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+---
+title: "Green's functions"
+date: 2021-11-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+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 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 **time-ordered** or **causal Green's function** $G_{\nu \nu'}$ is as follows,
+where $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/),
+$\nu$ and $\nu'$ are single-particle states,
+and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.:
+
+$$\begin{aligned}
+ \boxed{
+ G_{\nu \nu'}(t, t')
+ \equiv -\frac{i}{\hbar} \Expval{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t') \Big\}}
+ }
+\end{aligned}$$
+
+The expectation value $\Expval{}$ is
+with respect to thermodynamic equilibrium.
+This is sometimes in the [canonical ensemble](/know/concept/canonical-ensemble/)
+(for some two-particle Green's functions, see below),
+but usually 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 $\hat{H}$.
+Explicitly, for a complete set of many-particle states $\Ket{\Psi_n}$, we have:
+
+$$\begin{aligned}
+ G_{\nu \nu'}(t, t')
+ &= -\frac{i}{\hbar Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}} \Big)
+ \\
+ &= -\frac{i}{\hbar Z} \sum_{n}
+ \Matrixel{\Psi_n}{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}}}{\Psi_n}
+\end{aligned}$$
+
+Arguably more prevalent are
+the **retarded Green's function** $G_{\nu \nu'}^R$
+and the **advanced Green's function** $G_{\nu \nu'}^A$
+which are defined like so:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ G_{\nu \nu'}^R(t, t')
+ &\equiv -\frac{i}{\hbar} \Theta(t - t') \Expval{\comm{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+ \\
+ G_{\nu \nu'}^A(t, t')
+ &\equiv \frac{i}{\hbar} \Theta(t' - t) \Expval{\comm{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+ \end{aligned}
+ }
+\end{aligned}$$
+
+Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/),
+and $[,]_{\mp}$ is a commutator for bosons,
+and an anticommutator for fermions.
+Depending on the context,
+we could either be in the [Heisenberg picture](/know/concept/heisenberg-picture/)
+or in the [interaction picture](/know/concept/interaction-picture/),
+hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent.
+
+Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$
+and **lesser Green's function** $G_{\nu \nu'}^<$ are:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ G_{\nu \nu'}^>(t, t')
+ &\equiv -\frac{i}{\hbar} \Expval{\hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')}
+ \\
+ G_{\nu \nu'}^<(t, t')
+ &\equiv \mp \frac{i}{\hbar} \Expval{\hat{c}_{\nu'}^\dagger(t') \: \hat{c}_{\nu}(t)}
+ \end{aligned}
+ }
+\end{aligned}$$
+
+Where $-$ is for bosons, and $+$ for fermions.
+With this, the causal, retarded and advanced Green's functions
+can thus be expressed as follows:
+
+$$\begin{aligned}
+ G_{\nu \nu'}(t, t')
+ &= \Theta(t - t') \: G_{\nu \nu'}^>(t, t') + \Theta(t' - t) \: G_{\nu \nu'}^<(t, t')
+ \\
+ G_{\nu \nu'}^R(t, t')
+ &= \Theta(t - t') \big( G_{\nu \nu'}^>(t, t') - G_{\nu \nu'}^<(t, t') \big)
+ \\
+ G_{\nu \nu'}^A(t, t')
+ &= \Theta(t' - t) \big( G_{\nu \nu'}^<(t, t') - G_{\nu \nu'}^>(t, t') \big)
+\end{aligned}$$
+
+If the Hamiltonian involves interactions,
+it might be more natural to use quantum field operators $\hat{\Psi}(\vb{r}, t)$
+instead of choosing a basis of single-particle states $\psi_\nu$.
+In that case, instead of a label $\nu$,
+we use the spin $s$ and position $\vb{r}$, leading to:
+
+$$\begin{aligned}
+ G_{ss'}(\vb{r}, t; \vb{r}', t')
+ &= -\frac{i}{\hbar} \Theta(t - t') \Expval{\mathcal{T}\Big\{ \hat{\Psi}_{s}(\vb{r}, t) \hat{\Psi}_{s'}^\dagger(\vb{r}', t') \Big\}}
+ \\
+ &= \sum_{\nu \nu'} \psi_\nu(\vb{r}) \: \psi^*_{\nu'}(\vb{r}') \: G_{\nu \nu'}(t, t')
+\end{aligned}$$
+
+And analogously for $G_{ss'}^R$, $G_{ss'}^A$, $G_{ss'}^>$ and $G_{ss'}^<$.
