immediately relevant for NMR, but recall that the (free) Hamiltonian in NMR is simply $-\gamma B_0 \hat{I}_z$. So, the eigenstates of $\hat{I}_z$, $|\alpha\rangle$ and $|\beta\rangle$, are also the eigenstates of the Hamiltonian. If we never had superposition eigenstates, then the spins would simply sit there forever and do nothing.*


* As long as the Hamiltonian doesn't change. But if you change the Hamiltonian, say from $\hat{H}_0$ to $\hat{H}_1$, then the eigenstates change, and so your system (which was originally an eigenstate of $\hat{H}_0$ is no longer an eigenstate of $\hat{H}_1$). Indeed, it would be in a superposition of eigenstates of $\hat{H}_1$, and would evolve in time according to the TDSE, so it doesn't invalidate any of the preceding discussion.

$\endgroup$
4
  • $\begingroup$ I think your answer (and mine) might benefit from inclusion of a description of how the observed angle $\theta$ depends on the quantum state (or coherence). $\endgroup$ – Buck Thorn Jan 11 '20 at 15:29
  • 1
    $\begingroup$ @BuckThorn, I considered going into it but felt too lazy...! Imo it would require describing density matrices, Pauli matrices and the Bloch sphere (I may be overcomplicating it, of course). NMR theory uses these tools regularly to describe bulk magnetisation, less so for a single spin. But quantum computing uses Bloch spheres for single spins. Maybe if I have time... $\endgroup$ – orthocresol Jan 11 '20 at 15:33
  • $\begingroup$ @orthocresol I appreciate the technical correctness of your response, but I must admit that I am not particularly au fait with QM. From what I understand of your answer, the angle theta is constant for an angular momentum vector of a spin with a certain magnetic quantum number only when it's observed. When not being observed it can take any value of theta. This is essentially the Copenhagen interpretation. Is this correct? The bulk magnetization vector can take many values of theta when observed as it's an average of all the thetas of each individual spin. Is this also correct? $\endgroup$ – TSA Jan 12 '20 at 16:54
  • $\begingroup$ @ETS, that’s a pretty accurate summary. $\endgroup$ – orthocresol Jan 12 '20 at 19:43
0
$\begingroup$

The magnetization is composed of the sum of dipole moments associated with individual spins:

$$\vec{M}=\sum_i \vec{\mu}_i$$

and the component along the magnetizing field (assumed to lie along the z axis) is

$$\begin{align}M_z&= \left( \sum_i \vec{\mu}_i\right)_z \\ &= \sum_i \mu_{iz}\end{align}$$

Since the individual dipole moment components along z take on discrete values (are quantized, ie for the homonuclear case $\vec{\mu}_{iz}=\gamma\hbar m_z $) we can write

$$M_z = \gamma\hbar\sum_i m_{i} \tag{1}$$

where the symbols have their usual meanings. Since the sum extends over a large number of spins $i$, the resulting magnetization can for practical purposes be regarded as taking on a continuous range of values, even if the z-component of individual spins is discretely valued.

$\endgroup$
10
  • $\begingroup$ You are correct, except that individual spins need not always take quantised values of $m_z$. The bulk magnetisation is a valuable point to bring up though, especially in the context of NMR. $\endgroup$ – orthocresol Jan 11 '20 at 15:07
  • $\begingroup$ @orthocresol in the presence of an external field in an NMR experiment the z component of each spin is assumed quantized, as far as I am aware. $\endgroup$ – Buck Thorn Jan 11 '20 at 15:07
  • $\begingroup$ I've gone into more detail about that in my answer; it's not true that individual spins must be either up or down. This is something that is always taught at the introductory level (probably to make things easier), but is inconsistent with QM, where superposition is a key concept. $\endgroup$ – orthocresol Jan 11 '20 at 15:09
  • $\begingroup$ @orthocresol that's a good point but it brings to mind the confusing distinction between superpositions and coherences. In general measurement is associated with quantization along a measurement axis. Note that since I limit discussion to $M_z$ afaik equation 1 is general. Also, I am trying to explain how an observable macroscopic property could have a continuum of values while made up of discrete entities, but thanks for bringing this to my attention, I will have to reread my copy of Cavanagh. $\endgroup$ – Buck Thorn Jan 11 '20 at 15:26
  • $\begingroup$ "how an observable macroscopic property could have a continuum of values" I understand, and that point is entirely valid (hence why I said it's valuable!). It's just that OP is likely confused over the single-spin case. Modern NMR textbooks, like that of Keeler (and also Levitt's Spin Dynamics), point out that superpositions are allowed and that individual spins only being in up / down states is mostly a myth. I'm not sure if Cavanagh mentions it (I've only perused small sections of the book), but I wouldn't be surprised if he does, too. $\endgroup$ – orthocresol Jan 11 '20 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.