Expand knowledge base

5 files changed, 463 insertions, 26 deletions

diff --git a/content/know/concept/berry-phase/index.pdc b/content/know/concept/berry-phase/index.pdc
new file mode 100644
index 0000000..339599b
--- /dev/null
+++ b/content/know/concept/berry-phase/index.pdc
@@ -0,0 +1,219 @@
+---
+title: "Berry phase"
+firstLetter: "B"
+publishDate: 2021-11-29
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-11-25T20:42:45+01:00
+draft: false
+markup: pandoc
+---
+
+# Berry phase
+
+Consider a Hamiltonian $\hat{H}$ that does not explicitly depend on time,
+but does depend on a given parameter $\vb{R}$.
+The Schrödinger equations then read:
+
+$$\begin{aligned}
+    i \hbar \dv{t} \ket{\Psi_n(t)}
+    &= \hat{H}(\vb{R}) \ket{\Psi_n(t)}
+    \\
+    \hat{H}(\vb{R}) \ket{\psi_n(\vb{R})}
+    &= E_n(\vb{R}) \ket{\psi_n(\vb{R})}
+\end{aligned}$$
+
+The general full solution $\ket{\Psi_n}$ has the following form,
+where we allow $\vb{R}$ to evolve in time,
+and we have abbreviated the traditional phase of the "wiggle factor" as $L_n$:
+
+$$\begin{aligned}
+    \ket{\Psi_n(t)}
+    = \exp\!(i \gamma_n(t)) \exp\!(-i L_n(t) / \hbar) \: \ket{\psi_n(\vb{R}(t))}
+    \qquad
+    L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'}
+\end{aligned}$$
+
+The **geometric phase** $\gamma_n(t)$ is more interesting.
+It is not included in $\ket{\psi_n}$,
+because it depends on the path $\vb{R}(t)$
+rather than only the present $\vb{R}$ and $t$.
+Its dynamics can be found by inserting the above $\ket{\Psi_n}$
+into the time-dependent Schrödinger equation:
+
+$$\begin{aligned}
+    \dv{t} \ket{\Psi_n}
+    &= i \dv{\gamma_n}{t} \ket{\Psi_n} - \frac{i}{\hbar} \dv{L_n}{t} \ket{\Psi_n}
+    + \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \dv{t} \ket{\psi_n}
+    \\
+    &= i \dv{\gamma_n}{t} \ket{\Psi_n} + \frac{1}{i \hbar} E_n \ket{\Psi_n}
+    + \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \: \ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
+    \\
+    &= i \dv{\gamma_n}{t} \ket{\Psi_n} + \frac{1}{i \hbar} \hat{H} \ket{\Psi_n}
+    + \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \: \ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
+\end{aligned}$$
+
+Here we recognize the Schrödinger equation, so those terms cancel.
+We are then left with:
+
+$$\begin{aligned}
+    - i \dv{\gamma_n}{t} \ket{\Psi_n}
+    &= \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \: \ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
+\end{aligned}$$
+
+Front-multiplying by $i \bra{\Psi_n}$ gives us
+the equation of motion of the geometric phase $\gamma_n$:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{\gamma_n}{t}
+        = - \vb{A}_n(\vb{R}) \cdot \dv{\vb{R}}{t}
+    }
+\end{aligned}$$
+
+Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{A}_n(\vb{R})
