Shuai Aris Chen

Supercurrent from a mean-field theory in superconductivity

2023-11-30T00:00:00+00:00

One of the remarkable features of a superconductor is the Meissner effect, which states that a current can flow through the sample without any dissipation. This phenomenon gives rise to what we call supercurrents, which occur due to a coherent effect. In contrast to the normal phase, or the Fermi liquid, where the current is more like a local property resulting from electron scattering with disorder or impurities, superconductors exhibit a different behavior. Interestingly, in the case of a simple s-wave superconductor, the normal phase and superconducting phase share the same current operator.

We consider a spinful Fermi gas with density-density interaction,

\[\begin{align} H & =\sum_{k\sigma}\xi(k)c_{k\sigma}^{\dagger}c_{k\sigma}-\int drUc_{\uparrow}^{\dagger}(r)c_{\downarrow}^{\dagger}(r)c_{\downarrow}(r)c_{\uparrow}(r)\\ & =\sum_{k\sigma}\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}-\frac{U}{V}\sum_{kk^{\prime}q}c_{\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}+\mathbf{q}\downarrow}^{\dagger}c_{-\mathbf{k}^{\prime}+\mathbf{q}\downarrow}c_{\mathbf{k}\prime\uparrow} \end{align}\]

The interaction is a uniform density-density interaction, and we do not expect a current from it. Therefore, regardless of the real phase, the current operator for electrons always have the form as

\[\mathbf{j}(\mathbf{k})=\sum_{\sigma}\nabla\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}\]

One should not confuse the current of an electron with the current of a quasiparticle. From the Landau’s Fermi liquid theory, we can learn that the current of quasiparticles can get renormalized by interaction or orbital hybridization.

We can apply the mean-field theory to obtain an ansatz for a ground state. We can make a mean-field ansatz by assuming

\[\Delta=-\frac{1}{V}\int dr\langle c_{\downarrow}(r)c_{\uparrow}(r)\rangle=-\frac{1}{V}\sum_k\langle c_{-\mathbf{k}\downarrow}c_{\mathbf{k}\uparrow}\rangle\]

Then we obtain a mean-field Hamiltonian $H_{MF}$

\[H_{MF}(\mathbf{k})=\sum_{\sigma}\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}+\Delta\left[c_{\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger}+c_{-\mathbf{k}\downarrow}c_{\mathbf{k}\uparrow}\right]\]

where we make a proper gauge choice such that the order parameter $\Delta$ be a real number. We can diagonalize the mean-field Hamiltonian

\[H_{MF}(\mathbf{k})=E(\mathbf{k})\left[\gamma_{\mathbf{k}\uparrow}^{\dagger}\gamma_{\mathbf{k}\uparrow}+\gamma_{\mathbf{k}\downarrow}^{\dagger}\gamma_{\mathbf{k}\downarrow}\right]\]

by the Bogoliubov transformation

\[\begin{aligned} & \left[\begin{array}{c} \gamma_{\boldsymbol{k},\uparrow}\\ \gamma_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right]=\left[\begin{array}{cc} u_{\boldsymbol{k}} & -v_{\boldsymbol{k}}\\ v_{\boldsymbol{k}} & u_{\boldsymbol{k}} \end{array}\right]\left[\begin{array}{c} c_{\boldsymbol{k},\uparrow}\\ c_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right],\\ & \left[\begin{array}{c} c_{\boldsymbol{k},\uparrow}\\ c_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right]=\left[\begin{array}{cc} u_{\boldsymbol{k}} & v_{\boldsymbol{k}}\\ -v_{\boldsymbol{k}} & u_{\boldsymbol{k}} \end{array}\right]\left[\begin{array}{c} \gamma_{\boldsymbol{k},\uparrow}\\ \gamma_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right]. \end{aligned}\]

with the Bogolibov coefficients

\[\begin{align} u(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1+\frac{\xi(\mathbf{k})}{E(\mathbf{k})}\right)}\\ v(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1-\frac{\xi(\mathbf{k})}{E(\mathbf{k})}\right)} \end{align}\]

and the dispersion of quasiparticle $E(\mathbf{k})=\sqrt{\xi^{2}(\mathbf{k})+\Delta^{2}}$. Accordingly the ground state is

\[\vert\mathrm{BCS}\rangle=\prod_{\mathbf{k}}(u(\mathbf{k})+v(\mathbf{k})c_{\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger})\vert0\rangle\]

It is easy to check $\gamma_{\mathbf{k}\sigma}\vert\mathrm{BCS}\rangle=0$. The BCS ground state represents a condensation of Cooper pairs. The net momentum of a Cooper pair vanishes and thus we expect the current vanishes

\[\begin{align} \langle\mathrm{BCS}\vert J\vert\mathrm{BCS}\rangle & =\sum_{\mathbf{k}}\langle\mathrm{BCS}\vert\mathbf{j}(\mathbf{k})\vert\mathrm{BCS}\rangle\\ & =\sum_{\mathbf{k}}v^{2}(\mathbf{k})(\nabla\xi(\mathbf{k})-\nabla\xi(\mathbf{k}))=0 \end{align}\]

We need an excited state to show the supercurrents. There are two types of excitations. The first one is the Fermionic Bogolibov quasiparticle $\gamma_{\mathbf{k}\sigma}^{\dagger}\vert\mathrm{BCS}\rangle$ with an energy $E=E_{GS}+E(\mathbf{k})$. It has a finite energy gap. In fact, we have the second type of excitation, that is the gapless Bosonic Goldstone mode. By contrast it will cost at least energy $2\Delta$ to create a Cooper pair state $\gamma_{\mathbf{k}\uparrow}^{\dagger}\gamma_{\mathbf{-k}\downarrow}^{\dagger}\vert\mathrm{BCS}\rangle$, which can be taken as local excitations (in momentum space). Instead, to pursue the Goldstone mode, we can consider a new mean-field ansatz

\[\Delta_{\mathbf{q}}=\frac{1}{V}\sum_k\langle c_{-\mathbf{k}\downarrow}c_{\mathbf{k}+\mathbf{q}\uparrow}\rangle\]

or in the real space

\[\begin{align} \Delta_{\mathbf{q}} & =\int dre^{i\mathbf{q}\cdot\mathbf{r}}\langle c_{\downarrow}(\mathbf{r})c_{\uparrow}(\mathbf{r})\rangle \end{align}\]

Then we have the mean-field Hamiltonian

\[H_{\mathbf{q}}(\mathbf{k})=\xi(\mathbf{k}+\mathbf{q})c_{\mathbf{k}+\mathbf{q}\uparrow}^{\dagger}c_{\mathbf{k}+\mathbf{q}\uparrow}+\xi(\mathbf{k})c_{-\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}\uparrow}+\Delta_{\mathbf{q}}^{*}c_{\mathbf{k}+\mathbf{q}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger}+\Delta_{\mathbf{q}}c_{-\mathbf{k}\downarrow}c_{\mathbf{k}+\mathbf{q}\uparrow}\]

Also we can act the Bogoliubov transformation

\[\begin{align} u_{\mathbf{q}}(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1+\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2\epsilon_{\mathbf{q}}(\mathbf{k})}\right)}\\ v_{\mathbf{q}}(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1-\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2\epsilon_{\mathbf{q}}(\mathbf{k})}\right)} \end{align}\]

with a quasiparticle dispersion $E_{\mathbf{q\pm}}(\mathbf{k})$

\[\begin{align} E_{\mathbf{q\pm}}(\mathbf{k}) & =\sqrt{\left(\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2}\right)^{2}+\vert\Delta_{\mathbf{q}}\vert^{2}}\pm\frac{\xi(\mathbf{k}+\mathbf{q})-\xi(\mathbf{k})}{2}\\ & \equiv\epsilon_{\mathbf{q}}(\mathbf{k})\pm\frac{\xi(\mathbf{k}+\mathbf{q})-\xi(\mathbf{k})}{2} \end{align}\]

Similarly, we have the ground state

\[\vert\mathrm{BCS}(\mathbf{q})\rangle=\prod_{\mathbf{k}}(u_{\mathbf{q}}(\mathbf{k})+v_{\mathbf{q}}(\mathbf{k})c_{\mathbf{k}+\mathbf{q}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger})\vert0\rangle\]

When $\mathbf{q}$ approaches $0$, the state $\vert\mathrm{BCS}(\mathbf{q})\rangle$ reduces to $\vert\mathrm{BCS}\rangle$.Now let us evaluate the current operator

\[\begin{align} \langle\mathrm{BCS(\mathbf{q})}\vert\mathbf{J}\vert\mathrm{BCS(\mathbf{q})}\rangle & =\sum_{\mathbf{k}}\langle\mathrm{BCS(\mathbf{q})}\vert\mathbf{j}(\mathbf{k})\vert\mathrm{BCS(\mathbf{q})}\rangle\\ & =\sum_{\mathbf{k}}\langle\mathrm{BCS(\mathbf{q})}\vert\sum_{\sigma}\nabla\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}\vert\mathrm{BCS(\mathbf{q})}\rangle\\ & =\sum_{\mathbf{k}}\left(\nabla\xi(\mathbf{k}+\mathbf{q})-\nabla\xi(\mathbf{k})\right)v_{\mathbf{q}}^{2}(\mathbf{k}) \end{align}\]

If the $\mathbf{q}$ is vanishingly small, we can assume $\Delta_{\mathbf{q}}=\Delta$ or

\[\Delta_{\mathbf{q}}(\mathbf{r})=\Delta e^{i\mathbf{q}\cdot\mathbf{r}}\]

By noting

\[\begin{align} \epsilon_{\mathbf{q}}(\mathbf{k}) & =\sqrt{\left(\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2}\right)^{2}+\vert\Delta_{\mathbf{q}}\vert^{2}}\\ v_{\mathbf{q}}^{2}(\mathbf{k}) & =v^{2}(\mathbf{k})+\mathcal{O}(q) \end{align}\]

we can approximate

\[\langle\mathrm{BCS(\mathbf{q})}\vert J_{a}\vert\mathrm{BCS(\mathbf{q})}\rangle=\sum_{\mathbf{k}}\sum_{b}q_{b}\partial_{b}\partial_{a}\xi(\mathbf{k})v^{2}(\mathbf{k})\] \[\begin{align} \langle\mathrm{BCS(\mathbf{q})}\vert j_{a}\vert\mathrm{BCS(\mathbf{q})}\rangle & =\sum_{\mathbf{k}}\left(\nabla\xi(\mathbf{k}+\mathbf{q})-\nabla\xi(\mathbf{k})\right)v_{\mathbf{q}}^{2}(\mathbf{k})\\ & =\frac{q_{a}}{m}\sum_{\mathbf{k}}v^{2}(\mathbf{k})=\frac{q_{a}}{m}n_{s} \end{align}\]

where $n_{s}$ is the superfluid stiffness,

\[n_{s}=\sum_{\mathbf{k}}v^{2}(\mathbf{k})\]

and the superfluid weight is

\[D_{s}=\frac{n_{s}}{m}\]

The supercurrent in a superconductor is considered global because it involves the integral over all momentum states. In other words, the state represented by $\vert \mathrm{BCS}(\mathbf q)\rangle$, which consists of Bogoliubov quasiparticles, undergoes a change in its wave function throughout the entire system, indicating a global excitation.

