Expand knowledge base

8 files changed, 559 insertions, 16 deletions

diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc
index c5280e5..576863c 100644
--- a/content/know/concept/calculus-of-variations/index.pdc
+++ b/content/know/concept/calculus-of-variations/index.pdc
@@ -234,3 +234,107 @@ $$\begin{aligned}
         0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big)
     }
 \end{aligned}$$
+
+
+## Constraints
+
+So far, for multiple functions $f_1, ... f_N$,
+we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$.
+Suppose that we now have $M < N$ constraints $\phi_m$
+that all $f_n$ need to obey, introducing implicit dependencies between them.
+
+Let us consider constraints $\phi_m$ of the two forms below.
+It is important that they are **holonomic**,
+meaning they do not depend on any derivatives of any $f_n(x)$:
+
+$$\begin{aligned}
+    \phi_m(f_1, ..., f_N, x) = 0
+    \qquad
+    \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m
+\end{aligned}$$
+
+Where $C_m$ is a constant.
+Note that the first form can also be used for $\phi_m = C_m \neq 0$,
+by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.
+
+To solve this constrained optimization problem for $f_n(x)$,
+we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$.
+In the former case $\lambda_m(x)$ is a function of all $x$, while in the
+latter case $\lambda_m$ is constant:
+
+$$\begin{aligned}
+    \int \lambda_m(x_i) \: \phi_m(\{f_n\}, x) \dd{x} = 0
+    \qquad
+    \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m
+\end{aligned}$$
+
+The reason for this distinction in $\lambda_m$
+is that we need to find the stationary points with respect to $\varepsilon$
+of both constraint types. Written in the variational form, this is:
+
+$$\begin{aligned}
+    \delta \int \lambda_m \: \phi_m \dd{x} = 0
+\end{aligned}$$
+
+From this, we define a new Lagrangian $\Lambda$ for the functional $J$,
+with the contraints built in:
+
+$$\begin{aligned}
+    J[f_n]
+    &= \int \Lambda(f_1, ..., f_N; f_1', ..., f_N'; \lambda_1, ..., \lambda_M; x) \dd{x}
+    \\
+    &= \int L + \sum_{m} \lambda_m \phi_m \dd{x}
+\end{aligned}$$
+
+Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$:
+
+$$\begin{aligned}
+    0
+    &= \delta \int \Lambda \dd{x}
+    = \int \pdv{\Lambda}{\varepsilon} \dd{x}
+    = \int \sum_n \Big( \pdv{\Lambda}{f_n} \pdv{f_n}{\varepsilon} + \pdv{\Lambda}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x}
+    \\
+    &= \int \sum_n \Big( \pdv{\Lambda}{f_n} \eta_n + \pdv{\Lambda}{f_n'} \eta_n' \Big) \dd{x}
+    \\
+    &= \Big[ \sum_n \pdv{\Lambda}{f_n'} \eta_n \Big]_{x_0}^{x_1}
+    + \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{x} \Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$:
+
+$$\begin{aligned}
+    0
+    = \pdv{\Lambda}{f_n} - \dv{x} \Big( \pdv{\Lambda}{f_n'} \Big)
+\end{aligned}$$
+
+By inserting the definition of $\Lambda$, we then get the following.
+Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{L}{f_n} - \dv{x} \Big( \pdv{L}{f_n'} \Big) + \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
+    }
+\end{aligned}$$
+
+These are **Lagrange's equations of the first kind**,
+with their second-kind counterparts being the earlier Euler-Lagrange equations.
+Note that there are $N$ separate equations, one for each $f_n$.
+
+Due to the constraints $\phi_m$, the functions $f_n$ are not independent.
+This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold,
+i.e. solving a system of $M$ equations for $\lambda_m$:
+
+$$\begin{aligned}
+    \dv{x} \Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n}
+    = \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
+\end{aligned}$$
+
+And then the remaining $N - M$ equations can be solved in the normal unconstrained way.
+
+
+
+## References
+1.  G.B. Arfken, H.J. Weber,
+    *Mathematical methods for physicists*, 6th edition, 2005,
+    Elsevier.
diff --git a/content/know/concept/density-operator/index.pdc b/content/know/concept/density-operator/index.pdc
index 84c2d74..39c2e85 100644
--- a/content/know/concept/density-operator/index.pdc
+++ b/content/know/concept/density-operator/index.pdc
@@ -45,11 +45,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 However, from the special case where $\ket{\Psi_n}$ are indeed basis vectors,
-we can conclude that $\hat{\rho}$ is Hermitian,
+we can conclude that $\hat{\rho}$ is positive semidefinite and Hermitian,
 and that its trace (i.e. the total probability) is 100%:
 