+        \equiv -i \braket{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})}
+    }
+\end{aligned}$$
+
+Importantly, note that $\vb{A}_n$ is real,
+provided that $\ket{\psi_n}$ is always normalized for all $\vb{R}$.
+To prove this, we start from the fact that $\nabla_\vb{R} 1 = 0$:
+
+$$\begin{aligned}
+    0
+    &= \nabla_\vb{R} \braket{\psi_n}{\psi_n}
+    = \braket{\nabla_\vb{R} \psi_n}{\psi_n} + \braket{\psi_n}{\nabla_\vb{R} \psi_n}
+    \\
+    &= \braket{\psi_n}{\nabla_\vb{R} \psi_n}^* + \braket{\psi_n}{\nabla_\vb{R} \psi_n}
+    = 2 \Re\{ - i \vb{A}_n \}
+    = 2 \Im\{ \vb{A}_n \}
+\end{aligned}$$
+
+Consequently, $\vb{A}_n = \Im \braket{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real,
+because $\braket{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary.
+
+Suppose now that the parameter $\vb{R}(t)$ is changed adiabatically
+(i.e. so slow that the system stays in the same eigenstate)
+for $t \in [0, T]$, along a circuit $C$ with $\vb{R}(0) \!=\! \vb{R}(T)$.
+Integrating the phase $\gamma_n(t)$ over this contour $C$ then yields
+the **Berry phase** $\gamma_n(C)$:
+
+$$\begin{aligned}
+    \boxed{
+        \gamma_n(C)
+        = - \oint_C \vb{A}_n(\vb{R}) \cdot \dd{\vb{R}}
+    }
+\end{aligned}$$
+
+But we have a problem: $\vb{A}_n$ is not unique!
+Due to the Schrödinger equation's gauge invariance,
+any function $f(\vb{R}(t))$ can be added to $\gamma_n(t)$
+without making an immediate physical difference to the state.
+Consider the following general gauge transformation:
+
+$$\begin{aligned}
+    \ket*{\tilde{\psi}_n(\vb{R})}
+    \equiv \exp\!(i f(\vb{R})) \: \ket{\psi_n(\vb{R})}
+\end{aligned}$$
+
+To find $\vb{A}_n$ for a particular choice of $f$,
+we need to evaluate the inner product
+$\braket*{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:
+
+$$\begin{aligned}
+    \braket*{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}
+    &= \exp\!(i f) \Big( i \nabla_\vb{R} f \: \braket*{\tilde{\psi}_n}{\psi_n} + \braket*{\tilde{\psi}_n}{\nabla_\vb{R} \psi_n} \Big)
+    \\
+    &= i \nabla_\vb{R} f \: \braket*{\psi_n}{\psi_n} + \braket*{\psi_n}{\nabla_\vb{R} \psi_n}
+    \\
+    &= i \nabla_\vb{R} f + \braket*{\psi_n}{\nabla_\vb{R} \psi_n}
+\end{aligned}$$
+
+Unfortunately, $f$ does not vanish as we would have liked,
+so $\vb{A}_n$ depends on our choice of $f$.
+
+However, the curl of a gradient is always zero,
+so although $\vb{A}_n$ is not unique,
+its curl $\nabla_\vb{R} \cross \vb{A}_n$ is guaranteed to be.
+Conveniently, we can introduce a curl in the definition of $\gamma_n(C)$
+by applying Stokes' theorem, under the assumption
+that $\vb{A}_n$ has no singularities in the area enclosed by $C$
+(fortunately, $\vb{A}_n$ can always be chosen to satisfy this):
+
+$$\begin{aligned}
+    \boxed{
+        \gamma_n(C)
