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.

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Figure 1. Nuclear spin 1 produces the magnetic field that is sensed by a spin 2. Similarly, nuclear spin 2 produces the magnetic field that is sensed by a spin 1.

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?

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Figure 2. Two nuclear spins, 1 and 2, are separated by the distance r\theta is defined as an angle between z-direction of the coordinate system and the vector connecting the two spins. If there is an external magnetic field B0, it is usually defined along the z-axis.

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).

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Figure 3. If written in Zeeman basis (αααβ, βα and ββ), dipole-dipole Hamiltonian can be split into the following 6 terms of the “Dipolar Alphabet”.

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.

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Figure 4. Energy level structure of the Zeeman Hamiltonian: lowest energy level (E1) corresponds to the state αα, energy levels E2 and E3 correspond to the states αβ and βα. The highest energy level (E4) corresponds to the nuclear spin state ββ. The essence of secular approximation is to make the to-be-simplified Hamiltonian match the eigenvalue structure of the main Hamiltonian. One can see from Figure 3 that the first two terms of the DD Hamiltonian will match the structure of the Zeeman Hamiltonian.

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.

 

 

Parahydrogen

To begin my blog, let’s introduce parahydrogen. Lately, this little molecule has been attracting a lot of attention in the magnetic resonance community due to tremendous opportunities it brings for NMR/MRI signal enhancement. I will explain a bit later how this parahydrogen-based NMR signal enhancement works. But first, let’s talk about physical origins of parahydrogen!

ortho_para_H2-01

Parahydrogen (para-H2) is a nuclear spin isomer of a hydrogen molecule. Nuclear spin isomerism is a very special form of isomerism. Unlike “traditional” molecular isomers (molecules having the same atomic composition but different chemical structure) and isotopologues (isomers that differ only in their isotopic composition), nuclear spin isomers are chemically identical: they have exactly the same atomic (and even isotopic) structure. However, nuclear spin isomers differ in the nuclear spin state of their atoms. It turns out that this tiny change (energy difference associated with nuclear spin transitions is only ~0.1 J/mol) may lead to different thermodynamic and spectroscopic properties of molecules. So, how does this work?

Unfortunately (or fortunately), we will have to use rules of quantum mechanics and some math. In quantum mechanics, in order to describe properties of quantum systems (atoms, molecules, etc.), physicists use wavefunctions. By knowing a wavefunction one will be able to calculate probabilities to find a quantum system in different states (namely, a squared modulus of the wavefunction determines the probability to find a system in a given state). Let’s look how it works taking as an example hydrogen molecule.

Hydrogen molecule consists of two hydrogen atoms (H) and is denoted as Н2. Each atom has a nucleus – a proton which is a spin-1/2 particle. Physicists say that hydrogen molecule has several degrees of freedom: translational, rotational, vibrational, etc., and these degrees of freedom can be considered independent. In other words, rotation of the hydrogen molecule does not depend on how and where the molecule is moving and how it is vibrating. Each degree of freedom has a wavefunction associated with it. I will use different colors to describe electron and nuclear wavefunctions. A position of the molecule in space, as well as its rotation and vibrations, are determined by the position and movements of nuclei, therefore, these degrees of freedom are described by translational (ψtr), rotational (ψrot), vibrational (ψvib), and nuclear spin (ψspin) wavefunctions. Atomic nuclei are surrounded by electrons which provides the bonding between the nuclei. The wavefunction describing movements of electrons is called orbital wavefunction ψorb, and state of the electron spins is described by the electron spin wavefunction ψspin.

Since probabilities of independent events are multiplied, the total wavefunction is a product of the above-mentioned wavefunctions:

ψtot = ψtr·ψrot·ψvib·ψspin·ψorb·ψspin

However, rules of quantum mechanics are trickier than they may sound. According to Pauli’s principle, the total wavefunction of the system of spins-1/2 particles has to be antisymmetric with respect to the exchange (also called permutation) of two identical particles. What does this mean?

