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diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md new file mode 100644 index 0000000..6e395b4 --- /dev/null +++ b/source/know/concept/jellium/index.md @@ -0,0 +1,416 @@ +--- +title: "Jellium" +date: 2021-11-23 +categories: +- Physics +- Quantum mechanics +- Perturbation +layout: "concept" +--- + +**Jellium**, also called the **uniform** or **homogeneous electron gas**, +is a theoretical material where all electrons are free, +and the ions' positive charge is smeared into a uniform background "jelly". +This simple model lets us study electron interactions easily. + + +## Without interactions + +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 $V$, +and impose periodic boundary conditions on it, +such that the single-particle orbitals are (suppressing spin): + +$$\begin{aligned} + \Inprod{\vb{r}}{\psi_{\vb{k}}} + = \psi_{\vb{k}}(\vb{r}) + = \frac{1}{\sqrt{V}} \exp(i \vb{k} \cdot \vb{r}) + \qquad \quad + \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z) +\end{aligned}$$ + +Where $n_x, n_y, n_z \in \mathbb{Z}$. +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: + +$$\begin{aligned} + \hat{H}_0 + = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}} +\end{aligned}$$ + +Assuming that the temperature $T = 0$, +the $N$-electron ground state of this Hamiltonian +is known as the **Fermi sea** or **Fermi sphere** $\Ket{\mathrm{FS}}$, +and is constructed by filling up the single-electron states +starting from the lowest energy: + +$$\begin{aligned} + \Ket{\mathrm{FS}} + = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0} +\end{aligned}$$ + +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** $k_F$, +and obey the expected kinetic energy relation: + +$$\begin{aligned} + \boxed{ + \epsilon_F + = \frac{\hbar^2}{2 m} k_F^2 + } +\end{aligned}$$ + +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. + +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: + +$$\begin{aligned} + N + = \sum_{s} \sum_{\vb{k}} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} + = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}} +\end{aligned}$$ + +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. + +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/): + +$$\begin{aligned} + N + = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}} + = 2 \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}} +\end{aligned}$$ + +Where we realized that spin does not matter, +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/) +$(|\vb{k}|, \theta, \varphi)$: + +$$\begin{aligned} + N + &= \frac{V}{4 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi} + \\ + &= \frac{V}{4 \pi^3} 4 \pi \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} + = \frac{V}{\pi^2} \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_0^{k_F} + = \frac{V}{3 \pi^2} k_F^3 +\end{aligned}$$ + +Using that the electron density $n = N/V$, +we thus arrive at the following relation: + +$$\begin{aligned} + \boxed{ + k_F^3 + = 3 \pi^2 n + } +\end{aligned}$$ + +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, +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 $E^{(0)}$? + +$$\begin{aligned} + E^{(0)} + = \matrixel{\mathrm{FS}}{\hat{H}_0}{\mathrm{FS}} + = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} +\end{aligned}$$ + +Once again, we turn the sum over $\vb{k}$ into an integral, +and recognize the spin's irrelevance: + +$$\begin{aligned} + E^{(0)} + &= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m} + \matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}} + \\ + &= \frac{\hbar^2 V}{8 \pi^3 m} \int_{-\infty}^\infty |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \dd{\vb{k}} +\end{aligned}$$ + +In spherical coordinates, +we evaluate the integral and find that $E^{(0)}$ is proportional to $k_F^5$: + +$$\begin{aligned} + E^{(0)} + &= \frac{\hbar^2 V}{8 \pi^3 m} \int_0^{2 \pi} + \int_0^\pi \int_0^\infty \Big( |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \Big) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi} + \\ + &= \frac{\hbar^2 V}{8 \pi^3 m} 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|} + = \frac{\hbar^2 V}{2 \pi^2 m} \bigg[ \frac{|\vb{k}|^5}{5} \bigg]_0^{k_F} + = \frac{\hbar^2 V}{10 \pi^2 m} k_F^5 +\end{aligned}$$ + +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 $k_F^3 = 3 \pi^2 n$: + +$$\begin{aligned} + \boxed{ + \frac{E^{(0)}}{N} + = \frac{3 \hbar^2}{10 m} k_F^2 + = \frac{3}{5} \epsilon_F + } + \:\sim\: n^{2/3} +\end{aligned}$$ + +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 $a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$: + +$$\begin{aligned} + \frac{4 \pi}{3} (a_0 r_s)^3 + = \frac{1}{n} + = \frac{3 \pi^2}{k_F^3} + \quad \implies \quad + r_s + = \Big( \frac{3}{4 \pi a_0^3 n} \Big)^{1/3} + = \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 k_F} +\end{aligned}$$ + +Such that the ground state energy can be rewritten in Rydberg units of energy like so: + +$$\begin{aligned} + \frac{E^{(0)}}{N} + = \frac{3 \hbar^2}{10 m} \frac{4 \pi \varepsilon_0 e^2}{4 \pi \varepsilon_0 e^2} \frac{a_0^2 k_F^2}{a_0^2} + = \frac{3 e^2}{40 \pi \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{2/3} \frac{1}{a_0 r_s^2} + \approx \frac{2.21}{r_s^2} \; \mathrm{Ry} +\end{aligned}$$ + + +## With interactions + +To include Coulomb interactions, let us try +[time-independent pertubation theory](/know/concept/time-independent-perturbation-theory/). +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 $n^{2/3}$, +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 $n \to \infty$. + +The two-body Coulomb interaction operator $\hat{W}$ +is as follows in second-quantized form: + +$$\begin{aligned} + \hat{W} + = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e^2}{\varepsilon_0 |\vb{q}|^2} + \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} +\end{aligned}$$ + +The first-order correction $E^{(1)}$ to the ground state (i.e. Fermi sea) energy +is then given by: + +$$\begin{aligned} + E^{(1)} + = \matrixel{\mathrm{FS}}{\hat{W}}{\mathrm{FS}} + = \frac{e^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} + \matrixel{\mathrm{FS}}{ + \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} + }{\mathrm{FS}} +\end{aligned}$$ + +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 $\vb{k}_1 \!