 $$\begin{gathered}
     \boxed{
+        \hat{\rho} \ge 0
+    }
+    \qquad \qquad
+    \boxed{
         \hat{\rho}^\dagger = \hat{\rho}
     }
     \qquad \qquad
diff --git a/content/know/concept/holomorphic-function/index.pdc b/content/know/concept/holomorphic-function/index.pdc
index 3e7a91e..1077060 100644
--- a/content/know/concept/holomorphic-function/index.pdc
+++ b/content/know/concept/holomorphic-function/index.pdc
@@ -196,7 +196,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 **Cauchy's residue theorem** generalizes Cauchy's integral theorem
-to meromorphic functions, and states that the integral of a contour $C$,
+to meromorphic functions, and states that the integral of a contour $C$
 depends on the simple poles $p$ it encloses:
 
 $$\begin{aligned}
@@ -206,7 +206,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 *__Proof__*. *From the definition of a meromorphic function,
-we know that we can decompose $f(z)$ as follows,
+we know that we can decompose $f(z)$ like so,
 where $h(z)$ is holomorphic and $p$ are all its poles:*
 
 $$\begin{aligned}
@@ -228,5 +228,5 @@ This theorem might not seem very useful,
 but in fact, thanks to some clever mathematical magic,
 it allows us to evaluate many integrals along the real axis,
 most notably [Fourier transforms](/know/concept/fourier-transform/).
-It can also be used to derive the Kramers-Kronig relations.
+It can also be used to derive the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations).
 