Let’s take for example ψspin. A system consisting of two spins-1/2 can be described as α1α2, β1β2 or combinations α1β2+β1α2, α1β2β1α2. Here α and β denote the projection of nuclear spin angular momentum along the quantization axis (more on this stuff later, for now, one can imagine the state α as a magnetic moment – spin – pointing up along the external magnetic field and the state β as a magnetic moment pointing down, opposite to the field). Indexes 1 and 2 say to which nucleus the spin belongs. For example, the state α1α2 means that both nuclear spins point along the field while the state β1β2 means that both spins point opposite to the field. The combination states α1β2+β1α2 and α1β2β1α2 are more interesting. Neither of spins points along or opposite to the field but if we take one spin and determine its orientation, the second spin will take the opposite orientation. We can see now that two spins are correlated: the state of the second spin depends on the state of the first one.

Now let’s look what happens if we exchange (permute) particles. Mathematically, permutation simply means interchange of indexes (1→2, 2→1). One can see that upon permutation of indexes the first three states do not change: α2α1= α1α2β2β1 = β1β2, (α2β1+β2α1) = (α1β2+β1α2), but the last state changes the sign: (α2β1β2α1) = –(α1β2β1α2). Therefore, the first three states are called symmetric wavefunctions and the last one – antisymmetric with respect to permutation of particles.

So, our hydrogen molecule contains four spin-1/2 particles: two electrons and two nuclei). Permutation of electrons can only affect ψorb and ψspin. The first wave function, corresponding to the electronic ground state, is symmetric with respect to the electrons, the second, the electron spin wavefunction, is antisymmetric, and the rest are independent of the electrons’ variables and, thus, symmetric. Therefore, Pauli’s principle is fulfilled for electrons: the total wavefunction is antisymmetric with respect to permutation of electrons, thanks to antisymmetric ψspin. Permutation of nuclei can affect two wavefunctions: ψspin (as we just saw above) and ψrot. A mathematical expression for ψrot is rather complicated but it is not necessary to know its full form to understand the symmetry properties.

RotE_ortho_para_H2-01
Figure 1. Schematic energy diagram of rotational levels of the hydrogen molecule.

This is because rotating diatomic molecules possess a set of stable rotational states, which can be described by only one parameter – the rotational quantum number J. This number can take integer values 0, 1, 2, 3, … This means that molecule can be in a stable state with J = 0, J = 1, J = 2, etc. (Figure 1). It turns out that the symmetry (with respect to permutation of nuclei) of the rotational wavefunction can be described as

P12·ψrot = (-1)J·ψrot

 

where P12 represents the permutation operator that interchanges the nuclei’s positions (indexes). This means that the rotational wavefunction is symmetric for even rotational states (J = 0, 2, 4, …) and antisymmetric for odd rotational states (J = 1, 3, 5, ).

Coming back to Pauli’s principle, permutation of nuclei should lead to the change of sign of the total wavefunction. Since only ψspin and ψrot can change sign upon such permutation, these two wavefunctions become connected: even (symmetric) rotational wavefunctions must be combined with the antisymmetric nuclear wavefunction (α1β2β1α2), whereas each antisymmetric rotational wavefunction has to be associated with one of the three symmetric spin functions. All this is required to yield a total wavefunction being antisymmetric with respect to the exchange of the nuclei. This is where two hydrogen spin isomers come from. One is called parahydrogen (para-H2), having an antisymmetric nuclear spin wavefunction 1β2β1α2) and existing only in even rotational states, and the other called orthohydrogen (ortho-H2), having a symmetric nuclear spin wavefunction and existing only in the odd rotational states.

It follows from the Pauli’s principle that nuclear spin state and rotational state of the hydrogen molecule are strictly correlated. This is remarkable, because the notion of independence (which allowed us to write a wavefunction as a product of individual wavefunctions) has led to complete dependence of these degrees of freedom from each other!

Remarkably, parahydrogen and orthohydrogen can be seen as two individual gases because their thermodynamic properties (boiling point, heat capacity, etc.) are slightly different. This is not surprising taking into account the fact that molecules constituting these two gases are rotating differently!