+\! \vb{q}$ +must be inside the Fermi sphere, +to avoid annihilating an empty orbital. +Let $s = s_1$ and $\vb{k} = \vb{k}_1$: + +$$\begin{aligned} + E^{(1)} + &= \frac{e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} + \matrixel{\mathrm{FS}}{ + \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}} \hat{c}_{s, \vb{k}} + }{\mathrm{FS}} + \\ + &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} + \matrixel{\mathrm{FS}}{ + \big( \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}}\big) \big(\hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k}}\big) + }{\mathrm{FS}} + \\ + &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} + \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|) +\end{aligned}$$ + +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, +so the largest possible jump +goes from a point on the Fermi surface to the opposite point, +and thus has length $2 k_F$. +This yields the integration limit, and therefore leads to: + +$$\begin{aligned} + E^{(1)} + &= \frac{- e^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}} + \int_0^{2 \pi} \!\!\int_0^\pi \!\!\int_0^\infty \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \frac{|\vb{q}|^2}{|\vb{q}|^2} + \sin(\theta_q) \dd{|\vb{q}|} \dd{\theta_q} \dd{\varphi_q} + \\ + &= \frac{- e^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}} + \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|} +\end{aligned}$$ + +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 $\vb{k}$: + +$$\begin{aligned} + E^{(1)} + &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty + \!\Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) + \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} + \\ + &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F} + \!\Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) + \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} +\end{aligned}$$ + +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 $|\vb{k} \!+\! \vb{q}|$. +The *law of cosines* then gives the following relation: + +$$\begin{aligned} + |\vb{k}|^2 + = |\vb{q}|^2 + |\vb{k} \!+\! \vb{q}|^2 - 2 |\vb{q}| |\vb{k} \!+\! \vb{q}| \cos(\theta_k) +\end{aligned}$$ + +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: + +$$\begin{aligned} + \cos(\theta_k) + = \frac{|\vb{k} \!+\! \vb{q}|^2 + |\vb{q}|^2 - |\vb{k}|^2}{2 |\vb{k} \!+\! \vb{q}| |\vb{q}|} + &\:\:\underset{|\vb{q}| \to 0}{>}\:\:\: \frac{k_F^2 + |\vb{q}|^2 - k_F^2}{2 k_F |\vb{q}|} + = \frac{|\vb{q}|}{2 k_F} + \\ + &\underset{|\vb{q}| \to 2 k_F}{<}\:\: \frac{k_F^2 + 4 k_F ^2 - k_F^2}{2 k_F 2 k_F} + = 1 +\end{aligned}$$ + +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 $|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$. +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. + +Armed with these integration limits, we return to calculating $E^{(1)}$, +substituting $\xi \equiv \cos(\theta_k)$: + +$$\begin{aligned} + E^{(1)} + &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}} + \!\!\int_{|\vb{q}|/(2 \cos{\theta_k})}^{k_F} |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} + \\ + &= \frac{e^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)} + \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \frac{\sin(\theta_k)}{\sin(\theta_k)} \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} + \\ + &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 + \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} +\end{aligned}$$ + +Where we have used that $\varphi_k$ does not appear in the integrand. +Evaluating these integrals: + +$$\begin{aligned} + E^{(1)} + &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 + \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|} + \\ + &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 + \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|} + \\ + &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} + \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|} + \\ + &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} + \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|} + \\ + &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F} + \\ + &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4 + = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4 + = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F +\end{aligned}$$ + +Per particle, the first-order energy correction $E^{(1)}$ +is therefore found to be as follows: + +$$\begin{aligned} + \boxed{ + \frac{E^{(1)}}{N} + = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} k_F + } +\end{aligned}$$ + +This can also be written using the parameter $r_s$ introduced above, leading to: + +$$\begin{aligned} + \frac{E^{(1)}}{N} + = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0} + = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s} +\end{aligned}$$ + +Consequently, for sufficiently high densities $n$, +the total energy $E$ per particle is given by: + +$$\begin{aligned} + \boxed{ + \frac{E}{N} + \approx \bigg( \frac{2.21}{r_s^2} - \frac{0.92}{r_s} \bigg) \; \mathrm{Ry} + } +\end{aligned}$$ + +Unfortunately, this is as far as we can go. +In theory, the second-order energy correction $E^{(2)}$ is as shown below, +but it turns out that it (and all higher orders) diverge: + +$$\begin{aligned} + E^{(2)} + = \sum_{\Psi_n \neq \mathrm{FS}} \frac{\big| \matrixel{\mathrm{FS}}{\hat{W}}{\Psi_n} \big|^2}{E^{(0)} - E_n} +\end{aligned}$$ + +The only cure for this is to go to infinite order, +where all the infinities add up to a finite result. + + + +## References +1. H. Bruus, K. Flensberg, + *Many-body quantum theory in condensed matter physics*, + 2016, Oxford. |