Expand knowledge base, add /sources/

6 files changed, 249 insertions, 15 deletions

diff --git a/content/know/concept/cauchy-strain-tensor/index.pdc b/content/know/concept/cauchy-strain-tensor/index.pdc
index f150723..cb48377 100644
--- a/content/know/concept/cauchy-strain-tensor/index.pdc
+++ b/content/know/concept/cauchy-strain-tensor/index.pdc
@@ -82,7 +82,7 @@ we expand the middle term to first order in $\va{a}$:
 $$\begin{aligned}
     \va{u}(\va{x} + \va{a})
     \approx \va{u}(\va{x}) + a_x \pdv{\va{u}}{x} + a_y \pdv{\va{u}}{y} + a_z \pdv{\va{u}}{z}
-    = \va{u}(\va{x}) + \va{a} \cdot \nabla \va{u}(\va{x})
+    = \va{u}(\va{x}) + (\va{a} \cdot \nabla) \va{u}(\va{x})
 \end{aligned}$$
 
 With this, we can now define the "shift" $\delta\va{a}$
@@ -91,7 +91,7 @@ as the difference between $\va{a}$ and $\va{A}$ like so:
 $$\begin{aligned}
     \delta{\va{a}}
     \equiv \va{a} - \va{A}
-    = \va{a} \cdot \nabla \va{u}(\va{x})
+    = (\va{a} \cdot \nabla) \va{u}(\va{x})
 \end{aligned}$$
 
 In index notation, we write this expression as follows,
@@ -245,8 +245,8 @@ is easy to express using the displacement field $\va{u}$:
 $$\begin{aligned}
     \boxed{
         \delta(\dd{\va{l}})
-        = \dd{\va{l}} \cdot \nabla \va{u}
-        = (\nabla \vec{u})^\top \cdot \dd{\va{l}}
+        = (\dd{\va{l}} \cdot \nabla) \va{u}
+        %= (\nabla \vec{u})^\top \cdot \dd{\va{l}}
     }
 \end{aligned}$$
 
@@ -259,9 +259,9 @@ $$\begin{aligned}
     = \delta(\va{a} \cross \va{b} \cdot \va{c})
     &= \delta\va{a} \cross \va{b} \cdot \va{c} + \va{a} \cross \delta\va{b} \cdot \va{c} + \va{a} \cross \va{b} \cdot \delta\va{c}
     \\
-    &= (\va{a} \cdot \nabla\va{u}) \cross \va{b} \cdot \va{c}
-    + \va{a} \cross (\va{b} \cdot \nabla\va{u}) \cdot \va{c}
-    + \va{a} \cross \va{b} \cdot (\va{c} \cdot \nabla\va{u})
+    &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c}
+    + \va{a} \cross (\va{b} \cdot \nabla )\va{u} \cdot \va{c}
+    + \va{a} \cross \va{b} \cdot (\va{c} \cdot \nabla) \va{u}
 \end{aligned}$$
 
 We can reorder the factors like so
@@ -303,7 +303,7 @@ $$\begin{aligned}
     \delta(\dd{V})
     = \delta(\va{c} \cdot \dd{\va{S}})
     = \delta\va{c} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
-    = (\va{c} \cdot \nabla\va{u}) \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
+    = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
 \end{aligned}$$
 
 By comparing this to the previous result for $\delta(\dd{V})$,
@@ -311,7 +311,7 @@ we arrive at the following equation:
 
 $$\begin{aligned}
     \nabla \cdot \va{u} (\va{c} \cdot \dd{\va{S}})
-    = (\va{c} \cdot \nabla\va{u}) \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
+    = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
 \end{aligned}$$
 
