Construct a matrix operator that translates (moves) each bead clockwise one “bead slot” along the ring.
We want to move each bead once clockwise, or one down the column vector, so we need to have a matrix with only non-zero entries on the spaces just above the main diagonal (plus one in the off corner since we have a periodic boundary condition). So we have:
\begin{align*} S^\downarrow =\begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{bmatrix} \end{align*}
I can test this on a arbitrary column vector:
\begin{align} S^\downarrow \begin{bmatrix} a\\b\\c \end{bmatrix} = \begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{bmatrix} \begin{bmatrix} a\\b\\c \end{bmatrix} = \begin{bmatrix}c\\a\\b \end{bmatrix} \end{align}
Yep it moves each element one space down!
Write down an eigenvalue equation for this operator.
\begin{align*} S^\downarrow \vec{s} = \lambda \vec{s} \end{align*}
Find eigenvalues and eigenvectors of this operator.
If I operate this matrix on a column vector 3 times, we we get the exact same vector back again:
\begin{align} (S^\downarrow)^{3}\vec{s}=1\vec{s} \end{align}
But this isn't an eigenvalue equation for \(S^\downarrow\), it is an eigevalue equation for \((S^\downarrow)^{3}\). However, if I use \(S^\downarrow\) on a column vector 3 times, I will get the eigenvalue of 1, and since I know I need the same eigenvalue for \(S^\downarrow\) each time, I can take the 3th root of 1 and use it as the eigenvalue of \(S^\downarrow\).
\begin{align} \lambda_n = \sqrt[3]{1}=\left(1\right)^{1/3}=\left(e^{i2\pi n}\right)^{1/3}=e^{i\frac{2\pi}{3}n} \end{align}
Where I used the exponential complex representation of 1 with a integer n which goes from 1 to 3 (because I have 3 beads). So my eigenvalues are, in exponential and retanglar complex numbers:
n 1 2 3 \(e^{i\frac{2\pi}{3}}\) \(e^{i\frac{4\pi}{3}}\) \(e^{i\frac{6\pi}{3}}=1\) Now I want the eigenvectors.
I'll start with thinking about what \(S^{\downarrow}\) with a vector:
\begin{align*} S^{\downarrow} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}z\\x\\y\end{bmatrix} \\ \end{align*}
For eigenvalue \(\lambda_3 =1\), the eigenvalue equation is:
\begin{align*} S^{\downarrow} \begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}x\\y\\z\end{bmatrix} \\ \end{align*}
This means that:
\begin{align*} \begin{bmatrix}z\\x\\y\end{bmatrix} &= \begin{bmatrix}x\\y\\z\end{bmatrix}\\ z &= x \\ x &= y \\ y &= z \end{align*}
This only works if \(x = y = z\), and the unnormalized eigenvector is:
\begin{align*} \left|{1}\right\rangle = \begin{bmatrix}1\\1\\1\end{bmatrix} \end{align*}
For eigenvalue \(\lambda_1 =e^{i\frac{2\pi}{3}}\), the eigenvalue equation is:
\begin{align*} S^{\downarrow} \begin{bmatrix}x\\y\\z\end{bmatrix} &= e^{i\frac{2\pi}{3}}\begin{bmatrix}x\\y\\z\end{bmatrix} \\ \end{align*}
This means that:
\begin{align*} \begin{bmatrix}z\\x\\y\end{bmatrix} &= e^{i\frac{2\pi}{3}}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ z &= e^{i\frac{2\pi}{3}}x \\ x &= e^{i\frac{2\pi}{3}}y \\ y &= e^{i\frac{2\pi}{3}}z \end{align*}
This works if:
\begin{align*} \left|{e^{i\frac{2\pi}{3}}}\right\rangle = \begin{bmatrix}1\\e^{i\frac{4\pi}{3}}\\e^{i\frac{2\pi}{3}}\end{bmatrix} \end{align*}
Using the same reasoning for \(\lambda_2 =e^{i\frac{4\pi}{3}}\):
\begin{align*} \left|{e^{i\frac{4\pi}{3}}}\right\rangle &= \begin{bmatrix}1\\e^{i\frac{8\pi}{3}}\\e^{i\frac{4\pi}{3}}\end{bmatrix} &= \begin{bmatrix}1\\e^{i\frac{2\pi}{3}}\\e^{i\frac{4\pi}{3}}\end{bmatrix} \end{align*}
Go ahead and check that these vectors solve the eigenvalue equations for their corresponding eigenvalue.
