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diff --git a/content/know/concept/equation-of-motion-theory/index.pdc b/content/know/concept/equation-of-motion-theory/index.pdc new file mode 100644 index 0000000..e7b1120 --- /dev/null +++ b/content/know/concept/equation-of-motion-theory/index.pdc @@ -0,0 +1,203 @@ +--- +title: "Equation-of-motion theory" +firstLetter: "E" +publishDate: 2021-11-08 +categories: +- Physics +- Quantum mechanics + +date: 2021-11-08T18:09:29+01:00 +draft: false +markup: pandoc +--- + +# Equation-of-motion theory + +In many-body quantum theory, **equation-of-motion theory** +is a method to calculate the time evolution of a system's properties +using [Green's functions](/know/concept/greens-functions/). + +Starting from the definition of +the retarded single-particle Green's function $G_{\nu \nu'}^R(t, t')$, +we simply take the $t$-derivative +(we could do the same with the advanced function $G_{\nu \nu'}^A$): + +$$\begin{aligned} + i \hbar \pdv{G^R_{\nu \nu'}(t, t')}{t} + &= \pdv{\Theta(t \!-\! t')}{t} \expval{\comm*{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + + \Theta(t \!-\! t') \pdv{t}\expval{\comm*{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + \\ + &= \delta(t \!-\! t') \expval{\comm*{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + + \Theta(t \!-\! t') \expval{\comm{\dv{\hat{c}_\nu(t)}{t}}{\hat{c}_{\nu'}^\dagger(t)}_{\mp}} +\end{aligned}$$ + +Where we have used that the derivative +of a [Heaviside step function](/know/concept/heaviside-step-function/) $\Theta$ +is a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$. +Also, from the [second quantization](/know/concept/second-quantization/), +$\expval**{\comm*{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$ +for $t = t'$ is zero when $\nu \neq \nu'$. + +Since we are in the [Heisenberg picture](/know/concept/heisenberg-picture/), +we know the equation of motion of $\hat{c}_\nu(t)$: + +$$\begin{aligned} + \dv{\hat{c}_\nu(t)}{t} + = \frac{i}{\hbar} \comm*{\hat{H}_0(t)}{\hat{c}_\nu(t)} + \frac{i}{\hbar} \comm*{\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)} +\end{aligned}$$ + +Where the single-particle part of the Hamiltonian $\hat{H}_0$ +and the interaction part $\hat{H}_\mathrm{int}$ +are assumed to be time-independent in the Schrödinger picture. +We thus get: + +$$\begin{aligned} + i \hbar \pdv{G^R_{\nu \nu'}}{t} + &= \delta_{\nu \nu'} \delta(t \!-\! t')+ \frac{i}{\hbar} \Theta(t \!-\! t') + \expval{\comm{\comm*{\hat{H}_0}{\hat{c}_\nu} + \comm*{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} +\end{aligned}$$ + +The most general form of $\hat{H}_0$, for any basis, +is as follows, where $u_{\nu' \nu''}$ are constants: + +$$\begin{aligned} + \hat{H}_0 + = \sum_{\nu' \nu''} u_{\nu' \nu''} \hat{c}_{\nu'}^\dagger \hat{c}_{\nu''} + \quad \implies \quad + \comm*{\hat{H}_0}{\hat{c}_\nu} + = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''} +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-commH0"/> +<label for="proof-commH0">Proof</label> +<div class="hidden"> +<label for="proof-commH0">Proof.</label> +Using the commutator identity for $\comm*{A B}{C}$, +we decompose it like so: + +$$\begin{aligned} + \comm*{\hat{H}_0}{\hat{c}_\nu} + &= \sum_{\nu' \nu''} u_{\nu \nu''} \comm*{\hat{c}_{\nu'}^\dagger \hat{c}_{\nu''}}{\hat{c}_\nu} + = \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{c}_{\nu'}^\dagger \comm*{\hat{c}_{\nu''}}{\hat{c}_\nu} + + \comm*{\hat{c}_{\nu'}^\dagger}{\hat{c}_\nu} \hat{c}_{\nu''} \Big) +\end{aligned}$$ + +Bosons have well-known commutation relations, +so the result follows directly: + +$$\begin{aligned} + \comm*{\hat{H}_0}{\hat{b}_\nu} + &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{b}_{\nu'}^\dagger \comm*{\hat{b}_{\nu''}}{\hat{b}_\nu} + + \comm*{\hat{b}_{\nu'}^\dagger}{\hat{b}_\nu} \hat{b}_{\nu''} \Big) + = - \sum_{\nu''} u_{\nu \nu''} \hat{b}_{\nu''} +\end{aligned}$$ + +Fermions only have anticommutation relations, +so a bit more work is necessary: + +$$\begin{aligned} + \comm*{\hat{H}_0}{\hat{f}_{\!\nu}} + &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm*{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}} + + \comm*{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) + \\ + &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm*{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}} + - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''} + + \acomm*{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} + - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big) + \\ + &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''} + - 2 \acomm*{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) + = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''} +\end{aligned}$$ +</div> +</div> + +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 $t - t'$: $$\begin{aligned} + G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t') + \qquad \quad + G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t') + \\ G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t') \qquad \quad G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t') \end{aligned}$$ +<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}$$ diff --git a/content/know/concept/kubo-formula/index.pdc b/content/know/concept/kubo-formula/index.pdc index f0208da..f5430da 100644 --- a/content/know/concept/kubo-formula/index.pdc +++ b/content/know/concept/kubo-formula/index.pdc @@ -35,7 +35,7 @@ $$\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}$$ -<div class="accordion"> -<input type="checkbox" id="proof-time-difference"/> -<label for="proof-time-difference">Proof</label> -<div class="hidden"> -<label for="proof-time-difference">Proof.</label> -This is trivial for $\Theta(t\!-\!t')$, -so the challenge is to prove that -$\expval*{\comm*{\hat{A}_I(t)}{B_I(t')}}$ -depends only on the time difference $t - t'$. - -Suppose that the system started in thermodynamic equilibrium -(see [canonical ensemble](/know/concept/canonical-ensemble/)), -so that its (unnormalized) [density operator](/know/concept/density-operator/) -$\hat{\rho}$ was as follows before $\hat{H}_{1,I}$ was applied: - -$$\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}$$ - -As expected, this clearly only depends on the time difference $t - t'$, -because $\hat{H}_{0,S}$ is time-independent by assumption. -</div> -</div> - With this, the Kubo formula can be written as follows, where we have set $t_0 = - \infty$: diff --git a/content/know/concept/lehmann-representation/index.pdc b/content/know/concept/lehmann-representation/index.pdc index f38f803..5808934 100644 --- a/content/know/concept/lehmann-representation/index.pdc +++ b/content/know/concept/lehmann-representation/index.pdc @@ -18,9 +18,8 @@ is an alternative way to write the [Green's functions](/know/concept/greens-func obtained by expanding in the many-particle eigenstates under the assumption of a time-independent Hamiltonian $\hat{H}$. -We start by writing out the -greater Green's function $G_{\nu \nu'}(t, t')$, -and then expanding its thermal expectation value $\expval{}$ +First, we write out the greater Green's function $G_{\nu \nu'}(t, t')$, +and then expand its expected value $\expval{}$ (at thermodynamic equilibrium) into a sum of many-particle basis states $\ket{n}$: $$\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) } |