+        = - \iint_{S(C)} \vb{B}_n(\vb{R}) \cdot \dd{\vb{S}}
+    }
+\end{aligned}$$
+
+Where we defined $\vb{B}_n$ as the curl of $\vb{A}_n$.
+Now $\gamma_n(C)$ is guaranteed to be unique.
+Note that $\vb{B}_n$ is analogous to a magnetic field,
+and $\vb{A}_n$ to a magnetic vector potential:
+
+$$\begin{aligned}
+    \vb{B}_n(\vb{R})
+    \equiv \nabla_\vb{R} \cross \vb{A}_n(\vb{R})
+    = \Im\Big\{ \nabla_\vb{R} \cross \braket{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\}
+\end{aligned}$$
+
+Unfortunately, $\nabla_\vb{R} \psi_n$ is difficult to evaluate explicitly,
+so we would like to rewrite $\vb{B}_n$ such that it does not enter.
+We do this as follows, inserting $1 = \sum_{m} \ket{\psi_m} \bra{\psi_m}$ along the way:
+
+$$\begin{aligned}
+    i \vb{B}_n
+    = \nabla_\vb{R} \cross \braket{\psi_n}{\nabla_\vb{R} \psi_n}
+    &= \braket{\psi_n}{\nabla_\vb{R} \cross \nabla_\vb{R} \psi_n} + \bra{\nabla_\vb{R} \psi_n} \cross \ket{\nabla_\vb{R} \psi_n}
+    \\
+    &= \sum_{m} \braket{\nabla_\vb{R} \psi_n}{\psi_m} \cross \braket{\psi_m}{\nabla_\vb{R} \psi_n}
+\end{aligned}$$
+
+The fact that $\braket{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary
+means it is parallel to its complex conjugate,
+and thus the cross product vanishes, so we exclude $n$ from the sum:
+
+$$\begin{aligned}
+    \vb{B}_n
+    &= \sum_{m \neq n} \braket{\nabla_\vb{R} \psi_n}{\psi_m} \cross \braket{\psi_m}{\nabla_\vb{R} \psi_n}
+\end{aligned}$$
+
+From the [Hellmann-Feynman theorem](/know/concept/hellmann-feynman-theorem/),
+we know that the inner products can be rewritten:
+
+$$\begin{aligned}
+    \braket{\psi_m}{\nabla_\vb{R} \psi_n}
+    = \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m}}{E_n - E_m}
+\end{aligned}$$
+
+Where we have assumed that there is no degeneracy.
+This leads to the following result:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{B}_n
+        = \Im \sum_{m \neq n}
+        \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m} \cross \matrixel{\psi_m}{\nabla_\vb{R} \hat{H}}{\psi_n}}{(E_n - E_m)^2}
+    }
+\end{aligned}$$
+
+Which only involves $\nabla_\vb{R} \hat{H}$,
+and is therefore easier to evaluate than any $\ket{\nabla_\vb{R} \psi_n}$.
+
+
+
+## References
+1.  M.V. Berry,
+    [Quantal phase factors accompanying adiabatic changes](https://doi.org/10.1098/rspa.1984.0023),
+    1984, Royal Society.
+2.  G. Grosso, G.P. Parravicini,
+    *Solid state physics*,
+    2nd edition, Elsevier.
diff --git a/content/know/concept/conditional-expectation/index.pdc b/content/know/concept/conditional-expectation/index.pdc
index 50120e8..0384222 100644
--- a/content/know/concept/conditional-expectation/index.pdc
+++ b/content/know/concept/conditional-expectation/index.pdc
@@ -6,6 +6,7 @@ categories:
 - Mathematics
 - Statistics
 - Measure theory
+- Stochastic analysis
 
