Expand knowledge base, reorganize Green's functions

1 files changed, 253 insertions, 38 deletions

diff --git a/content/know/concept/greens-functions/index.pdc b/content/know/concept/greens-functions/index.pdc
index 10ab09b..2f86e63 100644
--- a/content/know/concept/greens-functions/index.pdc
+++ b/content/know/concept/greens-functions/index.pdc
@@ -13,38 +13,52 @@ markup: pandoc
 
 # Green's functions
 
-In many-body quantum theory, **Green's functions**
-are correlation functions between particle creation/annihilation operators.
+In many-body quantum theory, a **Green's function**
+can be any correlation function between two given operators,
+although it is usually used to refer to the special case
+where the operators are particle creation/annihilation operators
+from the [second quantization](/know/concept/second-quantization/).
+
 They are somewhat related to
-[fundamental solution](/know/concept/fundamental-solution/) functions,
-which are also often called *Green's functions*.
+[fundamental solutions](/know/concept/fundamental-solution/),
+which are also called *Green's functions*,
+but in general they are not the same,
+except in a special case, see below.
+
+
+## Single-particle functions
+
+If the two operators are single-particle creation/annihilation operators,
+then we get the **single-particle Green's functions**,
+for which the symbol $G$ is used.
 
 The **retarded Green's function** $G_{\nu \nu'}^R$
 and the **advanced Green's function** $G_{\nu \nu'}^A$
 are defined like so,
 where the expectation value $\expval{}$ is
-with respect to thermal equilibrium,
-$\nu$ and $\nu'$ are labels of single-particle states that may include spin,
-and $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are annihilation/creation operators
-from the [second quantization](/know/concept/second-quantization/):
+with respect to thermodynamic equilibrium,
+$\nu$ and $\nu'$ are labels of single-particle states,
+and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.:
 
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
             G_{\nu \nu'}^R(t, t')
-            &\equiv -\frac{i}{\hbar} \Theta(t - t') \expval{\comm{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}}
+            &\equiv -\frac{i}{\hbar} \Theta(t - t') \expval{\comm*{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
             \\
             G_{\nu \nu'}^A(t, t')
-            &\equiv \frac{i}{\hbar} \Theta(t' - t) \expval{\comm{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}}
+            &\equiv \frac{i}{\hbar} \Theta(t' - t) \expval{\comm*{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
         \end{aligned}
     }
 \end{aligned}$$
 
-Where $\Theta$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
-This is for bosons; for fermions the commutator
-must be replaced by an anticommutator, as usual.
-Notice that $G^R_{\nu \nu'}$ has the same form as the correlation function
-from the [Kubo formula](/know/concept/kubo-formula/).
+Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/),
+and $[,]_{\mp}$ is a commutator for bosons,
+and an anticommutator for fermions.
+We are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent,
+but keep in mind that time-dependent Hamiltonians are allowed,
+so it might not be trivial.
 
 Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$
 and **lesser Green's function** $G_{\nu \nu'}^<$ are:
@@ -53,10 +67,10 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             G_{\nu \nu'}^>(t, t')
-            &\equiv -\frac{i}{\hbar} \expval{\hat{c}_{\nu}(t) \hat{c}_{\nu'}^\dagger(t')}
+            &\equiv -\frac{i}{\hbar} \expval{\hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')}
             \\
             G_{\nu \nu'}^<(t, t')
-            &\equiv \mp \frac{i}{\hbar} \expval{\hat{c}_{\nu'}^\dagger(t') \hat{c}_{\nu}(t)}
+            &\equiv \mp \frac{i}{\hbar} \expval{\hat{c}_{\nu'}^\dagger(t') \: \hat{c}_{\nu}(t)}
         \end{aligned}
     }
 \end{aligned}$$
@@ -80,46 +94,94 @@ we use the spin $s$ and position $\vb{r}$, leading to:
 