diff --git a/content/know/concept/quantum-entanglement/index.pdc b/content/know/concept/quantum-entanglement/index.pdc
new file mode 100644
index 0000000..f8d0b76
--- /dev/null
+++ b/content/know/concept/quantum-entanglement/index.pdc
@@ -0,0 +1,152 @@
+---
+title: "Quantum entanglement"
+firstLetter: "Q"
+publishDate: 2021-03-07
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-07T15:36:15+01:00
+draft: false
+markup: pandoc
+---
+
+# Quantum entanglement
+
+Consider a composite quantum system which consists of two subsystems $A$ and $B$,
+respectively with basis states $\ket{a_n}$ and $\ket{b_n}$.
+All accessible states of the sytem $\ket{\Psi}$ lie in
+the tensor product of the subsystems'
+[Hilbert spaces](/know/concept/hilbert-space/) $\mathbb{H}_A$ and $\mathbb{H}_B$:
+
+$$\begin{aligned}
+    \ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B
+\end{aligned}$$
+
+A subset of these states can be written as the tensor product (i.e. Kronecker product in a basis)
+of a state $\ket{\alpha}$ in $A$ and a state $\ket{\beta}$ in $B$,
+often abbreviated as $\ket{\alpha} \ket{\beta}$:
+
+$$\begin{aligned}
+    \ket{\Psi}
+    = \ket{\alpha} \ket{\beta}
+    = \ket{\alpha} \otimes \ket{\beta}
+\end{aligned}$$
+
+The states that can be written in this way are called **separable**,
+and states that cannot are called **entangled**.
+Therefore, we are dealing with **quantum entanglement**
+if the state of subsystem $A$ cannot be fully described
+independently of the state of subsystem $B$, and vice versa.
+
+To detect and quantify entanglement,
+we can use the [density operator](/know/concept/density-operator/) $\hat{\rho}$.
+For a pure ensemble in a given (possibly entangled) state $\ket{\Psi}$,
+$\hat{\rho}$ is given by:
+
+$$\begin{aligned}
+    \hat{\rho} = \ket{\Psi} \bra{\Psi}
+\end{aligned}$$
+
+From this, we would like to extract the corresponding state of subsystem $A$.
+For that purpose, we define the **reduced density operator** $\hat{\rho}_A$ of subsystem $A$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}_A
+        = \Tr_B(\hat{\rho})
+        = \sum_m \bra{b_m} \Big( \hat{\rho} \Big) \ket{b_m}
+    }
+\end{aligned}$$
+
+Where $\Tr_B(\hat{\rho})$ is called the **partial trace** of $\hat{\rho}$,
+which basically eliminates subsystem $B$ from $\hat{\rho}$.
+For a pure composite state $\ket{\Psi}$,
+the resulting $\hat{\rho}_A$ describes a pure state in $A$ if $\ket{\Psi}$ is separable,
+else, if $\ket{\Psi}$ is entangled, it describes a mixed state in $A$.
+In the former case we simply find:
+
+$$\begin{aligned}
+    \boxed{
+        \ket{\Psi} = \ket{\alpha} \otimes \ket{\beta}
+        \quad \implies \quad
+        \hat{\rho}_A = \ket{\alpha} \bra{\alpha}
+    }
+\end{aligned}$$
+
+We call $\ket{\Psi}$ **maximally entangled**
+if its reduced density operators are **maximally mixed**,
+where $N$ is the dimension of $\mathbb{H}_A$ and $\hat{I}$ is the identity matrix:
+
+$$\begin{aligned}
+    \hat{\rho}_A
+    = \frac{1}{N} \hat{I}
+\end{aligned}$$
+
+Suppose that we are given an entangled pure state
+$\ket{\Psi} \neq \ket{\alpha} \otimes \ket{\beta}$.
+Then the partial traces $\hat{\rho}_A$ and $\hat{\rho}_B$
+of $\hat{\rho} = \ket{\Psi} \bra{\Psi}$ are mixed states with the same probabilities $p_n$
+(assuming $\mathbb{H}_A$ and $\mathbb{H}_B$ have the same dimensions,
+which is usually the case):
+
+$$\begin{aligned}
+    \hat{\rho}_A
+    = \Tr_B(\hat{\rho})
+    = \sum_n p_n \ket{a_n} \bra{a_n}
+    \qquad \quad
+    \hat{\rho}_B
+    = \Tr_A(\hat{\rho})
+    = \sum_n p_n \ket{b_n} \bra{b_n}
+\end{aligned}$$
+
+There exists an orthonormal choice
+of the subsystem basis states $\ket{a_n}$ and $\ket{b_n}$,
+such that $\ket{\Psi}$ can be written as follows,
+where $p_n$ are the probabilities in the reduced density operators:
+
+$$\begin{aligned}
+    \ket{\Psi}
+    = \sum_n \sqrt{p_n} \Big( \ket{a_n} \otimes \ket{b_n} \Big)
+\end{aligned}$$
