# Rotations and Irreducible Tensor Operators

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## Overview

Irreducible Tensor Operators are extremely valuable to help reduce complex mathematical problems commonly found in NMR. This page will be devoted to deriving the tensors and showing how to use the tensors in calculations.

## Cartesian Rotations

Irreducible Tensor Operators operate on angular momenta, which is a fairly abstract concept to understand. We begin this discussion by investigating rotations in 2 and 3 dimensions so the reader understands a physical picture before moving into a complex quantum mechanical treatment.

2D Rotations

Let's begin with the simple example of the hand of a clock. The clock hand is a vector which has a magnitude (the length of the hand) and a direction (the direction the arrow is pointing). The vector is rotating in a 2D plane (the clock face). We define our cartesian axes that the y axis lies along 12 and the x axis is along 3 on the clock. Initally, lets say it is 3:00 (t=0) and the vector lies along the x-axis and rotates counter-clockwise, then a some time, (t=2.2hrs) later the vector points along some arbitrary angle theta from the inital x-axis. Taking a snapshot of the vector at this time, we can see that vector a can now be described by

$\vec{r}=a\hat{i}+b{\hat{j}=r\hat{i}\cos(\theta) \hat{j} + r\sin(\theta}) \hat{j}$

insert picture

We can actually describe this vectors position using any arbitrary axis (time) for x and y as long as x and y remain orthogonal. If for example the axes are rotated by $$\phi$$ then the vector a time=2.2hrs is

$\vec{r}=c\hat{i}+d{\hat{j}=r\hat{i}\cos(\theta-\phi) \hat{j} + r\sin(\theta-\phi}) \hat{j}$

insert picture

Using the following triognometric identities, we can solve for a ,b, c, and d.

$\cos(\theta-\phi)=\cos\theta \cos\phi + \sin\theta \sin\phi$

$\sin(\theta-\phi)=\sin\theta \cos\phi - \cos\theta \sin\phi$

the it follows that

$a=r\cos\theta$

$b=r\sin\theta$

$c=r\cos(\theta-\phi)=r\cos\theta \cos\phi+r\sin\theta \sin\phi= a\cos\phi +b\sin\phi$

$d=r\sin(\theta-\phi)=r\sin\theta \cos\phi- r\cos\theta \sin\phi= b\cos\phi -a\sin\phi$

From inspection we can see that this can be readily transformed into a matrix

$\begin{pmatrix} c\\d \end{pmatrix}=\begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{pmatrix}\begin{pmatrix} a\\b \end{pmatrix}$

In other word we can rotate r into r' by applying a rotation trnasformation R or

$r'=Rr$

where R has the property

$RR^{\dagger}=1$

$\begin{pmatrix} a\\b \end{pmatrix}=\begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix}\begin{pmatrix} c\\d \end{pmatrix}$

### 3D Rotations

In 3 dimensions rotations are slightly more complex. We need to specify another dimension in which the object can be rotated. In order to do this we need to employ three angles $$\alpha, \beta, \gamma$$, commonly referred to as the euler angles. $$\alpha$$ is rotation about the original z-axis, $$beta$$ is rotation about y' and $$\gamma$$ is rotation about z'. How the angle correspond to rotations are shown below.

insert figure

Our vector, r, may now be described as

$\vec{r}=a\hat{i} +b\hat{j} + c\hat{k}$

We can now describe rotations R when one of the 3 axes is fixed If Z axis is fixed,

$R_z=\begin{pmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha &\cos\alpha &0 \\ 0&0&1\end{pmatrix}$

When Y axis is fixed

$R_y=\begin{pmatrix} \cos\beta & 0 & -\sin\beta \\ 0 &1 &0 \\ \sin\beta &0&\cos\beta\end{pmatrix}$

When X axis is fixed

$R_z=\begin{pmatrix} \cos\gamma & \sin\gamma & 0 \\ -\sin\gamma &\cos\gamma &0 \\ 0&0&1\end{pmatrix}$

Then rotation of the original axis system to the new axis system is then

$R=R_xR_yR_z=\begin{pmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha &\cos\alpha &0 \\ 0&0&1\end{pmatrix} \begin{pmatrix} \cos\beta & 0 & -\sin\beta \\ 0 &1 &0 \\ \sin\beta &0&\cos\beta\end{pmatrix} \begin{pmatrix} \cos\gamma & \sin\gamma & 0 \\ -\sin\gamma &\cos\gamma &0 \\ 0&0&1\end{pmatrix}$

## Angular Momentum Rotations

### J=1/2

In a similar manner we can rotated angular momentum states. Like we developed a rotation operator, R for Cartesian space, we can define a new rotation operators which operates on angular momentum states using the Euler relations we made for 3D rotations.

