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10 files changed, 524 insertions, 16 deletions

diff --git a/content/know/concept/bell-state/index.pdc b/content/know/concept/bell-state/index.pdc
new file mode 100644
index 0000000..5e147e2
--- /dev/null
+++ b/content/know/concept/bell-state/index.pdc
@@ -0,0 +1,93 @@
+---
+title: "Bell state"
+firstLetter: "B"
+publishDate: 2021-03-09
+categories:
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-09T17:31:29+01:00
+draft: false
+markup: pandoc
+---
+
+# Bell state
+
+In quantum information, the **Bell states** are a set of four two-qubit states
+which are simple and useful examples of [quantum entanglement](/know/concept/quantum-entanglement/).
+They are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \ket*{\Phi^{\pm}}
+            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{0}_B \pm \ket{1}_A \ket{1}_B \Big)
+            \\
+            \ket*{\Psi^{\pm}}
+            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{1}_B \pm \ket{1}_A \ket{0}_B \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where e.g. $\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$
+is the tensor product of qubit $A$ in state $\ket{0}$ and $B$ in $\ket{1}$.
+These states form an orthonormal basis for the two-qubit
+[Hilbert space](/know/concept/hilbert-space/).
+
+More importantly, however,
+is that the Bell states are maximally entangled,
+which we prove here for $\ket*{\Phi^{+}}$.
+Consider the following pure [density operator](/know/concept/density-operator/):
+
+$$\begin{aligned}
+    \hat{\rho}
+    = \ket*{\Phi^{+}} \bra*{\Phi^{+}}
+    &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big)
+\end{aligned}$$
+
+The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows:
+
+$$\begin{aligned}
+    \hat{\rho}_A
+    &= \Tr_B(\hat{\rho})
+    = \sum_{b = 0, 1} \bra{b}_B \Big( \ket*{\Phi^{+}} \bra*{\Phi^{+}} \Big) \ket{b}_B
+    \\
+    &= \sum_{b = 0, 1} \Big( \ket{0}_A \braket{b}{0}_B + \ket{1}_A \braket{b}{1}_B \Big)
+    \Big( \bra{0}_A \braket{0}{b}_B + \bra{1}_A \braket{1}{b}_B \Big)
+    \\
+    &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big)
+    = \frac{1}{2} \hat{I}
+\end{aligned}$$
+
+This result is maximally mixed, therefore $\ket*{\Phi^{+}}$ is maximally entangled.
+The same holds for the other three Bell states,
+and is equally true for qubit $B$.
+
+This means that a measurement of qubit $A$
+has a 50-50 chance to yield $\ket{0}$ or $\ket{1}$.
+However, due to the entanglement,
+measuring $A$ also has consequences for qubit $B$:
+
+$$\begin{aligned}
+    \big| \bra{0}_A \bra{0}_B \: \ket*{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \braket{0}{0}_A \braket{0}{0}_B + \braket{0}{1}_A \braket{0}{1}_B \Big)^2
+    = \frac{1}{2}
+    \\
+    \big| \bra{0}_A \bra{1}_B \: \ket*{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \braket{0}{0}_A \braket{1}{0}_B + \braket{0}{1}_A \braket{1}{1}_B \Big)^2
+    = 0
+    \\
+    \big| \bra{1}_A \bra{0}_B \: \ket*{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \braket{1}{0}_A \braket{0}{0}_B + \braket{1}{1}_A \braket{0}{1}_B \Big)^2
+    = 0
+    \\
+    \big| \bra{1}_A \bra{1}_B \: \ket*{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \braket{1}{0}_A \braket{1}{0}_B + \braket{1}{1}_A \braket{1}{1}_B \Big)^2
+    = \frac{1}{2}
+\end{aligned}$$
+
+As an example, if $A$ collapses into $\ket{0}$ due to a measurement,
+then $B$ instantly also collapses into $\ket{0}$, never $\ket{1}$,
+even if it was not measured.
+This was a specific example for $\ket*{\Phi^{+}}$,
+but analogous results can be found for the other Bell states.
diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc
index 68dec8c..70cc897 100644
--- a/content/know/concept/binomial-distribution/index.pdc
+++ b/content/know/concept/binomial-distribution/index.pdc
@@ -191,3 +191,9 @@ $$\begin{aligned}
         \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big)
     }
 \end{aligned}$$
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/bloch-sphere/bloch-small.jpg b/content/know/concept/bloch-sphere/bloch-small.jpg
new file mode 100644