 Since $\va{c}$ is dot-multiplied at the front of each term,
diff --git a/content/know/concept/deutsch-jozsa-algorithm/deutsch-circuit.png b/content/know/concept/deutsch-jozsa-algorithm/deutsch-circuit.png
new file mode 100644
index 0000000..04f81fe
--- /dev/null
+++ b/content/know/concept/deutsch-jozsa-algorithm/deutsch-circuit.png
diff --git a/content/know/concept/deutsch-jozsa-algorithm/deutsch-jozsa-circuit.png b/content/know/concept/deutsch-jozsa-algorithm/deutsch-jozsa-circuit.png
new file mode 100644
index 0000000..a43ae6a
--- /dev/null
+++ b/content/know/concept/deutsch-jozsa-algorithm/deutsch-jozsa-circuit.png
diff --git a/content/know/concept/deutsch-jozsa-algorithm/index.pdc b/content/know/concept/deutsch-jozsa-algorithm/index.pdc
new file mode 100644
index 0000000..b2b3d98
--- /dev/null
+++ b/content/know/concept/deutsch-jozsa-algorithm/index.pdc
@@ -0,0 +1,234 @@
+---
+title: "Deutsch-Jozsa algorithm"
+firstLetter: "D"
+publishDate: 2021-04-08
+categories:
+- Quantum information
+
+date: 2021-04-08T10:31:45+02:00
+draft: false
+markup: pandoc
+---
+
+# Deutsch-Jozsa algorithm
+
+The **Deutsch algorithm** and its extension, the **Deutsch-Jozsa algorithm**,
+were first to prove that quantum computers can
+solve certain problems more efficiently
+than any classical system.
+
+Given an unknown "black box" binary function $f(x)$ of one or more bits $x$,
+the goal is determine whether $f$ is
+**constant** (i.e. $f(x)$ is the same for all $x$)
+or **balanced** (i.e. exactly 50\% of all $x$-values yield $f(x) = 0$,
+and the other 50\% yield $f(x) = 1$).
+We can query $f$ as many times as we want with inputs of our choice,
+but we want to solve the problem using as few queries as possible.
+
+The problem is extremely artificial and of no practical use,
+but quantum computers can solve it with a single query,
+while classical computers need up to $2^{N - 1} + 1$ queries
+for an $N$-bit $x$.
+
+
+## Deutsch algorithm
+
+The Deutsch algorithm handles the simplest case,
+where $x$ is only a single bit.
+Only four $f$ exist:
+
++ **Constant**: $(f(0) = f(1) = 0)$ or $(f(0) = f(1) = 1)$.
++ **Balanced**: $(f(0) = 0, f(1) = 1)$, or $(f(0) = 1, f(1) = 0)$.
+
+In other words, we only need to determine if $f(0) = f(1)$ or $f(0) \neq f(1)$.
+To do this, we use the following quantum circuit,
+where $U_f$ is the oracle we query:
+
+<a href="deutsch-circuit.png">
+<img src="deutsch-circuit.png" style="width:50%;display:block;margin:auto;">
+</a>
+
+Due to unitarity constraints,
+the action of $U_f$ is defined to be as follows,
+with $\oplus$ meaning XOR:
+
+$$\begin{aligned}
+    \ket{x} \ket{y}
+    \quad \to \boxed{U_f} \to \quad
+    \ket{x} \ket{y \oplus f(x)}
+\end{aligned}$$
+
+Starting on the left from two qubits $\ket{0}$ and $\ket{1}$,
+we apply the Hadamard gate $H$ to both:
+
+$$\begin{aligned}
+    \ket{0} \ket{1}
+    \quad \to \boxed{H^{\otimes 2}} \to \quad
+    \ket{+} \ket{-}
+    = \frac{1}{2} \Big( \ket{0} + \ket{1} \Big) \Big( \ket{0} - \ket{1} \Big)
+\end{aligned}$$
+
+Feeding this result into the oracle $U_f$ then leads us to:
+
+$$\begin{aligned}
+    \to \boxed{U_f} \to \quad
+    \frac{1}{2} \ket{0} \Big( \ket{0 \oplus f(0)} - \ket{1 \oplus f(0)} \Big)
+    + \frac{1}{2} \ket{1} \Big( \ket{0 \oplus f(1)} - \ket{1 \oplus f(1)} \Big)
+\end{aligned}$$
+
+The parenthesized superpositions can be reduced.
+Assuming that $f(b) = 0$, we notice:
+
+$$\begin{aligned}
+    \ket{0 \oplus f(b)} - \ket{1 \oplus f(b)}
+    = \ket{0 \oplus 0} - \ket{1 \oplus 0}
+    = \ket{0} - \ket{1}
+\end{aligned}$$
+
+On the other hand, if we assume that $f(b) = 1$,
+we get the opposite result:
+
+$$\begin{aligned}
+    \ket{0 \oplus f(b)} - \ket{1 \oplus f(b)}
+    = \ket{0 \oplus 1} - \ket{1 \oplus 1}
+    = - \big(\ket{0} - \ket{1}\big)
+\end{aligned}$$
+
+We can thus combine both cases, $f(b) = 0$ or $f(b) = 1$,
+into the following single expression:
+
+$$\begin{aligned}
+    \ket{0 \oplus f(b)} - \ket{1 \oplus f(b)}
+    = (-1)^{f(b)} \big(\ket{0} - \ket{1}\big)
+\end{aligned}$$
+
+Using this, we rewrite the intermediate state of the quantum circuit like so:
+
+$$\begin{aligned}
+    \ket{0} \ket{1}
+    \quad \to \boxed{H^{\otimes 2}} \to \boxed{U_f} \to \quad
+    \frac{1}{2} \Big( (-1)^{f(0)} \ket{0} + (-1)^{f(1)} \ket{1} \Big) \Big( \ket{0} - \ket{1} \Big)
+\end{aligned}$$
+
+The second qubit in state $\ket{-}$ is garbage; it is no longer of interest.
+The first qubit is given by:
+
+$$\begin{aligned}
+    \frac{1}{\sqrt{2}} \Big( (-1)^{f(0)} \ket{0} + (-1)^{f(1)} \ket{1} \Big)
+    = \frac{(-1)^{f(0)}}{\sqrt{2}} \Big( \ket{0} + (-1)^{f(0) \oplus f(1)} \ket{1} \Big)
+\end{aligned}$$
+
+If $f$ is constant, then $f(0) \oplus f(1) = 0$,
+meaning this state is $(-1)^{f(0)} \ket{+}$.
+On the other hand, if $f$ is balanced, then $f(0) \oplus f(1) = 1$,