Consider linear operators \(\hat{A}\) and \(\hat{B}\) share a complete set of eigenvectors but with different eigenvalues.
\begin{align*} \hat{A}\left|{ab}\right\rangle &= a\left|{ab}\right\rangle \\ \hat{B}\left|{ab}\right\rangle &= b\left|{ab}\right\rangle \end{align*}
Now consider an operator \(\hat{C}\) that is a linear combination of \(\hat{A}\) and \(\hat{B}\).
Show that \(\left|{ab}\right\rangle \) is also an eigenvector of \(\hat{C}\) and find it's eigenvalues.
Let \(\hat{C} = \alpha \hat{A} + \beta \hat{B}\), then:
\begin{align*} \hat{C}\left|{ab}\right\rangle &= (\alpha \hat{A} + \beta \hat{B}) \left|{ab}\right\rangle \\[6pt] &=\alpha \hat{A}\left|{ab}\right\rangle + \beta \hat{B}\left|{ab}\right\rangle \\[6pt] &=\alpha a\left|{ab}\right\rangle + \beta b\left|{ab}\right\rangle \\[6pt] &=(\alpha a + \beta b)\left|{ab}\right\rangle \\ \end{align*}
Therefore \(\hat{C}\) shares the eigenvectors of \(\hat{A}\) and \(\hat{B}\) and its eigenvalues are the same linear combination of the eigenvalues of \(\hat{A}\) and \(\hat{B}\).
Show that \(\hat{S^\uparrow}\) and \(\hat{S^\downarrow}\) are inverses of each other.
If these operators are inverses of each other, than I should get the identity operator if I multiply them together:
\begin{align*} \hat{S^\uparrow}\hat{S^\downarrow} &\doteq \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{bmatrix}\begin{bmatrix} 0&0&1\\1&0&0\\0&1&0 \end{bmatrix} \\[6pt] &= \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} \end{align*}
I can also see if they undo the transformation of each other:
\begin{align*} \hat{S^\uparrow}\hat{S^\downarrow} \begin{bmatrix}x\\y\\z\end{bmatrix} &\doteq \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{bmatrix} \left(\begin{bmatrix} 0&0&1\\1&0&0\\0&1&0 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}\right) \\[6pt] &= \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{bmatrix} \begin{bmatrix}z\\x\\y\end{bmatrix}\\[6pt] &= \begin{bmatrix}x\\y\\z\end{bmatrix} \end{align*}
Therefore, \(\hat{S^\uparrow}\) and \(\hat{S^\downarrow}\) are inverses.
Show that the eigenvectors of \(\hat{S^\downarrow}\) are also eigenvectors of \(\hat{S^\uparrow}\).
The easiest way to show this is to let \(\hat{S^\uparrow}\) act on the eigenvectors of \(\hat{S^\downarrow}\). Remember that the kets are labeled with the \(\hat{S^\downarrow}\) eigenvalues.
