## Magnetic resonance without magnets: explained

Have you ever done MRI? If yes, you probably remember, it is not the most pleasant experience. MRI is claustrophobic, loud, and takes a long time. The main reason for these problems — high magnetic fields. Would it be cool to avoid them? Why can’t we simply do MRI without magnetic fields?

It turns out, we can do it though it is not that simple…

To understand why high magnetic fields are needed to obtain magnetic resonance images, we need to understand the principles of signal detection and the concept of polarization.

Some atomic nuclei possess a property called spin and, therefore, produce magnetic fields. These nuclei can simply be visualized as tiny magnets (magnetic dipoles) swirling in space as their hosts — molecules — move, rotate or vibrate depending on the physical state of the chemical system they comprise (solid, liquid or gas).

In the absence of external fields, nuclear spin orientations are random:

By placing the spins in high magnetic fields, one can align them in the same direction, creating “polarization”:

The level to which spins are polarized in the picture above is extremely exaggerated. In reality, even in the very high magnetic fields of modern magnetic resonance spectrometers and imaging scanners (thousands of times higher than the Earth’s magnetic field), nuclear spin polarization is only a fraction of percent. Complete (100%) polarization is also possible. However, instead of placing spins in high fields, it is practically more feasible to use a different approach. Physicists and chemists came up with several creative techniques to create extremely high polarization or hyperpolarization. Techniques based on chemical reactions (such as parahydrogen-induced polarization) are my favorite ones since they don’t require magnetic fields at all (in reality, polarization in such techniques can still be magnetic-field-dependent but the requirements are very modest).

Now about detection. Nuclear magnetism can be detected in several ways. I will talk about two ways and both of them are named after a brilliant physicist Michael Faraday.

The first detection technique is the so-called inductive detection. If spins are polarized and brought to the magnetic field perpendicular to the polarization axis, collective nuclear magnetization will start oscillating about the field axis. This phenomenon is called Larmor precession and it is a general result of classical physics (no quantum mechanics is necessary to explain it but quantum mechanics brings more fun). Oscillating magnetization can be picked up by a coil (solenoid) and a voltage across the coil terminals (electromotive force) will be generated according to Faraday’s law of induction:

The problem is that it is really hard to construct magnets that produce high magnetic fields. Moreover, these fields need to be very uniform to be suitable for magnetic resonance purposes. This is why bores of MRI scanners are usually small – this way engineers achieve higher magnetic field homogeneity over the active volume.

The conventional logic of MRI engineers is straightforward: by using higher magnetic fields (1) we achieve higher nuclear spin polarization (indeed, polarization scales linearly with the applied field B0) and (2) we increase sensitivity of the detection (the voltage generated in the coil by Faraday’s law is linearly proportional to the oscillation frequency which, in turn, proportional to B0). Therefore, conventional MRI signal scales as a square of the magnetic field. Higher the field is — bigger the magnet should be, and we already reached the limits of our engineering capabilities to make the highest magnetic fields that are stable and uniform.

Now imagine that we already have high polarization produced without magnetic fields, for example, via chemical reactions mentioned above. Therefore, the signal is only linearly proportional to the magnetic field strength and even low magnetic fields can be used to produce high-quality images.

Recently my students and I played with MRI at the Earth’s magnetic field (compare 50 micro-Tesla vs. several Tesla of conventional MRI scanners). In our experiments, we simply bubbled a gas through a solution containing the required ingredients and picked up the signal by a coil:

The gas that we used is parahydrogen (para-H2) and the enhanced nuclear polarization in solution is the result of Signal Amplification By Reversible Exchange (SABRE) effect. This effect is a remarkable consequence of the interplay between chemical kinetics and nuclear spin dynamics. Unfortunately, explaining it here, in this post, would require many more words and even some math 🙂 But the main point is simple: bubbling the gas through the solution results in high nuclear polarization and, therefore, high MRI contrast. Without bubbling the gas, magnetic nuclei are oriented randomly and the signal is too low to be detected.