+meaning this state is $(-1)^{f(0)} \ket{-}$.
+Taking the Hadamard transform of this qubit therefore yields:
+
+$$\begin{aligned}
+    \to \boxed{H} \to \quad
+    (-1)^{f(0)} \ket{f(0) \oplus f(1)}
+\end{aligned}$$
+
+Depending on whether $f$ is constant or balanced,
+the mearurement outcome of this state will be $\ket{0}$ or $\ket{1}$
+with 100\% probability. We have solved the problem!
+
+Note that we only consulted the oracle (i.e. applied $U_f$) once.
+A classical computer would need to query it twice,
+once with input $x = 0$, and again with $x = 1$.
+
+
+## Full Deutsch-Jozsa algorithm
+
+The Deutsch-Jozsa algorithm generalizes the above to $N$-bit inputs $x$.
+We are promised that $f(x)$ is either constant or balanced;
+other possibilities are assumed to be impossible.
+This algorithm is then implemented by the following quantum circuit:
+
+<a href="deutsch-jozsa-circuit.png">
+<img src="deutsch-jozsa-circuit.png" style="width:50%;display:block;margin:auto;">
+</a>
+
+There are $N$ qubits in initial state $\ket{0}$, and one in $\ket{1}$.
+For clarity, the oracle $U_f$ works like so:
+
+$$\begin{aligned}
+    \ket{x_1} \ket{x_2} \cdots \ket{x_N} \ket{y}
+    \quad \to \boxed{U_f} \to \quad
+    \ket{x_1} \cdots \ket{x_N} \ket{y \oplus f(x_1, ..., x_N)}
+\end{aligned}$$
+
+Applying the $N + 1$ Hadamard gates to the initial state
+yields the following superposition:
+
+$$\begin{aligned}
+    \ket{0}^{\otimes N} \ket{1}
+    \quad \to \boxed{H^{\otimes N + 1}} \to \quad
+    \ket{+}^{\otimes N} \ket{-}
+    = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \ket{x} \ket{-}
+\end{aligned}$$
+
+Where $\ket{x} = \ket{x_1} \cdots \ket{x_N}$ denotes a classical binary state.
+For example, if $x = 5 = 2^0 + 2^2$ in the summation,
+then $\ket{x} = \ket{1} \ket{0} \ket{1} \ket{0}^{\otimes N-3}$
+(from least to most significant).
+
+We give this state to the oracle,
+and, by the same logic as for the Deutsch algorithm,
+get back:
+
+$$\begin{aligned}
+    \to \boxed{U_f} \to \quad
+    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \ket{x} \ket{-}
+\end{aligned}$$
+
+The last qubit $\ket{-}$ is garbage.
+Next, applying the Hadamard transform to the other $N$ gives:
+
+$$\begin{aligned}
+    \to \boxed{H^{\otimes N}} \to \quad
+    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)}
+    \bigg( \frac{1}{\sqrt{2^N}} \sum_{y = 0}^{2^N - 1} (-1)^{x \cdot y} \ket{y} \bigg)
+\end{aligned}$$
+
+Where $x \cdot y$ is the bitwise dot product of the binary representations of $x$ and $y$,
+so, for example, if $N = 2$, then $x \cdot y = x_1 y_1 + x_2 y_2$.
+Note that the above expression has not been reduced at all;
+it follows from the definition of the Hadamard transform.
+We can rewrite it like so:
+
+$$\begin{aligned}
+    \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} \sum_{y = 0}^{2^N - 1} (-1)^{f(x) + x \cdot y} \ket{y}
+    = \sum_{y = 0}^{2^N - 1} \bigg( \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x) + x \cdot y} \bigg) \ket{y}
+    = \sum_{y = 0}^{2^N - 1} c_y \ket{y}
+\end{aligned}$$
+
+The parenthesized expression can be interpreted as the coefficients
+of a superposition of several $y$-values.
+Therefore, the probability that a measurement yields $y = 0$,
+i.e. $\ket{y} = \ket{0}^{\otimes N}$, is:
+
+$$\begin{aligned}
+    |c_0|^2
+    = \bigg| \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \bigg|^2
+\end{aligned}$$
+
+The summation always contains an even number of terms, for all values of $N$.
+Consequently, if $f$ is constant, then $|c_0|^2 = |\!\pm\! 2^N / 2^N|^2 = 1$.
+Otherwise, if $f$ is balanced, all the terms cancel out, so we are left with $|c_0|^2 = 0$.
+In other words, we reach the same result as the Deutsch algorithm:
+we only need to measure the $N$ qubits once;
+$f$ is constant if and only if all are zero.
+
+The Deutsch-Jozsa algorithm needs only one oracle query to give an error-free result,
+whereas a classical computer needs $2^{N-1} + 1$ queries in the worst case;
+a revolutionary discovery.
+
+
+## References
+1.  J.S. Neergaard-Nielsen,
+    *Quantum information: lectures notes*,
+    2021, unpublished.
+2.  S. Aaronson,
+    *Introduction to quantum information science: lecture notes*,
+    2018, unpublished.
diff --git a/content/know/concept/euler-equations/index.pdc b/content/know/concept/euler-equations/index.pdc
index 37d2fea..cedfd93 100644
--- a/content/know/concept/euler-equations/index.pdc
+++ b/content/know/concept/euler-equations/index.pdc
@@ -149,7 +149,7 @@ to which we apply a vector identity:
 $$\begin{aligned}
     0
     = \dv{\rho}{t} + \nabla \cdot (\rho \va{v})
-    = \dv{\rho}{t} + \va{v} \cdot \nabla \rho + \rho (\nabla \cdot \va{v})
+    = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho + \rho (\nabla \cdot \va{v})
 \end{aligned}$$
 
