Update week7.do.txt

Author: mhjensen

doc/src/week7/week7.do.txt

@@ -730,10 +730,16 @@ These are the gates that are relevant for the simpler two-qubit Hamiltonian of p
 For a two qubit system we list here the possible transformations 
 !bt
 \begin{align*}
-\bm{Z}\otimes\bm{I}\hspace{1cm} & \bm{U}=\bm{I}\otimes\bm{I}\\
-\bm{I}\otimes\bm{Z}\hspace{1cm} & \bm{U}=\text{SWAP}\\
-\bm{Z}\otimes\bm{Z}\hspace{1cm} & \bm{U}=CX_{10}\\
-\bm{X}\otimes\bm{X}\hspace{1cm} & \bm{U}=CX_{10}(\bm{H}\otimes\bm{H})\\
+\bm{Y}\otimes\bm{I}\hspace{1cm} & \bm{U}=\bm{H}\bm{S}^{\dagger}\otimes\bm{I}\\
+\bm{I}\otimes\bm{X}\hspace{1cm} & \bm{U}=(\bm{H}\otimes\bm{I})\text{SWAP}\\
+\bm{I}\otimes\bm{X}\hspace{1cm} & \bm{U}=(\bm{H}\otimes\bm{I})\text{SWAP}\\
+\bm{I}\otimes\bm{X}\hspace{1cm} & \bm{U}=(\bm{H}\bm{S}^{\dagger}\otimes\bm{I})\text{SWAP}\\
+\bm{X}\otimes\bm{Z}\hspace{1cm} & \bm{U}=CX_{10}(\bm{H}\otimes\bm{I})\\
+\bm{Y}\otimes\bm{Z}\hspace{1cm} & \bm{U}=CX_{10}(\bm{H}\bm{S}^{\dagger}\otimes\bm{I})\\
+\bm{Z}\otimes\bm{Y}\hspace{1cm} & \bm{U}=CX_{10}(\bm{I}\otimes\bm{H}\bm{S}^{\dagger})\\
+\bm{Y}\otimes\bm{X}\hspace{1cm} & \bm{U}=CX_{10}(\bm{H}\bm{S}^{\dagger}\otimes\bm{H})\\
+\bm{X}\otimes\bm{Y}\hspace{1cm} & \bm{U}=CX_{10}(\bm{H}\otimes\bm{H}\bm{S}^{\dagger})\\
+\bm{Y}\otimes\bm{Y}\hspace{1cm} & \bm{U}=CX_{10}(\bm{H}\bm{S}^{\dagger}\otimes\bm{H}\bm{S}^{\dagger})\\
 \end{align*}
 !et
 
@@ -742,20 +748,24 @@ For a two qubit system we list here the possible transformations
 ===== Additional remarks =====
 
 Here, the CNOT (CX10) operation appears for the following reason. Each Pauli
-measurement that does not include the matrix is equivalent up to a
-unitary to by the earlier reasoning. The eigenvalues of only depend on
+measurement that does not include the identity matrix is equivalent up to a
+unitary to $\bm{Z}\otimes\bm{Z}$. The eigenvalues $\bm{Z}\otimes\bm{Z}$ of only depend on
 the parity of the qubits that comprise each computational basis
 vector, and the controlled-not operations serve to compute this parity
 and store it in the first bit. Then once the first bit is measured,
 you can recover the identity of the resultant half-space, which is
 equivalent to measuring the Pauli operator.
 
-Also, while it can be tempting to assume that measuring is the same as
-sequentially measuring 𝟙 and then 𝟙 , this assumption would be
-false. The reason is that measuring projects the quantum state into
-either the or eigenstate of these operators. Measuring 𝟙 and then 𝟙
-projects the quantum state vector first onto a half space of 𝟙 and
-then onto a half space of 𝟙 . As there are four computational basis
+!split
+===== Reducing space =====
+
+Also, while it can be tempting to assume that measuring $\bm{Z}\otimes\bm{Z}$ is the same as
+sequentially measuring $\bm{Z}\otimes\bm{I}$ and then $\bm{I}\otimes\bm{Z}$, this assumption would be
+false. The reason is that measuring $\bm{Z}\otimes\bm{Z}$ projects the quantum state into
+either the $+1$ or $-1$ eigenstate of these operators. Measuring $\bm{Z}\otimes\bm{I}$ and then
+$\bm{I}\otimes\bm{Z}$
+projects the quantum state vector first onto a half space of $\bm{Z}\otimes\bm{I}$ and
+then onto a half space of $\bm{I}\otimes\bm{Z}$. As there are four computational basis
 vectors, performing both measurements reduces the state to a
 quarter-space and hence reduces it to a single computational basis
 vector.