+
+Substituting this into $G_{\nu \nu'}^R$'s equation of motion,
+we recognize another Green's function $G_{\nu'' \nu'}^R$:
+
+$$\begin{aligned}
+ i \hbar \pdv{G^R_{\nu \nu'}}{t}
+ &= \delta_{\nu \nu'} \delta(t \!-\! t') + \frac{i}{\hbar} \Theta(t \!-\! t')
+ \bigg( \expval{\comm*{\comm*{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
+ - \sum_{\nu''} u_{\nu \nu''} \expval{\comm*{\hat{c}_{\nu''}}{\hat{c}_{\nu'}^\dagger}_{\mp}} \bigg)
+ \\
+ &= \delta_{\nu \nu'} \delta(t \!-\! t')
+ + \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm*{\comm*{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
+ + \sum_{\nu''} u_{\nu \nu''} G_{\nu''\nu'}^R(t, t')
+\end{aligned}$$
+
+Rearranging this as follows yields the main result
+of equation-of-motion theory:
+
+$$\begin{aligned}
+ \boxed{
+ \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t, t')
+ = \delta_{\nu \nu'} \delta(t \!-\! t') + D_{\nu \nu'}^R(t, t')
+ }
+\end{aligned}$$
+
+Where $D_{\nu \nu'}^R$ represents a correction due to interactions $\hat{H}_\mathrm{int}$,
+and also has the form of a retarded Green's function,
+but with $\hat{c}_{\nu}$ replaced by $\comm*{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$:
+
+$$\begin{aligned}
+ \boxed{
+ D^R_{\nu'' \nu'}(t, t')
+ \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm*{\comm*{-\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+ }
+\end{aligned}$$
+
+Unfortunately, calculating $D_{\nu \nu'}^R$
+might still not be doable due to $\hat{H}_\mathrm{int}$.
+The key idea of equation-of-motion theory is to either approximate $D_{\nu \nu'}^R$ now,
+or to differentiate it again $i \hbar \dv*{D_{\nu \nu'}^R}{t}$,
+and try again for the resulting corrections,
+until a solvable equation is found.
+There is no guarantee that that will ever happen;
+if not, one of the corrections needs to be approximated.
+
+For non-interacting particles $\hat{H}_\mathrm{int} = 0$,
+so clearly $D_{\nu \nu'}^R$ trivially vanishes then.
+Let us assume that $\hat{H}_0$ is also time-independent,
+such that $G_{\nu'' \nu'}^R$ only depends on the difference $t - t'$:
+
+$$\begin{aligned}
+ \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t - t')
+ = \delta_{\nu \nu'} \delta(t - t')
+\end{aligned}$$
+
+We take the [Fourier transform](/know/concept/fourier-transform/)
+$(t \!-\! t') \to (\omega + i \eta)$, where $\eta \to 0^+$ ensures convergence:
+
+$$\begin{aligned}
+ \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
+ = \delta_{\nu \nu'}
+\end{aligned}$$
+
+If we assume a diagonal basis $u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$,
+this reduces to the following:
+
+$$\begin{aligned}
+ \delta_{\nu \nu'}
+ &= \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - \varepsilon_\nu \delta_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
+ \\
+ &= \Big( \hbar (\omega + i \eta) - \varepsilon_\nu \Big) G^R_{\nu \nu'}(\omega)
+\end{aligned}$$
+
+For a non-interacting, time-independent Hamiltonian,
+we therefore arrive at:
+
+$$\begin{aligned}
+ \boxed{
+ G^R_{\nu \nu'}(\omega)
+ = \frac{\delta_{\nu \nu'}}{\hbar (\omega + i \eta) - \varepsilon_\nu}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
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 $t - t'$:
$$\begin{aligned}
+ G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t')
+ \qquad \quad
+ G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t')
+ \\
G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
\qquad \quad
G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
\end{aligned}$$
+
+
+
+
+
+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.
