## 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.

## My “Dream Research” Project

If I was asked to identify the most challenging biological question, I would answer immediately. What is the nature of memory and thought? This question always fascinated me as a child. For a long time, I thought only biologists can figure that out. It took me 10 years deeply studying physics and chemistry, becoming a specialist in nuclear magnetic resonance (NMR) and magnetic resonance imaging (MRI), to realize that actually now we are close to start answering the question which captivated my childish mind.

With all its complexity, the end result of the genetic machinery is to affect the chemistry of the body. Small molecules, metabolites, serve as fingerprints of what is happening inside us. Studying metabolites is almost like looking at someone’s apartment and coming up with a story of their recent lives: we can make guesses about a lifestyle based on what we see! And the chemistry of thinking is not an exception – our thought processes are accompanied by myriads of chemical transformations, and leftover metabolites can tell us about the process behind them.

Studying metabolites is almost like looking at someone’s apartment and coming up with a story of their recent lives: we can make guesses about a lifestyle based on what we see!

Routinely, metabolites are measured through analytical techniques like NMR and mass spectrometry (MS). But a new astonishing era is emerging. With new sensitivity enhancement techniques (signals can be increased by more than 20,000 times [1-4]), MR imaging will become a new tool to study metabolism in vivo and will move beyond morphology onto a platform to visualize molecules. Being fundamentally a quantum mechanical technique, the full potential of MRI is yet to be discovered.

I see my “dream research” project as the development of a new experimental MRI-NMR/MS platform to study metabolomics of memory and thought in living creatures. By developing novel MRI pulse sequences (which will take into account quantum-mechanical nature of molecules) and by applying state-of-the-art signal enhancement techniques, we will be able to “light up” the regions of the brain to study chemistry in them with an unprecedented level of accuracy. I believe that once all new methodologies available today are combined, it will become possible to create functional MRI for metabolomics – a tool to study instant chemical changes in the brain associated with memory and thinking. This will not only revisit the known biochemical processes at a new quantitative level, it will allow unraveling unexpected secrets of metabolism. And it is not only a fun thing to do — understanding the biochemical reasons for making decisions will bring us much closer to a society in which everyone truly enjoys living.

[1] J. H. Ardenkjær-Larsen et al. Increase in signal-to-noise ratio of >10,000 times in liquid-state NMR. Proc. Nat. Acad. Sci., 2003, 100 (18), 10158-10163.

[2] R. W. Adams et al. Reversible Interactions with parahydrogen Enhance NMR Sensitivity by Polarization Transfer. Science, 2009, 323 (5922), 1708-1711.

[3] D. A. Barskiy et al. Over 20% 15N Hyperpolarization in Under One Minute for Metronidazole, an Antibiotic and Hypoxia Probe. J. Am. Chem. Soc., 2016, 138 (26), 8080–8083.

[4] D. A. Barskiy et al. NMR Hyperpolarization Techniques of Gases. Chem. Eur. J., 2017, 23 (4), 725-751.