index 0000000..e99c0e1
--- /dev/null
+++ b/content/know/concept/bloch-sphere/bloch-small.jpg
diff --git a/content/know/concept/bloch-sphere/bloch.jpg b/content/know/concept/bloch-sphere/bloch.jpg
new file mode 100644
index 0000000..9515d84
--- /dev/null
+++ b/content/know/concept/bloch-sphere/bloch.jpg
diff --git a/content/know/concept/bloch-sphere/index.pdc b/content/know/concept/bloch-sphere/index.pdc
new file mode 100644
index 0000000..843de1f
--- /dev/null
+++ b/content/know/concept/bloch-sphere/index.pdc
@@ -0,0 +1,129 @@
+---
+title: "Bloch sphere"
+firstLetter: "B"
+publishDate: 2021-03-09
+categories:
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-09T15:35:33+01:00
+draft: false
+markup: pandoc
+---
+
+# Bloch sphere
+
+In quantum mechanics, particularly quantum information,
+the **Bloch sphere** is an invaluable tool to visualize qubits.
+All pure qubit states are represented by a point on the sphere's surface:
+
+<a href="bloch.jpg">
+<img src="bloch-small.jpg" style="width:50%;display:block;margin:auto;">
+</a>
+
+The $x$, $y$ and $z$-axes represent the components of a spin-1/2-alike system,
+and their extremes are the eigenstates of the Pauli matrices:
+
+$$\begin{aligned}
+    \hat{\sigma}_z
+    \to \{\ket{0}, \ket{1}\}
+    \qquad
+    \hat{\sigma}_x
+    \to \{\ket{+}, \ket{-}\}
+    \qquad
+    \hat{\sigma}_y
+    \to \{\ket{+i}, \ket{-i}\}
+\end{aligned}$$
+
+Where the latter two states are expressed as follows in the conventional $z$-basis:
+
+$$\begin{aligned}
+    \ket{\pm}
+    = \frac{\ket{0} \pm \ket{1}}{\sqrt{2}}
+    \qquad \quad
+    \ket{\pm i}
+    = \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}}
+\end{aligned}$$
+
+More generally, every point on the surface of the sphere
+describes a pure qubit state in terms of the angles $\theta$ and $\varphi$,
+respectively the elevation and azimuth:
+
+$$\begin{aligned}
+    \ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \ket{1}
+\end{aligned}$$
+
+We can generalize this further by describing points using the **Bloch vector** $\vec{r}$,
+with radius $r \le 1$:
+
+$$\begin{aligned}
+    \boxed{
+        \vec{r}
+        = \begin{bmatrix} r_x \\ r_y \\ r_z \end{bmatrix}
+        = \begin{bmatrix} r \sin\theta \cos\varphi \\ r \sin\theta \sin\varphi \\ r \cos\theta \end{bmatrix}
+    }
+\end{aligned}$$
+
+Note that $\vec{r}$ is not actually a qubit state,
+but rather an implicit description of one,
+meaning that it does not need to be normalized.
+The main point of the Bloch vector is that it allows us
+to describe the qubit using a [density operator](/know/concept/density-operator/):
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}
+        = \frac{1}{2} \Big( \hat{I} + \vec{r} \cdot \vec{\sigma} \Big)
+    }
+\end{aligned}$$
+
+Where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$ is the Pauli "vector".
+Now, we know that $\hat{\rho}$ represents a pure ensemble
+if and only if it is idempotent, i.e. $\hat{\rho}^2 = \hat{\rho}$:
+
+$$\begin{aligned}
+    \hat{\rho}^2
+    &= \frac{1}{4} \Big( \hat{I}^2 + 2 \hat{I} (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
+    = \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
+\end{aligned}$$
+
+You can easily convince yourself that if $(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$,
+then we get $\hat{\rho}$ again, and the state is pure:
+
+$$\begin{aligned}
+    (\vec{r} \cdot \vec{\sigma})^2
+    &= (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z)^2
+    \\
+    &= r_x^2 \hat{\sigma}_x^2 + r_x r_y \hat{\sigma}_x \hat{\sigma}_y + r_x r_z \hat{\sigma}_x \hat{\sigma}_z
+    + r_x r_y \hat{\sigma}_y \hat{\sigma}_x + r_y^2 \hat{\sigma}_y^2
+    \\
+    &\quad + r_y r_z \hat{\sigma}_y \hat{\sigma}_z + r_x r_z \hat{\sigma}_z \hat{\sigma}_x
+    + r_y r_z \hat{\sigma}_z \hat{\sigma}_y + r_z^2 \hat{\sigma}_z^2
+    \\
+    &= r_x^2 \hat{I} + r_y^2 \hat{I} + r_z^2 \hat{I}
+    + r_x r_y \{ \hat{\sigma}_x,  \hat{\sigma}_y \}
+    + r_y r_z \{ \hat{\sigma}_y,  \hat{\sigma}_z \}
+    + r_x r_z \{ \hat{\sigma}_x,  \hat{\sigma}_z \}
+    \\
+    &= (r_x^2 + r_y^2 + r_z^2) \hat{I}
+    = r^2 \hat{I}
