diff options
Diffstat (limited to 'content/know')
| -rw-r--r-- | content/know/concept/convolution-theorem/index.pdc | 30 | ||||
| -rw-r--r-- | content/know/concept/coulomb-logarithm/index.pdc | 5 | ||||
| -rw-r--r-- | content/know/concept/dielectric-function/index.pdc | 144 | ||||
| -rw-r--r-- | content/know/concept/electric-field/index.pdc | 19 | ||||
| -rw-r--r-- | content/know/concept/greens-functions/index.pdc | 49 | ||||
| -rw-r--r-- | content/know/concept/laplace-transform/index.pdc | 12 | ||||
| -rw-r--r-- | content/know/concept/lindhard-function/index.pdc | 535 | ||||
| -rw-r--r-- | content/know/concept/maxwell-bloch-equations/index.pdc | 60 |
8 files changed, 490 insertions, 364 deletions
diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc index 2712c21..a14e960 100644 --- a/content/know/concept/convolution-theorem/index.pdc +++ b/content/know/concept/convolution-theorem/index.pdc @@ -26,7 +26,7 @@ and $A$ and $B$ are constants from its definition: $$\begin{aligned} \boxed{ \begin{aligned} - A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\ + A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\ B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\} \end{aligned} } @@ -41,12 +41,12 @@ We expand the right-hand side of the theorem and rearrange the integrals: $$\begin{aligned} - \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} - &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k} + \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} + &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp\!(i s k x') \dd{x'} \Big) \exp\!(-i s k x) \dd{k} \\ - &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'} + &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp\!(- i s k (x - x')) \dd{k} \Big) \dd{x'} \\ - &= A \int_{-\infty}^\infty g(x') f(x - x') \dd{x'} + &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'} = A \cdot (f * g)(x) \end{aligned}$$ @@ -55,11 +55,11 @@ this time starting from a product in the $x$-domain: $$\begin{aligned} \hat{\mathcal{F}}\{f(x) \: g(x)\} - &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x} + &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp\!(- i s x k') \dd{k'} \Big) \exp\!(i s k x) \dd{x} \\ - &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'} + &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp\!(i s x (k - k')) \dd{x} \Big) \dd{k'} \\ - &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'} + &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'} = B \cdot (\tilde{f} * \tilde{g})(k) \end{aligned}$$ </div> @@ -73,10 +73,10 @@ the convolution theorem can also be stated using the [Laplace transform](/know/concept/laplace-transform/): $$\begin{aligned} - \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} + \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} \end{aligned}$$ -Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is +Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is unpleasant, the theorem is often stated using the forward transform instead: @@ -95,20 +95,20 @@ because we set both $f(t)$ and $g(t)$ to zero for $t < 0$: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t} + &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp\!(- s t) \dd{t} \\ - &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp\!(- s t) \dd{t} \Big) g(t') \dd{t'} \end{aligned}$$ Then we define a new integration variable $\tau = t - t'$, yielding: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp\!(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'} \\ - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp\!(- s \tau) \dd{\tau} \Big) g(t') \exp\!(- s t') \dd{t'} \\ - &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'} + &= \int_0^\infty \tilde{f}(s) \: g(t') \exp\!(- s t') \dd{t'} = \tilde{f}(s) \: \tilde{g}(s) \end{aligned}$$ </div> diff --git a/content/know/concept/coulomb-logarithm/index.pdc b/content/know/concept/coulomb-logarithm/index.pdc index 71b13a8..422f73b 100644 --- a/content/know/concept/coulomb-logarithm/index.pdc +++ b/content/know/concept/coulomb-logarithm/index.pdc @@ -84,8 +84,9 @@ and is thus barely affected by this inconsistency. You can easily convince yourself that the average time $\tau$ between "collisions" -is related as follows to the cross-section $\sigma$, -the density $n$, and relative velocity $|\vb{v}|$: +is related like so to the cross-section $\sigma$, +the total density $n$ of charged particles, +and the relative velocity $|\vb{v}|$: $$\begin{aligned} \frac{1}{\tau} diff --git a/content/know/concept/dielectric-function/index.pdc b/content/know/concept/dielectric-function/index.pdc new file mode 100644 index 0000000..cd2a0b7 --- /dev/null +++ b/content/know/concept/dielectric-function/index.pdc @@ -0,0 +1,144 @@ +--- +title: "Dielectric function" +firstLetter: "D" +publishDate: 2022-01-24 +categories: +- Physics +- Electromagnetism +- Quantum mechanics + +date: 2022-01-20T22:04:13+01:00 +draft: false +markup: pandoc +--- + +# Dielectric function + +The **dielectric function** or **relative permittivity** $\varepsilon_r$ +is a measure of how strongly a given medium counteracts +[electric fields](/know/concept/electric-field/) compared to a vacuum. +Let $\vb{D}$ be the applied external field, +and $\vb{E}$ the effective field inside the material: + +$$\begin{aligned} + \boxed{ + \vb{D} = \varepsilon_0 \varepsilon_r \vb{E} + } +\end{aligned}$$ + +If $\varepsilon_r$ is large, then $\vb{D}$ is strongly suppressed, +because the material's electrons and nuclei move to create an opposing field. +In order for $\varepsilon_r$ to be well defined, we only consider linear media, +where the induced polarization $\vb{P}$ is proportional to $\vb{E}$. + +We would like to find an alternative definition of $\varepsilon_r$. +Consider that the usual electric fields $\vb{E}$, $\vb{D}$, and $\vb{P}$ +can each be written as the gradient of an electrostatic potential like so, +where $\Phi_\mathrm{tot}$, $\Phi_\mathrm{ext}$ and $\Phi_\mathrm{ind}$ +are the