In contrast, in a Fermi liquid theory, when an electron is excited from the Fermi surface, it carries charge currents that involve only a single momentum point on the Fermi surface. Therefore, we can consider this type of current as local, as it is confined to a specific region or point in the system.

Formula reference

2023-11-29T00:00:00+00:00

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Bogoliubov transformation

Basics for bosons

Bogoliubov transformations are linear transformation of creation/annihilation operators that preserve the algebraic relations among them. In the following, we deal with bosons that satisfies canonical communication.

As a fact, generally, a Bogoliubov transformation $U$ is not unitary $U^\dagger\neq U^{-1}$. Let us consider a general model with
\[H= a^\dagger a +b^\dagger b +\gamma a^\dagger b^\dagger +\gamma a b\]
where operators $a$ and $b$ are bosonic operators that satisfy canonical communication relation
\[[a,a^\dagger]=1,[b,b^\dagger]=1\]
Our aim is to find new quasiparticles $\alpha,\beta$ under which $H$ is in a diagonal form. The first piece of requirement is to preserve the bosonic relation. Suppose
\[\begin{align} \alpha& =M_{11}a+M_{12}b^\dagger \\ \beta^\dagger&=M_{21}a+M_{22}b^\dagger \end{align}\]
Then
\[\begin{align} [\alpha,\alpha^\dagger]& =[M_{11}a,M_{11}^*a^\dagger]+[M_{12}b^\dagger,M_{12}^*b]=|M_{11}|^2-|M_{12}|^2=1 \\ [\beta,\beta^\dagger]&=|M_{22}|^2-|M_{11}|^2=1 \end{align}\]
when formulated into matrix form
\[h= \begin{pmatrix} \epsilon_0 & \Delta \\ \Delta^* & \epsilon_0 \end{pmatrix}\]

The eigenenergy appears as a solution to the equation

\[\mathrm{Det}\left[\begin{pmatrix}\lambda & 0 \\ 0 & -\lambda \end{pmatrix} -h\sigma_z\right]=0\]

which yields the spectrum function for the canonical modes $\alpha,\beta$

\[\lambda_{1,2}= \sqrt{\epsilon^2-|\Delta|^2}\]

Further, one may write down the Bogoliubov transformation

\[\begin{align} \alpha & = \sqrt{\frac{\epsilon_0-\lambda}{2\epsilon_0}} a +e^{-\theta}\sqrt{\frac{\lambda+\epsilon_0}{2\epsilon_0}}b^\dagger \\ \beta & = e^{i\theta}\sqrt{\frac{\lambda+\epsilon_0}{2\epsilon_0}} a + \sqrt{\frac{\epsilon_0-\lambda}{2\epsilon_0}}b^\dagger \end{align}\]

where $\Delta=\vert\Delta\vert e^{i\theta}$. As matter of fact, the case we often meet is that $a_{k},a_{-k}^\dagger$ in the original Hamiltonian, in which, we have

\[\begin{align} \alpha_k & = \sqrt{\frac{\epsilon_0-\lambda}{2\epsilon_0}} a_k +e^{-\theta}\sqrt{\frac{\lambda+\epsilon_0}{2\epsilon_0}}a_{-k}^\dagger \\ \alpha^\dagger_{-k} & = e^{i\theta}\sqrt{\frac{\lambda+\epsilon_0}{2\epsilon_0}} a_{k} + \sqrt{\frac{\epsilon_0-\lambda}{2\epsilon_0}}a_{-k}^\dagger \end{align}\]

==Remark==

For a quantum field theory, the diagnolization procedures have some sublattices compared with the case above. For example, a scalar field with Hamiltonian

\[\mathcal L = \frac{1}{2}\partial_0 \phi \partial _0\phi-\frac{1}{2}\nabla\phi\nabla\phi\]

has a quadratic time derivative. However, in such a system, a unitary transformation cannot diagonalize it. Of course, one may introduce canonical creation and annihilation operators with the time derivative of order 1. We first derive the canonical quantization relation

\[\pi =\frac{\delta\mathcal L}{\delta \partial_0\phi}= \partial_0\phi\]

which the Hamiltonian

\[H=\pi \partial_0\phi-L=\frac{1}{2}(\pi^2+\nabla\phi\nabla\phi)\]

We introduce the raising and lowering operators

\[\begin{align} a^\dagger &=\frac{1}{\sqrt{2|\mathbf k|}}(\pi+i|\mathbf k|\phi)\\ a &=\frac{1}{\sqrt{2|\mathbf k|}}(\pi-i|\mathbf k|\phi) \end{align}\]

which gives the Hamiltonian

\[H=|\mathbf k|a^\dagger(\mathbf k)a(\mathbf k)+\frac{1}{2}|\mathbf k|\]

Another example is the QED for which one has to introduce

\[A_i(k)= \frac{\xi_i(k)}{\sqrt{\mathbf k^2}}(a_\mathbf k^\dagger+ia_\mathbf k)\]

with $\xi_i(\mathbf k)$ being the polarized vector.

Basic Fermion

The case for a system of fermions are much simple and we can just apply the unitary transformation which perserves the anticommunication relation. Here we present the transformations. Consider a BdG Hamiltonian (Please double check the sign before the order parameter!!!)

\[\begin{align} \mathcal{H}_{\mathrm{MF}} &=\sum_{k, \alpha} \xi_{k} c_{k, \alpha}^{\dagger} c_{k, \alpha}-\sum_{k}^{\prime}\left[\Delta e^{-i \varphi} c_{k, \uparrow} c_{-k, \downarrow}+\Delta e^{i \varphi} c_{-k, \downarrow}^{\dagger} c_{k, \uparrow}^{\dagger}\right]+\frac{V_{\mathrm{vol}} \Delta^{2}}{g} \\ &=\sum_{k}\left[c_{k, \uparrow}^{\dagger}, c_{-k, \downarrow}\right]\left[\begin{array}{cc} \xi_{k} & \Delta e^{i \varphi} \\ \Delta e^{-i \varphi} & -\xi_{k} \end{array}\right]\left[\begin{array}{c} c_{k, \uparrow} \\ c_{-k, \downarrow}^{\dagger} \end{array}\right]+\frac{V_{\mathrm{vol}} \Delta^{2}}{g} \end{align}\]

Then

\[E_{k}=\sqrt{\xi_{k}^{2}+\Delta^{2}}, \quad u_{k}=\sqrt{\frac{1}{2}\left(1+\frac{\xi_{k}}{E_{k}}\right)}, \quad v_{k}=\sqrt{\frac{1}{2}\left(1-\frac{\xi_{k}}{E_{k}}\right)}\]

with

\[\begin{align} &{\left[\begin{array}{c} \gamma_{\boldsymbol{k}, \uparrow} \\ \gamma_{-\boldsymbol{k}, \downarrow}^{\dagger} \end{array}\right]=\left[\begin{array}{cc} u_{\boldsymbol{k}} & -v_{\boldsymbol{k}} e^{i \varphi} \\ v_{\boldsymbol{k}} e^{-i \varphi} & u_{\boldsymbol{k}} \end{array}\right]\left[\begin{array}{c} c_{\boldsymbol{k}, \uparrow} \\ c_{-\boldsymbol{k}, \downarrow}^{\dagger} \end{array}\right],} \\ &{\left[\begin{array}{c} c_{\boldsymbol{k}, \uparrow} \\ c_{-\boldsymbol{k}, \downarrow}^{\dagger} \end{array}\right]=\left[\begin{array}{cc} u_{\boldsymbol{k}} & v_{\boldsymbol{k}} e^{i \varphi} \\ -v_{\boldsymbol{k}} e^{-i \varphi} & u_{\boldsymbol{k}} \end{array}\right]\left[\begin{array}{c} \gamma_{\boldsymbol{k}, \uparrow} \\ \gamma_{-\boldsymbol{k}, \downarrow}^{\dagger} \end{array}\right] .} \end{align}\]

Then we obtain the diagonalized Hamiltonian

\[\mathcal H_\mathrm{MF} = \sum_k (E_k\gamma_{k,\uparrow}^\dagger\gamma_{k,\uparrow }-E_k\gamma_{-k,\downarrow}\gamma_{-k,\downarrow}^\dagger)+\frac{V_\mathrm{vol}\Delta^2}{g}\]

and the according BCS can be constructed as

\[|\mathrm{BCS}\rangle = \prod_k\gamma_{k,\uparrow}\gamma_{k\downarrow}|\Omega\rangle\sim\prod_{k}(u_k+v_k c_{k,\uparrow}^\dagger c_{-k\downarrow}^\dagger)|\Omega\rangle\]

with $\vert\Omega\rangle$ being the vacuum i.g. a Fermi sea with a Fermi surface. We can expect it for any $k,\sigma$, with

\[\gamma_{k,\sigma}|\mathrm{BCS}\rangle =0\]

With the BCS ground state, one can evaluate the expectation value

\[\begin{align} & \langle c^\dagger_{k,\uparrow}c_{k,\uparrow}\rangle=\langle c^\dagger_{k,\downarrow}c_{k,\downarrow} \rangle = \upsilon_k^2 \\ & \langle c^\dagger_{k,\uparrow}c^\dagger_{-k,\downarrow}\rangle = u_k v_k \end{align}\]

Matsubara Summation

Pair Correlation function $\chi_{n,\mathbf q}^c$

A building block enters the calculation of the Cooper pair propagator

\[\begin{align} \chi_{n, \mathbf{q}}^{\mathrm{c}} \equiv-\frac{T}{L^{d}} \sum_{m, \mathbf{p}} G_{0}\left(\mathbf{p}, i \omega_{m}\right) G_{0}\left(-\mathbf{p}+\mathbf{q},-i \omega_{m}+i \omega_{n}\right) \\ =\frac{1}{L^{d}} \sum_{\mathbf{p}} \frac{1-n_{\mathrm{F}}\left(\xi_{\mathbf{p}}\right)-n_{\mathrm{F}}\left(\xi_{-\mathbf{p}+\mathbf{q}}\right)}{i \omega_{n}-\xi_{\mathbf{p}}-\xi_{-\mathbf{p}+\mathbf{q}}} \end{align}\]

where (around the transition point)

\[G_0(\mathbf p,i\omega_m)=\frac{1}{i\omega_m-\xi_\mathbf p}\]

and $\omega_m = (2m+1)\pi T$ are fermionic Matsubara frequencies, while $\omega_n =2\pi nT$ is a bosonic Matsubara frequency.

