The Other Bell Basis and 24 | M.E. Irizarry-Gelpí

Wed 11 April 2018
Physics
#8.370.2x

The following $2 \times 2$ matrix is very interesting:

$\begin{equation*} J = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & i \\ 1 & -i \end{bmatrix} \end{equation*}$

It satisfies the following identities:

$\begin{align*} J^{3} &= \frac{1}{\sqrt{2}}(1 + i) I & J^{6} &= i I & J^{9} &= -\frac{1}{\sqrt{2}}(1 - i) I & J^{12} &= -I \\ J^{15} &= -\frac{1}{\sqrt{2}}(1 + i) I & J^{18} &= -i I & J^{21} &= \frac{1}{\sqrt{2}}(1 - i) I & J^{24} &= +I \end{align*}$

Note that $J$ is unitary:

$\begin{equation*} J^{\dagger} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -i & i \end{bmatrix} = J^{23} = J^{-1} \end{equation*}$

Thus

$\begin{align*} J^{4} &= \frac{1}{\sqrt{2}}(1 + i) J & J^{7} &= i J & J^{10} &= -\frac{1}{\sqrt{2}}(1 - i) J & J^{13} &= -J \\ J^{16} &= -\frac{1}{\sqrt{2}}(1 + i) J & J^{19} &= -i J & J^{22} &= \frac{1}{\sqrt{2}}(1 - i) J \end{align*}$

and

$\begin{align*} J^{2} &= \frac{1}{\sqrt{2}}(1 + i) J^{\dagger} & J^{5} &= i J^{\dagger} & J^{8} &= -\frac{1}{\sqrt{2}}(1 - i) J^{\dagger} & J^{11} &= -J^{\dagger} \\ J^{14} &= -\frac{1}{\sqrt{2}}(1 + i) J^{\dagger} & J^{17} &= -i J^{\dagger} & J^{20} &= \frac{1}{\sqrt{2}}(1 - i) J^{\dagger} \end{align*}$

Note that the coefficients correspond to the 8th-roots of unity, reflecting the fact that the cyclic group $\mathbb{Z}_{24}$ is the product group $\mathbb{Z}_{3} \times \mathbb{Z}_{8}$ .

If the kets $|0 \rangle$ and $|1 \rangle$ are defined via

$\begin{align*} |0 \rangle &= \begin{bmatrix} 1 \\ 0 \end{bmatrix} & |1 \rangle &= \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{align*}$

and the kets $|Y{-}\rangle$ and $|Y{+}\rangle$ are defined via

$\begin{align*} |Y{-}\rangle &= \frac{1}{\sqrt{2}} \left( |0 \rangle - i | 1 \rangle \right) & |Y{+}\rangle &= \frac{1}{\sqrt{2}} \left( |0 \rangle + i | 1 \rangle \right) \end{align*}$

then $J$ and $J^{\dagger}$ transform between these two ortho-normal bases:

$\begin{align*} J^{\dagger} |0 \rangle &= |Y{-} \rangle & J^{\dagger} |1 \rangle &= |Y{+} \rangle \end{align*}$

Note that the kets $|Y{-}\rangle$ and $|Y{+}\rangle$ are eigen-vectors of the unitary hermitian matrix $Y$ given by

$\begin{align*} Y |Y{-}\rangle &= -|Y{-}\rangle & Y |Y{+}\rangle &= |Y{+}\rangle & Y &= \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \end{align*}$

This is one of the Pauli matrices.

Let $J_{2} = J$ . It follows that

$\begin{equation*} (J_{2})^{24} = I_{2} \end{equation*}$

Note that $J_{2}$ correspond to a 1-qubit gate. Consider the Kronecker product:

$\begin{equation*} J_{4} = J_{2} \otimes J_{2} \end{equation*}$

It follows that

$\begin{equation*} (J_{4})^{12} = I_{4} \end{equation*}$

Note that $J_{4}$ correspond to a 2-qubit gate. Now consider the Kronecker product:

$\begin{equation*} J_{16} = J_{4} \otimes J_{4} \end{equation*}$

It follows that

$\begin{equation*} (J_{16})^{6} = I_{16} \end{equation*}$

Note that $J_{16}$ correspond to a 4-qubit gate. Now consider the Kronecker product:

$\begin{equation*} J_{256} = J_{16} \otimes J_{16} \end{equation*}$

It follows that

$\begin{equation*} (J_{256})^{3} = I_{256} \end{equation*}$

Note that $J_{256}$ correspond to an 8-qubit gate.