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5 files changed, 445 insertions, 20 deletions

diff --git a/content/know/concept/canonical-ensemble/index.pdc b/content/know/concept/canonical-ensemble/index.pdc
new file mode 100644
index 0000000..9f75fc8
--- /dev/null
+++ b/content/know/concept/canonical-ensemble/index.pdc
@@ -0,0 +1,248 @@
+---
+title: "Canonical ensemble"
+firstLetter: "C"
+publishDate: 2021-07-10
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+
+date: 2021-07-08T11:01:02+02:00
+draft: false
+markup: pandoc
+---
+
+# Canonical ensemble
+
+The **canonical ensemble** or **NVT ensemble** builds on
+the [microcanonical ensemble](/know/concept/microcanonical-ensemble/),
+by allowing the system to exchange energy with a very large heat bath,
+such that its temperature $T$ remains constant,
+but internal energy $U$ does not.
+The conserved state functions are
+the temperature $T$, the volume $V$, and the particle count $N$.
+
+We refer to the system of interest as $A$, and the heat bath as $B$.
+The combination $A\!+\!B$ forms a microcanonical ensemble,
+i.e. it has a fixed total energy $U$,
+and eventually reaches an equilibrium
+with a uniform temperature $T$ in both $A$ and $B$.
+
+Assuming that this equilibrium has been reached,
+we want to know which microstates $A$ prefers in that case.
+Specifically, if $A$ has energy $U_A$, and $B$ has $U_B$,
+which $U_A$ does $A$ prefer?
+
+Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
+Then the probability that $A$ is in a specific microstate $s_A$ is as follows,
+where $U_A(s_A)$ is the resulting energy:
+
+$$\begin{aligned}
+    p(s_A)
+    = \frac{c_B(U - U_A(s_A))}{D}
+    \qquad \quad
+    D \equiv \sum_{s_A} c_B(U - U_A(s_A))
+\end{aligned}$$
+
+In other words, we choose an $s_A$,
+and count the number $c_B$ of compatible $B$-microstates.
+
+Since the heat bath is large, let us assume that $U_B \gg U_A$.
+We thus approximate $\ln{p(s_A)}$ by
+Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$:
+
+$$\begin{aligned}
+    \ln{p(s_A)}
+    &= -\ln{D} + \ln\!\big(c_B(U - U_A(s_A))\big)
+    \\
+    &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A)
+\end{aligned}$$
+
+Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$,
+and that its $U_B$-derivative is $1/T$:
+
+$$\begin{aligned}
+    \ln{p(s_A)}
+    &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k} \Big( \pdv{S_B}{U_B} \Big)
+    \\
+    &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T}
+\end{aligned}$$
+
+We now define the **partition function** or **Zustandssumme** $Z$ as follows,
+which will act as a normalization factor for the probability:
+
+$$\begin{aligned}
+    \boxed{
+        Z
+        \equiv \sum_{s_A}^{} \exp\!(- \beta U_A(s_A))
+    }
+    = \frac{D}{c_B(U)}
+\end{aligned}$$
+
+Where $\beta \equiv 1/ (k T)$.
+The probability of finding $A$ in a microstate $s_A$ is thus given by:
+
+$$\begin{aligned}
+    \boxed{
+        p(s_A) = \frac{1}{Z} \exp\!(- \beta U_A(s_A))
+    }
+\end{aligned}$$
+
+This is the **Boltzmann distribution**,
+which, as it turns out, maximizes the entropy $S_A$
+for a fixed value of the average energy $\expval{U_A}$,
+i.e. a fixed $T$ and set of microstates $s_A$.
+
+Because $A\!+\!B$ is a microcanonical ensemble,
+we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/)
+is the entropy $S$.
+But what about the canonical ensemble, just $A$?
+
+The solution is a bit backwards.