 date: 2021-10-22T15:19:23+02:00
 draft: false
diff --git a/content/know/concept/hellmann-feynman-theorem/index.pdc b/content/know/concept/hellmann-feynman-theorem/index.pdc
new file mode 100644
index 0000000..3b88cd8
--- /dev/null
+++ b/content/know/concept/hellmann-feynman-theorem/index.pdc
@@ -0,0 +1,97 @@
+---
+title: "Hellmann-Feynman theorem"
+firstLetter: "H"
+publishDate: 2021-11-29
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-11-29T10:30:37+01:00
+draft: false
+markup: pandoc
+---
+
+# Hellmann-Feynman theorem
+
+Consider the time-independent Schrödinger equation,
+where the Hamiltonian $\hat{H}$ depends on a general parameter $\lambda$,
+whose meaning or type we will not specify:
+
+$$\begin{aligned}
+    \hat{H}(\lambda) \ket{\psi_n(\lambda)}
+    = E_n(\lambda) \ket{\psi_n(\lambda)}
+\end{aligned}$$
+
+Assuming all eigenstates $\ket{\psi_n}$ are normalized,
+this gives us the following basic relation:
+
+$$\begin{aligned}
+    \matrixel{\psi_m}{\hat{H}}{\psi_n}
+    = E_n \braket{\psi_m}{\psi_n}
+    = \delta_{mn} E_n
+\end{aligned}$$
+
+We differentiate this with respect to $\lambda$,
+which could be a scalar or a vector.
+This yields:
+
+$$\begin{aligned}
+    \delta_{mn} \nabla_\lambda E_n
+    &= \nabla_\lambda \matrixel{\psi_m}{\hat{H}}{\psi_n}
+    \\
+    &= \matrixel{\nabla_\lambda \psi_m}{\hat{H}}{\psi_n}
+    + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    + \matrixel{\psi_m}{\hat{H}}{\nabla_\lambda \psi_n}
+    \\
+    &= E_m \braket{\psi_m}{\nabla_\lambda \psi_n} + E_n \braket{\nabla_\lambda \psi_m}{\psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+\end{aligned}$$
+
+In order to simplify this,
+we differentiate the orthogonality relation
+$\braket{\psi_m}{\psi_n} = \delta_{mn}$,
+which ends up telling us that
+$\braket{\nabla_\lambda \psi_m}{\psi_n} = - \braket{\psi_m}{\nabla_\lambda \psi_n}$:
+
+$$\begin{aligned}
+    0
+    = \nabla_\lambda \delta_{mn}
+    = \nabla_\lambda \braket{\psi_m}{\psi_n}
+    = \braket{\nabla_\lambda \psi_m}{\psi_n} + \braket{\psi_m}{\nabla_\lambda \psi_n}
+\end{aligned}$$
+
+Using this result to replace $\braket{\nabla_\lambda \psi_m}{\psi_n}$
+in the previous equation leads to:
+
+$$\begin{aligned}
+    \delta_{mn} \nabla_\lambda E_n
+    &= (E_m - E_n) \braket{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+\end{aligned}$$
+
+For $m = n$, we therefore arrive at the **Hellmann-Feynman theorem**,
+which is useful when doing numerical calculations
+to minimize energies with respect to $\lambda$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla_\lambda E_n
+        = \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    }
+\end{aligned}$$
+
+While for $m \neq n$, we get the **Epstein generalization**
+of the Hellmann-Feynman theorem, which is for example relevant for
+the [Berry phase](/know/concept/berry-phase/):
+
+$$\begin{aligned}
+    \boxed{
+        (E_n - E_m) \braket{\psi_m}{\nabla_\lambda \psi_n}
+        = \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  G. Grosso, G.P. Parravicini,
+    *Solid state physics*,
+    2nd edition, Elsevier.
diff --git a/content/know/concept/self-energy/index.pdc b/content/know/concept/self-energy/index.pdc
index d2908eb..c86f8c5 100644
--- a/content/know/concept/self-energy/index.pdc
+++ b/content/know/concept/self-energy/index.pdc
@@ -50,8 +50,6 @@ and $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
 