\begin{align*} \hat{S^\uparrow}\left|{1}\right\rangle &= \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{bmatrix} \begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix} \\[12pt] \hat{S^\uparrow}\left|{e^{i\frac{2\pi}{3}}}\right\rangle &= \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{bmatrix} \begin{bmatrix}1\\e^{i\frac{4\pi}{3}}\\e^{i\frac{2\pi}{3}}\end{bmatrix} = \begin{bmatrix}e^{i\frac{4\pi}{3}}\\e^{i\frac{2\pi}{3}}\\1\end{bmatrix} = e^{i\frac{4\pi}{3}}\begin{bmatrix}1\\e^{i\frac{4\pi}{3}}\\e^{i\frac{2\pi}{3}}\end{bmatrix} \\[12pt] \hat{S^\uparrow}\left|{e^{i\frac{4\pi}{3}}}\right\rangle &= \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{bmatrix} \begin{bmatrix}1\\e^{i\frac{2\pi}{3}}\\e^{i\frac{4\pi}{3}}\end{bmatrix} = \begin{bmatrix}e^{i\frac{2\pi}{3}}\\e^{i\frac{4\pi}{3}}\\1\end{bmatrix} = e^{i\frac{2\pi}{3}} \begin{bmatrix}1\\e^{i\frac{2\pi}{3}}\\e^{i\frac{4\pi}{3}}\end{bmatrix} \end{align*}
I see that the eigenvectors of \(\hat{S^\downarrow}\) are also eigenvectors of \(\hat{S^\uparrow}\), but with different eigenvalues.
Since \(\hat{S^\uparrow}\) and \(\hat{S^\downarrow}\) are inverses and undo each other's action, the product of their eigenvalues for each eigenvector must be equal to 1.
\begin{align*} \lambda^\uparrow_1\lambda^\downarrow_1 = (e^{i\frac{4\pi}{3}}) (e^{i\frac{2\pi}{3}}) = 1 \\[6pt] \lambda^\uparrow_2\lambda^\downarrow_2 = (e^{i\frac{2\pi}{3}}) (e^{i\frac{4\pi}{3}}) = 1 \\[6pt] \lambda^\uparrow_3\lambda^\downarrow_3 = (e^{i\frac{6\pi}{3}}) (e^{i\frac{6\pi}{3}}) = 1 \\[6pt] \end{align*}
where \(\lambda^\uparrow_n\) are the eigenvalues of \(\hat{S^\uparrow}\) and \(\lambda^\downarrow_n\) are the eigenvalues of \(\hat{S^\downarrow}\). In other words, if
\[\lambda^\uparrow_n = e^{i\frac{2\pi}{N}n}\]
then
\[\lambda^\downarrow_n = e^{-i\frac{2\pi}{N}n}\]
where \(m\) is the number of beads.
Find the three eigenvalues of the operator (\(\hat{S^\uparrow}+\hat{S^\downarrow}\)) and show that at least one of them checks out.
Since \(\hat{S^\uparrow}\) and \(\hat{S^\downarrow}\) share a common set of eigenvectors, the eigenvalues of the operator that is their sum must be the sum of the eigenvalues:
\begin{align*} (\hat{S^\uparrow}+\hat{S^\downarrow})\left|{n}\right\rangle &= \hat{S^\uparrow}\left|{n}\right\rangle +\hat{S^\downarrow}\left|{n}\right\rangle \\[6pt] &= \lambda^\uparrow\left|{n}\right\rangle +\lambda^\downarrow\left|{n}\right\rangle \\[6pt] &= (\lambda^\uparrow + \lambda^\downarrow)\left|{n}\right\rangle \\[6pt] &= (e^{i\frac{2\pi}{3}n}+ e^{-i\frac{2\pi}{3}n})\left|{n}\right\rangle \end{align*}
For example, for \(n=1\):
\begin{align*} (\hat{S^\uparrow}+\hat{S^\downarrow})\left|{1}\right\rangle &= \begin{bmatrix} 0&1&1\\1&0&1\\1&1&0 \end{bmatrix} \begin{bmatrix} 1\\e^{i\frac{2\pi}{3}}\\e^{i\frac{4\pi}{3}}\end{bmatrix}\\ &= \begin{bmatrix} e^{i\frac{2\pi}{3}}+e^{i\frac{4\pi}{3}}\\1+e^{i\frac{4\pi}{3}}\\1+e^{i\frac{2\pi}{3}}\end{bmatrix}\\ &= (e^{i\frac{2\pi}{3}}+ e^{-i\frac{2\pi}{3}})\begin{bmatrix} 1\\e^{i\frac{2\pi}{3}}\\e^{i\frac{4\pi}{3}}\end{bmatrix}\quad \checkmark \end{align*}