tons of typos

Author: mhjensen

doc/pub/week6/html/week6-bs.html

@@ -250,34 +250,34 @@ <h2 id="single-qubit-gates">Single qubit gates </h2>
 </section>
 
 <section>
-<h2 id="pauli-x-gate">Pauli-\( X \) gate </h2>
+<h2 id="pauli-boldsymbol-x-gate">Pauli-\( \boldsymbol{X} \) gate </h2>
 
-<p>The Pauli-\( X \) gate is also known as the <b>NOT</b> gate, which flips the state of the qubit.</p>
+<p>The Pauli-\( \boldsymbol{X} \) gate is also known as the <b>NOT</b> gate, which flips the state of the qubit.</p>
 <p>&nbsp;<br>
 $$
 \begin{align*}
-	X\vert 0\rangle &= \vert 1\rangle, \\
-	X\vert 1\rangle &= \vert 0\rangle.	
+	\boldsymbol{X}\vert 0\rangle &= \vert 1\rangle, \\
+	\boldsymbol{X}\vert 1\rangle &= \vert 0\rangle.	
 \end{align*}
 $$
 <p>&nbsp;<br>
 
-<p>The Pauli-\( Y \) gate flips the bit and multiplies the phase by $ i $. </p>
+<p>The Pauli-\( \boldsymbol{Y} \) gate flips the bit and multiplies the phase by $ i $. </p>
 <p>&nbsp;<br>
 $$
 \begin{align*}
-	Y\vert 0\rangle &= i\vert 1\rangle, \\
-	Y\vert 1\rangle &= -i\vert 0\rangle.
+	\boldsymbol{Y}\vert 0\rangle &= i\vert 1\rangle, \\
+	\boldsymbol{Y}\vert 1\rangle &= -i\vert 0\rangle.
 \end{align*}
 $$
 <p>&nbsp;<br>
 
-<p>The Pauli-\( Z \) gate multiplies only the phase of \( \vert 1\rangle \) by $ -1 \( .</p>
+<p>The Pauli-\( \boldsymbol{Z} \) gate multiplies only the phase of \( \vert 1\rangle \) by $ -1 \( .</p>
 <p>&nbsp;<br>
 $$
 \begin{align*}
-	Z\vert 0\rangle &= \vert 0\rangle, \\
-	Z\vert 1\rangle &= -\vert 1\rangle.
+	\boldsymbol{Z}\vert 0\rangle &= \vert 0\rangle, \\
+	\boldsymbol{Z}\vert 1\rangle &= -\vert 1\rangle.
 \end{align*}
 $$
 <p>&nbsp;<br>
@@ -289,7 +289,7 @@ <h2 id="hadamard-gate">Hadamard gate </h2>
 <p>The Hadamard gate is defined as</p>
 <p>&nbsp;<br>
 $$
-	H = \frac{1}{\sqrt{2}} \begin{pmatrix}
+	\boldsymbol{H} = \frac{1}{\sqrt{2}} \begin{pmatrix}
 		1 & 1 \\
 		1 & -1
 	\end{pmatrix}.
@@ -300,9 +300,9 @@ <h2 id="hadamard-gate">Hadamard gate </h2>
 <p>&nbsp;<br>
 $$
 \begin{align}
-	H\vert 0\rangle &= \frac{1}{\sqrt{2}} \left( \vert 0\rangle + \vert 1\rangle \right), 
+	\boldsymbol{H}\vert 0\rangle &= \frac{1}{\sqrt{2}} \left( \vert 0\rangle + \vert 1\rangle \right), 
 \tag{1}\\
-	H\vert 1\rangle &= \frac{1}{\sqrt{2}} \left( \vert 0\rangle - \vert 1\rangle \right).
+	\boldsymbol{H}\vert 1\rangle &= \frac{1}{\sqrt{2}} \left( \vert 0\rangle - \vert 1\rangle \right).
 \tag{2}
 \end{align}
 $$
@@ -316,7 +316,7 @@ <h2 id="phase-gates">Phase Gates </h2>
 <p>The phase gate is usually denoted as \( S \) and is defined as</p>
 <p>&nbsp;<br>
 $$
-	S = \begin{pmatrix}
+	\boldsymbol{S} = \begin{pmatrix}
 		1 & 0 \\
 		0 & i
 	\end{pmatrix}.
@@ -327,32 +327,32 @@ <h2 id="phase-gates">Phase Gates </h2>
 <p>&nbsp;<br>
 $$
 \begin{align*}
-	S\vert 0\rangle &= \vert 0\rangle, \\
-	S\vert 1\rangle &= i\vert 1\rangle.
+	\boldsymbol{S}\vert 0\rangle &= \vert 0\rangle, \\
+	\boldsymbol{S}\vert 1\rangle &= i\vert 1\rangle.
 \end{align*}
 $$
 <p>&nbsp;<br>
 </section>
 