+
+
+
+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}$$
+
+
+
+
+
+
+In the second quantization,
+the Hamiltonian $\hat{H}_0$ is written like so:
+
+$$\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}$$
+
+
+
+After substituting this into the equation of motion,
+we recognize $G^R(\vb{r}, t; \vb{r}', t')$ itself:
+
+$$\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}$$
diff --git a/content/know/concept/kubo-formula/index.pdc b/content/know/concept/kubo-formula/index.pdc
index f0208da..f5430da 100644
--- a/content/know/concept/kubo-formula/index.pdc
+++ b/content/know/concept/kubo-formula/index.pdc
@@ -35,7 +35,7 @@ $$\begin{aligned}
= \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)}
&= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)}
\\
- &= \matrixel{\psi_I(t_0)}{\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)}{\psi_I(t_0)}
+ &= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)}
\end{aligned}$$
Where the time evolution operator $\hat{K}_I(t, t_0)$ is as follows,
@@ -116,9 +116,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-This result applies to bosonic operators,
-whereas for fermionic operators
-the commutator would be replaced by an anticommutator.
+Note that observables are bosonic,
+because in the [second quantization](/know/concept/second-quantization/)
+they consist of products of even numbers
+of particle creation/annihiliation operators.
+Therefore, this correlation function
+is a two-particle [Green's function](/know/concept/greens-functions/).
A common situation is that $\hat{H}_1$ consists of
a time-independent operator $\hat{B}$
@@ -133,67 +136,16 @@ $$\begin{aligned}
= C^R_{A B}(t, t') f(t')
\end{aligned}$$
-Conveniently, it can be shown that in this case
-$C^R_{AB}$ only depends on the difference $t - t'$,
-if we assume that the system was initially in thermodynamic equilibrium:
+Since $C_{AB}^R$ is a Green's function,
+we know that it only depends on the difference $t - t'$,
+as long as the system was initially in thermodynamic equilibrium,
+and $\hat{H}_{0,S}$ is time-independent:
$$\begin{aligned}
C^R_{A B}(t, t')
= C^R_{A B}(t - t')
\end{aligned}$$
-
-
-
-
-
-This is trivial for $\Theta(t\!-\!t')$,
-so the challenge is to prove that
-$\expval*{\comm*{\hat{A}_I(t)}{B_I(t')}}$
-depends only on the time difference $t - t'$.
-
-Suppose that the system started in thermodynamic equilibrium
-(see [canonical ensemble](/know/concept/canonical-ensemble/)),
-so that its (unnormalized) [density operator](/know/concept/density-operator/)
-$\hat{\rho}$ was as follows before $\hat{H}_{1,I}$ was applied:
-
-$$\begin{aligned}
- \hat{\rho} = \exp\!(- \beta \hat{H}_{0,S})
-\end{aligned}$$
-
-Let us assume that the perturbation $\hat{H}_{1,I}$
-does not affect the distribution of states,
-but only their individual evolutions in time.
-Note that, in general, this is not equilibrium.
-
-In that case, the expectation value of the product
-of two time-independent observables $\hat{A}$ and $\hat{B}$
-can be calculated as follows,
-where $Z_0 \equiv \Tr\!(\hat{\rho})$ is the partition function:
-
-$$\begin{aligned}
- \expval*{\hat{A} \hat{B}}
- = \frac{1}{Z_0} \Tr\!\big( \hat{\rho} \hat{A}_I \hat{H}_{1,I} \big)
- = \frac{1}{Z_0} \Tr\!\Big( e^{-\beta \hat{H}_0} e^{i t \hat{H}_0 / \hbar} \hat{A}_S e^{-i t \hat{H}_0 / \hbar}
- e^{i t' \hat{H}_0 / \hbar} \hat{B}_S e^{-i t' \hat{H}_0 / \hbar} \Big)
-\end{aligned}$$
-
-Where $\hat{H}_0 \equiv \hat{H}_{0,S}$ for brevity.
-Using that the trace $\Tr$ is invariant
-under cyclic permutations of its argument,
-and that functions of $\hat{H}_{0,S}$ always commute, we find:
-
-$$\begin{aligned}
- \expval*{\hat{A} \hat{B}}
- = \frac{1}{Z_0} \Tr\!\Big( e^{-\beta \hat{H}_0} e^{i (t - t') \hat{H}_0 / \hbar} \hat{A}_S
- e^{-i (t - t') \hat{H}_0 / \hbar} \hat{B}_S \Big)
-\end{aligned}$$
-
-As expected, this clearly only depends on the time difference $t - t'$,
-because $\hat{H}_{0,S}$ is time-independent by assumption.