This idea of avoiding magnetic fields can be extended even further. Indeed, if we don’t need a magnetic field to generate polarization, can we detect nuclear spin signals completely without magnets? This somewhat weird idea (magnetic resonance without magnets) is, in fact, one of the most outstanding achievements in magnetic resonance of the last decade. It was pioneered by chemists and physicists from UC Berkeley (laboratories of Alex Pines and Dmitry Budker).

It turns out that very weak magnetic signals can be detected by atomic magnetometers. These magnetometers use vapor cells, glass containers filled with the vapor of alkali metal. Atoms of alkali metals have unpaired valence electrons. Since electrons are tiny magnets in the same way as their nuclear brothers, they can also be polarized, this time by lasers. Properties of a laser beam passing through such polarized “electronic gas” can be modulated if electrons are subjected to additional magnetic fields. This is called the Faraday effect and it is a second way of detecting magnetic resonance signals. In a nutshell, electrons “feel” the magnetic fields around them in a very sensitive way, even if these fields are generated by nuclear spins from a sample placed on top of a vapor cell:

The end result of this signal detection scheme is the same as before – the voltage is generated and picked up, this time by a photodetector. If magnetic fields coming from the sample are time-dependent, the voltage is time-dependent as well and it can be converted into a spectrum containing useful chemical information about the sample.

You might wonder where this all can be applied? Well, it is worth mentioning Faraday here again:

I personally feel that MRI without magnets is not only possible but necessary. However, it will obviously not replace the conventional approach but will rather complement it. One may imagine sensitive magnetometers and Earth’s-field MRI applied to study chemicals at a large scale, to investigate processes taking place inside big industrial chemical reactors. Indeed, the Earth is the best magnet for these purposes – it is extremely homogeneous on the scale that we need. Nullifying magnetic fields (to produce zero field) is another option – remarkably, the sensitivity of atomic magnetometers is maximized at the near-zero-field conditions. It is hard to predict the direction that future will take but one thing is clear – magnetic resonance is entering a new era. So you better stay tuned! 😉

https://chemrxiv.org/articles/Zero-Field_Nuclear_Magnetic_Resonance_of_Chemically_Exchanging_Systems/7658372

https://aip.scitation.org/doi/abs/10.1063/1.5003347

https://www.nature.com/articles/nphys1986

https://onlinelibrary.wiley.com/doi/abs/10.1002/cphc.201402607

https://pubs.acs.org/doi/abs/10.1021/ac501638p

https://www.sciencedirect.com/science/article/abs/pii/S1090780717301659

## Dipole-dipole interactions in NMR: explained

Interactions between spins are fundamental for understanding magnetic resonance. One of the most important ones is the magnetic dipole-dipole interaction. Spins act as tiny magnets, thus, they can interact with each other directly through space, pretty much the same way as classical magnetic dipoles (Figure 1). In NMR, dipole-dipole interactions very often determine lineshapes of solid-state samples and relaxation rates of nuclei in the liquid state.

In many NMR textbooks, you can find the following expression for dipole-dipole (DD) Hamiltonian between two spins 1 and 2:

$\hat{{H}}_{\rm DD} = d_{12} \left( 3 \hat{I}_{1z} \hat{I}_{2z} - \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right)$

where $d_{12} = - \displaystyle\frac{\mu_0}{4 \pi} \displaystyle\frac{\gamma_1 \gamma_2 \hbar}{r^3} \frac{\left( 3 \cos^2{\theta} - 1 \right)}{2}$, $\gamma_1$ and $\gamma_2$ are gyromagnetic ratios of the spins, $r$ is the distance between them.  Meaning of the angle $\theta$ can be seen from the Figure 2. But how was this expression derived?