Density-Density response function $\chi^d_{\mathbf q,\omega_n}$

\[\begin{align} & \chi_{\mathbf{q}, \omega_{n}}^{\mathrm{d}} \equiv-\frac{T}{L^{d}} \sum_{\mathbf{p}, \omega_{m}} G_{0}\left(\mathbf{p}, i \omega_{m}\right) G_{0}\left(\mathbf{p}+\mathbf{q}, i \omega_{m}+i \omega_{n}\right) \\ = & -\frac{1}{L^{d}} \sum_{\mathbf{p}} \frac{n_{\mathrm{F}}\left(\xi_{\mathbf{p}}\right)-n_{\mathrm{F}}\left(\xi_{\mathbf{p}+\mathbf{q}}\right)}{i \omega_{n}+\xi_{\mathbf{p}}-\xi_{\mathbf{p}+\mathbf{q}}} \end{align}\]

In real calculation, we often take the static limit. That is

\[(\mathbf q,\omega_n)\rightarrow (\mathbf q,0)\]

It is convenient to change from an explicit matrix representation of the Gorkov Green function to an expression in terms of the Pauli matrices

\[\mathcal G_{0,p}=\frac{1}{i\sigma_0\omega_n-\sigma_3 \xi_p+\sigma_1\Delta_0}=\frac{-i\sigma_0\omega_n-\sigma_3\xi_p+\sigma_1\Delta_0}{\omega_n^2+\xi_p^2+\Delta_0^2}\]

where the Gorkov Green function takes the form as

\[\mathcal G^{-1}=\begin{pmatrix}-\partial_\tau-i\phi-\frac{1}{2m}(-i\nabla-\mathbf A)^2+\mu & \Delta_0e^{2i\theta} \\ \Delta_0e^{-2i\theta} & -\partial_\tau +i\phi +\frac{1}{2m}(-i\nabla-\mathbf A)-\mu \end{pmatrix}\]

On the other hand we have the expansion

\[\begin{align} \mathcal G^{-1} & =-\sigma_0\partial_\tau-\sigma_3 (i\phi+\frac{1}{2m}(-i\nabla-\sigma_3\mathbf A)^2-\mu)+\sigma_1\Delta_0 \\ & =-\sigma_0\partial_\tau -\sigma_3 (-\frac{\nabla^2}{2m}-\mu)+\sigma_1\Delta_0-i\sigma_3\phi+\frac{i}{2m}\sigma_0[\nabla,\mathbf A]_+-\sigma_3\frac{1}{2m}\mathbf A^2 \end{align}\]

Gorkov Green function
\[\begin{align} G_{0,p} & =\frac{u_k^2}{\omega - \epsilon_k}+\frac{v_k^2}{\omega+\epsilon_k}\\ F_{0,p} & =u_kv_k\frac{2\omega}{\omega^2+\epsilon_k^2} \end{align}\]

Conventions for Fourier transformation

Fermionic operator (N is the lattice number)
\[\begin{align} c(r)&=\frac{1}{\sqrt{N}}\sum_{k}c_{k}e^{ik\cdot r}\\ c(k)&=\frac{1}{\sqrt{N}}\sum_{r}c(r)e^{-ik\cdot r} \end{align}\]
Order parameter formed by two Fermion operator

For the superconducting order parameter $\Delta(r)=c_\uparrow(r)c_\downarrow(r)$
\[\Delta (k) = \sum_r \Delta(r)e^{ik\cdot r}\]
Function
\[\begin{align} f(r) &=\sum_{k}f(k)e^{ik\cdot r} \\ f(k) &=\frac{1}{N}\sum_{r}f(r)e^{-ik\cdot r} \end{align}\]

Quantum geometry: a view from the adiabatic theorem

2023-11-20T00:00:00+00:00

Riemann structure of Hilbert space
Quantum geometry tensor and adiabatic theorem

Riemann structure of Hilbert space

Given a parameter dependent Hamiltonian $H(\lambda_{a})$ with $\lambda_{a},a=1,2,\cdots$, we have a family of grand states

\[H(\lambda)\vert\psi(\lambda)\rangle=E(\lambda)\vert\psi(\lambda)\rangle.\]

The manifold is formed by the parameter-dependent ground wave functions. In their seminal paper, the authors introduced a Riemannian differential structure in the Hilbert spacei ¹. Specifically, the Riemann structure is embedded in the Quantum Geometry Tensor (QGT) denoted as $\mathfrak{G}$. This tensor can be expressed using the following formula:

\[\mathfrak{G}_{ab}(\lambda)=\langle\partial_{a}\psi(\lambda)\vert\partial_{b}\psi(\lambda)\vert-\langle\partial_{a}\psi(\lambda)\vert\psi(\lambda)\rangle\langle\psi(\lambda)\vert\partial_{b}\psi(\lambda)\rangle.\]

The Quantum Geometry Tensor captures the structural information of the Hilbert space. It consists of two components: the quantum metric, denoted as $\mathcal{G}$, and the Berry curvature, denoted as $\mathcal{B}$. The quantum metric corresponds to the real part of the QGT, while the Berry curvature corresponds to its imaginary part. These quantities provide essential geometric properties that describe the quantum system under consideration.

\[\mathfrak{G}=\mathcal{G}-i\mathcal{B}/2\]

In differential geometry, the curvature of a manifold provides information about its intrinsic geometric properties, such as how the manifold curves and bends. It is related to the concept of parallel transport and describes how vectors change as they are transported along different paths on the manifold. On the other hand, the metric of a manifold determines the notion of distance and angle between points. It defines the geometric properties of length and angle measurement on the manifold. The Riemannian structure asserts that the curvature and metric of a manifold are intimately related. Specifically, the metric tensor determines the curvature tensor, which encodes the curvature properties of the manifold. The compatibility between the curvature and metric is captured by the Riemann curvature tensor, which provides a systematic way to measure and quantify the curvature of the manifold. This relationship between curvature and metric is a fundamental aspect of Riemannian geometry. Further insights and explanations regarding the compatibility between curvature and metric can be found in subsequent blog posts on the topic.

Quantum geometry tensor and adiabatic theorem

In quantum mechanics, the Berry curvature is closely tied to the Berry phase and can be understood within the framework of the adiabatic approximation.

Consider a time-dependent Hamiltonian $H(\lambda)$, which relies on a set of varying parameters. Our goal is to solve the time-dependent Schrödinger equation:

\[i\frac{\partial}{\partial t}\vert\Psi(t)\rangle = H(\lambda)\vert\Psi(t)\rangle.\]

In general, solving this equation is challenging compared to the case of a static equation where we can separate time using a dynamical phase. However, the adiabatic theorem comes to our aid. It states that if there exists a significant energy gap between the non-degenerate ground state and the excited states, we can approximately treat the time and spatial coordinates as separate entities. According to the adiabatic approximation, we make an assumption, known as an ansatz, that the ground state only acquires phase factors without undergoing transitions to the excited states.

Thus, we express the wave function in the adiabatic approximation as:

\[\vert\Psi(\lambda(t))\rangle = e^{-i\gamma(t)}e^{-i\int^{t}E(\lambda)}\vert\psi(\lambda)\rangle,\]

where $\vert\psi(\lambda)\rangle$ represents the simultaneous eigenstate of the Hamiltonian $H(\lambda)$:

\[H(\lambda)\vert\psi(\lambda)\rangle = E(\lambda)\vert\psi(\lambda)\rangle.\]

The first phase factor, $e^{-i\gamma(t)}$, corresponds to the well-known Berry phase. It can be related to the Berry phase by considering the phase factor accumulated along a closed path $\Omega$ in the Hilbert space:

\[e^{-i\gamma_{\Omega}(t)}e^{-i\int_{\Omega}A\cdot d\lambda} = \prod_{\lambda_{i}}\frac{\langle\psi(\lambda(t_{i}))\vert\psi(\lambda(t_{i+1})\rangle}{\vert\langle\psi(\lambda(t_{i}))\vert\psi(\lambda(t_{i+1})\rangle\vert}, \label{eq:Berry}\]

where $\lambda_{i}$ represents the discrete points along the closed path. When no phase is accumulated along a path, it appears featureless. In the definition of the Berry phase, we introduce a normalization factor, specifically the absolute value of the inner product between the states $\langle\psi(\lambda(t_{i}))\vert\psi(\lambda(t_{i+1})\rangle$, as shown in Eq.$~(\ref{eq:Berry})$. Consequently, the adiabatic theorem captures only the curvature structure. To obtain the complete Riemannian structure, we need to go beyond the adiabatic approximation.

