Finding the generator of rotations for a 3-state triangle

AI Thread Summary
The discussion focuses on finding the generator of rotations for a 3-state triangle using the operator ##\hat{T}##, which has been computed in the ##a,b,c## basis. The eigenvalues and eigenvectors of this operator are identified, and the need to solve for the eigenvalues of the generator of rotations ##\hat{J}_{z}## is emphasized. The participants suggest that the vertices of the triangle correspond to specific vectors in a normalized basis, and a rotation of ##2\pi/3## should be applied about a specific axis. A change of basis is necessary to derive the rotation matrix, which leads to the conclusion that the generator can be found using the transformation properties of the matrix. The conversation highlights the importance of proper basis selection and transformation in solving the problem.
PhysicsKush
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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
 
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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.
 
pasmith said:
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 ?
 
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 <br /> \exp(M^{-1} A M) = M^{-1} \exp(A) M.
 
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