total, external and induced potentials, respectively: + +$$\begin{aligned} + \vb{E} + = -\nabla \Phi_\mathrm{tot} + \qquad \qquad + \vb{D} + = - \varepsilon_0 \nabla \Phi_\mathrm{ext} + \qquad \qquad + \vb{P} + = \varepsilon_0 \nabla \Phi_\mathrm{ind} +\end{aligned}$$ + +Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$. +Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$ +then suggests defining: + +$$\begin{aligned} + \boxed{ + \varepsilon_r + \equiv \frac{\Phi_\mathrm{ext}}{\Phi_\mathrm{tot}} + } +\end{aligned}$$ + + +## From induced charge density + +A common way to calculate $\varepsilon_r$ is from +the induced charge density $\rho_\mathrm{ind}$, +i.e. the offset caused by the material's particles responding to the field. +We start from [Gauss' law](/know/concept/maxwells-equations/) for $\vb{P}$: + +$$\begin{aligned} + \nabla \cdot \vb{P} + = \varepsilon_0 \nabla^2 \Phi_\mathrm{ind}(\vb{r}) + = - \rho_\mathrm{ind}(\vb{r}) +\end{aligned}$$ + +This is Poisson's equation, which has the following well-known +[Fourier transform](/know/concept/fourier-transform/): + +$$\begin{aligned} + \Phi_\mathrm{ind}(\vb{q}) + = \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2} + = V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q}) +\end{aligned}$$ + +Where $V(\vb{q})$ represents Coulomb interactions, +and $V(0) = 0$ to ensure overall neutrality: + +$$\begin{aligned} + V(\vb{q}) + = \frac{1}{\varepsilon_0 |\vb{q}|^2} + \qquad \implies \qquad + V(\vb{r} - \vb{r}') + = \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|} +\end{aligned}$$ + +The [convolution theorem](/know/concept/convolution-theorem/) +then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain: + +$$\begin{aligned} + \Phi_\mathrm{ind}(\vb{r}) + = (V * \rho_\mathrm{ind})(\vb{r}) + = \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'} +\end{aligned}$$ + +To proceed, we need to find an expression for $\rho_\mathrm{ind}$ +that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$, +or some linear combination thereof. +Such an expression must exist for a linear material. + +Suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$, +for some $C_\mathrm{ext}$, which may depend on $\vb{q}$. Then: + +$$\begin{aligned} + \Phi_\mathrm{tot} + = (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext} + \quad \implies \quad + \boxed{ + \varepsilon_r(\vb{q}) + = \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})} + } +\end{aligned}$$ + +Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$, +for some quantity $C_\mathrm{tot}$, then: + +$$\begin{aligned} + \Phi_\mathrm{ext} + = (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot} + \quad \implies \quad + \boxed{ + \varepsilon_r(\vb{q}) + = 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q}) + } +\end{aligned}$$ + + + +## References +1. H. Bruus, K. Flensberg, + *Many-body quantum theory in condensed matter physics*, + 2016, Oxford. +2. M. Fox, + *Optical properties of solids*, 2nd edition, + Oxford. diff --git a/content/know/concept/electric-field/index.pdc b/content/know/concept/electric-field/index.pdc index 6162e0b..62ce1f5 100644 --- a/content/know/concept/electric-field/index.pdc +++ b/content/know/concept/electric-field/index.pdc @@ -31,7 +31,7 @@ since opposite charges attracts and like charges repel. If two opposite point charges with magnitude $q$ are observed from far away, they can be treated as a single object called a **dipole**, -which has an **electric dipole moment** $\vb{p}$ defined as follows, +which has an **electric dipole moment** $\vb{p}$ defined like so, where $\vb{d}$ is the vector going from the negative to the positive charge (opposite direction of $\vb{E}$): @@ -88,7 +88,7 @@ and that $\vb{M}$ has the opposite sign of $\vb{P}$. The polarization $\vb{P}$ is a function of $\vb{E}$. In addition to the inherent polarity of the material $\vb{P}_0$ (zero in most cases), -there is a possibly nonlinear response +there is a (possibly nonlinear) response to the applied $\vb{E}$-field: $$\begin{aligned} @@ -101,10 +101,7 @@ $$\begin{aligned} Where the $\chi_e^{(n)}$ are the **electric susceptibilities** of the medium. For simplicity, we often assume that only the $n\!=\!1$ term is nonzero, which is the linear response to $\vb{E}$. -In that case, we define -the **relative permittivity** $\varepsilon_r \equiv 1 + \chi_e^{(1)}$ -and the **absolute permittivity** $\varepsilon \equiv \varepsilon_r \varepsilon_0$, -so that: +In that case, we define the **absolute permittivity** $\varepsilon$ so that: $$\begin{aligned} \vb{D} @@ -114,14 +111,20 @@ $$\begin{aligned} = \varepsilon \vb{E} \end{aligned}$$ +I.e. $\varepsilon \equiv \varepsilon_r \varepsilon_0$, +where $\varepsilon_r \equiv 1 + \chi_e^{(1)}$ is +the [**dielectric function**](/know/concept/dielectric-function/) +or **relative permittivity**, +whose calculation is of great interest in physics. + In reality, a material cannot respond instantly to $\vb{E}$, meaning that $\chi_e^{(1)}$ is a function of time, and that $\vb{P}$ is the convolution of $\chi_e^{(1)}(t)$ and $\vb{E}(t)$: $$\begin{aligned} \vb{P}(t) - = (\chi_e^{(1)} * \vb{E})(t) - = \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau + = \varepsilon_0 \big(\chi_e^{(1)} * \vb{E}\big)(t) + = \varepsilon_0 \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau \end{aligned}$$ Note that this definition requires $\chi_e^{(1)}(t) = 0$ for $t < 0$ diff --git a/content/know/concept/greens-functions/index.pdc b/content/know/concept/greens-functions/index.pdc index b3c9ede..92f0fcf 100644 --- a/content/know/concept/greens-functions/index.pdc +++ b/content/know/concept/greens-functions/index.pdc @@ -32,12 +32,9 @@ If the two operators are single-particle creation/annihilation operators, then we get the **single-particle Green's functions**, for which the symbol $G$ is used. -The **time-ordered** or **causal Green's function** $G_{\nu \nu'}$ -is defined as follows, +The **time-ordered** or **causal Green's function** $G_{\nu \nu'}$ is as follows, where $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/), -the expectation value $\expval{}$ is -with respect to thermodynamic equilibrium, -$\nu$ and $\nu'$ are labels of single-particle states, +$\nu$ and $\nu'$ are single-particle states, and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.: $$\begin{aligned} @@ -47,6 +44,24 @@ $$\begin{aligned} } \end{aligned}$$ +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** $G_{\nu \nu'}^A$ @@ -67,10 +82,10 @@ $$\begin{aligned} Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/), and $[,]_{\mp}$ is a commutator for bosons, and an anticommutator for fermions. -We are in the [Heisenberg picture](/know/concept/heisenberg-picture/), -hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent, -but keep in mind that time-dependent Hamiltonians are allowed, -so it might not be trivial. +Depending on the context, +we could either be in the [Heisenberg picture](/know/concept/heisenberg-picture/) +or in the [interaction picture](/know/concept/interaction-picture/), +hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent. Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$ and **lesser Green's function** $G_{\nu \nu'}^<$ are: @@ -146,16 +161,7 @@ $\expval*{\hat{A}(t) \hat{B}(t')}$ only depends on $t - t'$ for arbitrary $\hat{A}$ and $\hat{B}$, and it trivially follows that the Green's functions do too. -Suppose that the system started in thermodynamic equilibrium. -This could sometimes be in the [canonical ensemble](/know/concept/canonical-ensemble/) -(for two-particle Green's functions, see below), -but usually it will be in the -[grand canonical ensemble](/know/concept/grand-canonical-ensemble/), -since we are adding/removing particles. -In the latter case, we assume that the chemical potential $\mu$ -is already included in the Hamiltonian. - -In any case, at equilibrium, we know that the +In (grand) canonical equilibrium, we know that the [density operator](/know/concept/density-operator/) $\hat{\rho}$ is as follows: @@ -163,9 +169,8 @@ $$\begin{aligned} \hat{\rho} = \frac{1}{Z} \exp\!(- \beta \hat{H}) \end{aligned}$$ -Where $Z \equiv \Tr\!(\exp\!(- \beta \hat{H}))$ is the partition function. -In that case, the expected value of the product -of the time-independent operators $\hat{A}$ and $\hat{B}$ is calculated like so: +The expected value of the product +of the time-independent operators $\hat{A}$ and $\hat{B}$ is then: $$\begin{aligned} \expval*{\hat{A}(t) \hat{B}(t')} diff --git a/content/know/concept/laplace-transform/index.pdc b/content/know/concept/laplace-transform/index.pdc index bd7673b..5e91a04 100644 --- a/content/know/concept/laplace-transform/index.pdc +++ b/content/know/concept/laplace-transform/index.pdc @@ -33,6 +33,12 @@ This is solved by restricting the domain of $\tilde{f}(s)$ to $s$ where $\mathrm{Re}\{s\} > s_0$, for an $s_0$ large enough to compensate for the growth of $f(t)$. +The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration, +and is therefore a lot more difficult to calculate. +Fortunately, it is usually avoidable by rewriting a given $s$-space expression +using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/), +and then looking up the individual terms. + ## Derivatives @@ -42,7 +48,7 @@ This is especially useful for transforming ODEs with variable coefficients: $$\begin{aligned} \boxed{ - \tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\} + \tilde{f}{}'(s) = - \hat{\mathcal{L}}\{t f(t)\} } \end{aligned}$$ @@ -107,9 +113,9 @@ $$\begin{aligned} \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\} &= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t} \\ - &= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t} + &= \Big[ f^{(n - 1)}(t) \exp\!(- s t) \Big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t} \\ - &= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t} + &= - f^{(n - 1)}(0) + s \Big[ f^{(n - 2)}(t) \exp\!(- s t) \Big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t} \end{aligned}$$ And so on. diff --git a/content/know/concept/lindhard-function/index.pdc b/content/know/concept/lindhard-function/index.pdc index 96244c9..aedf0f5 100644 --- a/content/know/concept/lindhard-function/index.pdc +++ b/content/know/concept/lindhard-function/index.pdc @@ -1,7 +1,7 @@ --- title: "Lindhard function" firstLetter: "L" -publishDate: 2021-10-12 +publishDate: 2022-01-24 # Originally 2021-10-12, major rewrite categories: - Physics - Quantum mechanics @@ -13,9 +13,9 @@ markup: pandoc # Lindhard function -The **Lindhard function** describes the response of an electron gas -to an external perturbation, -and can be regarded as a quantum-mechanical +The **Lindhard function** describes the response of +[jellium](/know/concept/jellium) (i.e. a free electron gas) +to an external perturbation, and is a quantum-mechanical alternative to the [Drude model](/know/concept/drude-model/). We start from the [Kubo formula](/know/concept/kubo-formula/) @@ -28,389 +28,370 @@ $$\begin{aligned} = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \expval{\comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'} \end{aligned}$$ -Where $\Theta$ is the [Heaviside step function](/know/concept/heaviside-step-function/), -and the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/). -Notice from the limits that the perturbation is switched on at $t = -\infty$. -Now, let us consider the following harmonic $\hat{H}_1$ in the Schrödinger picture: +Where the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/), +and the expectation $\expval{}_0$ is for +a thermal equilibrium before the perturbation was applied. +Now consider a harmonic $\hat{H}_1$: $$\begin{aligned} \hat{H}_{1,S}(t) - = g(t) \: \hat{V}_S - \qquad - g(t) - \equiv \exp\!(- i \omega t) \exp\!