 <section>
-<h2 id="the-inverse-of-the-s-gate">The  inverse of the \( S \)-gate </h2>
+<h2 id="the-inverse-of-the-boldsymbol-s-gate">The  inverse of the \( \boldsymbol{S} \)-gate </h2>
 
 <p>The inverse</p>
 <p>&nbsp;<br>
 $$
-	S^\dagger = \begin{pmatrix}
+	\boldsymbol{S}^\dagger = \begin{pmatrix}
 		1 & 0 \\
 		0 & -i
 	\end{pmatrix}
 $$
 <p>&nbsp;<br>
 
-<p>is known as the $ S^\dagger$ gate which applies an \( \imath \) phase shift to \( \vert 1\rangle \).</p>
+<p>is known as the $ \boldsymbol{S}^\dagger$ gate which applies an \( \imath \) phase shift to \( \vert 1\rangle \).</p>
 <p>&nbsp;<br>
 $$
 \begin{align*}
-	S^\dagger\vert 0\rangle &= \vert 0\rangle, \\
-	S^\dagger\vert 1\rangle &= -i\vert 1\rangle.
+	\boldsymbol{S}^\dagger\vert 0\rangle &= \vert 0\rangle, \\
+	\boldsymbol{S}^\dagger\vert 1\rangle &= -i\vert 1\rangle.
 \end{align*}
 $$
 <p>&nbsp;<br>
@@ -415,12 +415,12 @@ <h2 id="the-swap-gate">The SWAP gate </h2>
 <section>
 <h2 id="pauli-strings">Pauli Strings </h2>
 
-<p>A Pauli string, such as $ XIYZ $ is a tensor product of Pauli matrices acting on different qubits.
-The Pauli string $ XIYZ $ is defined as (from qubit one to qubit four, from left to right)
+<p>A Pauli string, such as \( \boldsymbol{XIYZ} \) is a tensor product of Pauli matrices acting on different qubits.
+The Pauli string \( \boldsymbol{XIYZ} \) is defined as (from qubit one to qubit four, from left to right)
 </p>
 <p>&nbsp;<br>
 $$
-	XIYZ \equiv X_0 \otimes I_1 \otimes Y_2 \otimes Z_3.
+	\boldsymbol{XIYZ} \equiv \boldsymbol{X}_0 \otimes \boldsymbol{I}_1 \otimes \boldsymbol{Y}_2 \otimes \boldsymbol{Z}_3.
 $$
 <p>&nbsp;<br>
 
@@ -730,33 +730,33 @@ <h2 id="expectation-values">Expectation values </h2>
 by post-processing measurements of quantum circuits in different
 basis sets. To rotate bases, one uses the basis rotator \( R_\sigma \) which is
 defined for each Pauli gate \( \sigma \) to be (using the Hadamard rotation \( H \) and Phase rotation \( S \))
-for a Pauli-\( X \) matrix
+for a Pauli-\( \boldsymbol{X} \) matrix
 </p>
 <p>&nbsp;<br>
 $$
-X=R_{\sigma}ZR_{\sigma} = HZH
+\boldsymbol{X}=R_{\sigma}\boldsymbol{Z}R_{\sigma} = HZH
 $$
 <p>&nbsp;<br>
 