In physics, I rarely struggled with imagining abstract things and concepts, but rather, I was often lazy to do math thoroughly and derive equations from the beginning to the end. That’s why I decided to write here a full derivation, from the beginning to the end, for the Hamiltonian of interacting nuclear spins. We will start from a classical expression of the magnetic field produced by the dipole and finish by the analysis of a truncated dipole-dipole Hamiltonian. Usually, you don’t see such a full derivation in textbooks. Maybe, this is because textbooks have limited space… Luckily, here we do not have such limitations, thus, we can have some fun here! Let’s go then! 🙂

Classical equation describing the magnetic field $\vec{\boldsymbol{B}}_{\rm dip} (\vec{\boldsymbol{r}})$ produced by a magnetic dipole moment $\vec{\boldsymbol{\mu}}$ is

$\vec{\boldsymbol{B}}_{\rm dip} (\vec{\boldsymbol{r}}) = \displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \left( \vec{\boldsymbol{\mu}} \cdot \hat{\boldsymbol{r}} \right) \hat{\boldsymbol{r}} - \vec{\boldsymbol{\mu}} \right)$

where $\hat{\boldsymbol{r}} = \displaystyle\frac{\vec{\boldsymbol{r}}}{|\vec{\boldsymbol{r}}|}$ is a unit vector. It is easy to show that $\left( \vec{\boldsymbol{\mu}} \cdot \vec{\boldsymbol{r}} \right) \vec{\boldsymbol{r}} = \left( \vec{\boldsymbol{r}} \cdot \vec{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}$.

Indeed,

$\left( \vec{\boldsymbol{\mu}} \cdot \vec{\boldsymbol{r}} \right) \vec{\boldsymbol{r}} = \left( \mu_x r_x + \mu_y r_y + \mu_z r_z \right) \cdot \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} = \begin{pmatrix} \mu_{x} r_{x} r_{x} + \mu_{y} r_{y} r_{x} + \mu_{z} r_{z} r_{x} \\ \mu_{x} r_{x} r_{y} + \mu_{y} r_{y} r_{y} + \mu_{z} r_{z} r_{y} \\ \mu_{x} r_{x} r_{z} + \mu_{y} r_{y} r_{z} + \mu_{z} r_{z} r_{z} \end{pmatrix}$

which is the same as

$\left( \vec{\boldsymbol{r}} \cdot \vec{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}} = \begin{pmatrix} r_{x} r_{x} & r_{x} r_{y} & r_{x} r_{z} \\ r_{y} r_{x} & r_{y} r_{y} & r_{y} r_{z} \\ r_{z} r_{x} & r_{z} r_{y} & r_{z} r_{z} \end{pmatrix} \cdot \begin{pmatrix} \mu_x \\ \mu_y \\ \mu_z \end{pmatrix} = \begin{pmatrix} r_{x} r_{x} \mu_{x} + r_{x} r_{y} \mu_{y} + r_{x} r_{z} \mu_{z} \\ r_{y} r_{x} \mu_{x} + r_{y} r_{y} \mu_{y} + r_{y} r_{z} \mu_{z} \\ r_{z} r_{x} \mu_{x} + r_{z} r_{y} \mu_{y} + r_{z} r_{z} \mu_{z} \end{pmatrix}$

So, we write the magnetic field produced by a magnetic dipole $\vec{\boldsymbol{\mu}}_2$ as

$\vec{\boldsymbol{B}}_{\mu_2} = \displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}_2 - \vec{\boldsymbol{\mu}}_2 \right)$

The energy of a magnetic dipole $\vec{\boldsymbol{\mu}}_1$ interacting with the magnetic field $\vec{\boldsymbol{B}}_{\mu_2}$ produced by a magnetic dipole $\vec{\boldsymbol{\mu}}_2$ (dipole-dipole interaction) is therefore

$E_{\rm DD} = - \left( \vec{\boldsymbol{\mu}}_1 \cdot \vec{\boldsymbol{B}}_{\mu_2} \right) = -\displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \cdot \vec{\boldsymbol{\mu}}_1 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}_2 - \left( \vec{\boldsymbol{\mu}}_1 \cdot \vec{\boldsymbol{\mu}}_2 \right) \right)$

The transition from classical to quantum mechanics is realized by substituting the measurable quantities by corresponding quantum mechanical operators:

$E_{\rm DD} \rightarrow \hat{H}_{\rm DD} \quad \vec{\boldsymbol{\mu}}_1 \rightarrow \gamma_1 \hbar \hat{\mathbf{I}}_1\quad \vec{\boldsymbol{\mu}}_2 \rightarrow \gamma_2 \hbar \hat{\mathbf{I}}_2$

$\label{Eq_Hdd} \hat{H}_{\rm DD} = - \displaystyle\frac{\mu_0}{4 \pi } \displaystyle\frac{\gamma_1 \gamma_1 \hbar}{r^3} \left( 3 \cdot \hat{\mathbf{I}}_1 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \hat{\mathbf{I}}_2 - \left( \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right) \right) = b_{12} \hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2$

here $b_{12}$ is a factor which depends only on the types of the nuclear spins and the distance between them, and a tensor of dipole-dipole interactions contains information about the mutual orientation of two spins:

$\hat{\mathbf{D}} = 3 \cdot \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) - \hat{1}$

here $\hat{1}$ is a unit matrix. Note that we write Hamiltonian $\hat{H}_{\rm DD}$ in units of [rad/s], that is why one $\hbar$ is missing. In spherical coordinates:

$\hat{\boldsymbol{r}} = \begin{pmatrix} \sin{\theta} \cos{\phi} \\ \sin{\theta} \sin{\phi} \\ \cos{\theta} \end{pmatrix}$

Therefore,

$\hat{\mathbf{D}} = \begin{pmatrix} 3 \sin^2{\theta} \cos^2{\phi} - 1 & 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin{\theta} \cos{\theta} \cos{\phi} \\ 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin^2{\theta} \sin^2{\phi} - 1 & 3 \sin{\theta} \cos{\theta} \sin{\phi} \\ 3 \sin{\theta} \cos{\theta} \cos{\phi} & 3 \sin{\theta} \cos{\theta} \sin{\phi} & 3 \cos^2{\theta} - 1 \end{pmatrix}$

Looks good, doesn’t it? Now, let’s evaluate the product $\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2$:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \\ \begin{pmatrix} \hat{I}_{1x} & \hat{I}_{1y} & \hat{I}_{1z} \end{pmatrix} \begin{pmatrix} 3 \sin^2{\theta} \cos^2{\phi} - 1 & 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin{\theta} \cos{\theta} \cos{\phi} \\ 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin^2{\theta} \sin^2{\phi} - 1 & 3 \sin{\theta} \cos{\theta} \sin{\phi} \\ 3 \sin{\theta} \cos{\theta} \cos{\phi} & 3 \sin{\theta} \cos{\theta} \sin{\phi} & 3 \cos^2{\theta} - 1 \end{pmatrix} \begin{pmatrix} \hat{I}_{2x} \\ \hat{I}_{2y} \\ \hat{I}_{2z} \end{pmatrix} =$

$\begin{pmatrix} \hat{I}_{1x} & \hat{I}_{1y} & \hat{I}_{1z} \end{pmatrix} \begin{pmatrix} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) + \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) \\ \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) + \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) \\ \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) \end{pmatrix} =$

$= \hat{I}_{1x} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) + \hat{I}_{1x} \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{1x} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + + \hat{I}_{1y} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{1y} \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) + \hat{I}_{1y} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + + \hat{I}_{1z} \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{1z} \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

Let’s color terms to make it easier grouping them:

$\hat{I}_{1x} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) +$ $\hat{I}_{1x} \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) +$ $\hat{I}_{1x} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) +$ $\hat{I}_{1y} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) +$ $\hat{I}_{1y} \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) +$ $\hat{I}_{1y} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) +$ $\hat{I}_{1z} \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{1z} \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) +$ $\hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

Groupling the red terms gives

$\left( \hat{I}_{1x} \hat{I}_{2x} \cos^2{\phi} + \hat{I}_{1y} \hat{I}_{2y} \sin^2{\phi} \right) 3 \sin^2{\theta} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{1y} \hat{I}_{2y} \right)$