The leading correction to the adiabatic approximation takes into account the rate at which the particle transitions to excited states with a certain probability.

\[\begin{align} P & =1-\vert\langle\psi(\lambda(t_{i}))\vert\psi(\lambda(t_{i+1})\vert^{2}=1-2\langle\psi(\lambda(t_{i}))\vert\frac{d}{dt}\vert\psi(\lambda(t_{i+1}))\rangle dt, \end{align}\]

which is nothing more than the quantum metric (or the Bures distance which measures the distance between two quantum states in the sense of the time evolution). Accordingly, the ground state ansatz shall be modified

\[\vert\Psi(\lambda)\rangle=e^{-\mathbb{g}(t)-i\gamma(t)}e^{-i\int^{t}E(\lambda)}\vert\psi(\lambda)\rangle\]

For a closed path, we have

\[\begin{align} e^{-\mathbb{g}_{\Omega}-i\gamma_{\Omega}} & =e^{-\int\vert\langle\lambda(t_{i})\vert\frac{d}{dt}\vert\lambda(t_{i})\rangle\vert dt-i\int_{\Omega}A\cdot d\lambda}\\ & =\prod_{\lambda_{i}}\langle\psi(\lambda_{i})\vert\psi(\lambda_{i+1})\rangle\\ & =\prod_{\lambda_{i}}\frac{\langle\psi(\lambda_{i})\vert\psi(\lambda_{i+1})\rangle}{\vert\langle\psi(\lambda_{i})\vert\psi(\lambda_{i+1})\rangle\vert}\vert\langle\psi(\lambda_{i})\vert\psi(\lambda_{i+1})\rangle\vert \end{align}\]

where $\lambda_i\equiv \lambda(t_i)$. Currently, it is evident that the amplitude of the term $e^{-\mathbb{g}{\Omega}-i\gamma{\Omega}}$ can decay over time or along a given path. The Berry phase, denoted as $e^{-i\gamma_{\Omega}(t)}$, relies on the path’s homotopy. Conversely, the term $e^{-\mathbb{g}{\Omega}}$, which corresponds to the quantum metric, is sensitive to the specific details of the path. For instance, when the particle traverses the same path multiple times, say $n$ times, the probability of it remaining in the original state becomes $P=e^{-n \mathbb{g}_{\Omega}}$.

Locally, we can expand the distance for a infinitesimal change of the parameters

\[D=1-\vert\langle\psi(\lambda)\vert\psi(\lambda+d\lambda)\rangle\vert^{2}=\mathcal{G}_{ab}d\lambda_{a}d\lambda_{b}+\mathcal{O}((d\lambda)^{3})\]

with

\[\mathcal{G}_{ab}=\mathrm{Re}[\langle\partial_{a}\psi(\lambda)\vert\partial_{b}\psi(\lambda)\vert-\langle\partial_{a}\psi(\lambda)\vert\psi(\lambda)\rangle\langle\psi(\lambda)\vert\partial_{b}\psi(\lambda)\rangle].\]

Riemannian structure on manifolds of quantum states, by J. P. Provost and G. Vallee ↩

Peierls Substitution

2023-11-16T00:00:00+00:00

Peierls substitution

In condensed matter, we need to solve the Shroedinger equation in the presence of a periodic potential. One may start with a continuous model

\[H=-\frac{\nabla^{2}}{2m}+V(\mathbf{r}) \label{eq:Seq}\]

with a periodic potential $V(\mathbf{r}+\mathbf{a}_{i})=V(\mathbf{r})$. The periodic potential defines a lattice $\mathbf{R}\in\{\sum_{i}n_{i}\mathbf{a}_{i}\}$. The famous Bloch theorem tells us that a general solution

\[H\psi_{\mathbf{k}}(\mathbf{r})=E(\mathbf{k})\psi_{\mathbf{k}}(\mathbf{r})\]

to the Schroedinger equation in Eq. $(\ref{eq:Seq})$ can be written in the form as

\[\begin{align} \psi_{\mathbf{k}}(\mathbf{r}) & =e^{i\mathbf{k}\cdot\mathbf{r}}u_{\mathbf{k}}(\mathbf{r}), \end{align}\]

where $N$ is the number of unit cells and $\mathbf{k}$ is the crystal momentum related to the lattice. To consider the minimal coupling to the gauge field, it is better to separate the crystal momentum dependence from $u_{\mathbf{k}}(\mathbf{r})$.It can be achieved by by the Wannier function $W_{\mathbf{R}}(\mathbf{r})$ that appears as the Fourier transformation of the periodic part of the Bloch wave,

\[\psi_{\mathbf{k}}(\mathbf{r})=\frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{i\mathbf{k}\cdot\mathbf{R}}W_{\mathbf{R}}(\mathbf{r}).\label{eq:Wann}\]

Then we can calculate the eigenvalue $E(\mathbf{k})$ , by calculating the matrix element

\[\begin{align} E(\mathbf{k}) & =\int d\mathbf{r}\psi_{\mathbf{k}}^{*}(\mathbf{r})H\psi_{\mathbf{k}}(\mathbf{r})=\int d\mathbf{r}\frac{1}{N}\sum_{\mathbf{R}^{\prime}\mathbf{R}}e^{i\mathbf{k}\cdot(\mathbf{R}^{\prime}-\mathbf{R})}W_{\mathbf{R}}^{*}(\mathbf{r})HW_{\mathbf{R}^{\prime}}(\mathbf{r})\\ & =\frac{1}{N}\sum_{\mathbf{R}^{\prime}\mathbf{R}}e^{i\mathbf{k}\cdot(\mathbf{R}^{\prime}-\mathbf{R})}t(\mathbf{R}^{\prime}-\mathbf{R}) \end{align}\]

with $t(\mathbf{R}^{\prime}-\mathbf{R})$ is the hopping integral

\[\begin{align} t(\mathbf{R}^{\prime}-\mathbf{R}) & =\int d\mathbf{r}W_{\mathbf{R}}^{*}(\mathbf{r})HW_{\mathbf{R}^{\prime}}(\mathbf{r}). \end{align}\]

Therefore, we can have a tight-binding model corresponding to the continuous model in Eq.~$(\ref{eq:Seq})$

\[\begin{align} H_{\mathrm{tb}} & =\frac{1}{N}\sum_{\mathbf{R}^{\prime}\mathbf{R}}e^{i\mathbf{k}\cdot(\mathbf{R}^{\prime}-\mathbf{R})}t(\mathbf{R}^{\prime}-\mathbf{R})c_{\mathbf{R}}^{\dagger}c_{\mathbf{R}^{\prime}}\\ & =\sum_{\mathbf{k}}E(\mathbf{k})c_{\mathbf{k}}^{\dagger}c_{\mathbf{k}}. \end{align}\]

One of the most striking features of Eq.$~(\ref{eq:Wann})$ is the independence of Wannier function on the crystal momentum. With the preparation above, we can proceed to consider a minimal coupling to a gauge field, with the Hamiltonian

\[H(\mathbf{A})=\frac{(-i\mathbf{\nabla-\mathbf{A}}(\mathbf{r},t))^{2}}{2m}+V(\mathbf{r}).\]

Our aim is to obtain the guiding principles directing minimal coupling in the tight-binding model to the gauge field. That is, we want to express physical quantities in terms of the crystal momentum. We can consider the modified Wannier functions by the gauging principle for a continuous model

\[W_{\mathbf{R}}(\mathbf{r},\mathbf{A})=\exp\left(i\int_{\mathbf{R}}^{\mathbf{r}}\mathbf{A}(\mathbf{r}^{\prime},t)\right)W_{\mathbf{R}}(\mathbf{r}).\]

Obviously, $W_{\mathbf{R}}(\mathbf{r})=W_{\mathbf{R}}(\mathbf{r},\mathbf{A}\rightarrow0)$. We then have a new solution to $H(\mathbf{A})$

\[\psi_{\mathbf{k}}(\mathbf{r},\mathbf{A})=\frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{i\mathbf{k}\cdot\mathbf{R}}W_{\mathbf{R}}(\mathbf{r},\mathbf{A})\]

We can check this by

\[\begin{align} & H(\mathbf{A})\psi_{\mathbf{k}}(\mathbf{r},\mathbf{A}) \\ =& \left[\frac{(-i\mathbf{\nabla-\mathbf{A}}(\mathbf{r},t))^{2}}{2m}+V(\mathbf{r})\right]\frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{i\mathbf{k}\cdot\mathbf{R}}W_{\mathbf{R}}(\mathbf{r},\mathbf{A})\\ =& \frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{i\mathbf{k}\cdot\mathbf{R}}\left[\frac{(-i\mathbf{\nabla-\mathbf{A}}(\mathbf{r},t))^{2}}{2m}+V(\mathbf{r})\right]\left[\exp\left(i\int_{\mathbf{R}}^{\mathbf{r}}\mathbf{A}(\mathbf{r}^{\prime},t)\right)W_{\mathbf{R}}(\mathbf{r})\right]\\ =& \frac{1}{\sqrt{N}}\sum_{\mathbf{R}}\exp\left(i\int_{\mathbf{R}}^{\mathbf{r}}\mathbf{A}(\mathbf{r}^{\prime},t)\right)e^{i\mathbf{k}\cdot\mathbf{R}}\left[\frac{(-i\mathbf{\nabla-\mathbf{A}}(\mathbf{r},t)+\mathbf{A}(\mathbf{r},t))^{2}}{2m}+V(\mathbf{r})\right]W_{\mathbf{R}}(\mathbf{r})\\ =& \frac{1}{\sqrt{N}}\sum_{\mathbf{R}}\exp\left(i\int_{\mathbf{R}}^{\mathbf{r}}\mathbf{A}(\mathbf{r}^{\prime},t)\right)e^{i\mathbf{k}\cdot\mathbf{R}}\left[\frac{(-i\nabla)^{2}}{2m}+V(\mathbf{r})\right]W_{\mathbf{R}}(\mathbf{r}) \end{align}\]

When calculating the hopping integral, we have

\[\begin{align} t_{\mathbf R\mathbf R^{\prime}}(\mathbf{A}) & =\int d\mathbf{r}W_{\mathbf{R}}^{*}(\mathbf{r},\mathbf{A})HW_{\mathbf{R}^{\prime}}(\mathbf{r},\mathbf{A})\\ & =\exp\left[i\int_{\mathbf{R}^{\prime}}^{\mathbf{R}}\mathbf{A}\cdot d\mathbf{r}^{\prime}\right]\int d\mathbf{r}e^{i\Phi_{R^{\prime}rR}}HW_{\mathbf{R}^{\prime}}(\mathbf{r}) \end{align}\]

where $\Phi_{R^{\prime}rR}=\oint_{R^{\prime}\rightarrow r\rightarrow R\rightarrow R^{\prime}}\mathbf{A}(\mathbf{r}^{\prime})\cdot d\mathbf{r}^{\prime}$, is the flux through the triangular made by three position arguments.Since the $\mathbf{A}$ is approximately uniform at the lattice scale,— the scale at which the Wannier states are localized to the positions $\mathbf{R}$, we can than approximate $\Phi_{R^{\prime}rR}=0$, yielding the desired result.

\[t_{\mathbf R\mathbf R^{\prime}}(\mathbf{A})=\exp\left[i\int_{\mathbf{R}^{\prime}}^{\mathbf{R}}\mathbf{A}\cdot d\mathbf{r}^{\prime}\right]t(\mathbf{R}^{\prime}-\mathbf{R})\]

Therefore, the matrix elements are the same as in the case without magnetic field, apart from the extra phase factor picked up, which is denoted by the Peierls phase factor.