-
-
-
With this, the Kubo formula can be written as follows,
where we have set $t_0 = - \infty$:
diff --git a/content/know/concept/lehmann-representation/index.pdc b/content/know/concept/lehmann-representation/index.pdc
index f38f803..5808934 100644
--- a/content/know/concept/lehmann-representation/index.pdc
+++ b/content/know/concept/lehmann-representation/index.pdc
@@ -18,9 +18,8 @@ is an alternative way to write the [Green's functions](/know/concept/greens-func
obtained by expanding in the many-particle eigenstates
under the assumption of a time-independent Hamiltonian $\hat{H}$.
-We start by writing out the
-greater Green's function $G_{\nu \nu'}(t, t')$,
-and then expanding its thermal expectation value $\expval{}$
+First, we write out the greater Green's function $G_{\nu \nu'}(t, t')$,
+and then expand its expected value $\expval{}$ (at thermodynamic equilibrium)
into a sum of many-particle basis states $\ket{n}$:
$$\begin{aligned}
@@ -29,9 +28,9 @@ $$\begin{aligned}
&= - \frac{i}{\hbar Z} \sum_{n} \matrixel**{n}{\hat{c}_\nu(t) \hat{c}_{\nu'}^\dagger(t') e^{-\beta \hat{H}}}{n}
\end{aligned}$$
-Where $\beta = 1 / (k_B T)$, and $Z$ is the partition function
-(see [canonical ensemble](/know/concept/canonical-ensemble/));
-the operator $e^{\beta \hat{H}}$ gives the weight of each term at thermal equilibrium.
+Where $\beta = 1 / (k_B T)$, and $Z$ is the grand partition function
+(see [grand canonical ensemble](/know/concept/grand-canonical-ensemble/));
+the operator $e^{\beta \hat{H}}$ gives the weight of each term at equilibrium.
Since $\ket{n}$ is an eigenstate of $\hat{H}$ with energy $E_n$,
this gives us a factor of $e^{\beta E_n}$.
Furthermore, we are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
diff --git a/content/know/concept/second-quantization/index.pdc b/content/know/concept/second-quantization/index.pdc
index b8d9a18..b4d920a 100644
--- a/content/know/concept/second-quantization/index.pdc
+++ b/content/know/concept/second-quantization/index.pdc
@@ -317,10 +317,10 @@ which create or destroy a particle at a given position $\vec{r}$:
$$\begin{aligned}
\boxed{
- \hat{\psi}^\dagger(\vec{r})
+ \hat{\Psi}^\dagger(\vec{r})
= \sum_{\alpha} \braket{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger
\qquad \quad
- \hat{\psi}(\vec{r})
+ \hat{\Psi}(\vec{r})
= \sum_{\alpha} \braket{\vec{r}}{\alpha} \hat{c}_\alpha
}
\end{aligned}$$
diff --git a/content/know/concept/time-ordered-product/index.pdc b/content/know/concept/time-ordered-product/index.pdc
index 82c9d0f..7af5acc 100644
--- a/content/know/concept/time-ordered-product/index.pdc
+++ b/content/know/concept/time-ordered-product/index.pdc
@@ -110,7 +110,7 @@ The general definition of $\mathcal{T}$ is:
$$\begin{aligned}
\boxed{
\mathcal{T} \big\{ \hat{a}_1 \cdots \hat{a}_N \big\}
- \equiv \sum_{p \in P_N}^{} (-1)^p
+ \equiv \sum_{p \in P_N}^{} (\pm 1)^p
\bigg( \prod_{j = 1}^{N-1} \Theta\big(t_{p_j} \!-\! t_{p_{j+1}}\big) \bigg)
\bigg( \prod_{k = 1}^N \hat{a}_{p_k}(t_{p_k}) \bigg)
}
--
cgit v1.2.3