+Note that the partition function $Z$ is not a constant;
+it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$).
+Using the same logic as for the microcanonical ensemble,
+we define "equilibrium" as the set of microstates $s_A$
+that $A$ is most likely to occupy,
+which must be the set (as a function of $T,V,N$) that maximizes $Z$.
+
+However, $T$, $V$ and $N$ are fixed,
+so how can we maximize $Z$?
+Well, as it turns out,
+the Boltzmann distribution has already done it for us!
+We will return to this point later.
+
+Still, $Z$ does not have a clear physical interpretation.
+To find one, we start by showing that the ensemble averages
+of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$
+can be calculated by differentiating $Z$.
+As preparation, note that:
+
+$$\begin{aligned}
+    \pdv{Z}{\beta} = - \sum_{s_A} U_A \exp\!(- \beta U_A)
+\end{aligned}$$
+
+With this, we can find the ensemble averages
+$\expval{U_A}$, $\expval{P_A}$ and $\expval{\mu_A}$ of the system:
+
+$$\begin{aligned}
+    \expval{U_A}
+    &= \sum_{s_A} p(s_A) \: U_A
+    = \frac{1}{Z} \sum_{s_A} U_A \exp\!(- \beta U_A)
+    = - \frac{1}{Z} \pdv{Z}{\beta}
+    \\
+    \expval{P_A}
+    &= - \sum_{s_A} p(s_A) \pdv{U_A}{V}
+    = - \frac{1}{Z} \sum_{s_A} \exp\!(- \beta U_A) \pdv{U_A}{V}
+    \\
+    &= \frac{1}{Z \beta} \pdv{V} \sum_{s_A} \exp\!(- \beta U_A)
+    = \frac{1}{Z \beta} \pdv{Z}{V}
+    \\
+    \expval{\mu_A}
+    &= \sum_{s_A} p(s_A) \pdv{U_A}{N}
+    = \frac{1}{Z} \pdv{N} \sum_{s_A} \exp\!(- \beta U_A) \pdv{U_A}{N}
+    \\
+    &= - \frac{1}{Z \beta} \pdv{N} \sum_{s_A} \exp\!(- \beta U_A)
+    = - \frac{1}{Z \beta} \pdv{Z}{N}
+\end{aligned}$$
+
+It will turn out more convenient to use derivatives of $\ln{Z}$ instead,
+in which case:
+
+$$\begin{aligned}
+    \expval{U_A}
+    = - \pdv{\ln{Z}}{\beta}
+    \qquad \quad
+    \expval{P_A}
+    = \frac{1}{\beta} \pdv{\ln{Z}}{V}
+    \qquad \quad
+    \expval{\mu_A}
+    = - \frac{1}{\beta} \pdv{\ln{Z}}{N}
+\end{aligned}$$
+
+Now, to find a physical interpretation for $Z$.
+Consider the quantity $F$, in units of energy,
+whose minimum corresponds to a maximum of $Z$:
+
+$$\begin{aligned}
+    F \equiv - k T \ln{Z}
+\end{aligned}$$
+
+We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element:
+
+$$\begin{aligned}
+    \dd{(\beta F)}
+    = - \dd{(\ln{Z})}
+    &= - \pdv{\ln{Z}}{\beta} \dd{\beta} - \pdv{\ln{Z}}{V} \dd{V} - \pdv{\ln{Z}}{N} \dd{N}
+    \\
+    &= \expval{U_A} \dd{\beta} - \beta \expval{P_A} \dd{V} + \beta \expval{\mu_A} \dd{N}
+    \\
+    &= \expval{U_A} \dd{\beta} + \beta \dd{\expval{U_A}} - \beta \dd{\expval{U_A}} - \beta \expval{P_A} \dd{V} + \beta \expval{\mu_A} \dd{N}
+    \\
+    &= \dd{(\beta \expval{U_A})} - \beta \: \big( \dd{\expval{U_A}} + \expval{P_A} \dd{V} - \expval{\mu_A} \dd{N} \big)
+\end{aligned}$$
+
+Rearranging and substituting
+the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
+then gives:
+
+$$\begin{aligned}
+    \dd{(\beta F - \beta \expval{U_A})}
+    &= - \beta \: \big( \dd{\expval{U_A}} + \expval{P_A} \dd{V} - \expval{\mu_A} \dd{N} \big)
+    = - \beta T \dd{S_A}
+\end{aligned}$$
+
+We integrate this and ignore the integration constant,
+leading us to the desired result:
+
+$$\begin{aligned}
+    - \beta T S_A
+    &= \beta F - \beta \expval{U_A}
+    \quad \implies \quad
+    F = \expval{U_A} -  T S_A
+\end{aligned}$$
+
+As was already suggested by our notation,