Importantly, conversion between the two states occurs extremely slowly because the transition between symmetric and antisymmetric nuclear spin states are forbidden by the selection rules of quantum mechanics. Therefore, after its production parahydrogen may be stored for long periods before use in a tank as an individual gas, as the relaxation rate of the parahydrogen back to room-temperature equilibrium can be on the order of months.

However, the use of paramagnetic catalysts (i.e., activated charcoal, nickel, hydrated iron(III) oxide) promotes the establishment of Boltzmann thermodynamic equilibrium between ortho-H2/para-H2 states for a given temperature at greatly accelerated rates. This happens because paramagnetic materials can create a strong inhomogeneous magnetic field on the atomic scale. In such fields the two hydrogen atoms are no longer equivalent, thus, spin-flip transitions between ortho-H2 and para-H2 are no longer forbidden. In practice, normal hydrogen gas (i.e., equilibrium ratio of spin isomers at room temperature) consisting of 75% ortho– and 25% para-hydrogen is passed through a chamber filled with paramagnetic catalyst and maintained at cryogenic temperatures, where the equilibration to the isomer ratio governed by the Boltzmann distribution occurs. For example, a parahydrogen generator operating at 77 K (obtained conveniently by a liquid-N2 bath) yields a mixture with ~50% parahydrogen, whereas the designs based on cryo-chillers (e.g. T~20 K) yield >99% parahydrogen (Figure 2). I should note that the enrichment of hydrogen with para-isomer happens so easily only because of the big energy gap between rotational spin states. This, in turn, is due to the small mass of molecular hydrogen (in general, the energy difference between rotational spin states is inversely proportional to the moment of inertia of a rotating molecule).

Figure_1-01
Figure 2. a) Equilibrium para-H2 fraction as a function of ortho-para conversion temperature. The plot shows that at temperatures near the boiling point (1), the equilibrium composition is almost pure parahydrogen, and at elevated temperatures, the composition asymptotically approaches a 3:1 ratio, (2) — liquid nitrogen temperature, (3) — room temperature. b) The population of rotational energy states (J = 1, 2, 3) of hydrogen as a function of temperature.

The existence of nuclear spin isomers of molecular hydrogen (which was experimentally confirmed by the early 1930s) was one of the first triumphs of quantum mechanics. Indeed, the citation of the Nobel Prize awarded to Werner Heisenberg in 1932 stated that he had “created quantum mechanics, the application of which led to the discovery of the two allotropic forms of hydrogen”!

Knowledge about ortho– to para-H2 conversion is important for the storage of liquid hydrogen (especially as a rocket fuel). The difference in energy associated with the different rotational levels means that energy is released when ortho-H2 converts to para-H2, and energy is absorbed in the reverse process. This phenomenon can be thought of as a latent heat of conversion. If one quickly liquefies normal hydrogen, it will still have 3:1 ortho:para composition which will eventually lead to the heat release. This can vaporize a significant portion of hydrogen and break the impermeability of the storage container. At the dawn of industrial liquid hydrogen production, this presented a major problem. Modern hydrogen liquefying processes now ensure that the liquid hydrogen has reached equilibrium concentration at 99.8% para-H2 before being transported and stored for use.

One may ask how can para-H2 be important for NMR? Indeed, this spin isomer has a zero total nuclear spin and, thus, it does not possess 1H NMR spectrum. However, para-H2 is a pure quantum mechanical state and a highly organized spin order which is readily achievable simply by cooling. Pure state means that all para-H2 molecules are described by the same wavefunction – 1β2β1α2). For comparison, ortho-H2 is a mixture of three wavefunctions, α1α2α1β2+β1α2 and β1β2 and, thus, it is not a pure state. It turns out that once you have a quantum mechanically pure state, you can manipulate it and transfer the spin order from one form to another. This is how parahydrogen-induced polarization (PHIP) and signal amplification by reversible exchange (SABRE) work: they transfer NMR-silent singlet spin order of para-H2 into observable nuclear magnetization.