 $$\begin{aligned}
     G_{ba}
-    %&= - \frac{\expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}}}{\expval{\hat{K}(\hbar \beta, 0)}}
-    %\\
     &= - \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
     \expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
     {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
@@ -141,12 +139,15 @@ $$\begin{aligned}
 
 These integrals over products of interactions and Green's functions
 are the perfect place to apply [Feynman diagrams](/know/concept/feynman-diagram/).
-Conveniently, it even turns out that the factor $(-1)^p$
-is exactly equivalent to the rule that each diagram is multiplied by $(-1)^F$,
+Conveniently, it turns out that the factor $(-1)^p$
+is equivalent to the rule that each diagram must be multiplied by $(-1)^F$,
 with $F$ the number of fermion loops.
+Keep in mind that fermion lines absorb a factor $-\hbar$ each (see above),
+and interactions $-1/\hbar$.
 
-The denominator thus turns into a sum of all possible diagrams for each total order $n$
-(the order of a diagram is the number of interaction lines it contains).
+The denominator turns into a sum of all possible diagrams
+(including equivalent ones) for each total order $n$
+(the order is the number of interaction lines).
 The endpoints $a$ and $b$ do not appear here,
 so we conclude that all those diagrams only have internal vertices;
 we will therefore refer to them as **internal diagrams**.
@@ -168,24 +169,25 @@ We thus find:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \bigg[ \sum_\mathrm{all\;ext}^{n} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m \!\le\! n}
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)} \bigg]}
-    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
+    \bigg[ \sum_{m = 0}^{n} \frac{n!}{m! (n \!-\! m)!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
 \end{aligned}$$
 
-Where the total order refers to the sum of the orders of all disconnected diagrams.
-Note that the external diagram does not directly depend on $n$.
-We can therefore reorganize:
+Where the total order is the sum of the orders of all considered diagrams,
+and the new factor is needed for all the possible choices
+of vertices to put in the external part.
+Note that the external diagram does not directly depend on $n$,
+so we reorganize:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_\mathrm{all\;ext}^{\infty} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
-    \bigg[ \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)} \bigg]}
-    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{2^{n-m} (n \!-\! m)!}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
 \end{aligned}$$
 
 Since both $n$ and $m$ start at zero,
@@ -194,18 +196,19 @@ we see that the second sum in the numerator does not actually depend on $m$:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_\mathrm{all\;ext} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
-    \bigg[ \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n} \bigg]}
-    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
     \\
-    &= \sum_\mathrm{all\;ext} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
+    &= \sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
 \end{aligned}$$
 
 In other words, all the disconnected diagrams simply cancel out,
 and we are left with a sum over all possible fully connected diagrams
-that contain $a$ and $b$. Let $G(b,a) = G_{ba}$:
+that contain $a$ and $b$. Furthermore, it can be shown using combinatorics
+that exactly $2^m m!$ diagrams at each order are topologically equivalent,
+so we are left with non-equivalent diagrams only.
+Let $G(b,a) = G_{ba}$:
 
 <a href="fullgf.png">
 <img src="fullgf.png" style="width:90%;display:block;margin:auto;">
diff --git a/content/know/concept/wiener-process/index.pdc b/content/know/concept/wiener-process/index.pdc
index 11c7a6e..1919284 100644
--- a/content/know/concept/wiener-process/index.pdc
+++ b/content/know/concept/wiener-process/index.pdc
@@ -68,6 +68,123 @@ since each increment has mean zero (so it is a martingale),
 and all increments are independent (so it is a Markov process).
 