-<p>for a Pauli-\( Y \) matrix</p>
+<p>for a Pauli-\( \boldsymbol{Y} \) matrix</p>
 <p>&nbsp;<br>
 $$
-Y=R_{\sigma}ZR_{\sigma}=HS^{\dagger}ZHS,
+\boldsymbol{Y}=R_{\sigma}\boldsymbol{Z}R_{\sigma}=\boldsymbol{HS}^{\dagger}\boldsymbol{ZHS},
 $$
 <p>&nbsp;<br>
 
 <p>and</p>
 <p>&nbsp;<br>
 $$
-Z=R_{\sigma}ZR_{\sigma}=\boldsymbol{I}Z\boldsymbol{I}=Z.
+\boldsymbol{Z}=R_{\sigma}ZR_{\sigma}=\boldsymbol{I}\boldsymbol{Z}\boldsymbol{I}=\boldsymbol{Z}.
 $$
 <p>&nbsp;<br>
 </section>
 
 <section>
 <h2 id="measurements-of-eigenvalues-of-the-pauli-operators">Measurements of eigenvalues of the Pauli operators </h2>
 
-<p>We can show that these rotations allow us to measure the eigenvalues of the Pauli operators. The eigenvectors of the Pauli \( X \) gate are</p>
+<p>We can show that these rotations allow us to measure the eigenvalues of the Pauli operators. The eigenvectors of the Pauli \( \boldsymbol{X} \) gate are</p>
 <p>&nbsp;<br>
 $$
 \vert\pm\rangle = \frac{\vert 0\rangle \pm \vert 1\rangle}{\sqrt{2}},
@@ -768,14 +768,14 @@ <h2 id="measurements-of-eigenvalues-of-the-pauli-operators">Measurements of eige
 </p>
 <p>&nbsp;<br>
 $$
-H\vert +\rangle = +1\vert 0\rangle,
+\boldsymbol{H}\vert +\rangle = +1\vert 0\rangle,
 $$
 <p>&nbsp;<br>
 
 <p>and</p>
 <p>&nbsp;<br>
 $$
-H\vert -\rangle = -1\vert 1\rangle.
+\boldsymbol{H}\vert -\rangle = -1\vert 1\rangle.
 $$
 <p>&nbsp;<br>
 </section>
@@ -790,7 +790,7 @@ <h2 id="single-qubit-states">Single-qubit states </h2>
 $$
 <p>&nbsp;<br>
 
-<p>We then have the following expectation value for the Pauli \( X \) operator</p>
+<p>We then have the following expectation value for the Pauli \( \boldsymbol{X} \) operator</p>
 <p>&nbsp;<br>
 $$
 \langle \vert X\vert \rangle = \langle \psi\vert X \vert \psi\rangle = |\alpha|^2 - |\beta|^2.
@@ -809,10 +809,10 @@ <h2 id="single-qubit-states">Single-qubit states </h2>
 <h2 id="interpretations">Interpretations </h2>
 
 <p>This tells us that we are able to estimate \( |\alpha|^2 \) and
-\( |\beta|^2 \) (and hence the expectation value of the Pauli \( X \)
+\( |\beta|^2 \) (and hence the expectation value of the Pauli \( \boldsymbol{X} \)
 operator) by using a rotation and measure the
 resulting state in the computational basis. We can show this for the
-Pauli \( Z \) and Pauli \( Y \) similarly.
+Pauli \( \boldsymbol{Z} \) and Pauli \( \boldsymbol{Y} \) similarly.
 </p>
 </section>
 
@@ -822,11 +822,11 @@ <h2 id="reminder-on-rotations">Reminder on rotations </h2>
 <p>Note the following identity of the basis rotator</p>
 <p>&nbsp;<br>
 $$
-R^\dagger_\sigma Z R_\sigma = \sigma,
+R^\dagger_\sigma \boldsymbol{Z} R_\sigma = \sigma,
 $$
 <p>&nbsp;<br>
 
