diff options
Diffstat (limited to 'source/know/concept/jellium')
| -rw-r--r-- | source/know/concept/jellium/index.md | 158 |
1 files changed, 79 insertions, 79 deletions
diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md index c951fd7..5c50f80 100644 --- a/source/know/concept/jellium/index.md +++ b/source/know/concept/jellium/index.md @@ -21,7 +21,7 @@ Let us start by neglecting electron-electron interactions. This is clearly a dubious assumption, but we will stick with it for now. For an infinitely large sample of jellium, the single-electron states are simply plane waves. -We consider an arbitrary cube of volume $V$, +We consider an arbitrary cube of volume $$V$$, and impose periodic boundary conditions on it, such that the single-particle orbitals are (suppressing spin): @@ -33,22 +33,22 @@ $$\begin{aligned} \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z) \end{aligned}$$ -Where $n_x, n_y, n_z \in \mathbb{Z}$. +Where $$n_x, n_y, n_z \in \mathbb{Z}$$. This is a discrete (but infinite) set of independent orbitals, so it is natural to use the [second quantization](/know/concept/second-quantization/) -to write the non-interacting Hamiltonian $\hat{H}_0$, -where $\hbar^2 |\vb{k}|^2 / (2 m)$ is the kinetic energy -of the orbital with wavevector $\vb{k}$, and $s$ is the spin: +to write the non-interacting Hamiltonian $$\hat{H}_0$$, +where $$\hbar^2 |\vb{k}|^2 / (2 m)$$ is the kinetic energy +of the orbital with wavevector $$\vb{k}$$, and $$s$$ is the spin: $$\begin{aligned} \hat{H}_0 = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}} \end{aligned}$$ -Assuming that the temperature $T = 0$, -the $N$-electron ground state of this Hamiltonian -is known as the **Fermi sea** or **Fermi sphere** $\Ket{\mathrm{FS}}$, +Assuming that the temperature $$T = 0$$, +the $$N$$-electron ground state of this Hamiltonian +is known as the **Fermi sea** or **Fermi sphere** $$\Ket{\mathrm{FS}}$$, and is constructed by filling up the single-electron states starting from the lowest energy: @@ -57,9 +57,9 @@ $$\begin{aligned} = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0} \end{aligned}$$ -Because $T = 0$, all the electrons stay in their assigned state. -The energy and wavenumber $|\vb{k}|$ of the highest filled orbital -are called the **Fermi energy** $\epsilon_F$ and **Fermi wavenumber** $k_F$, +Because $$T = 0$$, all the electrons stay in their assigned state. +The energy and wavenumber $$|\vb{k}|$$ of the highest filled orbital +are called the **Fermi energy** $$\epsilon_F$$ and **Fermi wavenumber** $$k_F$$, and obey the expected kinetic energy relation: $$\begin{aligned} @@ -69,13 +69,13 @@ $$\begin{aligned} } \end{aligned}$$ -The Fermi sea can be visualized in $\vb{k}$-space as a sphere with radius $k_F$. -Because $\vb{k}$ is discrete, the sphere's surface is not smooth, -but in the limit $V \to \infty$ it becomes perfect. +The Fermi sea can be visualized in $$\vb{k}$$-space as a sphere with radius $$k_F$$. +Because $$\vb{k}$$ is discrete, the sphere's surface is not smooth, +but in the limit $$V \to \infty$$ it becomes perfect. Now, we would like a relation between the system's parameters, -e.g. $N$ and $V$, and the resulting values of $\epsilon_F$ or $k_F$. -The total population $N$ must be given by: +e.g. $$N$$ and $$V$$, and the resulting values of $$\epsilon_F$$ or $$k_F$$. +The total population $$N$$ must be given by: $$\begin{aligned} N @@ -83,11 +83,11 @@ $$\begin{aligned} = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}} \end{aligned}$$ -Where we have turned the sum over $\vb{k}$ into an integral with a constant factor, -by using that each orbital exclusively occupies a volume $(2 \pi)^3 / V$ in $\vb{k}$-space. +Where we have turned the sum over $$\vb{k}$$ into an integral with a constant factor, +by using