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diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md new file mode 100644 index 0000000..48f1e76 --- /dev/null +++ b/source/know/concept/greens-functions/index.md @@ -0,0 +1,390 @@ +--- +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 $t - t'$: + +$$\begin{gathered} + G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t - t') + \\ + G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t') + \qquad \quad + G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t') + \\ + G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t') + \qquad \quad + G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t') +\end{gathered}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-time-diff"/> +<label for="proof-time-diff">Proof</label> +<div class="hidden" markdown="1"> +<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. + +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. +</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/). + +$$\begin{aligned} + G_{\nu \nu}^>(t, t') + &= -\frac{i}{\hbar} \Expval{\hat{c}_{\nu} \hat{c}_{\nu}^\dagger} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big) + \\ + &= -\frac{i}{\hbar} (1 - f_\nu) \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big) + \\ + G_{\nu \nu}^<(t, t') + &= \mp \frac{i}{\hbar} \Expval{\hat{c}_{\nu}^\dagger \hat{c}_{\nu}} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big) + \\ + &= \mp \frac{i}{\hbar} f_\nu \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big) +\end{aligned}$$ + + +## 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: + +$$\begin{aligned} + \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})} + = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}) +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-commH0"/> +<label for="proof-commH0">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-commH0">Proof.</label> +In the second quantization, +the Hamiltonian $\hat{H}_0$ is written like so: + +$$\begin{aligned} + \hat{H}_0 + &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \Inprod{\psi_\nu}{\nabla^2 \psi_{\nu'}} + \\ + &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'} + \\ + &= - \frac{\hbar^2}{2 m} + \int \Big( \sum_{\nu} \psi_\nu^*(\vb{r}') \hat{c}_\nu^\dagger \Big) \Big( \nabla^2 \sum_{\nu'} \psi_{\nu'}(\vb{r}') \hat{c}_{\nu'} \Big) \dd{\vb{r}'} + \\ + &= - \frac{\hbar^2}{2 m} + \int \hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})} + &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'} + \int \delta(\vb{r} - \vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'} + \\ + &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r}) + = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}) +\end{aligned}$$ +</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 + +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. |