(\eta t) - \qquad - \hat{V}_S - \equiv \int_{-\infty}^\infty V(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}} + = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}} \end{aligned}$$ -Where $\eta$ is a tiny positive number, -which represents a gradual switching-on of $\hat{H}_1$, -eliminating transient effects -and helping the convergence of an integral later. - -We assume that $V(\vb{r})$ varies slowly compared to the electrons' wavefunctions, -so we argue that $\hat{V}_S$ is practically time-independent, -because the total number of electrons is conserved, -and $\hat{n}$ is only weakly perturbed by $\hat{H}_1$. - -Because $\hat{H}_1$ starts at $t = -\infty$, -we can always shift the time axis such that the point of interest is at $t = 0$. -We thus have, without loss of generality: +Where $S$ is the Schrödinger picture, +$\eta$ is a positive infinitesimal to ensure convergence later, +and $U(\vb{r})$ is an arbitrary potential function. +The Kubo formula becomes: $$\begin{aligned} - \delta\expval{{\hat{n}}}(\vb{r}) - = \delta\expval{{\hat{n}}}(\vb{r}, 0) - &= -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(- t') - \Big( \expval{\hat{n}_I \hat{V}_I}_0 - \expval{\hat{V}_I \hat{n}_I}_0 \Big) g(t') \dd{t'} + \delta\expval{{\hat{n}}}(\vb{r}, t) + = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'} \end{aligned}$$ -The expectation values $\expval{}_0$ are calculated for $\ket{0}$, -which was the state at $t = -\infty$. -Note that if $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$, -there is no difference which picture (Schrödinger or interaction) $\ket{0}$ is in, -because any operator $\hat{A}$ then satisfies: +Here, $\chi$ is the density-density correlation function, +i.e. a two-particle [Green's function](/know/concept/greens-functions/): $$\begin{aligned} - \matrixel{0_I}{\hat{A}_I}{0_I} - &= \matrixel**{0_S}{\exp\!(-i \hat{H}_{0,S} t / \hbar) \:\: \hat{A}_I \: \exp\!(i \hat{H}_{0,S} t / \hbar)}{0_S} - \\ - &= \matrixel{0_S}{\hat{A}_I}{0_S} \:\exp\!\big(i (E_0\!-\!E_0) t / \hbar\big) - = \matrixel{0_S}{\hat{A}_I}{0_S} + \chi(\vb{r}, \vb{r}'; t, t') + \equiv - \frac{i}{\hbar} \Theta(t - t') \expval{\comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0 \end{aligned}$$ -Therefore, we will assume that $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$. -Next, we insert the identity operator $\hat{I} = \sum_{j} \ket{j} \bra{j}$, -where $\ket{j}$ are all the eigenstates of $\hat{H}_{0,S}$: +Let us assume that the unperturbed system (i.e. without $U$) is spatially uniform, +so that $\chi$ only depends on the difference $\vb{r} - \vb{r}'$. +We then take its [Fourier transform](/know/concept/fourier-transform/) +$\vb{r}\!-\!\vb{r}' \to \vb{q}$: $$\begin{aligned} - \delta\expval{\hat{n}}(\vb{r}) - &= -\frac{i}{\hbar} \sum_{j} \int \Theta(-t') - \Big( \matrixel{0}{\hat{n}_I}{j} \matrixel{j}{\hat{V}_I}{0} - \matrixel{0}{\hat{V}_I}{j} \matrixel{j}{\hat{n}_I}{0} \Big) g(t') \dd{t'} + \chi(\vb{q}; t, t') + &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint + \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} \end{aligned}$$ -Using the fact that $\ket{0}$ and $\ket{j}$ -are eigenstates of $\hat{H}_{0,S}$, -and that we chose $t = 0$, we find: +Where both $\hat{n}_I$ have been written as inverse Fourier transforms, +giving a factor $(2 \pi)^{-2 D}$, with $D$ being the number of spatial dimensions. +We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$: $$\begin{aligned} - \matrixel{j}{\hat{V}_I(t')}{0} - &= \matrixel{j}{\hat{V}_S}{0} \:\exp\!\big(i (E_j \!-\! E_0) t' / \hbar\big) + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint + \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint + \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \\ - \matrixel{j}{\hat{n}_I(0)}{0} - &= \matrixel{j}{\hat{n}_S(0)}{0} \:\exp\!(0 - 0) - = \matrixel{j}{\hat{n}_S(0)}{0} + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int + \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \end{aligned}$$ -We define $\omega_{j0} \equiv (E_j \!-\! E_0)/\hbar$ -and insert the above expressions into $\delta\expval{\hat{n}}$, -yielding: +On the left, $\vb{r}'$ does not appear, so it must also disappear on the right. +If we choose an arbitrary (hyper)cube of volume $V$ in real space, +then clearly $\int_V \dd{\vb{r}'} = V$. Therefore: $$\begin{aligned} - \delta\expval{\hat{n}}(\vb{r}) - &= -\frac{i}{\hbar} \sum_{j} \bigg( - \matrixel{0}{\hat{n}_S}{j} \matrixel{j}{\hat{V}_S}{0} \int \Theta(-t') \exp\!(i \omega_{j0} t') \: g(t') \dd{t'} - \\ - &\qquad\qquad\:\, - - \matrixel{0}{\hat{V}_S}{j} \matrixel{j}{\hat{n}_S}{0} \int \Theta(-t') \exp\!(-i \omega_{j0} t') \: g(t') \dd{t'} \bigg) + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty + \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'} \end{aligned}$$ -These integrals are [Fourier transforms](/know/concept/fourier-transform/), -and are straightforward to evaluate. The first is: +For $V \to \infty$ we get a Dirac delta function, +but in fact the conclusion holds for finite $V$ too: $$\begin{aligned} - \int_{-\infty}^\infty \Theta(-t') \exp\!(i \omega_{j0} t') \: g(t') \dd{t'} - &= \int_{-\infty}^0 \exp\!\big(\!-\! i (\omega - \omega_{j0}) t' + \eta t' \big) \dd{t'} - \\ - &= \bigg[ \frac{\exp\!\big(\!-\! i (\omega - \omega_{j0}) t' + \eta t' \big)}{- i (\omega - \omega_{j0}) + \eta} \bigg]_{-\infty}^0 + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty + \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2} \\ - &= \frac{1}{- i (\omega - \omega_{j0}) + \eta} - = \frac{i}{\omega - \omega_{j0} + i \eta} + &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 \end{aligned}$$ -The other integral simply has the opposite sign in front of $\omega_{j0}$. -We thus arrive at: +Similarly, if the unperturbed Hamiltonian $\hat{H}_0$ is time-independent, +$\chi$ only depends on the time difference $t - t'$. +Note that $\delta{\expval{\hat{n}}}$ already has the form of a Fourier transform, +which gives us an opportunity to rewrite $\chi$ +in the [Lehmann representation](/know/concept/lehmann-representation/): $$\begin{aligned} - \delta\expval{\hat{n}}(\vb{r}, \omega) - &= \frac{1}{\hbar} \sum_{j} \bigg( - \frac{\matrixel{0}{\hat{n}_S}{j} \matrixel{j}{\hat{V}_S}{0}}{\omega - \omega_{j0} + i \eta} - - \frac{\matrixel{0}{\hat{V}_S}{j} \matrixel{j}{\hat{n}_S}{0}}{\omega + \omega_{j0} + i \eta} \bigg) + \chi(\vb{q}, \omega) + = \frac{1}{Z V} \sum_{\nu \nu'} + \frac{\matrixel{\nu}{\hat{n}_S(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}_S(-\vb{q})}{\nu}}{\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big) \end{aligned}$$ -Inserting the definition $\hat{V}_S = \int V(\vb{r}') \:\hat{n}(\vb{r}') \dd{\vb{r}'}$ -leads us to the following formula for $\delta\expval{\hat{n}}$, -which has the typical form of a linear response, -with response function $\chi$: +Where $\ket{\nu}$ and $\ket{\nu'}$ are many-electron eigenstates of $\hat{H}_0$, +and $Z$ is the [grand partition function](/know/concept/grand-canonical-ensemble/). +According to the [convolution theorem](/know/concept/convolution-theorem/) +$\delta{\expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$. +In anticipation, we swap $\nu$ and $\nu''$ in the second term, +so the general response function is written as: $$\begin{aligned} - \boxed{ - \begin{gathered} - \delta{n}(\vb{r}, \omega) - = \int_{-\infty}^\infty \chi(\vb{r}, \vb{r'}, \omega) \: V(\vb{r}') \dd{\vb{r}'} - \qquad\quad \mathrm{where} - \\ - \chi(\vb{r}, \vb{r'}, \omega) - = \sum_{j} \bigg( \frac{\matrixel{0}{\hat{n}_S(\vb{r})}{j} \matrixel{j}{\hat{n}_S(\vb{r}')}{0}}{(\omega + i \eta) \hbar - E_j + E_0} - - \frac{\matrixel{0}{\hat{n}_S(\vb{r}')}{j} \matrixel{j}{\hat{n}_S(\vb{r})}{0}}{(\omega + i \eta) \hbar + E_j - E_0} \bigg) - \end{gathered} - } + \chi(\vb{q}, \omega) + = \frac{1}{Z V} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}} + {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}} + {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -By definition, $\ket{j}$ are eigenstates -of the many-electron Hamiltonian $\hat{H}_{0,S}$, -which is only solvable if we crudely neglect -any and all electron-electron interactions. -Therefore, to continue, we neglect those interactions. -According to tradition, we then rename $\chi$ to $\chi_0$. +All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$. -The well-known ground state of a non-interacting electron gas -is the Fermi sea $\ket{\mathrm{FS}}$, given below, -together with $\hat{n}_S$ in the language of the -[second quantization](/know/concept/second-quantization/): +To proceed, we need to rewrite $\hat{n}(\vb{q})$ somehow. +If we neglect electron-electron interactions, +the single-particle states are simply plane waves, in which case: $$\begin{aligned} - \ket{\mathrm{FS}} - = \prod_\alpha \hat{c}_\alpha^\dagger \ket{0} - \qquad \quad - \hat{n}_S(\vb{r}) - = \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r}) - = \sum_{\alpha \beta} \psi_\alpha^*(\vb{r}) \psi_\beta(\vb{r})\: \hat{c}_\alpha^\dagger \hat{c}_\beta + \hat{n}(\vb{q}) + = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}} + \qquad \qquad + \hat{n}(-\vb{q}) + = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ -For now, we ignore thermal excitations, -i.e. we set the temperature $T = 0$. -In $\chi_0 = \chi$, we thus insert the above $\hat{n}_S$, -and replace $\ket{0}$ with $\ket{\mathrm{FS}}$, yielding: +<div class="accordion"> +<input type="checkbox" id="proof-density"/> +<label for="proof-density">Proof</label> +<div class="hidden"> +<label for="proof-density">Proof.</label> +Starting from the general definition of $\hat{n}$, +we write out the field operators $\hat{\Psi}(\vb{r})$, +and insert the known non-interacting single-electron orbitals +$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$: $$\begin{aligned} - \matrixel{0}{\hat{n}_S}{j} - \quad\longrightarrow\quad \matrixel**{\mathrm{FS}}{\hat{\Psi}{}^\dagger \hat{\Psi}}{j} - = \sum_{\alpha \beta} \psi_\alpha^* \psi_\beta \matrixel**{\mathrm{FS}}{\hat{c}_\alpha^\dagger \hat{c}_\beta}{j} + \hat{n}(\vb{r}) + \equiv \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r}) + = \sum_{\vb{k} \vb{k}'} \psi_{\vb{k}}^*(\vb{r}) \: \psi_{\vb{k}'}(\vb{r})\: \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} + = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \end{aligned}$$ -This inner product is only nonzero if -$\ket{j} = \hat{c}_a \hat{c}_b^\dagger \ket{\mathrm{FS}}$ -with $a = \alpha$ and $b = \beta$, -or in other words, -only if $\ket{j}$ is a single-electron excitation of $\ket{\mathrm{FS}}$. -Furthermore, in $\ket{\mathrm{FS}}$, -$\alpha$ must be filled, and $\beta$ must be empty. -Let $f_\alpha \in \{0,1\}$ be the occupation number of orbital $\alpha$, then: +Taking the Fourier transfom yields a Dirac delta function $\delta$: $$\begin{aligned} - \matrixel{0}{\hat{n}_S}{j} - \longrightarrow \sum_{\alpha \beta} \psi_\alpha^* \psi_\beta \: f_\alpha (1 - f_\beta) \: \delta_{a \alpha} \delta_{b \beta} + \hat{n}(\vb{q}) + = \frac{1}{V} \int_{-\infty}^\infty + \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}} + = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \end{aligned}$$ -In $\chi_0$, the sum over $j$ becomes a sum over $a$ and $b$ -(this implicitly eliminates all $\ket{j}$ that are not single-electron excitations), -and $E_j\!-\!E_0$ becomes the cost of the excitation $\epsilon_b \!-\! \epsilon_a$, -where $\epsilon_a$ is the energy of orbital $a$. -Therefore, we find: +If we impose periodic boundary conditions +on our $D$-dimensional hypercube of volume $V$, +then $\vb{k}$ becomes discrete, +with per-value spacing $2 \pi / V^{1/D}$ along each axis. + +Consequently, each orbital $\psi_\vb{k}$ uniquely occupies +a volume $(2 \pi)^D / V$ in $\vb{k}$-space, so we make the approximation +$\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}$. +This becomes exact for $V \to \infty$, +in which case $\vb{k}$ also becomes continuous again, +which is what we want for jellium. + +We apply this standard trick from condensed matter physics to $\hat{n}$, +and $V$ cancels out: $$\begin{aligned} - \chi_0 - &= \sum_{a b} \bigg( \sum_{\alpha \beta \kappa \mu} f_\alpha (1 \!