+$F$ turns out to be the **Helmholtz free energy**:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            F
+            &\equiv - k T \ln{Z}
+            \\
+            &= \expval{U_A} - T S_A
+        \end{aligned}
+    }
+\end{aligned}$$
+
+We can therefore reinterpret
+the partition function $Z$ and the Boltzmann distribution $p(s_A)$
+in the following "more physical" way:
+
+$$\begin{aligned}
+    Z
+    = \exp\!(- \beta F)
+    \qquad \quad
+    p(s_A)
+    = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big)
+\end{aligned}$$
+
+Finally, by rearranging the expressions for $F$,
+we find the entropy $S_A$ to be:
+
+$$\begin{aligned}
+    S_A
+    = k \ln{Z} + \frac{\expval{U_A}}{T}
+\end{aligned}$$
+
+This is why $Z$ is already maximized:
+the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\expval{U_A}$,
+leaving $Z$ as the only "variable".
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/fermis-golden-rule/index.pdc b/content/know/concept/fermis-golden-rule/index.pdc
new file mode 100644
index 0000000..5ed273e
--- /dev/null
+++ b/content/know/concept/fermis-golden-rule/index.pdc
@@ -0,0 +1,91 @@
+---
+title: "Fermi's golden rule"
+firstLetter: "F"
+publishDate: 2021-07-10
+categories:
+- Physics
+- Quantum mechanics
+- Optics
+
+date: 2021-07-03T14:41:11+02:00
+draft: false
+markup: pandoc
+---
+
+# Fermi's golden rule
+
+In quantum mechanics, **Fermi's golden rule** expresses
+the transition rate between two states of a system,
+when a sinusoidal perturbation is applied
+at the resonance frequency $\omega = E_g / \hbar$ of the
+energy gap $E_g$. The main conclusion is that the rate is independent of
+time.
+
+From [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
+we know that the transition probability
+for a particle in state $\ket{a}$ to go to $\ket{b}$
+is as follows for a periodic perturbation at frequency $\omega$:
+
+$$\begin{aligned}
+    P_{ab}
+    = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2}
+\end{aligned}$$
+
+Where $\omega_{ba} \equiv (E_b - E_a) / \hbar$.
+If we assume that $\ket{b}$ irreversibly absorbs an unlimited number of particles,
+then we can interpret $P_{ab}$ as the "amount" of the current particle
+that has transitioned since the last period $2 \pi n / (\omega_{ba} \!-\! \omega)$.
+
+For generality, let $E_b$ be the center
+of a state continuum with width $\Delta E$.
+In that case, $P_{ab}$ must be modified as follows,
+where $\rho(E_x)$ is the destination's
+[density of states](/know/concept/density-of-states/):
+
+$$\begin{aligned}
+    P_{ab}
+    &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - \Delta E / 2}^{E_b + \Delta E / 2}
+    \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
+\end{aligned}$$
+
+If $E_b$ is not in a continuum, then $\rho(E_x) = \delta(E_x - E_b)$.
+The integrand is a sharp sinc-function around $E_x$.
+For large $t$, it is so sharp that we can take out $\rho(E_x)$.
+In that case, we also simplify the integration limits.
+Then we substitute $x \equiv (\omega_{xa}\!-\!\omega) / 2$ to get:
+
+$$\begin{aligned}
+    P_{ab}
+    &\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx
+\end{aligned}$$
+
+This definite integral turns out to be $\pi |t|$,
+so we find, because clearly $t > 0$:
+
+$$\begin{aligned}
+    P_{ab}
+    &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t
+\end{aligned}$$
+
+The transition rate $R_{ab}$,
+i.e. the number of particles per unit time,
+then takes this form:
+