One can generalize the Peierls substitution in a multi-band system.

Wanner waves, tight-binding model and quantum geometry

2023-11-13T00:00:00+00:00

Wanner waves, tight-binding model, and quantum geometry

Wanner waves, tight-binding model, and quantum geometry

Wannier waves

We begin with the single-particle Schroedinger equation in $d$ spatial dimensions

\[H\vert\psi\rangle=\left[-\frac{(\hbar\nabla)^{2}}{2m}+V(\mathbf{r})\right]\vert\psi\rangle, \label{eq:Hcont}\]

where $V(\mathbf{r}+\mathbf{a}_{i})=V(\mathbf{r})$ represents a periodic potential, and $\mathbf{a}_{i}$ ($i=1,\cdots,d$) defines a lattice system. According to the Bloch theorem, the solutions, known as Bloch waves, for an energy band $n$ can be expressed as:

\[\psi_{n\mathbf{k}}(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}u_{n\mathbf{k}}(\mathbf{r}),\]

where $u_{n\mathbf{k}}(\mathbf{r})$ is a periodic Bloch function satisfying $u_{n\mathbf{k}}(\mathbf{r})=u_{n\mathbf{k}}(\mathbf{r}+\mathbf{a}_{i})$, and $\mathbf{k}$ is the Bloch wavevector. The normalization condition for $u_{n\mathbf{k}}(\mathbf{r})$ is given by:

\[\int_{\text{u.c.}}d^{d}\mathbf{r} \vert u_{n\mathbf{k}}(\mathbf{r})\vert^{2}=1,\]

where the integral is taken over one unit cell. Here, u.c. represents the unit cell with volume $\mathcal{A}_\mathrm{uc}$. The energy $\epsilon_n(\mathbf{k})$ satisfies periodicity with respect to the reciprocal lattice vectors $\mathbf{G}_i$, given by the condition $\mathbf{a}_i \cdot \mathbf{G}_j = 2\pi \delta_{ij}$. In other words, the energy is invariant under translations by the reciprocal lattice vectors.

We consider composite bands labeled by the band index $n$ in a specific subset $\mathcal{V}$, which is separated from other bands by sufficiently large band gaps. In this case, we can construct a set of Wannier basis states $\{\vert\mathbf{r}_i\alpha\rangle\}$ that span the same sub-Hilbert space as the Bloch waves corresponding to the bands with indices $n\in\mathcal{V}$. The Wannier basis states can be expressed as follows:

\[\begin{align} \vert\mathbf{r}_i\alpha\rangle &= \frac{\mathcal{A}_\mathrm{uc}}{(2\pi)^d}\int_\mathrm{BZ} d^d\mathbf{k}\, e^{i\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}_i)} \sum_{n\in\mathcal{V}} (\mathcal{U}_\mathbf{k})_{n,\alpha} \vert u_{n\mathbf{k}}\rangle, \label{eq:wannier_Bloch1} \\ \vert u_{n\mathbf{k}}\rangle &= \sum_{\mathbf{r}_i}\sum_\alpha e^{-i\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}_i)} (\mathcal{U}_\mathbf{k}^\dagger)_{\alpha,n} \vert\mathbf{r}_i\alpha\rangle. \label{eq:wannier_Bloch2} \end{align}\]

Here, $\mathcal{A}_\mathrm{uc}$ is the volume of the unit cell, and $\mathbf{r}_i$ represents a lattice site spanned by the lattice vectors $\mathbf{a}_i$ $(i=1,\cdots,d)$. The integration over momentum is performed over the first Brillouin zone (BZ). The unitary matrix $\mathcal{U}_\mathbf{k}$ is chosen to optimize the localization of the Wannier functions. The Wannier function $\langle\mathbf{r}\vert\mathbf{r}_i\alpha\rangle \equiv w_\alpha(\mathbf{r}-\mathbf{r}_i)$ is localized around the lattice site $\mathbf{r}_i$. It turns out to be the Fourier transformation of the corresponding Bloch wave and thus inherits the orthonormality properties of the Bloch functions.

Quantum metric: lower bound of quadratic spread

The unitary matrix $\mathcal{U}_{\mathbf{k}}$ is chosen to maximize the localization of Wannier functions by minimizing a localization functional, as introduced by Marzari and Vanderbilt in their seminal work ¹. The localization functional is given by

\[\begin{align} F & =\sum_{\alpha\in\mathcal{V}}\left[\langle\mathbf{0}\alpha\vert r^{2}\vert\mathbf{0}\alpha\rangle-\vert\langle\mathbf{0}\alpha\vert\mathbf{r}\vert\mathbf{0}\alpha\rangle\vert^{2}\right] =F_{I}+\delta F~. \end{align}\]

Both parts, $F_I$ and $\delta F$, are non-negative, where

\[\begin{align} F_{I} & =\sum_{\alpha\in\mathcal{V}}\left[\langle\mathbf{0}\alpha\vert r^{2}\vert\mathbf{0}\alpha\rangle-\sum_{\mathbf{r}_{i}}\sum_{\beta}\vert\langle\mathbf{r}_{i}\beta\vert\mathbf{r}\vert\mathbf{0}\alpha\rangle\vert^{2}\right],\\ \delta F & =\sum_{\mathbf{r}_{i}(\neq\mathbf{0})}\sum_{\beta(\neq\alpha)}\vert\langle\mathbf{r}_{i}\beta\vert\mathbf{r}\vert\mathbf{0}\alpha\rangle\vert^{2}. \end{align}\]

The optimization of the unitary matrix $\mathcal{U}_{\mathbf{k}}$ aims to minimize the localization functional $F$, leading to the construction of maximally localized Wannier functions. The term $F_I$ is independent of the unitary transformation $\mathcal{U}_{\mathbf{k}}$ and therefore gauge invariant. This allows us to choose $\mathcal{U}_{\mathbf{k}}$ as an identity matrix with components $(\mathcal{U}_{\mathbf{k}})_{\alpha,n}=\delta_{\alpha,n}$ when calculating $F_I$. Then from the relation in Eqs.$~(\ref{eq:wannier_Bloch1})$ and $(\ref{eq:wannier_Bloch2})$, we have

\[\begin{align} \langle u_{n\mathbf{k}}\vert u_{m\mathbf{k+q}}\rangle & =\sum_{\mathbf{r}_{i}}e^{-i\mathbf{k}\cdot\mathbf{r}_{i}}\langle\mathbf{r}_{i}n\vert e^{-i\mathbf{q}\cdot\mathbf{r}}\vert\mathbf{0}m\rangle ,\label{eq:uuq} \end{align}\]

By taking the derivative with respect to $\mathbf{q}$ on both sides of Eq.~$(\ref{eq:uuq})$, we obtain a series of relations in the limit $q\rightarrow 0$. For example, taking the first and second derivatives with respect to $\mathbf{q}$ gives

\[\begin{align} \langle u_{n\mathbf{k}}\vert\nabla_{\mathbf{k}}u_{m\mathbf{k}}\rangle & =-i\sum_{\mathbf{r}_{i}}e^{-i\mathbf{k}\cdot\mathbf{r}_{i}}\langle\mathbf{r}_{i}n\vert\mathbf{r}\vert\mathbf{0}m\rangle ,\\ \langle u_{n\mathbf{k}}\vert\nabla_{\mathbf{k}}^{2}u_{m\mathbf{k}}\rangle & =-\sum_{\mathbf{r}_{i}}e^{-i\mathbf{k}\cdot\mathbf{r}_{i}}\langle\mathbf{r}_{i}n\vert\mathbf{r}^{2}\vert\mathbf{0}m\rangle, \end{align}\]

Similarly, we can establish the converse relations

\[\begin{align} \langle\mathbf{r}_{i}n\vert\mathbf{r}\vert\mathbf{0}m\rangle & =i\frac{\mathcal{A}_{\mathrm{uc}}}{(2\pi)^{d}}\int_\mathrm{BZ} d^{d}\mathbf{k}e^{i\mathbf{k}\cdot\mathbf{r}_{i}}\langle u_{n\mathbf{k}}\vert\nabla_{\mathbf{k}}u_{m\mathbf{k}}\rangle ,\\ \langle\mathbf{r}_{i}n\vert\mathbf{r}^{2}\vert\mathbf{0}m\rangle & =\frac{\mathcal{A}_{\mathrm{uc}}}{(2\pi)^{d}}\int_\mathrm{BZ} d^{d}\mathbf{k}e^{i\mathbf{k}\cdot\mathbf{r}_{i}}\langle\nabla_{\mathbf{k}}u_{n\mathbf{k}}\vert\nabla_{\mathbf{k}}u_{m\mathbf{k}}\rangle. \end{align}\]

Therefore, we can simplify $F_{I}$ as

\[\begin{align} F_{I} & =\sum_{\alpha\in\mathcal{V}}\left[\langle\mathbf{0}\alpha\vert r^{2}\vert\mathbf{0}\alpha\rangle-\sum_{\mathbf{r}_{i}}\sum_{\beta}\vert\langle\mathbf{r}_{i}\beta\vert\mathbf{r}\vert\mathbf{0}\alpha\rangle\vert^{2}\right]\\ & =\frac{\mathcal{A}_{\mathrm{uc}}}{(2\pi)^{d}}\int_\mathrm{BZ}d^{d}\mathbf{k}\sum_{n\in\mathcal{V}}\mathrm{Re}\langle\nabla_{\mathbf{k}}u_{n\mathbf{k}}\vert(\mathbb{I}_{\mathcal{V}}-\vert u_{n\mathbf{k}}\rangle\langle u_{n\mathbf{k}}\vert)\vert\nabla_{\mathbf{k}}u_{n\mathbf{k}}\rangle~, \end{align}\]

where $\mathbb{I}_{\mathcal{V}}$ is the identity operator in the sub-Hilbert space spanned by bands carrying indices in $\mathcal{V}$. This expression clearly shows that $F_{I}$ is expressed in terms of the quantum metric. Since $\delta F\geq 0$, we have the inequality relation,

\[F\geq F_{I}~.\]

Hence, we can conclude that the quantum metric characterizes an obstruction to finding a complete set of exponentially localized Wannier functions. When $F_{I}$ is finite, it indicates that more bands need to be included in the composite bands in order to construct a complete set of exponentially localized Wannier functions.

