- #1
PhysicsKush
- 29
- 4
- 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
$$ \hat{T} = \begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{pmatrix}.$$
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