+$$\begin{aligned}
+    \boxed{
+        R_{ab}
+        = \pdv{P_{ab}}{t}
+        = \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b)
+    }
+\end{aligned}$$
+
+Note that the $t$-dependence has disappeared,
+and all that remains is a constant factor involving $E_b = E_a \!+\! \hbar \omega$,
+where $\omega$ is the resonance frequency.
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/content/know/concept/grand-canonical-ensemble/index.pdc b/content/know/concept/grand-canonical-ensemble/index.pdc
new file mode 100644
index 0000000..5853a5f
--- /dev/null
+++ b/content/know/concept/grand-canonical-ensemble/index.pdc
@@ -0,0 +1,83 @@
+---
+title: "Grand canonical ensemble"
+firstLetter: "G"
+publishDate: 2021-07-11
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+
+date: 2021-07-08T11:01:11+02:00
+draft: false
+markup: pandoc
+---
+
+# Grand canonical ensemble
+
+The **grand canonical ensemble** or **μVT ensemble**
+extends the [canonical ensemble](/know/concept/canonical-ensemble/)
+by allowing the exchange of both energy $U$ and particles $N$
+with an external reservoir,
+so that the conserved state functions are
+the temperature $T$, the volume $V$, and the chemical potential $\mu$.
+
+The derivation is practically identical to that of the canonical ensemble.
+We refer to the system of interest as $A$,
+and the reservoir as $B$.
+In total, $A\!+\!B$ has energy $U$ and population $N$.
+
+Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
+Then the probability that $A$ is in a specific microstate $s_A$ is as follows:
+
+$$\begin{aligned}
+    p(s)
+    = \frac{c_B\big(U - U_A(s_A), N - N_A(s_A)\big)}{\sum_{s_A} c_B\big(U \!-\! U_A(s_A), N \!-\! N_A(s_A)\big)}
+\end{aligned}$$
+
+Then, as for the canonical ensemble,
+we assume $U_B \gg U_A$ and $N_B \gg N_A$,
+and approximate $\ln{p(s_A)}$
+by Taylor-expanding $\ln{c_B}$ around $U_B = U$ and $N_B = N$.
+The resulting probability distribution is known as the **Gibbs distribution**,
+with $\beta \equiv 1/(kT)$:
+
+$$\begin{aligned}
+    \boxed{
+        p(s_A) = \frac{1}{\mathcal{Z}} \exp\!\Big(\!-\! \beta \: \big( U_A(s_A) \!-\! \mu N_A(s_A) \big) \Big)
+    }
+\end{aligned}$$
+
+Where the normalizing **grand partition function** $\mathcal{Z}(\mu, V, T)$ is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \mathcal{Z} \equiv \sum_{s_A}^{} \exp\!\Big(\!-\! \beta \: \big( U_A(s_A) - \mu N_A(s_A) \big) \Big)
+    }
+\end{aligned}$$
+
+In contrast to the canonical ensemble,
+whose [thermodynamic potential](/know/concept/thermodynamic-potential/)
+was the Helmholtz free energy $F$,
+the grand canonical ensemble instead
+minimizes the **grand potential** $\Omega$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \Omega(T, V, \mu)
+            &\equiv - k T \ln{\mathcal{Z}}
+            \\
+            &= \expval{U_A} - T S_A - \mu \expval{N_A}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+So $\mathcal{Z} = \exp\!(- \beta \Omega)$.
+This is proven in the same way as for $F$ in the canonical ensemble.
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/microcanonical-ensemble/index.pdc b/content/know/concept/microcanonical-ensemble/index.pdc
index 89d114b..16628b8 100644
--- a/content/know/concept/microcanonical-ensemble/index.pdc
+++ b/content/know/concept/microcanonical-ensemble/index.pdc
@@ -34,8 +34,8 @@ $$\begin{aligned}
     = \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A)
 \end{aligned}$$
 