-\! f_\beta) f_\kappa (1 \!-\! f_\mu) - \frac{\psi_\alpha^*(\vb{r}) \psi_\beta(\vb{r}) \psi_\kappa(\vb{r}') \psi_\mu^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \delta_{a \alpha} \delta_{b \beta} \delta_{a \kappa} \delta_{b \mu} - \\ - &\qquad\:\:\: - \sum_{\alpha \beta \kappa \mu} f_\alpha (1 \!-\! f_\beta) f_\kappa (1 \!-\! f_\mu) - \frac{\psi_\alpha^*(\vb{r'}) \psi_\beta(\vb{r'}) \psi_\kappa(\vb{r}) \psi_\mu^*(\vb{r})}{(\omega + i \eta) \hbar + \epsilon_b - \epsilon_a} - \delta_{a \alpha} \delta_{b \beta} \delta_{a \kappa} \delta_{b \mu} \bigg) - \\ - &= \sum_{a b} \bigg( f_a^2 (1 \!-\! f_b)^2 - \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \\ - &\qquad\:\: - f_a^2 (1 \!-\! f_b)^2 - \frac{\psi_a^*(\vb{r'}) \psi_b(\vb{r'}) \psi_a(\vb{r}) \psi_b^*(\vb{r})}{(\omega + i \eta) \hbar + \epsilon_b - \epsilon_a} - \bigg) + \hat{n}(\vb{q}) + &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty + \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'} + = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \end{aligned}$$ -Because $f_a, f_b \in \{0, 1\}$ when $T = 0$, we know that $f_a^2 (1 \!-\! f_b)^2 = f_a (1 \!-\! f_b)$. -We then swap the indices $a$ and $b$ in the second term, leading to: +For negated arguments, we simply define $\vb{k}' \equiv \vb{k} - \vb{q}$ +to show that $\hat{n}(-\vb{q}) = \hat{n}{}^\dagger(\vb{q})$, +which can also be understood as a consequence of $\hat{n}(\vb{r})$ being real: $$\begin{aligned} - \chi_0 - &= \sum_{a b} \Big( f_a (1 \!-\! f_b) - f_b (1 \!-\! f_a) \Big) - \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \\ - &= \sum_{a b} \Big( f_a - f_b \Big) - \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} + \hat{n}(-\vb{q}) + = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}} + = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'} + = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ -To proceed, we make the radical assumption that $\vb{H}_{0,S}$ -has continuous translational symmetry, -or in other words, that the external potential is uniform in space. -Clearly, this is not realistic, -so our conclusions will be more qualitative than quantitative. +The summation variable $\vb{k}$ has an associated spin $\sigma$, +and $\hat{n}$ does not carry any spin. +</div> +</div> -In that case, the wavefunction of a non-interacting particle is simply a plane wave, -so we insert $\psi_a(\vb{r}) = \exp\!(i \vb{k}_a \cdot \vb{r})$ -and $\psi_b(\vb{r}) = \exp\!(i \vb{k}_b \cdot \vb{r})$, yielding: +When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$. +We insert $\hat{n}$, suppressing spin: $$\begin{aligned} \chi_0 - &= \sum_{a b} \Big( f_a - f_b \Big) - \frac{\exp\!\big( \!-\! i \vb{k}_a \cdot \vb{r} + i \vb{k}_b \cdot \vb{r} + i \vb{k}_a \cdot \vb{r}' - i \vb{k}_b \cdot \vb{r}' \big)} - {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \\ - &= \sum_{a b} \Big( f_a - f_b \Big) - \frac{\exp\!\big( \!-\! i (\vb{k}_a \!-\! \vb{k}_b) \cdot (\vb{r} \!-\! \vb{r}')\big)} - {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} + &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}} + {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -Here, we see that $\chi_0$ only depends on the differences -$\vb{r}\!-\!\vb{r}'$ and $\vb{k}_a\!-\!\vb{k}_b$. -Therefore, we define $\vb{q}' \equiv \vb{k}_b\!-\!\vb{k}_a$ -and rename $\vb{k}_a \to \vb{k}$. -We thus have: - -$$\begin{aligned} - \chi_0(\vb{r}\!-\!\vb{r}') - &= \sum_{\vb{k} \vb{q}'} \Big( f_k - f_{k+q'} \Big) - \frac{\exp\!\big( i \vb{q'} \cdot (\vb{r} \!-\! \vb{r}')\big)} - {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} -\end{aligned}$$ +Here, $\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}$ +is only nonzero if $\ket{\nu'}$ is contructed from $\ket{\nu}$ +by moving an electron from $\vb{k}$ to $\vb{k} \!+\! \vb{q}$, +and analogously for the other inner products. +As a result, $\vb{k} = \vb{k}'$ (and $\sigma = \sigma'$). -The summation goes over all $\vb{k}$ and $\vb{q}'$ -where $\vb{k}$ is inside the Fermi sphere, and $\vb{k}\!+\!\vb{q}'$ is outside. -Let $k_F$ be the Fermi radius, -then we convert this sum into an integral, -which means introducing a factor of $1/(2 \pi)^{3}$ -as usual in solid state physics: +For the same reason, the energy difference $E_\nu \!-\! E_{\nu'}$ +can simply be replaced by the cost of the single-particle excitation +$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$, +where $\xi_{\vb{k}}$ is the energy of a $\vb{k}$-orbital. +Therefore: $$\begin{aligned} - \chi_0(\vb{r}) - &= \sum_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)} - {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} - \\ - &= \frac{1}{(2 \pi)^3} \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)} - {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -Fourier transforming the position $\vb{r}$ into the wavevector $\vb{q}$, -we recognize an integral that can be evaluated -to a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(\vb{q})$: +Notice that we have eliminated all dependence on $\ket{\nu'}$, +so we remove it by $\sum_{\nu} \ket{\nu} \bra{\nu} = 1$: $$\begin{aligned} - \chi_0(\vb{q}) - &= \frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \exp\!\big( i (\vb{q}' \!-\! \vb{q}) \cdot \vb{r} \big)}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} - \dd{\vb{k}} \dd{\vb{q}'} \dd{\vb{r}} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \\ - &= \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \:\delta(\vb{q}'\!-\!