Let’s not forget about intrinsic connections of spin angular momentum with raising and lowering operators:

$\hat{I}_{1x} \hat{I}_{2x} = \displaystyle\frac{\left( \hat{I}_{1+} + \hat{I}_{1-} \right)}{2}\displaystyle\frac{\left( \hat{I}_{2+} + \hat{I}_{2-} \right)}{2} = \displaystyle\frac{1}{4} \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right)$

$\hat{I}_{1y} \hat{I}_{2y} = \displaystyle\frac{\left( \hat{I}_{1+} - \hat{I}_{1-} \right)}{2 i}\displaystyle\frac{\left( \hat{I}_{2+} - \hat{I}_{2-} \right)}{2 i} = -\displaystyle\frac{1}{4} \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1+} \hat{I}_{2-} - \hat{I}_{1-} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right)$

Therefore, grouping the red terms gives

$\left( \hat{I}_{1x} \hat{I}_{2x} \cos^2{\phi} + \hat{I}_{1y} \hat{I}_{2y} \sin^2{\phi} \right) 3 \sin^2{\theta} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{1y} \hat{I}_{2y} \right) = \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \frac{3}{4} \sin^2{\theta} + \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right) \frac{3}{4} \sin^2{\theta} \cdot \left( \cos{ 2 \phi} \right) - \frac{1}{2} \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) = \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \frac{1}{4} \left(1 - 3 \cos^2{\theta} \right) + \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right) \frac{3}{4} \sin^2{\theta} \cdot \left( \cos{ 2 \phi} \right)$

Groupling the blue terms gives

$\left( \hat{I}_{1x} \hat{I}_{2y} + \hat{I}_{1y} \hat{I}_{2x} \right) 3 \sin^2{\theta} \cos{\phi} \sin{\phi} = \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)\displaystyle\frac{3}{4} \sin^2{\theta} \cdot \left( - i \sin{2 \phi} \right)$

where we took into consideration that

$\hat{I}_{1x} \hat{I}_{2y} =\displaystyle\frac{\left( \hat{I}_{1+} + \hat{I}_{1-} \right)}{2} \frac{\left( \hat{I}_{2+} - \hat{I}_{2-} \right)}{2 i} = \frac{1}{4 i} \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)$
$\hat{I}_{1y} \hat{I}_{2x} = \displaystyle\frac{\left( \hat{I}_{1+} - \hat{I}_{1-} \right)}{2 i} \frac{\left( \hat{I}_{2+} + \hat{I}_{2-} \right)}{2} = \frac{1}{4 i} \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1+} \hat{I}_{2-} - \hat{I}_{1-} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)$

Red and blue terms can be combined nicely to form

$\left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \displaystyle\frac{1}{4} \left(1 - 3 \cos^2{\theta} \right) + \hat{I}_{1+} \hat{I}_{2+} \left( \frac{3}{4} \sin^2{\theta} \cdot e^{- 2 i \phi} \right) + \hat{I}_{1-} \hat{I}_{2-} \left( \frac{3}{4} \sin^2{\theta} \cdot e^{+ 2 i \phi} \right)$

Now let’s focus on purple terms:

$\left( \hat{I}_{1x} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2x} \right) \left( \hat{I}_{1x} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2x} \right) \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \left( \hat{I}_{1y} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2y} \right) \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) = \left( \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} + \hat{I}_{1z}\hat{I}_{2-} \right) \cos{\phi} \right) \displaystyle\frac{3}{4} \sin{2 \theta} + \left( \left( -i \hat{I}_{1+} \hat{I}_{2z} + i \hat{I}_{1-} \hat{I}_{2z} -i \hat{I}_{1z} \hat{I}_{2+} + i \hat{I}_{1z} \hat{I}_{2-} \right) \sin{\phi} \right) \frac{3}{4} \sin{2 \theta} = \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} \right) \left( \frac{3}{4} \sin{2 \theta} \right) e^{-i \phi} + \left( \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2-} \right) \left( \frac{3}{4} \sin{2 \theta} \right) e^{+i \phi}$