Second perspective: quantum metric from multi-band models

Another perspective on the quantum metric arises from considering a multiband tight-binding model. Assuming we have already obtained a complete set of exponentially localized Wannier functions constructed from composite bands, we can approximate the continuum Hamiltonian in Eq.$~(\ref{eq:Hcont})$ with a tight-binding model. In the language of second quantization, the continuum model in Eq.$~(\ref{eq:Hcont})$ can be expressed as

\[H=\int d^{d}\mathbf{r}\psi^{\dagger}(\mathbf{r})\left[-\frac{(\hbar\nabla)^{2}}{2m}+V(\mathbf{r})\right]\psi(\mathbf{r})~. \label{eq:2ndH}\]

We then expand the field operator $\psi(\mathbf{r})$ in the basis of Wannier functions

\[\psi(\mathbf{r})=\sum_{\mathbf{r}_{i}}\sum_{\alpha\in\mathcal{V}}w_{\alpha}(\mathbf{r}-\mathbf{r}_{i})a_{i\alpha}+\sum_{\mathbf{r}_{i}}\sum_{\beta\in\mathcal{V}^{\perp}}w_{\beta}^{\perp}(\mathbf{r}-\mathbf{r}_{i})b_{i\beta}~,\]

where $w_{\beta}^{\perp}(\mathbf{r}-\mathbf{r}_{i})$ denotes Wannier functions associated with the complementary band set $\mathcal{V}^{\perp}$. By substituting the expansion into the Hamiltonian in Eq.~$(\ref{eq:2ndH})$, we can derive a tight-binding model defined on the lattice $\{\mathbf{r}_{i}\}$

\[\begin{align} H & =\sum_{\alpha\beta\in\mathcal{V}}\sum_{\mathbf{r}_{i},\mathbf{r}_{j}}\langle\mathbf{r}_{i}\alpha\vert H\vert\mathbf{r}_{j}\beta\rangle a_{i\alpha}^{\dagger}a_{j\beta}^{}+\sum_{\alpha^{\prime},\beta^{\prime}\in\mathcal{V}^{\perp}}\sum_{\mathbf{r}_{i},\mathbf{r}_{j}}\langle\mathbf{r}_{i}\alpha^{\prime}\vert H\vert\mathbf{r}_{j}\beta^{\prime}\rangle b_{i\alpha^{\prime}}^{\dagger}b_{j\beta^{\prime}}^{}\nonumber \\ & =\sum_{\alpha\beta\in\mathcal{V}}\sum_{\mathbf{r}_{i},\mathbf{r}_{j}}t_{ij,\alpha\beta}a_{i\alpha}^{\dagger}a_{j\beta}^{}+\sum_{\alpha^{\prime},\beta^{\prime}\in\mathcal{V}^{\perp}}\sum_{\mathbf{r}_{i},\mathbf{r}_{j}}t_{ij\alpha^{\prime}\beta^{\prime}}^{\perp}b_{i\alpha^{\prime}}^{\dagger}b_{j\beta^{\prime}}^{}~, \label{eq:tbhfull} \end{align}\]

where no mixing term between indices from $\mathcal{V}$ and $\mathcal{V}^{\perp}$. Up to this point, all the derivations have been rigorous, and the expression in Eq.~$(\ref{eq:tbhfull})$ includes all bands. However, since our interest lies solely in the bands belonging to $\mathcal{V}$, we can utilize a complete set of exponentially localized Wannier functions to approximate the Hamiltonian in Eq.$~(\ref{eq:Hcont})$ with a multi-band tight-binding model $H_{\mathrm{tb}}$ by disregarding the $t^\perp$ terms

\[\begin{align} H_{\mathrm{tb}} & =\sum_{\alpha\beta\in\mathcal{V}}\sum_{\mathbf{r}_{i},\mathbf{r}_{j}}t_{ij,\alpha\beta}a_{i\alpha}^{\dagger}a_{j\beta} =\sum_{\alpha\beta\in\mathcal{V}}\sum_{\mathbf{k}}h_{\alpha\beta}(\mathbf{k})a_{\mathbf{k}\alpha}^{\dagger}a_{\mathbf{k}\beta}^, \label{eq:tb} \end{align}\]

where $t_{ij,\alpha\beta}$ exponentially decays with the distance $\vert\mathbf{r}_{i}-\mathbf{r}_{j}\vert$. In Eq.~$(\ref{eq:tb})$, we have further introduced the Fourier transformation $a_{\mathbf{k}\alpha}^{\dagger}=\frac{1}{\sqrt{N}}\sum_{\mathbf{r}_{i}}a_{i\alpha}^{\dagger}e^{i\mathbf{k}\cdot\mathbf{r}_{i}}$, where $N$ represents the total number of lattice sites. In our specific setup, where there is a significant gap between the targeted band and the others, we can project onto the targeted band using the following expressions

\[a_{i\alpha}\rightarrow\frac{1}{\sqrt{N}}\sum_{\mathbf{k}}e^{i\mathbf{k}\cdot\mathbf{r}_{i}}u_{\mathbf{k}}^{*}(\alpha)c_{\mathbf{k}}~,\]

\[\psi({\mathbf r})\rightarrow\frac{1}{\sqrt{N}}\sum_{\mathbf{k}}e^{i\mathbf{k}\cdot\mathbf{r}_{i}}g_{\mathbf{k}}^{*}(\alpha)c_{\mathbf{k}}~,\]

where $u_{\mathbf{k}}$ represents an eigenvector of $h_{\alpha\beta}(\mathbf{k})$, and $c_{\mathbf{k}}$ annihilates an electron in the targeted band. It is important to note that the index $\alpha$ appearing in both $a_{i\alpha}$ and $g_{\mathbf{k}}(\alpha)$ arises from the realization of a multiband tight-binding model, which accounts for the nontrivial quantum metric or Wannier obstruction. This can be inferred from the quantum metric associated with $g_{\mathbf{k}}$.

Nicola Marzari and David Vanderbilt, Maximally localized generalized Wannier functions for composite energy bands, Phys. Rev. B 56, 12847 ↩

Optical theorem

2023-11-04T00:00:00+00:00

Optical Theorem

Optical Theorem

We consider Schroedinger equation with the incident $\vert\psi_{in}\rangle$ and emergent $\vert\psi_{out}\rangle$ states. The scattering matrix is then defined as

\[\vert\psi_{out}\rangle=S\vert\psi_{in}\rangle\]

The incident state $\vert\psi_{in}\rangle=\vert\mathbf{k}\rangle$ is an eigenstate of the free Hamiltonian $H_{0}=-\Delta$ of energy $E=k_{0}^{2}$ that is

\[H_{0}\vert\psi_{in}\rangle=E\vert\psi_{in}\rangle\]

The emergent state $\vert\psi_{out}\rangle$ is an eigenstate of the total Hamiltonian $H=H_{0}+V$ with the same energy, namely

\[H\vert\psi_{out}\rangle=E\vert\psi_{out}\rangle\]

At infinity, the two states $\vert\psi_{in}\rangle$ and $\vert\psi_{out}\rangle$ only differ by the scattered wave, whose amplitude tends to zero. As a consequence, the two states are identical up to a phase difference. We then can obtain the relation

\[VS=(E-H_{0})(S-I)\]

by the observation

\[\begin{align} H\vert\psi_{out}\rangle-H_{0}\vert\psi_{in}\rangle & =E(\vert\psi_{out}\rangle-\vert\psi_{in}\rangle)\\ HS-H_{0} & =E(S-I)\\ VS & =E(S-I)-H_{0}(S-I)=(E-H)(S-I) \end{align}\]

Without interaction $V=0$, $S=1$. Using the resolvent operator $G_{0}$ associated with the free problem $(E-H_{0})G=1$, we can have the Dyson equation

\[S=1+G_{0}VS\]

Projecting this equation on an incident state $\vert\mathbf{k}\rangle$ we have the relation

\[\vert\psi_{out}\rangle=\vert\mathbf{k}\rangle+G_{0}V\vert\psi_{out}\rangle\]

Now we can define the scattering operator

\[T=VS\]

which satisfies the Lippman-Schwinger equation

\[S=1+G_{0}T\]

For the emergent state, we have by taking $\vert\psi_{in}\rangle=\vert\mathbf{k}\rangle$

\[\vert\psi_{out}\rangle=\vert\mathbf{k}\rangle+G_{0}T\vert\mathbf{k}\rangle\]

whose projection on $\vert\mathbf{r}\rangle$ yields

\[\langle\mathbf{r}\vert\psi_{out}\rangle=e^{i\mathbf{k}\cdot\mathbf{r}}+\int d\mathbf{r}^{\prime}\langle\mathbf{r}\vert G_{0}\vert\mathbf{r}^{\prime}\rangle\langle\mathbf{r}^{\prime}\vert T\vert\mathbf{k}\rangle\]

With the asymptotic expansion,

\[\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}+\frac{e^{i\mathbf{k}_{0}\mathbf{r}}}{r}f(\mathbf{k},\mathbf{k}^{\prime})\]

we have

\[f(\mathbf{k},\mathbf{k}^{\prime})=-\frac{1}{4\pi}\langle\mathbf{k}^{\prime}\vert T\vert\mathbf{k}\rangle\]

To derive the optical theorem, we start with the Lippman-Schwinger equation,

\[\langle\mathbf{k}\vert T^{\dagger}\vert\psi_{out}\rangle=\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}\rangle+\langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle\] \[\mathrm{Im}\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}\rangle=-\mathrm{Im}\langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle\]