\vb{q})}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'} + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} + \frac{\matrixel{\nu}{\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} + {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -This delta functions eliminates the integral over $\vb{q}'$, -giving the following linear response $\chi_0$ -of a non-interacting electron gas in a uniform potential: +Where we recognized the commutator, +and eliminated $E_\nu$ using $\hat{H}_0 \ket{n} = E_\nu \ket{\nu}$. +The resulting expression has the form of a matrix trace $\Tr$ +and a thermal expectation $\expval{}_0$: $$\begin{aligned} - \boxed{ - \chi_0(\vb{q}, \omega) - = \int_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}| > k_F}} - \frac{f_k - f_{k+q}}{(\omega + i \eta) \hbar - \epsilon_{k+q} + \epsilon_k} \dd{\vb{k}} - } + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} + {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + = \frac{1}{V} \sum_{\vb{k}} + \frac{\expval*{\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -The resulting electron density change $\delta{\expval{\hat{n}}}$ is as follows, -where we use the [convolution theorem](/know/concept/convolution-theorem/) -to convert the convolution in $\vb{r}$-space into a product in $\vb{q}$-space: - -$$\begin{gathered} - \delta\expval{\hat{n}}(\vb{r}, \omega) - = \int_{-\infty}^\infty \chi_0(\vb{r}\!-\!\vb{r}', \omega) \: V(\vb{r}') \dd{r'} - \\ - \implies \qquad - \boxed{ - \delta\expval{\hat{n}}(\vb{q}, \omega) - = \chi_0(\vb{q}, \omega) \: V(\vb{q}) - } -\end{gathered}$$ - -So far, we have neglected electron-electron interactions, -but now we approximately correct this. -We split the effective potential $\vb{V}_\mathrm{eff}$ felt by the electrons -into the external potential $V_\mathrm{ext}$ -and the internal interactions $V_\mathrm{int}$, -such that: +This commutator can be evaluated, +and in this particular case it turns out to be: $$\begin{aligned} - V_\mathrm{eff}(\vb{r}) - = V_\mathrm{ext} + V_\mathrm{int} + \comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} + = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \end{aligned}$$ -We approximate $V_\mathrm{int}$ as follows, -where $V_{ee}$ represents electron-electron interactions: - +<div class="accordion"> +<input type="checkbox" id="proof-commutator"/> +<label for="proof-commutator">Proof</label> +<div class="hidden"> +<label for="proof-commutator">Proof.</label> +In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have: $$\begin{aligned} - V_\mathrm{int}(\vb{r}) - \approx \int_{-\infty}^\infty V_{ee}(\vb{r} \!-\! \vb{r}') \: \delta{n}(\vb{r}') \dd{\vb{r}'} - \qquad\quad - V_{ee}(\vb{r} \!-\! \vb{r}') - = \frac{e^2}{4 \pi \varepsilon_0} \frac{1}{|\vb{r} - \vb{r}'|} + \comm*{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} + &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n + \\ + &= \hat{c}_m^\dagger \big( \acomm*{\hat{c}_n}{\hat{c}_o^\dagger} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p + - \hat{c}_o^\dagger \big( \acomm*{\hat{c}_p}{\hat{c}_m^\dagger} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n \end{aligned}$$ -Consequently, $V_\mathrm{int}$ satisfies Poisson's equation, -which has a well-known Fourier transform: +Using the standard fermion anticommutation relations, this becomes: $$\begin{aligned} - \nabla^2 V_\mathrm{int}(\vb{r}) - = - \frac{\delta{n}(\vb{r})}{\varepsilon_0} - \quad \implies \quad - V_\mathrm{int}(\vb{q}) - = \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \delta{n}(\vb{q}) + \comm*{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} + &= \hat{c}_m^\dagger \big( \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p + - \hat{c}_o^\dagger \big( \delta_{pm} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n + \\ + &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_m^\dagger \hat{c}_o^\dagger \hat{c}_n \hat{c}_p + - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} + \hat{c}_o^\dagger \hat{c}_m^\dagger \hat{c}_p \hat{c}_n + \\ + &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} \end{aligned}$$ -Meanwhile, from all of the above calculations, -we can write $\delta{n}$ as follows, -where $\chi$ and $\chi_0$ are the -(unknown) interacting and (known) non-interacting response functions: +In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$, +so the Kronecker deltas are unnecessary. +</div> +</div> + +We substitute this result into $\chi_0$, +and reintroduce the spin index $\sigma$ associated with $\vb{k}$: $$\begin{aligned} - \delta{n}(\vb{q}) - = \chi V_\mathrm{ext} - \approx \chi_0 V_\mathrm{eff} + \chi_0(\vb{q}, \omega) + = \frac{1}{V} \sum_{\sigma \vb{k}} + \frac{\expval*{\hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k}} - \hat{c}_{\sigma,\vb{k}+\vb{q}}^\dagger \hat{c}_{\sigma,\vb{k}+\vb{q}}}_0} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -Keep in mind that we are treating interactions as a perturbation to $V_\mathrm{ext}$, -therefore $V_\mathrm{ext} \approx V_\mathrm{eff}$. -With this, $V_\mathrm{eff}$ becomes as follows in $\vb{q}$-space, -where we have used the convolution theorem -to get the product $\delta{n} (\vb{q}) V_{ee}(\vb{q})$: +The operator $\hat{c}_{\sigma.\vb{k}}^\dagger \hat{c}_{\sigma.\vb{k}}$ +simply counts the number of electrons in state $(\sigma, \vb{k})$, +which is given by the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) $n_F$. +This gives us the **Lindhard response function**: $$\begin{aligned} - V_\mathrm{eff}(\vb{q}) - = V_\mathrm{ext} + \chi_0 V_\mathrm{eff} V_{ee} - \qquad \quad - V_{ee}(\vb{q}) - = \frac{e^2}{\varepsilon_0 |\vb{q}|^2} + \boxed{ + \chi_0(\vb{q}, \omega) + = \frac{1}{V} \sum_{\sigma \vb{k}} + \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + } \end{aligned}$$ -Isolating this equation for $V_\mathrm{ext}$ -yields the definition of the relative permittivity $\varepsilon_r$: +From this, we would like to get the +[dielectric function](/know/concept/dielectric-function/) $\varepsilon_r$. +Recall its definition, where $U_\mathrm{tot}$, $U_\mathrm{ext}$, and $U_\mathrm{ind}$ +are the total, external and induced potentials, respectively: $$\begin{aligned} - V_\mathrm{ext} - = (1 - \chi_0 V_{ee}) V_\mathrm{eff} - \equiv \varepsilon_r V_\mathrm{eff} + U_\mathrm{tot} + = U_\mathrm{ext} + U_\mathrm{ind} + = \frac{U_\mathrm{ext}}{\varepsilon_r} \end{aligned}$$ -Therefore, by inserting all the above expressions, -we arrive at the Lindhard dielectric function $\varepsilon_r$ -for a non-interacting electron