 $$\begin{aligned}
     G_{ss'}^R(\vb{r}, t; \vb{r}', t')
-    &= -\frac{i}{\hbar} \Theta(t - t') \expval{\comm{\hat{\Psi}_{s}(\vb{r}, t)}{\hat{\Psi}_{s'}^\dagger(\vb{r}', t')}}
+    &= -\frac{i}{\hbar} \Theta(t - t') \expval{\comm*{\hat{\Psi}_{s}(\vb{r}, t)}{\hat{\Psi}_{s'}^\dagger(\vb{r}', t')}_{\mp}}
     \\
     &= \sum_{\nu \nu'} \psi_\nu(\vb{r}) \: \psi^*_{\nu'}(\vb{r}') \: G_{\nu \nu'}^R(t, t')
 \end{aligned}$$
 
 And analogously for $G_{ss'}^A$, $G_{ss'}^>$ and $G_{ss'}^<$.
 Note that the time-dependence is given to the old $G_{\nu \nu'}^R$,
-i.e. to $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$.
-In other words, we are using the
-[Heisenberg picture](/know/concept/heisenberg-picture/).
+i.e. to $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$,
+because we are in the Heisenberg picture.
 
 If the Hamiltonian is time-independent,
 then it can be shown that all the Green's functions
-only depend on the time-difference $t - t'$
-(for a proof, see [Kubo formula](/know/concept/kubo-formula/)):
+only depend on the time-difference $t - t'$:
 
 $$\begin{aligned}
+    G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t')
+    \qquad \quad
+    G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t')
+    \\
     G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
     \qquad \quad
     G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
 \end{aligned}$$
 
+<div class="accordion">
+<input type="checkbox" id="proof-time-diff"/>
+<label for="proof-time-diff">Proof</label>
+<div class="hidden">
+<label for="proof-time-diff">Proof.</label>
+We will prove that the thermal expectation value
+$\expval*{\hat{A}(t) \hat{B}(t')}$ only depends on $t - t'$
+for arbitrary $\hat{A}$ and $\hat{B}$,
+and it trivially follows that the Green's functions do too.
 
-If the Hamiltonian is both time-independent and non-interacting,
-then the time-dependence of $\hat{c}_\nu$
-can simply be factored out as follows:
+Suppose that the system started in thermodynamic equilibrium.
+This could sometimes be in the [canonical ensemble](/know/concept/canonical-ensemble/)
+(for two-particle Green's functions, see below),
+but usually it will be in the
+[grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
+since we are adding/removing particles.
+In the latter case, we assume that the chemical potential $\mu$
+is already included in the Hamiltonian.
+
+In any case, at equilibrium, we know that the
+[density operator](/know/concept/density-operator/)
+$\hat{\rho}$ is as follows:
+
+$$\begin{aligned}
+    \hat{\rho} = \frac{1}{Z} \exp\!(- \beta \hat{H})
+\end{aligned}$$
+
+Where $Z \equiv \Tr\!(\exp\!(- \beta \hat{H}))$ is the partition function.
+In that case, the expected value of the product
+of the time-independent operators $\hat{A}$ and $\hat{B}$ is calculated like so:
 
 $$\begin{aligned}
-    \hat{c}_\nu(t)
-    = \hat{c}_\nu \exp\!(- i \varepsilon_\nu t / \hbar)
+    \expval*{\hat{A}(t) \hat{B}(t')}
+    &= \frac{1}{Z} \Tr\!\big( \hat{\rho} \hat{A}(t) \hat{B}(t') \big)
+    \\
+    &= \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i t \hat{H} / \hbar} \hat{A} e^{-i t \hat{H} / \hbar}
+    e^{i t' \hat{H} / \hbar} \hat{B} e^{-i t' \hat{H} / \hbar} \Big)
 \end{aligned}$$
 