 
+## Recurrence
+
+An important question about the Wiener process
+is whether it is **recurrent** or **transient**:
+given a hypersphere (interval in 1D, circle in 2D, sphere in 3D)
+away from the origin, will $B_t$ visit it after a finite time $\tau\!<\!\infty$?
+It is *recurrent* if yes, i.e. $P(\tau \!<\! \infty) = 1$, or *transient* otherwise.
+The answer to this question turns out to depend on the number of dimenions.
+
+To demonstrate this, we model the $d$-dimensional Wiener process
+as an [Itō diffusion](/know/concept/ito-calculus/) $X_t$,
+which also allows us to shift the initial condition $X_0$
+(or resume a "paused" process):
+
+$$\begin{aligned}
+    X_t
+    = X_0 + \int_0^t \dd{B_s}
+\end{aligned}$$
+
+Consider two hyperspheres, the inner with radius $R_i$,
+and the outer with $R_o > R_i$.
+Let the initial condition $|X_0| \in \, ]R_i, R_o[$,
+then we define the stopping times $\tau_i$, $\tau_o$ and $\tau$ like so:
+
+$$\begin{aligned}
+    \tau_i
+    \equiv \inf\{ t : |X_t| \le R_i \}
+    \qquad
+    \tau_o
+    \equiv \inf\{ t : |X_t| \ge R_o \}
+    \qquad
+    \tau
+    \equiv \min\{\tau_i, \tau_o\}
+\end{aligned}$$
+
+We stop when the inner or outer hypersphere is touched by $X_t$,
+whichever happens first.
+
+[Dynkin's formula](/know/concept/dynkins-formula/)
+is applicable to this situation, if we define $h(x)$ as follows,
+where the *terminal reward* $\Gamma$ equals $1$ for $|X_\tau| = R_i$,
+and $0$ for $|X_\tau| = R_o$,
+such that $h(X_0)$ equals the probability
+that we touch $R_i$ before $R_o$ for a given $X_0$:
+
+$$\begin{aligned}
+    h(X_0)
+    = \mathbf{E}\Big[ \Gamma(X_\tau) \Big| X_0 \Big]
+    = P\Big[|X_\tau| \!=\! R_i \:\Big|\: X_0\Big]
+\end{aligned}$$
+
+Dynkin's formula then tells us that $h(x)$ is given by the following equation,
+with the boundary conditions $h(R_i) = 1$ and $h(R_o) = 0$:
+
+$$\begin{aligned}
+    0
+    = \hat{L}\{h(x)\}
+    = \frac{1}{2} \nabla^2 h(x)
+\end{aligned}$$
+
+Thanks to this problem's spherical symmetry,
+$h$ only depends on the radial coodinate $r$,
+so the Laplacian $\nabla^2$ can be written as follows
+in $d$-dimensional [spherical coordinates](/know/concept/spherical-coordinates/):
+
+$$\begin{aligned}
+    0
+    = \nabla^2 h(r)
+    = \pdv[2]{h}{r} + \frac{d - 1}{r} \pdv{h}{r}
+\end{aligned}$$
+
+For $d = 1$, the solution $h_1(r)$ is as follows,
+of which we take the limit for $R_o \to \infty$:
+
+$$\begin{aligned}
+    h_1(r)
+    = \frac{r - R_o}{R_i - R_o}
+    \quad\underset{R_o \to \infty}{\longrightarrow}\quad
+    1
+\end{aligned}$$
+
+The outer hypersphere becomes harder to reach for larger $R_o$,
+and for $R_o \to \infty$ we are left with
+the probability of hitting $R_i$ only.
+This turns out to be $1$, so in 1D the Wiener process is recurrent:
+it always comes close to the origin in finite time.
+
+For $d = 2$, the solution $h_2(r)$ is as follows,
+whose limit turns out to be $1$,
+so the Wiener process is also recurrent in 2D:
+
+$$\begin{aligned}
+    h_2(r)
+    = 1 - \frac{\log\!(r/R_i)}{\log\!(R_o/R_i)}
+    \quad\underset{R_o \to \infty}{\longrightarrow}\quad
+    1
+\end{aligned}$$
+
+However, for $d \ge 3$, the solution $h_d(r)$
+does not converge to $1$ for $R_o \to \infty$,
+meaning the Wiener process is transient in 3D or higher:
+
+$$\begin{aligned}
+    h_d(r)
+    = \frac{R_o^{2 - d} - r^{2 - d}}{R_o^{2 - d} - R_i^{2 - d}}
+    \quad\underset{R_o \to \infty}{\longrightarrow}\quad
+    \frac{R_i^{d - 2}}{r^{d - 2}}
+    < 1
+\end{aligned}$$
+
+This is a major qualitative difference. For example, consider a situation
+where some substance is diffusing from a localized infinite source:
+in 3D, the substance can escape and therefore a steady state can exist,
+while in 2D, the substance never strays far from the source,
+so no steady state is ever reached as long as the source continues to emit.
+
+
 
 ## References
 1.  U.H. Thygesen,