that each orbital exclusively occupies a volume $$(2 \pi)^3 / V$$ in $$\vb{k}$$-space. -At zero temperature, this inner product can only be $0$ or $1$, -depending on whether $\vb{k}$ is outside or inside the Fermi sphere. +At zero temperature, this inner product can only be $$0$$ or $$1$$, +depending on whether $$\vb{k}$$ is outside or inside the Fermi sphere. We can therefore rewrite using a [Heaviside step function](/know/concept/heaviside-step-function/): @@ -98,10 +98,10 @@ $$\begin{aligned} \end{aligned}$$ Where we realized that spin does not matter, -and replaced the sum over $s$ by a factor $2$. +and replaced the sum over $$s$$ by a factor $$2$$. In order to evaluate this 3D integral, we go to [spherical coordinates](/know/concept/spherical-coordinates/) -$(|\vb{k}|, \theta, \varphi)$: +$$(|\vb{k}|, \theta, \varphi)$$: $$\begin{aligned} N @@ -112,7 +112,7 @@ $$\begin{aligned} = \frac{V}{3 \pi^2} k_F^3 \end{aligned}$$ -Using that the electron density $n = N/V$, +Using that the electron density $$n = N/V$$, we thus arrive at the following relation: $$\begin{aligned} @@ -122,15 +122,15 @@ $$\begin{aligned} } \end{aligned}$$ -This result also justifies our assumption that $T = 0$: -we can accurately calculate the density $n$ for many conducting materials, -and this relation then gives $k_F$ and $\epsilon_F$. -It turns out that $\epsilon_F$ is usually very large -compared to the thermal energy $k_B T$ at reasonable temperatures, +This result also justifies our assumption that $$T = 0$$: +we can accurately calculate the density $$n$$ for many conducting materials, +and this relation then gives $$k_F$$ and $$\epsilon_F$$. +It turns out that $$\epsilon_F$$ is usually very large +compared to the thermal energy $$k_B T$$ at reasonable temperatures, so we can conclude that thermal fluctuations are negligible. -Now, $\epsilon_F$ is the highest single-electron energy, -but about the total $N$-particle energy $E^{(0)}$? +Now, $$\epsilon_F$$ is the highest single-electron energy, +but about the total $$N$$-particle energy $$E^{(0)}$$? $$\begin{aligned} E^{(0)} @@ -138,7 +138,7 @@ $$\begin{aligned} = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \end{aligned}$$ -Once again, we turn the sum over $\vb{k}$ into an integral, +Once again, we turn the sum over $$\vb{k}$$ into an integral, and recognize the spin's irrelevance: $$\begin{aligned} @@ -150,7 +150,7 @@ $$\begin{aligned} \end{aligned}$$ In spherical coordinates, -we evaluate the integral and find that $E^{(0)}$ is proportional to $k_F^5$: +we evaluate the integral and find that $$E^{(0)}$$ is proportional to $$k_F^5$$: $$\begin{aligned} E^{(0)} @@ -163,8 +163,8 @@ $$\begin{aligned} \end{aligned}$$ In general, it is more useful to consider -the average kinetic energy per electron $E^{(0)} / N$, -which we find to be as follows, using that $k_F^3 = 3 \pi^2 n$: +the average kinetic energy per electron $$E^{(0)} / N$$, +which we find to be as follows, using that $$k_F^3 = 3 \pi^2 n$$: $$\begin{aligned} \boxed{ @@ -175,9 +175,9 @@ $$\begin{aligned} \:\sim\: n^{2/3} \end{aligned}$$ -Traditionally, this is expressed using a dimensionless parameter $r_s$, +Traditionally, this is expressed using a dimensionless parameter $$r_s$$, defined as the radius of a sphere containing a single electron, -measured in Bohr radii $a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$: +measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$$: $$\begin{aligned} \frac{4 \pi}{3} (a_0 r_s)^3 @@ -206,14 +206,14 @@ To include Coulomb interactions, let us try Clearly, this will give better results when the interaction is relatively weak, if ever. The Coulomb potential is proportional to the inverse distance, -and the average electron spacing is roughly $n^{-1/3}$, -so the interaction energy $E_\mathrm{int}$ should scale as $n^{1/3}$. -We already know that the kinetic energy $E_\mathrm{kin} = E^{(0)}$ scales as $n^{2/3}$, +and the average electron spacing is roughly $$n^{-1/3}$$, +so the interaction energy $$E_\mathrm{int}$$ should scale as $$n^{1/3}$$. +We already know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$ scales as $$n^{2/3}$$, meaning perturbation theory should be reasonable -if $1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$, -so in the limit of high density $n \to \infty$. +if $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$$, +so in the limit of high density $$n \to \infty$$. -The two-body Coulomb interaction operator $\hat{W}$ +The two-body Coulomb interaction operator $$\hat{W}$$ is as follows in second-quantized form: $$\begin{aligned} @@ -222,7 +222,7 @@ $$\begin{aligned} \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} \end{aligned}$$ -The first-order correction $E^{(1)}$ to the ground state (i.e. Fermi sea) energy +The first-order correction $$E^{(1)}$$ to the ground state (i.e. Fermi sea) energy is then given by: $$\begin{aligned} @@ -235,14 +235,14 @@ $$\begin{aligned} \end{aligned}$$ This inner product can only be nonzero -if the two creation operators $\hat{c}^\dagger$ -are for the same orbitals as the two annihilation operators $\hat{c}$. -Since $\vb{q} \neq 0$, this means that $s_1 = s_2$, -and that momentum is conserved: $\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$. -And of course both $\vb{k}_1$ and $\vb{k}_1 \!+\! \vb{q}$ +if the two creation operators $$\hat{c}^\dagger$$ +are for the same orbitals as the two annihilation operators $$\hat{c}$$. +Since $$\vb{q} \neq 0$$, this means that $$s_1 = s_2$$, +and that momentum is conserved: $$\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$$. +And of course both $$\vb{k}_1$$ and $$\vb{k}_1 \!+\! \vb{q}$$ must be inside the Fermi sphere, to avoid annihilating an empty orbital. -Let $s = s_1$ and $\vb{k} = \vb{k}_1$: +Let $$s = s_1$$ and $$\vb{k} = \vb{k}_1$$: $$\begin{aligned} E^{(1)} @@ -260,11 +260,11 @@ $$\begin{aligned} \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|) \end{aligned}$$ -Next, we convert the sum over $\vb{q}$ into an integral in spherical coordinates. -Clearly, $\vb{q}$ is the "jump" made by an electron from one orbital to another, +Next, we convert the sum over $$\vb{q}$$ into an integral in spherical coordinates. +Clearly, $$\vb{q}$$ is the "jump" made by an electron from one orbital to another, so the largest possible jump goes from a point on the Fermi surface to the opposite point, -and thus has length $2 k_F$. +and thus has length $$2 k_F$$. This yields the integration limit, and therefore leads to: $$\begin{aligned} @@ -277,11 +277,11 @@ $$\begin{aligned} \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|} \end{aligned}$$ -Where we have used that the direction of $\vb{q}$, -i.e. $(\theta_q,\varphi_q)$, is irrelevant, -as long as we define $\theta_k$ as -the angle between $\vb{q}$ and $\vb{k} \!+\! \vb{q}$ -when we go to spherical coordinates $(|\vb{k}|, \theta_k, \varphi_k)$ for $\vb{k}$: +Where we have used that the direction of $$\vb{q}$$, +i.e. $$(\theta_q,\varphi_q)$$, is irrelevant, +as long as we define $$\theta_k$$ as +the angle between $$\vb{q}$$ and $$\vb{k} \!+\! \vb{q}$$ +when we go to spherical coordinates $$(|\vb{k}|, \theta_k, \varphi_k)$$ for $$\vb{k}$$: $$\begin{aligned} E^{(1)} @@ -296,10 +296,10 @@ $$\begin{aligned} Unfortunately, this last step function is less easy to translate into integration limits. In effect, we are trying to calculate the intersection volume of two spheres, -both with radius $k_F$, one centered on the origin (for $\vb{k}$), -and the other centered on $\vb{q}$ (for $\vb{k} \!+\! \vb{q}$). -Imagine a triangle with side lengths $|\vb{k}|$, $|\vb{q}|$ and $|\vb{k} \!