-<p>which follows from the fact that \( HZH=X \) and \( SXS^\dagger=Y \).</p>
+<p>which follows from the fact that \( \boldsymbol{HZH}=\boldsymbol{X} \) and \( \boldsymbol{SXS}^\dagger=\boldsymbol{Y} \).</p>
 </section>
 
 <section>
@@ -994,7 +994,7 @@ <h2 id="vqe-and-efficient-computations-of-gradients">VQE and efficient computati
 </p>
 
 <p>We start with a simple \( 2\times 2 \) Hamiltonian matrix expressed in
-terms of Pauli \( X \) and \( Z \) matrices, as discussed in the project text.
+terms of Pauli \( \boldsymbol{X} \) and \( \boldsymbol{Z} \) matrices, as discussed in the project text.
 </p>
 </section>
 
@@ -1312,8 +1312,8 @@ <h2 id="computational-basis">Computational basis </h2>
 <p>The above equations require that we can make measurements in the chosen basis sets.</p>
 
 <p>However, this may not be possible. The difficulty comes  from the fact that one may have the possibility
-to measure only in the \( Z \)-basis. To solve this difficulty we still do
-a \( Z \)-basis measurement, but, before that, we apply specific operators
+to measure only in the \( \boldsymbol{Z} \)-basis. To solve this difficulty we still do
+a \( \boldsymbol{Z} \)-basis measurement, but, before that, we apply specific operators
 to the \( \left| \psi \right\rangle \) state.
 </p>
 </section>
@@ -1338,7 +1338,7 @@ <h2 id="unitary-transformation-of-boldsymbol-x">Unitary transformation of \( \bo
 $$
 <p>&nbsp;<br>
 
-<p>The Hadamard gate/matrix is a unitary matrix with the property that \( \boldsymbol{H}^2=\boldsymbol{I} \).</p>
+<p>The Hadamard gate/matrix is a unitary matrix with the property that \( H^2=\boldsymbol{I} \).</p>
 </section>
 
 <section>
@@ -1362,7 +1362,7 @@ <h2 id="generalizing">Generalizing </h2>
 <p>where \( \mathcal{P} \) represents some combination of the Pauli matrices and
 the identity matrix, \( \boldsymbol{U} \) is a unitary matrix and \( \boldsymbol{M} \)
 represents the gate/matrix which performs the measurements, often
-represented by a Pauli \( \boldsymbol{Z} \) gate/matrix.
+represented by a Pauli-\( \boldsymbol{Z} \) gate/matrix.
 </p>
 </section>
 
@@ -1616,7 +1616,7 @@ <h2 id="rotations-again-and-again">Rotations again and again </h2>
 <p>&nbsp;<br>
 
 <p>with \( \boldsymbol{sigma}_i \), with \( \boldsymbol{\sigma}_i \) being any of the Pauli
-matrices \( X \), \( Y \) and \( Z \). The latter can be generalized to other
+matrices \( \boldsymbol{X} \), \( \boldsymbol{Y} \) and \( \boldsymbol{Z} \). The latter can be generalized to other
 unitary matrices as well.
 The derivative with respect to \( \phi \) gives
 </p>
@@ -1642,7 +1642,7 @@ <h2 id="bloch-sphere-math">Bloch sphere math </h2>
 
 <p>&nbsp;<br>
 $$
-\langle \psi \vert \hat{H}\vert \psi \rangle = \langle 0 \vert R_i(\phi)^{\dagger} \hat{H}R_i(\phi)\vert 0\rangle. 
+\langle \psi \vert \mathcal{H}\vert \psi \rangle = \langle 0 \vert R_i(\phi)^{\dagger} \mathcal{H}R_i(\phi)\vert 0\rangle. 
 $$
 <p>&nbsp;<br>
 </section>
@@ -1680,15 +1680,15 @@ <h2 id="rewriting">Rewriting </h2>
 <h2 id="final-manipulations">Final manipulations </h2>
 
