Coherent Electronic Transport in Periodic Crystals

Abstract

We consider independent electrons in a periodic crystal in their ground state, and turn on a uniform electric field at some prescribed time. We rigorously define the current per unit volume and study its properties using both linear response and adiabatic theory. Our results provide a unified framework for various phenomena such as the quantization of Hall conductivity of insulators with broken time-reversibility, the ballistic regime of electrons in metals, Bloch oscillations in the long-time response of metals, and the static conductivity of graphene. We identify explicitly the regime in which each holds.

Proofs of Two Technical Lemmata

1.1 Proof of Lemma 4.1

Proof

We replicate the proof of the Faris–Lavine Theorem given in [32], replacing the Laplacian by $\frac{1}{2}(-i\nabla + {\mathcal {A}})^2$. It consists in verifying the following two hypotheses of [32, Theorem X.37]. Let $A = \frac{1}{2}(-i\nabla +{\mathcal {A}})^2 + W+V$ and $N=A+2c|x|^2 +b$, where $b\in {{\mathbb {R}}}$ will be specified below:

$$\begin{aligned} {} \textit{there exists h, such that for any } \phi \in {\mathcal {C}}, \quad \Vert A\phi \Vert \le h \Vert N\phi \Vert ; \end{aligned}$$

(90)

$$\begin{aligned} \textit{for some} \ell ,\textit{ for any } \phi \in {\mathcal {C}},\quad |(A\phi ,N\phi ) - (N\phi ,A\phi )| \le \ell \Vert N^\frac{1}{2} \phi \Vert ^2. \end{aligned}$$

(91)

By hypothesis 3 in Lemma 4.1 and the conditions on W, it is possible to choose b so that $N\ge 1$. As quadratic forms on ${\mathcal {C}}$,

$$\begin{aligned} N^2 = (A+b)^2 + 4c \sum _{j=1}^d x_j (A+b+c|x|^2)x_j - 2cd. \end{aligned}$$

Hypotheses 1 and 3 guarantee that $A+b+c|x|^2$ is bounded below. Hence, increasing the value of b if necessary to make this operator positive, we have

$$\begin{aligned} \Vert (A+b)\phi \Vert ^2_{L^2} \le \Vert N\phi \Vert ^2_{L^2} +4cd\Vert \phi \Vert ^2_{L^2}, \end{aligned}$$

which proves (90).

For (91), we observe that

$$\begin{aligned}&\pm i [A,N] = \pm 2c(x \cdot (-i\nabla +{\mathcal {A}}) + (-i\nabla +{\mathcal {A}})\cdot x)\\&\le 2 c \left( (-i\nabla +{\mathcal {A}})^2 +|x|^2\right) \le \ell N, \end{aligned}$$

where we have used

$$\begin{aligned} (-i\nabla + {\mathcal {A}})^2 + |x|^2 \pm (x\cdot (-i\nabla + {\mathcal {A}}) + (-i\nabla + {\mathcal {A}})\cdot x) = (-i\nabla + {\mathcal {A}} \pm x)^2 \ge 0 \end{aligned}$$

and

$$\begin{aligned}&N = \left( \frac{a}{2} (-i\nabla + {\mathcal {A}})^2 + V\right) + (W+c|x|^2)\\&+ \frac{1-a}{2}(-i\nabla + {\mathcal {A}})^2 + c|x|^2 +b \ge e((-i\nabla + {\mathcal {A}})^2+|x|^2), \end{aligned}$$

where $e = \min (c,\frac{1-a}{2})>0$ and where b is chosen so that

$$\begin{aligned} b- f +\min \,\sigma \left( \frac{a}{2} (-i\nabla + {\mathcal {A}})^2 + V\right) \ge 0. \end{aligned}$$

This proves (91). Hence A is essentially self-adjoint on ${\mathcal {C}}$. $\square$

1.2 Proof of Lemma 4.2

Proof

By the Kato-Rellich theorem, for any $0\le t\le T$, H(t) is self-adjoint on $L^2_{\mathrm{per}}$ with domain $H^2_\mathrm{per}$, and bounded below. We will show that there exists $\mu >0$ so that the graph norm of $(H(t)+\mu )$ for any $0\le t \le T$ is equivalent to the $H^2_{\mathrm{per}}$-norm. This will prove Lemma 4.2 by Proposition 2.1 in [36] (see also Theorem X.70 in [32]).

We have for any $\mu >0$, $0\le t\le T$ and $\phi \in H^{2}_\mathrm{per}$,

$$\begin{aligned} \Vert (H(t)+\mu )\phi \Vert _{L^{2}_{\mathrm{per}}} \le (1+a)\Vert H_0 \phi \Vert _{L^{2}_{\mathrm{per}}} + (b+\mu ) \Vert \phi \Vert _{L^{2}_{\mathrm{per}}} \le (1+a+b+\mu ) \Vert \phi \Vert _{H^2_{\mathrm{per}}}, \end{aligned}$$

and so the graph norm is controlled by the $H^2_{\mathrm{per}}$-norm.

For the other inequality, we relate the resolvent of H(t) to that of $H_0$ by a bounded operator, with bounded inverse. Notice that, for any $\mu >0$, since $H_0$ is positive,

$$\begin{aligned} \forall \; 0\le t\le T, \quad (H(t)+\mu ) = (1+H_1(t)(H_0+\mu )^{-1})(H_0+\mu ). \end{aligned}$$

Furthermore,

$$\begin{aligned} \forall \; 0\le t\le T, \quad \Vert H_1(t)(H_0+\mu )^{-1}\Vert \le a\Vert H_0(H_0+\mu )^{-1}\Vert +b\Vert (H_0+\mu )^{-1}\Vert \le a +\frac{b}{\mu }. \end{aligned}$$

and so, for $\mu > \frac{b}{1-a}$, the operator $1+ H_1(t)(H_0+\mu )^{-1}$ is bounded and invertible with bounded inverse in $L^{2}_{\mathrm{per}}$. Therefore $(H(t)+\mu )^{-1}$ is bounded from $L^2_{\mathrm{per}}$ to $H^2_{\mathrm{per}}$, which means there exists $C>0$ such that, for any $\phi \in H^2_{\mathrm{per}}$ and $0\le t\le T$,

$$\begin{aligned} \Vert \phi \Vert _{H^2_{\mathrm{per}}} = \Vert (H(t)+\mu )^{-1}(H(t)+\mu )\phi \Vert _{H^2_{\mathrm{per}}} \le C\Vert (H(t)+\mu )\phi \Vert _{L^{2}_{\mathrm{per}}}, \end{aligned}$$

which concludes the proof. $\square$