+\! \vb{q}|^2$, -where $\theta_k$ is the angle between $|\vb{k}|$ and $|\vb{k} \!+\! \vb{q}|$. +both with radius $$k_F$$, one centered on the origin (for $$\vb{k}$$), +and the other centered on $$\vb{q}$$ (for $$\vb{k} \!+\! \vb{q}$$). +Imagine a triangle with side lengths $$|\vb{k}|$$, $$|\vb{q}|$$ and $$|\vb{k} \!+\! \vb{q}|^2$$, +where $$\theta_k$$ is the angle between $$|\vb{k}|$$ and $$|\vb{k} \!+\! \vb{q}|$$. The *law of cosines* then gives the following relation: $$\begin{aligned} @@ -307,10 +307,10 @@ $$\begin{aligned} = |\vb{q}|^2 + |\vb{k} \!+\! \vb{q}|^2 - 2 |\vb{q}| |\vb{k} \!+\! \vb{q}| \cos(\theta_k) \end{aligned}$$ -We already know that $|\vb{k}| < k_F$ and $0 < |\vb{q}| < 2 k_F$, -so by isolating for $\cos(\theta_k)$, -we can obtain bounds on $\theta_k$ and $|\vb{k}|$. -Let $|\vb{k}| \to k_F$ in both cases, then: +We already know that $$|\vb{k}| < k_F$$ and $$0 < |\vb{q}| < 2 k_F$$, +so by isolating for $$\cos(\theta_k)$$, +we can obtain bounds on $$\theta_k$$ and $$|\vb{k}|$$. +Let $$|\vb{k}| \to k_F$$ in both cases, then: $$\begin{aligned} \cos(\theta_k) @@ -322,16 +322,16 @@ $$\begin{aligned} = 1 \end{aligned}$$ -Meaning that $0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$. -To get a lower limit for $|\vb{k}|$, we "cheat" by artificially demanding -that $\vb{k}$ does not cross the halfway point between the spheres, -with the result that $|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$. +Meaning that $$0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$$. +To get a lower limit for $$|\vb{k}|$$, we "cheat" by artificially demanding +that $$\vb{k}$$ does not cross the halfway point between the spheres, +with the result that $$|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$$. Then, thanks to symmetry (both spheres have the same radius), -we just multiply the integral by $2$, -for $\vb{k}$ on the other side of the halfway point. +we just multiply the integral by $$2$$, +for $$\vb{k}$$ on the other side of the halfway point. -Armed with these integration limits, we return to calculating $E^{(1)}$, -substituting $\xi \equiv \cos(\theta_k)$: +Armed with these integration limits, we return to calculating $$E^{(1)}$$, +substituting $$\xi \equiv \cos(\theta_k)$$: $$\begin{aligned} E^{(1)} @@ -345,7 +345,7 @@ $$\begin{aligned} \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} \end{aligned}$$ -Where we have used that $\varphi_k$ does not appear in the integrand. +Where we have used that $$\varphi_k$$ does not appear in the integrand. Evaluating these integrals: $$\begin{aligned} @@ -369,7 +369,7 @@ $$\begin{aligned} = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F \end{aligned}$$ -Per particle, the first-order energy correction $E^{(1)}$ +Per particle, the first-order energy correction $$E^{(1)}$$ is therefore found to be as follows: $$\begin{aligned} @@ -379,7 +379,7 @@ $$\begin{aligned} } \end{aligned}$$ -This can also be written using the parameter $r_s$ introduced above, leading to: +This can also be written using the parameter $$r_s$$ introduced above, leading to: $$\begin{aligned} \frac{E^{(1)}}{N} @@ -387,8 +387,8 @@ $$\begin{aligned} = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s} \end{aligned}$$ -Consequently, for sufficiently high densities $n$, -the total energy $E$ per particle is given by: +Consequently, for sufficiently high densities $$n$$, +the total energy $$E$$ per particle is given by: $$\begin{aligned} \boxed{ @@ -398,7 +398,7 @@ $$\begin{aligned} \end{aligned}$$ Unfortunately, this is as far as we can go. -In theory, the second-order energy correction $E^{(2)}$ is as shown below, +In theory, the second-order energy correction $$E^{(2)}$$ is as shown below, but it turns out that it (and all higher orders) diverge: $$\begin{aligned} |