+\end{aligned}$$
+
+Therefore, if the radius $r = 1$, the ensemble is pure,
+else if $r < 1$ it is mixed.
+
+Another useful property of the Bloch vector
+is that the expectation value of the Pauli matrices
+are given by the corresponding component of $\vec{r}$,
+for example for $\hat{\sigma}_z$:
+
+$$\begin{aligned}
+    \expval{\hat{\sigma}_z}
+    &= \Tr(\hat{\rho} \hat{\sigma}_z)
+    = \frac{1}{2} \Tr\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
+    = \frac{1}{2} \Tr\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
+    \\
+    &= \frac{1}{2} \Tr\big( r_x \hat{\sigma}_x \hat{\sigma}_z + r_y \hat{\sigma}_y \hat{\sigma}_z + r_z \hat{\sigma}_z^2 \big)
+    = \frac{1}{2} \Tr\big( r_z \hat{I} \big)
+    = r_z
+\end{aligned}$$
diff --git a/content/know/concept/central-limit-theorem/index.pdc b/content/know/concept/central-limit-theorem/index.pdc
new file mode 100644
index 0000000..270bb0b
--- /dev/null
+++ b/content/know/concept/central-limit-theorem/index.pdc
@@ -0,0 +1,208 @@
+---
+title: "Central limit theorem"
+firstLetter: "C"
+publishDate: 2021-03-09
+categories:
+- Statistics
+
+date: 2021-03-09T20:39:38+01:00
+draft: false
+markup: pandoc
+---
+
+# Central limit theorem
+
+In statistics, the **central limit theorem** states that
+the sum of many independent variables tends towards a normal distribution,
+even if the individual variables $x_n$ follow different distributions.
+
+For example, by taking $M$ samples of size $N$ from a population,
+and calculating $M$ averages $\mu_m$ (which involves summing over $N$),
+the resulting means $\mu_m$ are normally distributed
+across the $M$ samples if $N$ is sufficiently large.
+
+More formally, for $N$ independent variables $x_n$ with probability distributions $p(x_n)$,
+the central limit theorem states the following,
+where we define the sum $S$:
+
+$$\begin{aligned}
+    S = \sum_{n = 1}^N x_n
+    \qquad
+    \mu_S = \sum_{n = 1}^N \mu_n
+    \qquad
+    \sigma_S^2 = \sum_{n = 1}^N \sigma_n^2
+\end{aligned}$$
+
+And crucially, it states that the probability distribution $p_N(S)$ of $S$ for $N$ variables
+will become a normal distribution when $N$ goes to infinity:
+
+$$\begin{aligned}
+    \boxed{
+        \lim_{N \to \infty} \!\big(p_N(S)\big)
+        = \frac{1}{\sigma_S \sqrt{2 \pi}} \: \exp\!\Big( -\frac{(\mu_S - S)^2}{2 \sigma_S^2} \Big)
+    }
+\end{aligned}$$
+
+We prove this below,
+but first we need to introduce some tools.
+Given a probability density $p(x)$, its [Fourier transform](/know/concept/fourier-transform/)
+is called the **characteristic function** $\phi(k)$:
+
+$$\begin{aligned}
+    \phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x}
+\end{aligned}$$
+
+Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$.
+We take its Taylor expansion in two separate ways,
+where an overline denotes the mean:
+
+$$\begin{aligned}
+    \phi(k)
+    = \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0)
+    \qquad
+    \phi(k)
+    = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
+\end{aligned}$$
+
+By comparing the coefficients of these two power series,
+we get a useful relation:
+
+$$\begin{aligned}
+    \phi^{(n)}(0) = i^n \: \overline{x^n}
+\end{aligned}$$
+
+Next, the **cumulants** $C^{(n)}$ are defined from the Taylor expansion of $\ln\!\big(\phi(k)\big)$:
+
+$$\begin{aligned}
+    \ln\!\big( \phi(k) \big)
+    = \sum_{n = 1}^\infty \frac{(ik)^n}{n!} C^{(n)}
+    \quad \mathrm{where} \quad
+    C^{(n)} = \frac{1}{i^n} \: \dv[n]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
+\end{aligned}$$
+
+The first two cumulants $C^{(1)}$ and $C^{(2)}$ are of particular interest,
+since they turn out to be the mean and the variance respectively,
+using our earlier relation:
+
+$$\begin{aligned}
+    C^{(1)}
+    &= - i \dv{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
+    = - i \frac{\phi'(0)}{\exp(0)}
+    = \overline{x}
+    \\
+    C^{(2)}
+    &= - \dv[2]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
+    = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)}
+    = - \overline{x}^2 + \overline{x^2} = \sigma^2
+\end{aligned}$$
+