$D(\alpha, \beta, \gamma)=e^{i\gamma J_z} e^{i\beta J_y} e^{i\alpha J_x}$

Lets examine what happens when D operates on |J,m>

$D(\alpha, \beta, \gamma)=\sum_{J',m'} |J',m'><J',m'|D(\alpha, \beta, \gamma)|J,m>$

The operator can only operate on a \single J value reducing the above equation to

$D(\alpha, \beta, \gamma)=\sum_{m'}<J,m'|D(\alpha, \beta, \gamma)|J,m>|J,m' = \sum_{m'} D_{m',m}^{(J)}(\alpha, \beta, \gamma)|J,m'>$

where

$D_{m',m}^{(J)}(\alpha, \beta, \gamma)=<J,m'|D(\alpha, \beta, \gamma)|J,m>$

which can be simplified to

$D_{m',m}^{(J)}(\alpha, \beta, \gamma)=\sum_{n',n'} <J,m'|e^{i \gamma J_z}|J,n><J,n|e^{i \beta J_y}|J,n'><J,n'|e^{i \alpha J_z}|J,m>$

It is then realized that

$e^{i\alpha J_z}|J,m>=[1+i\alpha J_z + \frac{(i\alpha J_z)^2}{2!}+...]|J,m> = [1+i\alpha m +\frac{(i \alpha m)^2}{2!}+...]|J,m=e^{1 \alpha m}|J,m$

Then

$D_{m',m}^{J}(\alpha, \beta, \gamma)= e^{im'\gamma}<J,m'|e^{i\beta J_y}|J,m>e^{i m \alpha}= e^{im'\gamma}d_{m',m}^{J} (\beta) e^{i m \alpha}$

where

$d_{m',m}^{J} (\beta) =<J,m'|e^{i\beta J_y}|J,m>$

We now must calculate $$d_{m',m}^{J} (\beta)$$ for a arbitrary J. Lets examine the case where J=1/2 for rotation operator $$d^{1/2}(\beta)$$

$d^{1/2}(\beta)=e^{i\beta J_y}$

Taking a Taylor expansion gives

$e^{i\beta J_y}=1+i\beta J_y -\frac{1}{2!}\beta ^2 J_y ^2 -\frac{1}{3!} \beta ^3 j_Y^3$

$=\begin{pmatrix} 1 &0 \\0&1\end{pmatrix} (1-\frac{\frac{\beta}{2}^2}{2!}+...)+\begin{pmatrix} 0&1\\-1&0\end{pmatrix}(\frac{\beta}{2}-\frac{\frac{\beta}{2}^3}{3!}+...)$

We then see that multiplying the Taylor expansion of $$e^{i \beta J_y}$$ by the respective matrices, we obtain the Taylor expansions of \cos and \sin which can be rewritten as:

$= \begin{pmatrix} \cos(\beta/2)&\sin(\beta/2)\\-\sin(\beta/2) & \cos(\beta /2)\end{pmatrix}$

Then we have obtained the values for d

$d_{1/2,1/2}^{(1/2)}=\cos(\beta /2)$

$d_{1/2,-1/2}^{(1/2)}=\sin(\beta/ 2)$

### J>1/2

Now we must calculate $$d_{m',m}^{J} (\beta)$$ for J>1/2. To do this we must first uncouple J and m'

$D_{m',m}^{J}(\alpha, \beta, \gamma)|J,m>=\sum_{m'} D_{m',m}^{J}(\alpha, \beta, \gamma) |J,m'>$

$=(-1)^{j_2-j_1-m} \sqrt{2J+1} \sum{m_1,m_2} D(\alpha, \beta, \gamma) |j_1,m_1> D(\alpha, \beta, \gamma) |j_2,m_2> \begin{pmatrix} j_1 &j_2 & J \\ m_1 & m_2 & -m \end{pmatrix}$

$=(-1)^{j_2-j_1-m} \sqrt{2J+1} \sum{m_1,m_2 , m'_1, m'_2} D(\alpha, \beta, \gamma)^{(j_1)}_{m'_1 m_1} D(\alpha, \beta, \gamma)^{(j_2)}_{m'_2 m_2} \begin{pmatrix} j_1 &j_2 & J \\ m_1 & m_2 & -m \end{pmatrix} | j_1m'_1;j_2,m'_2>$

Taking this result and left multiplying by <Jm'| and uncoupling J and m' to the other j and m states results in the expression for $$D(\alpha, \beta, \gamma)^{(J)}_{m',m}$$