 Thanks to incompressibility, the last term disappears,
@@ -159,7 +159,7 @@ $$\begin{aligned}
     \boxed{
         0
         = \frac{\mathrm{D} \rho}{\mathrm{D} t}
-        = \dv{\rho}{t} + \va{v} \cdot \nabla \rho
+        = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho
     }
 \end{aligned}$$
 
diff --git a/content/know/concept/material-derivative/index.pdc b/content/know/concept/material-derivative/index.pdc
index af65ca0..1c6bfdc 100644
--- a/content/know/concept/material-derivative/index.pdc
+++ b/content/know/concept/material-derivative/index.pdc
@@ -57,7 +57,7 @@ then we can rewrite this expression like so:
 
 $$\begin{aligned}
     \dv{t} f\big(x(t), y(t), z(t), t\big)
-    &= \pdv{f}{t} + \va{v} \cdot \nabla f
+    &= \pdv{f}{t} + (\va{v} \cdot \nabla) f
 \end{aligned}$$
 
 Note that $\va{v} = \va{v}(\va{r}, t)$,
@@ -73,7 +73,7 @@ and is known as the **material derivative** or **comoving derivative**:
 $$\begin{aligned}
     \boxed{
         \frac{\mathrm{D}f}{\mathrm{D}t}
-        \equiv \pdv{f}{t} + \va{v} \cdot \nabla f
+        \equiv \pdv{f}{t} + (\va{v} \cdot \nabla) f
     }
 \end{aligned}$$
 
@@ -89,14 +89,14 @@ but in fact the definition also works for vector fields $\va{U}(\va{r}, t)$:
 $$\begin{aligned}
     \boxed{
         \frac{\mathrm{D} \va{U}}{\mathrm{D}t}
-        \equiv \pdv{f}{t} + \va{v} \cdot \nabla \va{U}
+        \equiv \pdv{\va{U}}{t} + (\va{v} \cdot \nabla) \va{U}
     }
 \end{aligned}$$
 
 Where the advective term is to be evaluated in the following way in Cartesian coordinates:
 
 $$\begin{aligned}
-    \va{v} \cdot \nabla \va{U}
+    (\va{v} \cdot \nabla) \va{U}
     =
     \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix}
     \cdot