+Let us now define $S$ as the sum of $N$ independent variables $x_n$, in other words:
+
+$$\begin{aligned}
+    S = \sum_{n = 1}^N x_n = x_1 + x_2 + ... + x_N
+\end{aligned}$$
+
+The probability density of $S$ is then as follows, where $p(x_n)$ are
+the densities of all the individual variables and $\delta$ is
+the [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+    p(S)
+    &= \idotsint_{-\infty}^\infty \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( S - \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N}
+    \\
+    &= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(S)
+\end{aligned}$$
+
+In other words, the integrals pick out all combinations of $x_n$ which
+add up to the desired $S$-value, and multiply the probabilities
+$p(x_1) p(x_2) \cdots p(x_N)$ of each such case. This is a convolution,
+so the [convolution theorem](/know/concept/convolution-theorem/)
+states that it is a product in the Fourier domain:
+
+$$\begin{aligned}
+    \phi_S(k) = \prod_{n = 1}^N \phi_n(k)
+\end{aligned}$$
+
+By taking the logarithm of both sides, the product becomes a sum,
+which we further expand:
+
+$$\begin{aligned}
+    \ln\!\big(\phi_S(k)\big)
+    = \sum_{n = 1}^N \ln\!\big(\phi_n(k)\big)
+    = \sum_{n = 1}^N \sum_{m = 1}^{\infty} \frac{(ik)^m}{m!} C_n^{(m)}
+\end{aligned}$$
+
+Consequently, the cumulants $C^{(m)}$ stack additively for the sum $S$
+of independent variables $x_m$, and therefore
+the means $C^{(1)}$ and variances $C^{(2)}$ do too:
+
+$$\begin{aligned}
+    C_S^{(m)} = \sum_{n = 1}^N C_n^{(m)} = C_1^{(m)} + C_2^{(m)} + ... + C_N^{(m)}
+\end{aligned}$$
+
+We now introduce the scaled sum $z$ as the new combined variable:
+
+$$\begin{aligned}
+    z = \frac{S}{\sqrt{N}} = \frac{1}{\sqrt{N}} (x_1 + x_2 + ... + x_N)
+\end{aligned}$$
+
+Its characteristic function $\phi_z(k)$ is then as follows,
+with $\sqrt{N}$ appearing in the arguments of $\phi_n$:
+
+$$\begin{aligned}
+    \phi_z(k)
+    &= \idotsint
+    \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z)
+    \dd{x_1} \cdots \dd{x_N}
+    \\
+    &= \idotsint
+    \Big( \prod_{n = 1}^N p(x_n) \Big) \exp\!\Big( i \frac{k}{\sqrt{N}} \sum_{n = 1}^N x_n \Big)
+    \dd{x_1} \cdots \dd{x_N}
+    \\
+    &= \prod_{n = 1}^N \phi_n\Big(\frac{k}{\sqrt{N}}\Big)
+\end{aligned}$$
+
+By expanding $\ln\!\big(\phi_z(k)\big)$ in terms of its cumulants $C^{(m)}$
+and introducing $\kappa = k / \sqrt{N}$, we see that the higher-order terms
+become smaller for larger $N$:
+
+$$\begin{gathered}
+    \ln\!\big( \phi_z(k) \big)
+    = \sum_{m = 1}^\infty \frac{(ik)^m}{m!} C^{(m)}
+    \\
+    C^{(m)}
+    = \frac{1}{i^m} \dv[m]{k} \sum_{n = 1}^N \ln\!\bigg( \phi_n\Big(\frac{k}{\sqrt{N}}\Big) \bigg)
+    = \frac{1}{i^m N^{m/2}} \dv[m]{\kappa} \sum_{n = 1}^N \ln\!\big( \phi_n(\kappa) \big)
+\end{gathered}$$
+
+For sufficiently large $N$, we can therefore approximate it using just the first two terms:
+
+$$\begin{aligned}
+    \ln\!\big( \phi_z(k) \big)
+    &\approx i k C^{(1)} - \frac{k^2}{2} C^{(2)}
+    = i k \overline{z} - \frac{k^2}{2} \sigma_z^2
+    \\
+    \phi_z(k)
+    &\approx \exp(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2)
+\end{aligned}$$
+
+We take its inverse Fourier transform to get the density $p(z)$,
+which turns out to be a Gaussian normal distribution,
+which is even already normalized:
+
+$$\begin{aligned}
+    p(z)
+    = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\}
+    &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k}
+    \\
+    &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big)
+\end{aligned}$$
+
+Therefore, the sum of many independent variables tends to a normal distribution,
+regardless of the densities of the individual variables.
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/density-operator/index.pdc b/content/know/concept/density-operator/index.pdc
index 39c2e85..5126f31 100644
--- a/content/know/concept/density-operator/index.pdc
+++ b/content/know/concept/density-operator/index.pdc
@@ -81,15 +81,17 @@ $$\begin{aligned}
 This can be used to find out whether a given $\hat{\rho}$
 represents a pure or mixed ensemble.
 
