Finding the generator of rotations for a 3-state triangle

[PhysicsKush]

Homework Statement

Imagine a system with $3$ states. to make it more physical we can think of an electron in a $3$-atom molecule which can be thought of as being close to any one of them. We label the different state by the three positions $|a \rangle$, $|b\rangle$ and $|c\rangle$ such that the position operator $\hat{x}$ can give one of three eigen values $a$ , $b$ and $c$ (with $\hat{x}|a \rangle = a | a\rangle$, etc). We now define a 'triangle translation operator' $\hat{T}$ such that:
$$ \hat{T} |a \rangle = |b \rangle \quad \hat{T} |b \rangle = | c\rangle \quad \text{and} \quad \hat{T} |c \rangle = |a \rangle$$

The translation around the triangle can be also thought of as a rotation about the middle of the triangle (by what angle?) . Find the generator of this rotation $\hat{J}_{z}$. Hint" work in a convenient basis. Then write the generator of rotations in the $a, b,c$ basis and check that $\hat{T}$ eigen states you found in the previous section are also eigenstates of $\hat{J}_{z}$ with the correct eigen values.

Relevant Equations

$$ R_{z}(\varphi) = e^{\frac{-i}{\hbar} \hat{J}_{z} \varphi} $$

I first computed the operator $\hat{T}$ in the $a,b,c$ basis (assuming $a = (1 \ 0 \ 0 )^{T} , b = (0 \ 1 \ 0)^{T}$ and $c = (0 \ 0 \ 1)^{T}$) and found
$$ \hat{T} = \begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{pmatrix}.$$
The eigenvalues and eigenvectors corresponding to this matrix are
$$ \lambda_{1} = 1,\quad \lambda_{2} = \frac{1}{2}(-1 + i\sqrt{3}) , \quad \lambda_{3} = \frac{1}{2}(-1 - i \sqrt{3}),$$
$$ v_{1} = \begin{pmatrix} 1 \\ 1 \\ 1\end{pmatrix}, \quad v_{2} = \begin{pmatrix} -1 -i\sqrt{3} \\ -1+i\sqrt{3} \\ 2 \end{pmatrix}, \quad v_{3} = \begin{pmatrix}-1 +i\sqrt{3} \\ -1-i\sqrt{3} \\ 2 \end{pmatrix}.$$

A full circle is $2 \pi$ ,therefore each rotation about a state is a rotation of $2 \pi /3$. Then I'm not sure how to properly set my system of equation (I suppose I have to solve one) to then solve for the generator of rotations $\hat{J}_{z}$.
I believe first that
$$ \hat{J}_{z} |v_{1}\rangle = \alpha_{1} |v_{1} \rangle, \quad \hat{J}_{z} |v_{2}\rangle = \alpha_{2} |v_{2} \rangle, \quad \hat{J}_{z} |v_{3}\rangle = \alpha_{3} |v_{3} \rangle,$$
and I'm trying to find the eigenvalues $\alpha_{i}$.
\begin{align*}
\hat{R}_{z}\left(\frac{2\pi }{3}\right)|v_{1} \rangle &= \exp{\frac{- i 2 \pi }{3 \hbar} \hat{J_{z}}} = \lambda_{1} \begin{pmatrix} \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{1}} & & \\ & \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{2}} & \\ & & \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{3}} \end{pmatrix} v_{1} = \alpha_{2} v_{2}, \\
\hat{R}_{z}\left(\frac{2\pi }{3}\right)|v_{2} \rangle &= \exp{\frac{- i 2 \pi }{3 \hbar} \hat{J_{z}}} = \lambda_{2} \begin{pmatrix} \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{1}} & & \\ & \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{2}} & \\ & & \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{3}} \end{pmatrix} v_{2} = \alpha_{3} v_{3}, \\
\hat{R}_{z}\left(\frac{2\pi }{3}\right)|v_{3} \rangle &= \exp{\frac{- i 2 \pi }{3 \hbar} \hat{J_{z}}} = \lambda_{3} \begin{pmatrix} \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{1}} & & \\ & \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{2}} & \\ & & \exp{\frac{- i 2 \pi }{3 \hbar} \alpha_{3}} \end{pmatrix} v_{3} = \alpha_{1} v_{1}, \\
\end{align*}
and then solve for $\alpha_{i}$, but this seems wrong. I don't know what to do nor if I'm on the good track.

Any hints would be appreciated

 

[pasmith]

I think you want to say that $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$ are the vertices of an equilateral triangle lying in the plane $x + y + z = 1$. The center of the triangle is at $(1/3,1/3,1/3)^T$ so you probably want your new basis vectors (which should, of course, be normalized) to be in the direction of $\mathbf{b}_1 = (1,0,0)^T - (1/3,1/3,1/3)^T$, $\mathbf{b}_2 = \mathbf{b}_3 \times \mathbf{b}_1$ and $\mathbf{b}_3 = (1,1,1)^T$. Then in these coordinates you can apply a rotation through $2\pi/3$ about an axis in direction $\mathbf{b}_3$.

 

[PhysicsKush]

I think you want to say that $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$ are the vertices of an equilateral triangle lying in the plane $x + y + z = 1$. The center of the triangle is at $(1/3,1/3,1/3)^T$ so you probably want your new basis vectors (which should, of course, be normalized) to be in the direction of $\mathbf{b}_1 = (1,0,0)^T - (1/3,1/3,1/3)^T$, $\mathbf{b}_2 = \mathbf{b}_3 \times \mathbf{b}_1$ and $\mathbf{b}_3 = (1,1,1)^T$. Then in these coordinates you can apply a rotation through $2\pi/3$ about an axis in direction $\mathbf{b}_3$.

Interessting, but how does that help me find the matrix generator of rotation ?

 

[pasmith]

A rotation through an angle $\theta$ about the $z$-axis is given by $$R = \begin{pmatrix} \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$ This is generated by $$J = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ If you were dealing with an equilateral triangle centered at the origin in the $x$-$y$ plane you would be done. But you aren't, so you need to apply a change of basis in which the new components $(X,Y,Z)$ are given by $M(x,y,z)^T$ for some invertible matrix $M$.

Thus the rotation matrix is given by $M^{-1}RM$. Finding the generator of this is fairly straightforward, in view of the fact that $$\exp(M^{-1} A M) = M^{-1} \exp(A) M.$$