+
+This is the **Schmidt decomposition**,
+and the **Schmidt number** is the number of nonzero terms in the summation,
+which can be used to determine if the state $\ket{\Psi}$
+is entangled (greater than one) or separable (equal to one).
+
+By looking at the Schmidt decomposition, we can notice that,
+if $\hat{O}_A$ and $\hat{O}_B$ are the subsystem observables
+with basis eigenstates $\ket{a_n}$ and $\ket{b_n}$,
+then measurement results of these operators
+will be perfectly correlated across $A$ and $B$.
+This is a general property of entangled systems,
+but beware: correlation does not imply entanglement!
+
+But what if the composite system is in a mixed state $\hat{\rho}$?
+The state is separable if and only if:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}
+        = \sum_m p_m \Big( \hat{\rho}_A \otimes \hat{\rho}_B \Big)
+    }
+\end{aligned}$$
+
+Where $p_m$ are probabilities,
+and $\hat{\rho}_A$ and $\hat{\rho}_B$ can be any subsystem states.
+In reality, it is very hard to determine, using this criterium,
+whether an arbitrary given $\hat{\rho}$ is separable or not.
+
+As a final side note, the expectation value
+of an obervable $\hat{O}_A$ acting only on $A$ is given by:
+
+$$\begin{aligned}
+    \expval*{\hat{O}_A}
+    = \Tr\big(\hat{\rho} \hat{O}_A\big)
+    = \Tr_A\big(\Tr_B(\hat{\rho} \hat{O}_A)\big)
+    = \Tr_A\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big)
+    = \Tr_A\big(\hat{\rho}_A \hat{O}_A\big)
+\end{aligned}$$
+
diff --git a/content/know/concept/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc
new file mode 100644
index 0000000..3287544
--- /dev/null
+++ b/content/know/concept/quantum-teleportation/index.pdc
@@ -0,0 +1,145 @@
+---
+title: "Quantum teleportation"
+firstLetter: "Q"
+publishDate: 2021-03-07
+categories:
+- Quantum information
+
+date: 2021-03-07T20:30:30+01:00
+draft: false
+markup: pandoc
+---
+
+# Quantum teleportation
+
+**Quantum teleportation** is a method to transfer quantum information
+between systems without the use of a quantum channel.
+It is based on [quantum entanglement](/know/concept/quantum-entanglement/).
+
+Suppose that Alice has a qubit $\ket{q}_{A'}$ that she wants to send to Bob.
+Since she has not measured it yet, she does not know $\alpha$ or $\beta$;
+she just wants Bob to get the same qubit:
+
+$$\begin{aligned}
+    \ket{q}
+    = \alpha \ket{0}_{A'} + \beta \ket{1}_{A'}
+\end{aligned}$$
+
+She can only directly communicate with Bob over a classical channel.
+This is not enough: even if Alice did know $\alpha$ and $\beta$ exactly
+(which would need her having infinitely many copies to measure),
+sending an arbitrary real number requires an infinite amount of classical data.
+
+However, between them, she and Bob also have an entangled Bell state,
+e.g. $\ket*{\Phi^+}_{AB}$ (it does not matter which Bell state it is)
+The state of the composite system is then as follows,
+with $A'$ being Alice' qubit, $A$ her side of the Bell state, and $B$ Bob's side:
+
+$$\begin{aligned}
+    \ket{q}_{A'} \otimes \ket*{\Phi^+}_{AB}
+    &= \frac{1}{\sqrt{2}} \Big( \alpha \ket{0} + \beta \ket{1} \Big)_{A'} \Big( \ket{00} + \ket{11} \Big)_{AB}
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \alpha \ket{000} + \beta \ket{100}
+    + \alpha \ket{011} + \beta \ket{111} \Big)_{A'AB}
+\end{aligned}$$
+
+Now, observe that we can write any combination of $\ket{0}$ and $\ket{1}$
+in the Bell basis like so:
+
+$$\begin{aligned}
+    \ket{00}
+    &= \frac{\ket{\Phi^{+}} + \ket{\Phi^{-}}}{\sqrt{2}}
+    \qquad \quad
+    \ket{11}
+    = \frac{\ket{\Phi^{+}} - \ket{\Phi^{-}}}{\sqrt{2}}
+    \\
+    \ket{01}
+    &= \frac{\ket{\Psi^{+}} + \ket{\Psi^{-}}}{\sqrt{2}}
+    \qquad \quad
+    \ket{10}
+    = \frac{\ket{\Psi^{+}} - \ket{\Psi^{-}}}{\sqrt{2}}
+\end{aligned}$$
+
+Using this, we can rewrite our previous result in terms of the Bell states as follows:
+
+$$\begin{aligned}