-Next, we define the ensemble average $\expval*{\expval*{\hat{L}}}$
-as the mean of the expectation values for states in the ensemble,
-which can be calculated like so:
+Next, we define the ensemble average $\expval*{\hat{O}}$
+as the mean of the expectation values of $\hat{O}$ for states in the ensemble.
+We use the same notation as for the pure expectation value,
+since this is only a small extension of the concept to mixed ensembles.
+It is calculated like so:
 
 $$\begin{aligned}
     \boxed{
-        \expval*{\expval*{\hat{L}}}
-        = \sum_{n} p_n \matrixel{\Psi_n}{\hat{L}}{\Psi_n}
-        = \mathrm{Tr}(\hat{L} \hat{\rho})
+        \expval*{\hat{O}}
+        = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O}}{\Psi_n}
+        = \mathrm{Tr}(\hat{\rho} \hat{O})
     }
 \end{aligned}$$
 
@@ -97,13 +99,13 @@ To prove the latter,
 we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so:
 
 $$\begin{aligned}
-    \mathrm{Tr}(\hat{L} \hat{\rho})
-    &= \sum_{j} \matrixel{j}{\hat{L} \hat{\rho}}{j}
-    = \sum_{j} \sum_{n} p_n \matrixel{j}{\hat{L}}{\Psi_n} \braket{\Psi_n}{j}
+    \mathrm{Tr}(\hat{\rho} \hat{O})
+    &= \sum_{j} \matrixel{j}{\hat{\rho} \hat{O}}{j}
+    = \sum_{j} \sum_{n} p_n \braket{j}{\Psi_n} \matrixel{\Psi_n}{\hat{O}}{j}
     \\
-    &= \sum_{n} \sum_{j} p_n \braket{\Psi_n}{j} \matrixel{j}{\hat{L}}{\Psi_n}
-    = \sum_{n} p_n \matrixel{\Psi_n}{\hat{I} \hat{L}}{\Psi_n}
-    = \expval*{\expval*{\hat{L}}}
+    &= \sum_{n} \sum_{j} p_n\matrixel{\Psi_n}{\hat{O}}{j} \braket{j}{\Psi_n}
+    = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O} \hat{I}}{\Psi_n}
+    = \expval*{\hat{O}}
 \end{aligned}$$
 