 <p>If we identify these operators as \( \hat{A}=\boldsymbol{I} \), with
-\( \boldsymbol{I} \) being the unit operator, \( \hat{B}=\hat{H} \) our Hamiltonian,
+\( \boldsymbol{I} \) being the unit operator, \( \hat{B}=\mathcal{H} \) our Hamiltonian,
 and \( \hat{C}=-\imath \boldsymbol{\sigma}_i/2 \), we obtain the following
 expression for the expectation value of the derivative (excluding the hermitian conjugate)
 </p>
 
 <p>&nbsp;<br>
 $$
-\langle \psi \vert \boldsymbol{I}^{\dagger}\hat{H}(-\frac{\imath}{2}\boldsymbol{\sigma}_i\vert \psi \rangle = \frac{1}{2}\left[
-\langle \psi \vert (\boldsymbol{I}-\frac{\imath}{2} \boldsymbol{\sigma}_i)^{\dagger}\hat{H}(\boldsymbol{I}-\frac{\imath}{2} \boldsymbol{\sigma}_i)\vert \psi \rangle-\langle \psi \vert (\boldsymbol{I}+\frac{\imath}{2} \boldsymbol{\sigma}_i)^{\dagger}\hat{H}(\boldsymbol{I}+\frac{\imath}{2} \boldsymbol{\sigma}_i)\vert \psi \rangle\right].
+\langle \psi \vert \boldsymbol{I}^{\dagger}\mathcal{H}(-\frac{\imath}{2}\boldsymbol{\sigma}_i\vert \psi \rangle = \frac{1}{2}\left[
+\langle \psi \vert (\boldsymbol{I}-\frac{\imath}{2} \boldsymbol{\sigma}_i)^{\dagger}\mathcal{H}(\boldsymbol{I}-\frac{\imath}{2} \boldsymbol{\sigma}_i)\vert \psi \rangle-\langle \psi \vert (\boldsymbol{I}+\frac{\imath}{2} \boldsymbol{\sigma}_i)^{\dagger}\mathcal{H}(\boldsymbol{I}+\frac{\imath}{2} \boldsymbol{\sigma}_i)\vert \psi \rangle\right].
 $$
 <p>&nbsp;<br>
 </section>
@@ -1717,8 +1717,8 @@ <h2 id="final-expression">Final expression </h2>
 <p>This means that we can write</p>
 <p>&nbsp;<br>
 $$
-\langle \psi \vert \boldsymbol{I}^{\dagger}\hat{H}(-\frac{\imath}{2}\boldsymbol{\sigma}_i\vert \psi \rangle = \frac{1}{2}\left[
-\langle \psi \vert R_i(\frac{\pi}{2})^{\dagger}\hat{H}R_i(\frac{\pi}{2})\vert \psi \rangle-\langle \psi \vert R_i(-\frac{\pi}{2})^{\dagger}\hat{H}R_i(-\frac{\pi}{2})^{\dagger}\vert \psi \rangle\right]=\frac{1}{2}(\langle\hat{H}(\phi+\frac{\pi}{2})\rangle-\langle\hat{H}(\phi-\frac{\pi}{2})\rangle).
+\langle \psi \vert \boldsymbol{I}^{\dagger}\mathcal{H}(-\frac{\imath}{2}\boldsymbol{\sigma}_i\vert \psi \rangle = \frac{1}{2}\left[
+\langle \psi \vert R_i(\frac{\pi}{2})^{\dagger}\mathcal{H}R_i(\frac{\pi}{2})\vert \psi \rangle-\langle \psi \vert R_i(-\frac{\pi}{2})^{\dagger}\mathcal{H}R_i(-\frac{\pi}{2})^{\dagger}\vert \psi \rangle\right]=\frac{1}{2}(\langle\mathcal{H}(\phi+\frac{\pi}{2})\rangle-\langle\mathcal{H}(\phi-\frac{\pi}{2})\rangle).
 $$
 <p>&nbsp;<br>
 </section>