+    \ket{q}_{A'} \ket*{\Phi^+}_{AB}
+    &= \frac{\alpha}{2} \Big( \ket*{\Phi^{+}} + \ket*{\Phi^{-}} \Big)_{A'A} \ket{0}_B
+    + \frac{\beta}{2} \Big( \ket*{\Psi^{+}} - \ket*{\Psi^{-}} \Big)_{A'A} \ket{0}_B
+    \\
+    &+ \frac{\alpha}{2} \Big( \ket*{\Psi^{+}} + \ket*{\Psi^{-}} \Big)_{A'A} \ket{1}_B
+    + \frac{\beta}{2} \Big( \ket*{\Phi^{+}} - \ket*{\Phi^{-}} \Big)_{A'A} \ket{1}_B
+\end{aligned}$$
+
+If we group all terms according to the Bell states,
+we end up with an interesting expression:
+
+$$\begin{aligned}
+    \ket{q}_{A'} \ket*{\Phi^+}_{AB}
+    = \frac{1}{2} \bigg( &\ket*{\Phi^{+}}_{A'A} \Big( \alpha \ket{0} + \beta \ket{1} \Big)_{B}
+    + \ket*{\Phi^{-}}_{A'A} \Big( \alpha \ket{0} - \beta \ket{1} \Big)_{B}
+    \\
+    + &\ket*{\Psi^{+}}_{A'A} \Big( \alpha \ket{1} + \beta \ket{0} \Big)_{B}
+    + \ket*{\Psi^{-}}_{A'A} \Big( \alpha \ket{1} - \beta \ket{0} \Big)_{B} \bigg)
+\end{aligned}$$
+
+Thus, purely due to entanglement,
+Bob's qubit $B$ is in a superposition of the following states:
+
+$$\begin{aligned}
+    \ket{q}
+    &= \alpha \ket{0} + \beta \ket{1}
+    \qquad \quad
+    \quad \hat{\sigma}_z \ket{q}
+    = \alpha \ket{0} - \beta \ket{1}
+    \\
+    \hat{\sigma}_x \ket{q}
+    &= \alpha \ket{1} + \beta \ket{0}
+    \qquad \quad
+    \hat{\sigma}_x \hat{\sigma}_z \ket{q}
+    = \alpha \ket{0} - \beta \ket{1}
+\end{aligned}$$
+
+Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of)
+the following entangled three-qubit state:
+
+$$\begin{aligned}
+    \ket{q}_{A'} \ket*{\Phi^+}_{AB}
+    = \frac{1}{2} \bigg( &\ket*{\Phi^{+}}_{A'A} \Big( \ket{q} \Big)_B \quad\, + \ket*{\Phi^{-}}_{A'A} \Big( \hat{\sigma}_z \ket{q} \Big)_B
+    \\
+    + &\ket*{\Psi^{+}}_{A'A} \Big( \hat{\sigma}_x \ket{q} \Big)_B + \ket*{\Psi^{-}}_{A'A} \Big( \hat{\sigma}_x \hat{\sigma}_z \ket{q} \Big)_B \bigg)
+\end{aligned}$$
+
+The point is that, thanks to the initial entanglement between Alice and Bob,
+adding $\ket{q}_{A'}$ into the mix somehow "teleports" that information to Bob,
+although it is not in a usable form yet.
+
+To finish the process, Alice measures her side $A'A$ in the Bell basis.
+Consequently, $A'A$ collapses into one of
+$\ket*{\Phi^{+}}$, $\ket*{\Phi^{-}}$, $\ket*{\Psi^{+}}$, $\ket*{\Psi^{-}}$
+with equal probability, and she knows which.
+This collapse leaves Bob's side $B$ in $\ket{q}$, $\hat{\sigma}_z \ket{q}$,
+$\hat{\sigma}_x \ket{q}$, or $\hat{\sigma}_x \hat{\sigma}_z \ket{q}$, respectively.
+The entanglement between $A$ and $B$ is thus broken,
+and instead Alice has local entanglement between $A'$ and $A$.
+
+She then uses the classical channel to tell Bob her result,
+who then either does nothing (for $\ket{q}$),
+applies $\hat{\sigma}_z$ (for $\hat{\sigma}_z \ket{q}$),
+applies $\hat{\sigma}_x$ (for $\hat{\sigma}_x \ket{q}$),
+or applies $\hat{\sigma}_z \hat{\sigma}_x$ (for $\hat{\sigma}_x \hat{\sigma}_z \ket{q}$).
+Then, due to the fact that $\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$,
+he recovers $\ket{q}$ in his local qubit $B$.
+
+This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem)
+because Alice does not require any knowledge of $\ket{q}$,
+and after the measurement, her qubit $A'$ will no longer be in that state.
+In other words, quantum teleportation *moves* states,
+rather than copying them.
+
+Nor does this conflict with Einstein's relativity,
+since the information travels no faster than light:
+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 $\ket{q}$.
diff --git a/content/know/concept/time-dependent-perturbation-theory/index.pdc b/content/know/concept/time-dependent-perturbation-theory/index.pdc
new file mode 100644
index 0000000..fbb71b2
--- /dev/null
+++ b/content/know/concept/time-dependent-perturbation-theory/index.pdc
@@ -0,0 +1,122 @@
+---
+title: "Time-dependent perturbation theory"
+firstLetter: "T"
+publishDate: 2021-03-07
+categories:
+- Physics
+- Quantum mechanics
+- Perturbation
+