$D(\alpha, \beta, \gamma)^{(J)}_{m',m}=(-1)^{j_2-j_1-m} \sqrt{2J+1} \sum{m_1,m_2 , m'_1, m'_2} D(\alpha, \beta, \gamma)^{(j_1)}_{m'_1 m_1} D(\alpha, \beta, \gamma)^{(j_2)}_{m'_2 m_2} \begin{pmatrix} j_1 &j_2 & J \\ m_1 & m_2 & -m \end{pmatrix} <J,m'|| j_1m'_1;j_2,m'_2>$

$=(2J+1)\sum_{m_1,m_2} \begin{pmatrix} j_1 &j_2 & J \\ m_1 & m_2 & -m \end{pmatrix}\begin{pmatrix} j_1 &j_2 & J \\ m'_1 & m'_2 & -m' \end{pmatrix}D(\alpha, \beta, \gamma)^{(j_1)}_{m'_1 m_1} D(\alpha, \beta, \gamma)^{(j_2)}_{m'_2 m_2}$

From this result we can calculate any $$D(\alpha, \beta, \gamma)^{(J)}_{m',m}$$ for any J value. This is important since the observables we wish to calculate in NMR are simply rotations of the angular momentum. Looking at the time-dependant Schrodinger equation

$\frac{d}{dt}|\psi (t)>=-iH|\psi (t)>$

$|\psi (t)=e^{iHt}|\psi (t)>$

Lets now examine what we can do by rotating the angular momentum vector J,m to J m'. Looking at the z component of J we know

$<J,m'|J_z|J,m>=m\delta_{m',m}$

which results in a diagonal matriz for J_z with the digonal elements being the m quantum numbers.

Then J_z may be written as

$J_z=\sum_m m|J,m><J,m|$

which has the complex conjugate

need to figure this out!

Now we have created the operatore for J_z.. In principle we can create an operator for any combination of |J'm'><J,m|. Now lets see what happens weh nwe rotate the operator in which the states are weighted by the angular momentum characteristics. We choose to label the new operator $$T^{(1)}_{(2)}(3,4)$$ by the following indicie.

$k=J+J'$ The Total Angular Momentum

$Q=m'-m$ The difference in M values

$J$ The initial and

$J'$ the final angular momentum.

$T^{(k)}_{(Q)}(J,J')=\sqrt{2k+1}\sum_{m,m'}(-1)^{J'-m'} \begin{pmatrix} J'&J&k\\m'&-m&-Q\end{pmatrix}|J',m'><J,m|$

To rotate this operator we must left multiply by $$D(\alpha, \beta, \gamma)$$ and right multiple by $$D(\alpha, \beta, \gamma)^{\dagger}$$ . This corresponds the rotation of the final and initial states.

$D(\alpha, \beta, \gamma)T^{(k)}_{(Q)}(J,J')(D(\alpha, \beta, \gamma)^{\dagger}=\sqrt{2k+1}\sum_{m,m',n,n'}(-1)^{J'-m'} \begin{pmatrix} J'&J&k\\ m'&-m&-Q\end{pmatrix} D(\alpha, \beta, \gamma)^{(J)}_{n',m} D(\alpha, \beta, \gamma)^{(J)\dagger}_{m,n} |J',m'><J,m|$

where

$D^{J_1}(\alpha, \beta, \gamma) D^{J_2}(\alpha, \beta, \gamma)^{\dagger}=\sum_{J,m,m'} (2J+1) \begin{pmatrix} j_1&j_2&J\\m_1'&M_2'&-m'\end{pmatrix} \begin{pmatrix} j_1&j_2&J\\m_1&M_2&-m\end{pmatrix}(-1)^{m'-m} D(\alpha, \beta, \gamma)^{(J)}_{-m',m}$

which simplifies the previous equation to

$D(\alpha, \beta, \gamma)T^{(k)}_{(Q)}(J,J')(D(\alpha, \beta, \gamma)^{\dagger}=\sqrt{2k+1}\sum_{n',n,q}(-1)^{J'-n'} \begin{pmatrix} J'&J&k\\n'&-n&-q\end{pmatrix} D_{qQ}^{(k)}(\alpha, \beta, \gamma)|J',n'><J,n|$

which is really $$T^{(k)}_{(Q)}(J,J')$$ with Q replaced by q and m',m with n',n or

$D(\alpha, \beta, \gamma)T^{(k)}_{(Q)}(J,J')(D(\alpha, \beta, \gamma)^{\dagger}=\sum_q D_{qQ}^{(k)}(\alpha, \beta, \gamma)T^{(k)}_{(q)}(J,J')$

This is a very unique property of these operators! Rotation of one operator produces another operator that is weighted by $$D_{qQ}^{(k)}(\alpha, \beta, \gamma)$$. It is now trivial to create analytical solutions as $$D_{mm'}^{(J)}(\alpha, \beta, \gamma)$$ are known.