To proceed, we can ($\epsilon(k)=k^{2}$)

\[\begin{align} \langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle & =\int dk^{\prime}\langle\mathbf{k}\vert T^{\dagger}\vert k^{\prime}\rangle\langle k^{\prime}\vert G_{0}\vert k^{\prime}\rangle\langle k^{\prime}\vert T\vert\mathbf{k}\rangle\\ & =\pi\int dk^{\prime}\delta(E-H_{0}(\mathbf{k}^{\prime}))\langle\mathbf{k}\vert T^{\dagger}\vert k^{\prime}\rangle\langle k^{\prime}\vert G_{0}\vert k^{\prime}\rangle\langle k^{\prime}\vert T\vert\mathbf{k}\rangle\\ & =\frac{\pi}{2k_{0}}\int d\mathbf{k}^{\prime}\vert\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}^{\prime}\rangle\vert^{2}\delta(\mathbf{k}^{\prime}-\mathbf{k}) \end{align}\]

by observing

\[\begin{align} \langle k^{\prime}\vert G_ {0}\vert k^{\prime}\rangle & =\langle k^{\prime}\vert\frac{1}{E-H_{0}(k^{\prime})+i0^{+}}\vert k^{\prime}\rangle\\ & =i\pi\delta(E-H_{0}(k^{\prime}) \end{align}\]

In general, we have

\[\langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle=\pi\left[\frac{d\epsilon(k_{0})}{dk_{0}}\right]^{-1}\int d\mathbf{k}^{\prime}\vert\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}^{\prime}\rangle\vert^{2}\delta(\mathbf{k}^{\prime}-\mathbf{k})\]

With these basic understandings, one may think about the optical theorem in condensed matter systems. Namely, one shall consider a scattering process under the background of a periodic potential.

Bound state and Bloch wave, Wannier function

2023-10-20T00:00:00+00:00

Bound states under a periodic potential

Schroedinger equation and Bloch waves

We first recall the Schroedinger equation with a periodic potential

\[\left[-\frac{\nabla^{2}}{2m}+V(\mathbf{r})\right]\psi(\mathbf{r})=E\psi(\mathbf{r})\label{eq:bloch}\]

Here $V(\mathbf{r})$ is a periodic potential $V(\mathbf{r})=V(\mathbf{r}+\mathbf{a})$. The system obeys the translation symmetry and the Bloch theorem tells that a general solution has the form as

\[\psi_{n\mathbf{k}}(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}u_{n\mathbf{k}}(\mathbf{r})\label{eq:solution}\]

where $u_{n\mathbf{k}}(\mathbf{r})$ is also periodic with $u_{n\mathbf{k}}(\mathbf{r})=u_{n\mathbf{k}}(\mathbf{r}+\mathbf{a})$ and $n$ is the band index. The wave function $u_{n\mathbf{k}}(\mathbf{r})$ satisfies the equation

\[\left[-\frac{1}{2m}\frac{d^2}{dx^2}+ \frac{k^{2}}{2m}+V(\mathbf{r})\right]u_{n\mathbf{k}}(\mathbf{r})=Eu_{n\mathbf{k}}(\mathbf{r})\label{eq:uk}\]

Given a band $n$, we can define the Wannier functions

\[W_{n\mathbf{R}}(\mathbf{r})=\frac{1}{\sqrt{N}}e^{i\mathbf{k}\cdot\mathbf{r}}\sum_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{R}}u_{n\mathbf{k}}(\mathbf{r})\label{eq:Wannier}\]

Here $\mathbf{R}$ is lattice vector and $N$ is the number of primitive cells in the crystals. The Bloch functions can be written in terms of the Wannier functions as follows

\[u_{n\mathbf{k}}(\mathbf{r})=\frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{-i\mathbf{k}\cdot\mathbf{r}}e^{i\mathbf{k}\cdot\mathbf{R}}W_{n\mathbf{R}}(\mathbf{r})\label{eq:unk}\]

The Wannier functions in Eq. (\ref{eq:Wannier}) constructed by the waves $u_{n\mathbf{k}}(\mathbf{r})$ is not unique due to a gauge degree of freedom in $u_{n\mathbf{k}}(\mathbf{r})$. Namely, when a phase change in $u_{n\mathbf{k}}(\mathbf{r})\rightarrow u_{n\mathbf{k}}(\mathbf{r})e^{i\phi_{n\mathbf{k}}}$ in Eq.~(\ref{eq:solution}), $\psi_{n\mathbf{k}}(\mathbf{r})$ stays as an eigenstate to the Schroedinger equation in Eq.(\ref{eq:bloch}).We can find the optimally localized Wannier functions by minimizing the quadratic spread

\[\begin{align} \langle r^{2}\rangle-\langle r\rangle^{2} & \equiv\int_{\mathcal{A}_{uc}}d\mathbf{r}W_{\mathbf{0}}^{*}(\mathbf{r})r^{2}W_{\mathbf{0}}(\mathbf{r})-\left[\int_{\mathcal{A}}d\mathbf{r}W_{\mathbf{0}}^{*}(\mathbf{r})\mathbf{r}W_{\mathbf{0}}(\mathbf{r})\right]^{2}\\ & \geq\frac{\mathcal{A}_{uc}}{(2\pi)^{d}}\int_{\mathrm{BZ}}d\mathbf{k}\mathrm{Tr}g(\mathbf{k}) \end{align}\]

with $\mathcal{A}_{uc}$ denotes a unit cell. Here $g(\mathbf{k})$ is the quantum metric with components

\[g_{ab}(\mathbf{k})=\mathrm{Re}\langle\partial_{a}u_{n\mathbf{k}}\vert\left(1-\vert u_{n\mathbf{k}}\rangle\langle u_{n\mathbf{k}}\vert\right)\vert\partial_{b}u_{n\mathbf{k}}\rangle\]

The average quantum metric appears as the lower bound of the quadratic spread which is gauge-independent.

Bound state at impurity

Here we consider an impurity along with the periodic potential. We focus on one spatial dimensional system The Schroedinger equation is

\[\left[-\frac{1}{2m}\frac{d^{2}}{dx^{2}}+V(x)-V_{0}\delta(x)\right]\psi(x)=E\psi(x)\]

Here $m_{0}$ is the mass of the electron and one should not confuse it with an effective mass of an energy band.

We are interested in a bound state that is trapped by the impurity. To solve the bound state, we can consider the Schroedinger equation in Eq.~(\ref{eq:bloch}) with boundary condition

\[\psi(\infty)=0\]

We assume a wave function ansatz

\[\psi_{\lambda}(x)=e^{-\lambda x}u_{\lambda}(x)\]

It is easy to find that $u_{\lambda}(x)$ satisfies the equation

\[\left[-\frac{\hbar}{2m}\frac{d^2}{dx^2}-\frac{\lambda^{2}}{2m}+V(x)\right]u_{\lambda}(x)=E_{n\lambda}u_{\lambda}(x)\label{eq:enl}\]

which has the same form as in Eq. (\ref{eq:uk}). Therefore, we can apply the Bloch theorem

\[u_{n\lambda}(x)=\frac{1}{\sqrt{N}}\sum_{R}e^{-i\lambda x}e^{i\lambda R}W_{nR}(x)\label{eq:u_l}\]

Here we also introduce the index $n$. We have the one-to-one correspondence between Eqs. (\ref{eq:u_l}) and (\ref{eq:unk}). The eigen energy in Eq. (\ref{eq:enl}) is

\[\begin{align} E_{n\lambda} & =-\frac{\lambda^{2}}{2m}+\sum_{\delta}e^{-i\lambda\delta}t(\delta)\nonumber \\ & =-\frac{\lambda^{2}}{2m}+t(0)+2\mathrm{Re}\left[e^{-i\lambda a}t(a)\right]+2\mathrm{Re}\left[e^{-i2\lambda a}t(2a)\right]+\cdots \end{align}\]

with $\delta=0,\pm a,\pm2a,\cdots$ and

\[t(\delta)=\int_{0}^{a}dxW_{R}(x)V(x)W_{R+\delta}^{*}(x)\]

If the profile of $W_{R}(x)$ exponentially decays, the hopping integral $t(\delta)$ decays exponentially as $\delta$ increases. Noting that $E_{n\lambda=0}>E_{n\lambda>0}$, we can conclude that the bound states have the energy lower than $\sum_{\delta}e^{-i\lambda\delta}t(\delta)$ while a scattering scattering state is larger than $\sum_{\delta}e^{-i\lambda\delta}t(\delta)$ given a band $n$.

For the impurity problem, we can construct the wave function

\[\begin{align} \psi_{n\lambda}(x)= & \begin{cases} \frac{1}{\sqrt{N}}e^{-\lambda x}u_{n\lambda}(x) & x\geq0\\ \frac{1}{\sqrt{N}}Ae^{-\lambda\vert x\vert}u_{n\lambda}(x) & x<0 \end{cases} \end{align}\]

At the $x=0$, we have the condition

\[\begin{align} \psi_{n}(0^{+}) & =\psi_{n}(0^{-})\\ -\frac{1}{2m}\left[\psi^{\prime}(0^{+})-\psi^{\prime}(0^{-})\right] & =V_{0}\psi(0) \end{align}\]

which gives rise to

\[\begin{align} \lambda & =mV_{0}\\ A & =1 \end{align}\]

Therefore we have solution

\[\begin{align} \psi_{n\lambda}(x) & =\frac{1}{\sqrt{N}}e^{-\lambda\vert x\vert}u_{n\lambda}(x)\\ & =\frac{1}{\sqrt{N}}e^{-\lambda\vert x\vert}e^{-i\lambda\cdot x}\sum_{R}e^{i\lambda R}W_{R}(x)\label{eq:psi_n} \end{align}\]

where $N$ is the normalization factor and the factor $e^{-i\lambda\cdot x}$ is to ensure the orthogonality of the wave functions $\psi_{n\lambda}(x)$ of bound states.

The question arises: what Wannier basis should we choose? In fact, we can take the form in Eq. (\ref{eq:psi_n}) as a wave function ansatz given a parameter $\lambda$. The choice on the Wannier functions should minimize the energy of the bound state, or the energy $E_{n0}=\sum_{\delta}e^{-i\lambda\delta}t(\delta)=\sum_{\delta}e^{-i\lambda\delta}\int_{0}^{a}dxW_{R}(x)V(x)W_{R+\delta}^{*}(x)$. We can expect that the more localized the Wanner functions are, the lower energy the ansatz state possesses. As we mention above, the optimal localized Wannier function is related to the quantum metric which is gauge independent.