+Note that the time-dependence is given to the old $G_{\nu \nu'}$,
+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 $t - t'$:
+
+$$\begin{gathered}
+ G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t - t')
+ \\
+ G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t')
+ \qquad \quad
+ G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t')
+ \\
+ G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
+ \qquad \quad
+ G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
+\end{gathered}$$
+
+
+
+
+
+
+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.
+
+In (grand) canonical equilibrium, we know that the
+[density operator](/know/concept/density-operator/)
+$\hat{\rho}$ is as follows:
+
+$$\begin{aligned}
+ \hat{\rho} = \frac{1}{Z} \exp(- \beta \hat{H})
+\end{aligned}$$
+
+The expected value of the product
+of the time-independent operators $\hat{A}$ and $\hat{B}$ is then:
+
+$$\begin{aligned}
+ \expval{\hat{A}(t) \hat{B}(t')}
+ &= \frac{1}{Z} \Tr\!\big( \hat{\rho} \hat{A}(t) \hat{B}(t') \big)
+ \\
+ &= \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i t \hat{H} / \hbar} \hat{A} e^{-i t \hat{H} / \hbar}
+ e^{i t' \hat{H} / \hbar} \hat{B} e^{-i t' \hat{H} / \hbar} \Big)
+\end{aligned}$$
+
+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/).
+
+$$\begin{aligned}
+ G_{\nu \nu}^>(t, t')
+ &= -\frac{i}{\hbar} \Expval{\hat{c}_{\nu} \hat{c}_{\nu}^\dagger} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+ \\
+ &= -\frac{i}{\hbar} (1 - f_\nu) \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+ \\
+ G_{\nu \nu}^<(t, t')
+ &= \mp \frac{i}{\hbar} \Expval{\hat{c}_{\nu}^\dagger \hat{c}_{\nu}} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+ \\
+ &= \mp \frac{i}{\hbar} f_\nu \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+\end{aligned}$$
+
+
+## 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:
+
+$$\begin{aligned}
+ \hat{H}_0
+ &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \Inprod{\psi_\nu}{\nabla^2 \psi_{\nu'}}
+ \\
+ &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+ \\
+ &= - \frac{\hbar^2}{2 m}
+ \int \Big( \sum_{\nu} \psi_\nu^*(\vb{r}') \hat{c}_\nu^\dagger \Big) \Big( \nabla^2 \sum_{\nu'} \psi_{\nu'}(\vb{r}') \hat{c}_{\nu'} \Big) \dd{\vb{r}'}
+ \\
+ &= - \frac{\hbar^2}{2 m}
+ \int \hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+ &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
+ \int \delta(\vb{r} - \vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+ \\
+ &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
+ = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
+\end{aligned}$$
+
+
+
+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
+
+We generalize the above to two arbitrary operators $\hat{A}$ and $\hat{B}$,
+giving us the **two-particle Green's functions**,
+or just **correlation functions**.
+The **causal correlation function** $C_{AB}$,
+the **retarded correlation function** $C_{AB}^R$
+and the **advanced correlation function** $C_{AB}^A$ are defined as follows
+(in the Heisenberg picture):
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ C_{AB}(t, t')
+ &\equiv -\frac{i}{\hbar} \Expval{\mathcal{T}\Big\{\hat{A}(t) \hat{B}(t')\Big\}}
+ \\
+ C_{AB}^R(t, t')
+ &\equiv -\frac{i}{\hbar} \Theta(t - t') \Expval{\comm{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
+ \\
+ C_{AB}^A(t, t')
+ &\equiv \frac{i}{\hbar} \Theta(t' - t) \Expval{\comm{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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 the two-particle functions
+only depend on the time-difference $t - t'$:
+
+$$\begin{aligned}
+ G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t \!-\! t')
+ \qquad
+ G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^>(t \!-\! t')
+ \qquad
+ G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^<(t \!-\! t')
+\end{aligned}$$
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
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