-Where $c_A$ and $c_B$ are the number of microstates of
-the subsystems at the given energy levels.
+Where $c_A$ and $c_B$ are the numbers of subsystem microstates
+at the given energy levels.
 
 The core assumption of the microcanonical ensemble
 is that each of these microstates has the same probability $1 / c$.
@@ -43,18 +43,20 @@ Consequently, the probability of finding an energy $U_A$ in $A$ is:
 
 $$\begin{aligned}
     p_A(U_A)
-    = \frac{c_A(U_A) \:c_B(U - U_A)}{c(U)}
+    = \frac{c_A(U_A) \: c_B(U - U_A)}{c(U)}
 \end{aligned}$$
 
 If a certain $U_A$ has a higher probability,
 then there are more $A$-microstates with that energy,
-meaning that $U_A$ is "easier to reach" or "more comfortable" for the system.
-Note that $c(U)$ is a constant, because $U$ is given.
+so, statistically, for an *ensemble* of many boxes,
+we expect that $U_A$ is more common.
 
-After some time, the system will reach equilibrium,
-where both $A$ and $B$ have settled into a "comfortable" position.
+The maximum of $p_A$ will be the most common in the ensemble.
+Assuming that we have given the boxes enough time to settle,
+we go one step further,
+and refer to this maximum as "equilibrium".
 In other words, the subsystem microstates at equilibrium
-must be maxima of their probability distributions $p_A$ and $p_B$.
+are maxima of $p_A$ and $p_B$.
 
 We only need to look at $p_A$.
 Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$:
@@ -66,13 +68,13 @@ $$\begin{aligned}
 
 Here, in the quantity $\ln{c_A}$,
 we recognize the definition of
-the entropy $S_A \equiv k_B \ln{c_A}$,
-where $k_B$ is Boltzmann's constant.
-We thus multiply by $k_B$:
+the entropy $S_A \equiv k \ln{c_A}$,
+where $k$ is Boltzmann's constant.
+We thus multiply by $k$:
 
 $$\begin{aligned}
-    k_B \ln p_A(U_A)
-    = S_A(U_A) + S_B(U - U_A) - k_B \ln{c(U)}
+    k \ln p_A(U_A)
+    = S_A(U_A) + S_B(U - U_A) - k \ln{c(U)}
 \end{aligned}$$
 
 Since entropy is additive over subsystems,
@@ -87,7 +89,7 @@ more concrete, equilibrium condition:
 
 $$\begin{aligned}
     0
-    = k_B \dv{(\ln{p_A})}{U_A}
+    = k \dv{(\ln{p_A})}{U_A}
     = \pdv{S_A}{U_A} + \pdv{S_B}{U_A}
     = \pdv{S_A}{U_A} - \pdv{S_B}{U_B}
 \end{aligned}$$
@@ -107,13 +109,14 @@ $$\begin{aligned}
 
 Recall that our partitioning into $A$ and $B$ was arbitrary,
 meaning that, in fact, the temperature $T$ must be uniform in the whole box.
-
 We get this specific result because
 heat was the only thing that $A$ and $B$ could exchange.
-The key point, however,
-is that the total entropy $S$ must be maximized.
-We also would have reached that conclusion if our imaginary wall
-allowed changes in volume $V_A$ and particle count $N_A$.
+
+The point is that the most likely state of the box
+maximizes the total entropy $S$.
+We also would have reached that conclusion
+if our imaginary wall was permeable and flexible,
+i.e if it allowed changes in volume $V_A$ and particle count $N_A$.
 
 
 
diff --git a/content/know/concept/thermodynamic-potential/index.pdc b/content/know/concept/thermodynamic-potential/index.pdc
index b65891f..813f3b4 100644
--- a/content/know/concept/thermodynamic-potential/index.pdc
+++ b/content/know/concept/thermodynamic-potential/index.pdc
@@ -25,7 +25,7 @@ and is consequently assumed to be at the same temperature $T = T_{\mathrm{sur}}$
 then $T$ must be a natural variable.
 
 The link from natural variables to potentials
-is established by thermodynamic ensembles.
+is established by [thermodynamic ensembles](/know/category/thermodynamic-ensembles/).
 
 Once enough natural variables have been found,
 the appropriate potential can be selected from the list below.