+Using that the trace $\Tr$ is invariant
+under cyclic permutations of its argument,
+and that all functions of $\hat{H}$ commute, we find:
+
+$$\begin{aligned}
+    \expval*{\hat{A}(t) \hat{B}(t')}
+    = \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i (t - t') \hat{H} / \hbar} \hat{A} e^{-i (t - t') \hat{H} / \hbar} \hat{B} \Big)
+\end{aligned}$$
+
+As expected, this only depends on the time difference $t - t'$,
+because $\hat{H}$ is time-independent by assumption.
+Note that thermodynamic equilibrium is crucial:
+intuitively, if the system is not in equilibrium,
+then it evolves in some transient time-dependent way.
+</div>
+</div>
+
+If the Hamiltonian is both time-independent and non-interacting,
+then the time-dependence of $\hat{c}_\nu$
+can simply be factored out as
+$\hat{c}_\nu(t) = \hat{c}_\nu \exp\!(- i \varepsilon_\nu t / \hbar)$.
 Then the diagonal ($\nu = \nu'$) greater and lesser Green's functions
 can be written in the form below, where $f_\nu$ is either
 the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/)
 or the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/).
-Note that the off-diagonal ($\nu \neq \nu'$) functions vanish,
-because $\expval*{\hat{c}_{\nu} \hat{c}_{\nu'}^\dagger} = 0$ there,
-since the many-particle states are simply orthogonal
-[Slater determinants](/know/concept/slater-determinant/)/permanents:
 
 $$\begin{aligned}
     G_{\nu \nu}^>(t, t')
@@ -133,15 +195,168 @@ $$\begin{aligned}
     &= \mp \frac{i}{\hbar} f_\nu \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
 \end{aligned}$$
 
-The diagonal retarded and advanced Green's functions then reduce to
-the following, where $+$ applies to fermions, and $-$ to bosons:
+
+## As fundamental solutions
+
+In the absence of interactions,
+we know from the derivation of
+[equation-of-motion theory](/know/concept/equation-of-motion-theory/)
+that the equation of motion of $G^R(\vb{r}, t; \vb{r}', t')$
+is as follows (neglecting spin):
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R}{t}
+    = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+    + \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\comm*{\hat{H}_0}{\hat{\Psi}(\vb{r}, t)}}{\hat{\Psi}^\dagger(\vb{r}', t')}}
+\end{aligned}$$
+
+If $\hat{H}_0$ only contains kinetic energy,
+i.e. there is no external potential,
+it can be shown that:
+
+$$\begin{aligned}
+    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+    = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-commH0"/>
+<label for="proof-commH0">Proof</label>
+<div class="hidden">
+<label for="proof-commH0">Proof.</label>
+In the second quantization,
+the Hamiltonian $\hat{H}_0$ is written like so:
+
+$$\begin{aligned}
+    \hat{H}_0
+    &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \braket{\psi_\nu}{\nabla^2 \psi_{\nu'}}
+    \\
+    &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+    \\
+    &= - \frac{\hbar^2}{2 m}
+    \int \Big( \sum_{\nu} \psi_\nu^*(\vb{r}') \hat{c}_\nu^\dagger \Big) \Big( \nabla^2 \sum_{\nu'} \psi_{\nu'}(\vb{r}') \hat{c}_{\nu'} \Big) \dd{\vb{r}'}
+    \\
+    &= - \frac{\hbar^2}{2 m}
+    \int \hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+We then insert this into the commutator that we want to prove, yielding:
+
+$$\begin{aligned}
+    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+    &= - \frac{\hbar^2}{2 m} \int \comm{\hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}')}{\hat{\Psi}(\vb{r})} \dd{\vb{r}'}
+    \\
+    &= - \frac{\hbar^2}{2 m} \int \hat{\Psi}^\dagger(\vb{r}') \comm{\nabla^2 \hat{\Psi}(\vb{r}')}{\hat{\Psi}(\vb{r})}
+    + \comm{\hat{\Psi}^\dagger(\vb{r}')}{\hat{\Psi}(\vb{r})} \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
+    \\
+    &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu' \nu''}
+    \Big( \hat{c}_\nu^\dagger \comm*{\hat{c}_{\nu''}}{\hat{c}_{\nu'}} + \comm*{\hat{c}_\nu^\dagger}{\hat{c}_{\nu'}} \hat{c}_{\nu''} \Big)
+    \psi_{\nu'}(\vb{r}) \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu''}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+When deriving equation-of-motion theory,
+we already showed that the following identity
+holds for both bosons and fermions:
+
+$$\begin{aligned}
+    \hat{c}_\nu^\dagger \comm*{\hat{c}_{\nu''}}{\hat{c}_{\nu'}} + \comm*{\hat{c}_\nu^\dagger}{\hat{c}_{\nu'}} \hat{c}_{\nu''}
+    = - \delta_{\nu \nu'} \hat{c}_{\nu''}
+\end{aligned}$$
+
+Such that the commutator can be significantly simplified to:
+
+$$\begin{aligned}
+    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+    &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
+    \int \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r}) \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+We know that the $\psi_\nu$ form a *complete* basis,
+which implies (see [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)):
+
+$$\begin{aligned}
+    \sum_{\nu} \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r})
+    = \delta(\vb{r} - \vb{r}')
+\end{aligned}$$
+
+With this, the commutator can be reduced even further as follows:
 