+date: 2021-03-07T11:08:14+01:00
+draft: false
+markup: pandoc
+---
+
+# Time-dependent perturbation theory
+
+In quantum mechanics, **time-dependent perturbation theory** exists to deal
+with time-varying perturbations to the Schrödinger equation.
+This is in contrast to [time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/),
+where the perturbation is is stationary.
+
+Let $\hat{H}_0$ be the base time-independent
+Hamiltonian, and $\hat{H}_1$ be a time-varying perturbation, with
+"bookkeeping" parameter $\lambda$:
+
+$$\begin{aligned}
+    \hat{H}(t) = \hat{H}_0 + \lambda \hat{H}_1(t)
+\end{aligned}$$
+
+We assume that the unperturbed time-independent problem
+$\hat{H}_0 \ket{n} = E_n \ket{n}$ has already been solved, such that the
+full solution is:
+
+$$\begin{aligned}
+    \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+Since these $\ket{n}$ form a complete basis, the perturbed wave function
+can be written in the same form, but with time-dependent coefficients $c_n(t)$:
+
+$$\begin{aligned}
+    \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+We insert this ansatz in the time-dependent Schrödinger equation, and
+reduce it using the known unperturbed time-independent problem:
+
+$$\begin{aligned}
+    0
+    &= \hat{H}_0 \ket{\Psi(t)} + \lambda \hat{H}_1 \ket{\Psi(t)} - i \hbar \dv{t} \ket{\Psi(t)}
+    \\
+    &= \sum_{n}
+    \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar)
+    \\
+    &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+We then take the inner product with an arbitrary stationary basis state $\ket{m}$:
+
+$$\begin{aligned}
+    0
+    &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+Thanks to orthonormality, this removes the latter term from the summation:
+
+$$\begin{aligned}
+    i \hbar \frac{d c_m}{dt} \exp(- i E_m t / \hbar)
+    &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+We divide by the left-hand exponential and define
+$\omega_{mn} = (E_m - E_n) / \hbar$ to get:
+
+$$\begin{aligned}
+    \boxed{
+        i \hbar \frac{d c_m}{dt}
+        = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp(i \omega_{mn} t)
+    }
+\end{aligned}$$
+
+So far, we have not invoked any approximation,
+so we can analytically find $c_n(t)$ for some simple systems.
+Furthermore, it is useful to write this equation in integral form instead:
+
+$$\begin{aligned}
+    c_m(t)
+    = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+\end{aligned}$$
+
+If this cannot be solved exactly, we must approximate it. We expand
+$c_m(t)$ in the usual way, with the initial condition $c_m^{(j)}(0) = 0$
+for $j > 0$:
+
+$$\begin{aligned}
+    c_m(t) = c_m^{(0)} + \lambda c_m^{(1)}(t) + \lambda^2 c_m^{(2)}(t) + ...
+\end{aligned}$$
+
+We then insert this into the integral and collect the non-zero orders of $\lambda$:
+
+$$\begin{aligned}
+    c_m^{(1)}(t)
+    &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    \\
+    c_m^{(2)}(t)
+    &= - \frac{i}{\hbar} \sum_{n}
+    \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    \\
+    c_m^{(3)}(t)
+    &= - \frac{i}{\hbar} \sum_{n}
+    \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+\end{aligned}$$
+
+And so forth. The pattern here is clear: we can calculate the $(j\!+\!1)$th
+correction using only our previous result for the $j$th correction.
+We cannot go any further than this without considering a specific perturbation $\hat{H}_1(t)$.
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/content/know/concept/time-independent-perturbation-theory/index.pdc b/content/know/concept/time-independent-perturbation-theory/index.pdc
index 2035fc2..3be3cd5 100644
--- a/content/know/concept/time-independent-perturbation-theory/index.pdc
+++ b/content/know/concept/time-independent-perturbation-theory/index.pdc
@@ -159,22 +159,22 @@ $$\begin{aligned}
 Here it is clear why this is only valid in the non-degenerate case:
 otherwise we would divide by zero in the denominator.
 