gas in a uniform potential: +Note that these are all *energy* potentials: +this choice is justified because all energy potentials +are caused by electric fields in this case. +The *electric* potential is recoverable as +$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$, +where $q_e < 0$ is the charge of an electron. + +From the Lindhard response function $\chi_0$, +we get the induced particle density offset $\delta{\expval{\hat{n}}}$ +caused by a potential $U$. +The density $\delta{\expval{\hat{n}}}$ should be self-consistent, +implying $U = U_\mathrm{tot}$. +In other words, we have a linear relation +$\delta{\expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$, +so the standard formula for $\varepsilon_r$ gives: $$\begin{aligned} \boxed{ \varepsilon_r(\vb{q}, \omega) - = 1 - \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \sum_{k} \frac{f_{k-q} - f_k}{\hbar (\omega + i \eta) + E_{k-q} - E_k} + = 1 - \frac{U_{ee}(\vb{q})}{V} + \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} } \end{aligned}$$ +Where $U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$ +is Coulomb repulsion. +This is the **Lindhard dielectric function** of a free +non-interacting electron gas, +at any temperature and for any dimensionality. + ## References diff --git a/content/know/concept/maxwell-bloch-equations/index.pdc b/content/know/concept/maxwell-bloch-equations/index.pdc index 020a120..3f090a2 100644 --- a/content/know/concept/maxwell-bloch-equations/index.pdc +++ b/content/know/concept/maxwell-bloch-equations/index.pdc @@ -34,8 +34,8 @@ $$\begin{aligned} \hat{H}_1(t) = - \hat{\vb{p}} \cdot \vb{E}(t) \qquad \quad - \hat{\vb{p}} - \equiv q \hat{\vb{x}} + \vu{p} + \equiv q \vu{x} \qquad \quad \vb{E}(t) = \vb{E}_0 \cos\!(\omega t) @@ -49,10 +49,10 @@ can then be described by: $$\begin{aligned} \dv{c_g}{t} - &= i \frac{q \matrixel{g}{\hat{\vb{x}}}{e} \cdot \vb{E}_0}{2 \hbar} \exp\!\big( i \omega t \!-\! i \omega_0 t \big) \: c_e + &= i \frac{q \matrixel{g}{\vu{x}}{e} \cdot \vb{E}_0}{2 \hbar} \exp\!\big( i \omega t \!-\! i \omega_0 t \big) \: c_e \\ \dv{c_e}{t} - &= i \frac{q \matrixel{e}{\hat{\vb{x}}}{g} \cdot \vb{E}_0}{2 \hbar} \exp\!\big(\!-\! i \omega t \!+\! i \omega_0 t \big) \: c_g + &= i \frac{q \matrixel{e}{\vu{x}}{g} \cdot \vb{E}_0}{2 \hbar} \exp\!\big(\!-\! i \omega t \!+\! i \omega_0 t \big) \: c_g \end{aligned}$$ We want to rearrange these equations a bit. @@ -70,11 +70,11 @@ Similarly, we define the transition dipole moment $\vb{p}_0^{-}$: $$\begin{aligned} \vb{p}_0^{-} - \equiv q \matrixel{e}{\vb{x}}{g} + \equiv q \matrixel{e}{\vu{x}}{g} \qquad \quad \vb{p}_0^{+} \equiv (\vb{p}_0^{-})^* - = q \matrixel{g}{\vb{x}}{e} + = q \matrixel{g}{\vu{x}}{e} \end{aligned}$$ With these, the equations for $c_g$ and $c_e$ can be rewritten as shown below. @@ -238,10 +238,10 @@ $$\begin{aligned} \vb{p} &= \matrixel{\Psi}{\hat{\vb{p}}}{\Psi} \\ - &= q \Big( c_g c_g^* \matrixel{g}{\hat{\vb{x}}}{g} + c_e c_e^* \matrixel{e}{\hat{\vb{x}}}{e} - + c_g c_e^* \matrixel{e}{\hat{\vb{x}}}{g} \exp\!(i \omega_0 t) + c_e c_g^* \matrixel{g}{\hat{\vb{x}}}{e} \exp\!(-i \omega_0 t) \Big) + &= q \Big( c_g c_g^* \matrixel{g}{\vu{x}}{g} + c_e c_e^* \matrixel{e}{\vu{x}}{e} + + c_g c_e^* \matrixel{e}{\vu{x}}{g} \exp\!(i \omega_0 t) + c_e c_g^* \matrixel{g}{\vu{x}}{e} \exp\!(-i \omega_0 t) \Big) \\ - &= q \Big( \rho_{ge} \matrixel{e}{\hat{\vb{x}}}{g} + \rho_{eg} \matrixel{g}{\hat{\vb{x}}}{e} \Big) + &= q \Big( \rho_{ge} \matrixel{e}{\vu{x}}{g} + \rho_{eg} \matrixel{g}{\vu{x}}{e} \Big) = \vb{p}_0^{-} \rho_{ge}(t) + \vb{p}_0^{+} \rho_{eg}(t) \equiv \vb{p}^{-}(t) + \vb{p}^{+}(t) \end{aligned}$$ @@ -366,41 +366,27 @@ Inserting the definition $\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$ together with Ohm's law $\vb{J}_\mathrm{free} = \sigma \vb{E}$ yields: $$\begin{aligned} - \nabla \cross \big( \nabla \cross \vb{E} \big) - = - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} - \mu_0 \pdv[2]{\vb{P}}{t} + \boxed{ + \nabla \cross \big( \nabla \cross \vb{E} \big) + = - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} - \mu_0 \pdv[2]{\vb{P}}{t} + } \end{aligned}$$ Where $\sigma$ is the medium's conductivity, if any; many authors assume $\sigma = 0$. -Next, we rewrite the left side using a vector identity, -and assume no net charge $\nabla \cdot \vb{E} = 0$: - -$$\begin{aligned} - \nabla^2 \vb{E} - \nabla \big( \nabla \cdot \vb{E} \big) - = \nabla^2 \vb{E} - = \mu_0 \sigma \pdv{\vb{E}}{t} + \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} + \mu_0 \pdv[2]{\vb{P}}{t} -\end{aligned}$$ - -After some rearranging, -we arrive at a variant of the electromagnetic wave equation: - -$$\begin{aligned} - \nabla^2 \vb{E} - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} - &= \mu_0 \pdv[2]{\vb{P}}{t} -\end{aligned}$$ - It is trivial to show that $\vb{E}$ and $\vb{P}$ can be replaced by $\vb{E}^{+}$ and $\vb{P}^{+}$. -It is also simple to convert -the dipole $\vb{p}^{+}$ and inversion $d$ -into their macroscopic versions $\vb{P}^{+}$ and $D$: + +It is also simple to convert $\vb{p}^{+}$ and $d$ +into the macroscopic polarization $\vb{P}^{+}$ and total inversion $D$ +by summing over the atoms: $$\begin{aligned} - \vb{P}^{+}(\vb{r}, t) - = \sum_{n} \vb{p}^{+}_n \: \delta(\vb{r} \!-\! \vb{r}_n) - \qquad \quad - D(\vb{r}, t) - = \sum_{n} d_n \: \delta(\vb{r} \!-\! \vb{r}_n) + \vb{P}^{+}(\vb{x}, t) + &= \sum_{n} \vb{p}^{+}_n \: \delta(\vb{x} - \vb{x}_n) + \\ + D(\vb{x}, t) + &= \sum_{n} d_n \: \delta(\vb{x} - \vb{x}_n) \end{aligned}$$ We thus arrive at the **Maxwell-Bloch equations**, @@ -410,7 +396,7 @@ $$\begin{aligned} \boxed{ \begin{aligned} \mu_0 \pdv[2]{\vb{P}^{+}}{t} - &= \nabla^2 \vb{E}^{+} - \mu_0 \sigma \pdv{\vb{E}^{+}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}^{+}}{t} + &= - \nabla \cross \nabla \cross \vb{E}^{+} - \mu_0 \sigma \pdv{\vb{E}^{+}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}^{+}}{t} \\ \pdv{\vb{P}^{+}}{t} &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+} |