 $$\begin{aligned}
-    G_{\nu \nu}^R(t, t')
-    &= - \frac{i}{\hbar} \Theta(t - t') \big( 1 - f_\nu \pm f_\nu \big) \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+    &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
+    \int \delta(\vb{r} - \vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
     \\
-    G_{\nu \nu}^A(t, t')
-    &= \frac{i}{\hbar} \Theta(t - t') \big( 1 - f_\nu \pm f_\nu \big) \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+    &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
+    = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
+\end{aligned}$$
+</div>
+</div>
+
+After substituting this into the equation of motion,
+we recognize $G^R(\vb{r}, t; \vb{r}', t')$ itself:
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R}{t}
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+    + \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}, t)}{\hat{\Psi}^\dagger(\vb{r}', t')}}
+    \\
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t') - \frac{\hbar^2}{2 m} \nabla_\vb{r}^2
+    \Big( \!-\! \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\hat{\Psi}(\vb{r}, t)}{\hat{\Psi}^\dagger(\vb{r}', t')}} \Big)
+    \\
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+    - \frac{\hbar^2}{2 m} \nabla_\vb{r}^2 G^R(\vb{r}, t; \vb{r}', t')
+\end{aligned}$$
+
+Rearranging this leads to the following,
+which is the definition of a fundamental solution:
+
+$$\begin{aligned}
+    \Big( i \hbar \pdv{t} + \frac{\hbar^2}{2 m} \nabla_\vb{r}^2 \Big) G^R(\vb{r}, t; \vb{r}', t')
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+\end{aligned}$$
+
+Therefore, the retarded Green's function
+(and, it turns out, the advanced Green's function too)
+is a fundamental solution of the Schrödinger equation
+if there is no potential,
+i.e. the Hamiltonian only contains kinetic energy.
+
+
+## Two-particle functions
+
+The above can be generalized to two arbitrary operators $\hat{A}$ and $\hat{B}$,
+giving us the **two-particle Green's functions**,
+or just **correlation functions**.
+The **retarded correlation function** $C_{AB}^R$
+and the **advanced correlation function** $C_{AB}^A$ are defined as
+(in the Heisenberg picture):
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            C_{AB}^R(t, t')
+            &\equiv -\frac{i}{\hbar} \Theta(t - t') \expval{\comm*{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
+            \\
+            C_{AB}^A(t, t')
+            &\equiv \frac{i}{\hbar} \Theta(t' - t) \expval{\comm*{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where the expectation value $\expval{}$ is taken of thermodynamic equilibrium.
+The name *two-particle* comes from the fact that $\hat{A}$ and $\hat{B}$
+will often consist of a sum of products
+of two single-particle creation/annihilation operators.
+
+Like for the single-particle Green's functions,
+if the Hamiltonian is time-independent,
+then it can be shown that $C_{AB}^R$ and $C_{AB}^A$
+only depend on the time-difference $t - t'$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
+    \qquad \quad
+    G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
 \end{aligned}$$