-Next, to find the second-order correction to the energy $E_n^{(2)}$, we
-take the corresponding equation and put $\bra{n}$ in front of it:
+Next, to find the second-order energy correction $E_n^{(2)}$,
+we take the corresponding equation and put $\bra{n}$ in front of it:
 
 $$\begin{aligned}
-    \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}}
-    &= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket{n}{\psi_n^{(1)}} + \varepsilon_n \braket{n}{\psi_n^{(2)}}
+    \matrixel*{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel*{n}{\hat{H}_0}{\psi_n^{(2)}}
+    &= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket*{n}{\psi_n^{(1)}} + \varepsilon_n \braket*{n}{\psi_n^{(2)}}
 \end{aligned}$$
 
 Because $\hat{H}_0$ is Hermitian, we know that
-$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket{n}{\psi_n^{(2)}}$,
+$\matrixel*{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket*{n}{\psi_n^{(2)}}$,
 i.e. we apply it to the bra, which lets us eliminate two terms. Also,
 since $\ket{n}$ is normalized, we find:
 
 $$\begin{aligned}
     E_n^{(2)}
-    = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}}
+    = \matrixel*{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket*{n}{\psi_n^{(1)}}
 \end{aligned}$$
 
 We explicitly removed the $\ket{n}$-dependence of $\ket*{\psi_n^{(1)}}$,
@@ -221,8 +221,8 @@ take the equation at order $\lambda^1$ and prepend an arbitrary
 eigenspace basis vector $\bra{n, \delta}$:
 
 $$\begin{aligned}
-    \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
-    &= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket{n, \delta}{\psi_n^{(1)}}
+    \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel*{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
+    &= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket*{n, \delta}{\psi_n^{(1)}}
 \end{aligned}$$
 
 Since $\hat{H}_0$ is Hermitian, we use the same trick as before to
@@ -320,11 +320,18 @@ $\ket{n, d_1}$ and $\ket{n, d_2}$ have distinct eigenvalues
 $\ell_1 \neq \ell_2$ for $d_1 \neq d_2$:
 
 $$\begin{aligned}
-    \hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1}
+    \hat{L} \ket{n, d_1} = \ell_1 \ket{n, d_1}
     \qquad
-    \hat{L} \ket{n, b_2} = \ell_2 \ket{n, b_2}
+    \hat{L} \ket{n, d_2} = \ell_2 \ket{n, d_2}
 \end{aligned}$$
 
-When this holds for any orthogonal choice of $\ket{n, d_1}$ and
+When this condition holds for any orthogonal choice of $\ket{n, d_1}$ and
 $\ket{n, d_2}$, then these specific eigenvectors of $\hat{L}$ are the
 "good states", for any valid choice of $\hat{L}$.
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
index 482650e..cf44fc8 100644
--- a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
+++ b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
@@ -14,7 +14,7 @@ markup: pandoc
 # Wentzel-Kramers-Brillouin approximation
 
 In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
-approximation** is a method to approximate the wave function $\psi(x)$ of
+approximation** is a technique to approximate the wave function $\psi(x)$ of
 the one-dimensional time-independent Schrödinger equation. It is an example
 of a **semiclassical approximation**, because it tries to find a
 balance between classical and quantum physics.
@@ -196,3 +196,12 @@ In the classical region ($E > V$), the wave function oscillates, and
 in the quantum-mechanical region ($E < V$) it is exponential. Note that for
 $E \approx V$ the approximation breaks down, due to the appearance of
 $p(x)$ in the denominator.
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
+2.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.