Overall, we have split our dipolar Hamiltonian into 6 term, so-called “Dipolar Alphabet”:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \hat{A} + \hat{B} + \hat{C} + \hat{D} + \hat{E} + \hat{F}$

where

$\hat{A} \quad = \quad \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

$\quad\quad \hat{B} \quad = \quad \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \cdot \displaystyle\frac{\left(1 - 3 \cos^2{\theta} \right)}{4}$

$\quad\quad \hat{C} \quad = \quad \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} \right) \left( \displaystyle\frac{3}{4} \sin{2 \theta} \right) e^{-i \phi}$

$\quad\quad \hat{D} \quad = \quad \left( \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2-} \right) \left( \displaystyle\frac{3}{4} \sin{2 \theta} \right) e^{+i \phi}$

$\hat{E} \quad = \quad \hat{I}_{1+} \hat{I}_{2+} \left( \displaystyle\frac{3}{4} \sin^2{\theta} \right) e^{- 2 i \phi}$

$\quad\quad \hat{F} \quad = \quad \hat{I}_{1-} \hat{I}_{2-} \left( \displaystyle\frac{3}{4} \sin^2{\theta} \right) e^{+ 2 i \phi}$

To summarize, the Hamiltonian of two interacting spins is a $4 \times 4$ matrix composed of 6 operators. Each of the letters of the dipolar alphabet corresponds to certain matrix elements in the final Hamiltonian (Figure 3).

Without an externally imposed direction in space (for example, in the case of two equivalent spins in zero magnetic field), all of the terms of the dipole-dipole Hamiltonian need to be used for calculating an NMR spectrum. This is because all orientations in space are equivalent. However, in the presence of the external high magnetic field, the Hamiltonian can be simplified via the use of so-called “secular approximation”.

The secular approximation concerns the case where the Hamiltonian is the sum of two terms:

$\hat{H} = \hat{A} + \hat{B}$

where $\displaystyle\hat{A}$ is a “large” operator and $\hat{B}$ is a “small” operator. In our case, $\hat{A}$ can be an operator describing the interaction with the magnetic field (Zeeman Hamiltonian) and $\hat{B}$ is DD Hamiltonian. Eigenstates of the Zeeman Hamiltonian are familiar αααβ, βα, ββGenerally, $\hat{B}$ does not commute with $\hat{A}$, therefore, if written in the eigenbasis of $\hat{A}$, it has finite elements everywhere.

The secular approximation for $\hat{B}$ means that we leave only the blocks that correspond to the eigenvalue structure of the operator $\hat{A}$ (Figure 4) and disregard all other elements.

In general, we can omit a matrix element $b_{nm}$ that is much smaller than

$|b_{mn}| \ll |E_m - E_n|$

For homonuclear case (e.g., two interacting protons), this means that only the first two terms of the dipolar Alphabet will survive:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \hat{A} + \hat{B} =$

$= \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) + \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \cdot \displaystyle\frac{\left(1 - 3 \cos^2{\theta} \right)}{4} =$

$= \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{2y} \hat{I}_{2y} \right) \cdot \displaystyle\frac{\left(3 \cos^2{\theta} - 1\right)}{2} =$

$= \displaystyle\frac{\left( 3 \cos^2{\theta} - 1\right)}{2} \cdot \left( 2 \hat{I}_{1z} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2z} - \hat{I}_{1z} \hat{I}_{2z} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{2y} \hat{I}_{2y} \right) \right) =$

$= \displaystyle\frac{\left(3 \cos^2{\theta} - 1\right)}{2} \cdot \left( 3 \hat{I}_{1z} \hat{I}_{2z} - \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right)$

Overall, this is how you go from the classical description of the magnetic field of the dipole to the truncated form of the Hamiltonian in the high nagnetic field. In the next post I will show how this Hamiltonian leads to the characteristic lineshape of the NMR line for solids.