 In both the pure and mixed cases,
diff --git a/content/know/concept/dirac-delta-function/index.pdc b/content/know/concept/dirac-delta-function/index.pdc
index 3982afc..6c9f2b0 100644
--- a/content/know/concept/dirac-delta-function/index.pdc
+++ b/content/know/concept/dirac-delta-function/index.pdc
@@ -13,8 +13,10 @@ markup: pandoc
 
 # Dirac delta function
 
-The **Dirac delta function** $\delta(x)$, often just called the **delta function**,
-is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is
+The **Dirac delta function** $\delta(x)$, often just the **delta function**,
+is a function (or, more accurately, a [Schwartz distribution](/know/concept/schwartz-distribution/))
+that is commonly used in physics.
+It is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is
 defined to be 1:
 
 $$\begin{aligned}
@@ -29,7 +31,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-It is sometimes also called the **sampling function**, due to its most
+It is sometimes also called the **sampling function**, thanks to its most
 important property: the so-called **sampling property**:
 
 $$\begin{aligned}
diff --git a/content/know/concept/impulse-response/index.pdc b/content/know/concept/impulse-response/index.pdc
new file mode 100644
index 0000000..012a2c3
--- /dev/null
+++ b/content/know/concept/impulse-response/index.pdc
@@ -0,0 +1,64 @@
+---
+title: "Impulse response"
+firstLetter: "I"
+publishDate: 2021-03-09
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-09T20:34:38+01:00
+draft: false
+markup: pandoc
+---
+
+# Impulse response
+
+The **impulse response** $u_p(t)$ of a system whose behaviour is described
+by a linear operator $\hat{L}$, is defined as the reponse of the system
+when forced by the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{L} \{ u_p(t) \} = \delta(t)
+    }
+\end{aligned}$$
+
+This can be used to find the response $u(t)$ of $\hat{L}$ to
+*any* forcing function $f(t)$, i.e. not only $\delta(t)$,
+by simply taking the convolution with $u_p(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{L} \{ u(t) \} = f(t)
+        \quad \implies \quad
+        u(t) = (f * u_p)(t)
+    }
+\end{aligned}$$
+
+*__Proof.__ Starting from the definition of $u_p(t)$,
+we shift the argument by some constant $\tau$,
+and multiply both sides by the constant $f(\tau)$:*
+
+$$\begin{aligned}
+    \hat{L} \{ u_p(t - \tau) \} &= \delta(t - \tau)
+    \\
+    \hat{L} \{ f(\tau) \: u_p(t - \tau) \} &= f(\tau) \: \delta(t - \tau)
+\end{aligned}$$
+
+*Where $f(\tau)$ can be moved inside using the
+linearity of $\hat{L}$. Integrating over $\tau$ then gives us:*
+
+$$\begin{aligned}
+    \int_0^\infty \hat{L} \{ f(\tau) \: u_p(t - \tau) \} \dd{\tau}
+    &= \int_0^\infty f(\tau) \: \delta(t - \tau) \dd{\tau}
+    = f(t)
+\end{aligned}$$
+
+*The integral and $\hat{L}$ are operators of different variables, so we reorder them:*
+
+$$\begin{aligned}
+    \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
+    &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
+\end{aligned}$$
+
+*__Q.E.D.__*
diff --git a/content/know/concept/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc
index 3287544..b4394c9 100644
--- a/content/know/concept/quantum-teleportation/index.pdc
+++ b/content/know/concept/quantum-teleportation/index.pdc
@@ -97,7 +97,7 @@ $$\begin{aligned}
     &= \alpha \ket{1} + \beta \ket{0}
     \qquad \quad
     \hat{\sigma}_x \hat{\sigma}_z \ket{q}
-    = \alpha \ket{0} - \beta \ket{1}
+    = \alpha \ket{1} - \beta \ket{0}
 \end{aligned}$$
 
 Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of)
@@ -143,3 +143,7 @@ the entangled $\ket*{\Phi^{+}}_{AB}$ state must be distributed in advance,
 and Alice' declaration of her result is sent classically.
 Before receiving that, Bob only sees his side of the maximally entangled
 Bell state $\ket*{\Phi^{+}}_{AB}$, which contains nothing of $\ket{q}$.
+
+
+## References
+1.  N. Brunner, *Quantum information theory: lecture notes*, 2019.