Furthermore, due the to localized property of Wannier function, the leading contribution to the $\psi_{n\lambda}(x)$ in Eq. (\ref{eq:psi_n}) is

\[\psi_{n\lambda}(x)=\frac{1}{\sqrt{N}}e^{-\lambda\vert x\vert}e^{-i\lambda\cdot x}W_{0}(x)\]

with $\lambda=mV_{0}$.We can consider a process in which we can adiabatically weaken the impurity strength $V_{0}=0$, the bound state will exclusively be controlled by the Wannier function. It is consistent with the property of Wannier functions which can be taken as the bound states that depends on the local profile of the periodic potential.

We emphasize that $m$ here is the bare mass of the electron. An effective mass of the band $n$ can be extracted as

\[\begin{align} \frac{1}{m_{eff}} & =\frac{d^{2}}{dk^{2}}E_{nk}\vert_{k\rightarrow0}=\frac{1}{m}-\delta^{2}\sum_{\delta}t(\delta) \end{align}\]

We can consider the leading orders and for a flat band, we have the constraint

$\frac{1}{m_{eff}}=\frac{1}{m}-a^{2}\left[t(a)+t(-a)\right]=0$ which can be met by tuning the hopping integral $t(\delta)$.

Introduction to quantum geometry

2023-08-28T00:00:00+00:00

Consider a generic parameter-dependent quantum system described by a Hamiltonian $H(k)$ and Schroedinger equation

\[H(x) |\psi_\alpha(x)\rangle = \epsilon_\alpha(x)|\psi_\alpha(x)\rangle\]

With eigenvalue $\epsilon_\alpha(x)$ that forms a band structure in parameter space, and corresponding eigenstates $\vert\psi_\alpha(x)\rangle$. We remark that $x$ indeed represents a general parameter such as momentum, or interaction parameter. As the name geometry indicates, we will define so-called quantum geometry that encodes structural information of the Hilbert spaces $\{\mathcal H_x\}$ when the parameter $x$ varies.

It is often the case that perturbation occurs (in reality or thinking experiments) $x\rightarrow x+\Delta x$. Then the system shows response to the external perturbation and for weak perturbation, the response can be expressed in the unperturbed Hamiltonian $H(x)$. There are two aspects of changes. First, the shift of the energy and we may call it the spectrum physics (I cook the name, not serious). More precisely, the information shall be encoded by the velocity

\[\mathbf v_\alpha(x) = \nabla _x \epsilon_\alpha(x).\]

The second part is the wave function (in parallel, we name it wavefunction physics), and similarly, the information is encoded in $\vert\partial_x \psi_\alpha(x)\rangle$. While the role of the energy bands and band velocities in such quantities is for the most part well known, considerable attention has in recent decades shifted to the eigenstates.

The quantum geometry works to provide a quantitative theory of the information encoded in the geometry $x$ dependence of eigenstates and to describe how this geometry is unveiled in the physical properties of the system of interest.

Quantum geometry tensor

We can regard the Hilbert space as a Riemannian manifold and then we can define the curvature and metric. Physically, we can define the quantum geometry tensor

\[T_{ij}(x) = \langle \partial_i \psi(x)|[1-|\psi(x)\rangle\langle\psi(x)|]|\partial_j\psi(x)\rangle\]

It is related to the Berry curvature and quantum metric by

\[T = g-\frac{i}{2}\Omega\]

The QGT unifies the quantum metric as the real part and the Berry curvature as the imaginary part in a single complex gauge-invariant tensor field.

The origin of quantum geometry

The quantum geometry represents structural information of the Hilbert space. To see its originality, we can consider a variation of parameter for the Hamiltonian

\[H(x+\Delta x) = H(x)+\nabla_xH\cdot\Delta x\]

When the perturbative term does not commute with $H(x)$ with

\[\langle \psi_\alpha|\nabla_xH\cdot\Delta x|\psi_\beta\rangle\neq 0\quad \exist\beta\neq \alpha\]

This means the eigenstate changes accordingly by means of the interband coupling mediated by the gradients of the Hamiltonian. To make this idea more quantitative, it is instructive to study the evolution of a single eigenstate when the parameters are varied by considering

\[\langle \psi(x)|\psi(x+\Delta x)\rangle = \mathcal Fe^{-i\Delta \varphi}\]

With $\mathcal F=\vert \langle \psi(x)\vert \psi(x+\Delta x)\rangle\vert$. Such an overlap is non-zero whenever the states do not form an orthonormal basis in $x$-space. Separating into real and imaginary parts, one will find the expression.

Asymptotic expansion

2023-01-02T00:00:00+00:00

Fundamental concept
- Something basic
- Examples
Laplace Type Integrals
Laplace’s method

The solution of a large class of physically important problems can be represented in terms of definite integrals and further can be expressed in terms of special functions. However, exact solutions fail to tell the true content. In order to decipher the main mathematical and physical features of these solutions, it is useful to study their behavior for large $x$ and $t$. Frequently, one particular class of the integral has a form as

\[I(k)= \int_a^b f(t) \exp(t\phi(k)) dt \quad \mbox{with}\quad k \rightarrow \infty.\]

In this note, we introduce appropriate mathematical techniques for evaluating the behavior of certain integrals containing a large parameter. Historically speaking, the development of these techniques was motivated by concrete physical problems. The main attention is paid to three methods: Laplace’s method, the method of stationary phase, and the steepest descent method.

Fundamental concept

The first step is to make a definition of asymptotic expansion under the condition that the parameter $\epsilon$ is real and small.

Something basic

==Def==

a) The notation
\[f(k)=O(g(k)),\quad k\rightarrow k_0\]
Means that there is a finite constant $M$ and a neighborhood of $k_0$ where $|f|\leq M|g|$. b) THe notation
\[f(k)\leq g(k),\quad k\rightarrow k_0\]
or $f(k)=o(g(k)),\quad k\rightarrow k_0$ means
\[\lim_{k\rightarrow k_0} |\frac{f(k)}{g(k)}|=0\]
c) We shall say that $f(k)$ is an approximation to $I(k)$ valid to order $\delta(k)$ as $k\rightarrow k_0$ if
\[\lim_{k\rightarrow k_0}\frac{I(k)-f(k)}{\delta(k)}=0\]

A typical purpose is to find a sequence of functions $\{\delta_j(k)\}$ such that

\[\delta_{j+1}(k)\ll \delta_j(k),\quad k\rightarrow k_0\]

In the sense of a notation, we claim the following two statements are equivalent.

\[I(k) \sum_{j=1}^m a_j\delta_j(k)+O(\delta_{m+1}(k)),\quad k\rightarrow k_0, m=1,2,\cdots N\]

and

\[I(k) \sim \sum_{j=1}^N a_j\delta_j(k),\quad k\rightarrow k_0\]

with $\sim$ meaning

\[\lim_ {k\rightarrow k_0}\left|\frac{I(k)}{\sum_{j=1}^Na_j\delta_j(k)} \right |=1\]

Examples

The first example given here is to find the value of the integral

\[I(\epsilon) =\int_0^\infty \frac{e^{-t}}{1+\epsilon t}d t,\quad \epsilon >0\]

For a sufficiently small real positive value of $\epsilon$. Since the value at the boundary is the largest, we can successively apply the partial integral. One integration by parts yields,

\[I(\epsilon)= 1- \epsilon\int_0^\infty\frac{e^{-t}}{(1+\epsilon t)^2}\]

Repeating this process $N$ more times yields,

\[I(\epsilon) = 1-\epsilon +2!\epsilon^2+\cdots +(-1)^N!\epsilon^N+(-1)^{N+1}(N+1)!\epsilon^{N+1}\int_0^\infty\frac{e^{-t}}{(1+\epsilon t)^{N+2}}dt\]

Therefore, the aymptotic series are $\{\epsilon^i,i=1,\cdots N\}$, which coincides with the Tayler expansion. The validatity of the expansion requires the error $R_{N+1}$ term vanishes as $o(\epsilon^N)$ as $\epsilon\rightarrow 0$,

\[|R_{N+1}| \leq (N+1)! \epsilon ^{N+1}\int_0^\infty e^{-t}dt =(N+1)!\epsilon^{N+1}\]

As a matter of fact, $R_{N+1}$ diverges with a fixed $\epsilon$ when $N$ goes to $\infty$. As a consequence, we cannot take too many terms in the series. In principle, one can find the ‘optimal’ value of $N$ for fixed $\epsilon$ for which the remainder is smallest.

There are cases where no sophisticated asymptotic methods are needed. The first class involves the evaluation of the integral

\[\int_a^b f(k,t)dt \quad f(k,t)\sim f_0(t), k\rightarrow k_0.\]

The second class is

\[\int_k ^b f(t)dt ,\quad k\rightarrow \infty.\]

To determine the leading behavior of integrals of the first class, we use the following fact. Assume that $f(k,t) \sim f_0(t)$ as $k\rightarrow k_0$ uniformly for $t$ in $[a,b]$, and that $\int_a^b f_0(t)dt$ is finite and nonzero. The limit as $k\rightarrow k_0$ and the integral can be interchanged, hence

\[\int_a^b f(k,t)dt\sim \int_a^b f_0(t)dt,\quad k\rightarrow k_0\]

One example follows as

\[I(k)=\int_0^1 \frac{\sin kt }{t}dt\sim \int_0^1(k-\frac{t^2k^3}{3!}\cdots)=k-\frac{k^3}{3\cdot 3!}.\]

The second class can be exemplified as with $k\rightarrow 0$

\[I(k)=\int_k^\infty e^{-t^2}dt = \int_0^\infty e^{-t^2}dt-\int_0^k e^{-t^2}dt\]

The second term can be asymptotically expanded.

A much nontrivial example is

\[E_1(k)=\int_k^\infty \frac{e^{-t}}{t}dt, \quad k\rightarrow 0^+~.\]

But now the integrand has a logarithmic singularity at $t=0$. If we subtract $\frac{1}{t(t+1)}$, the difficulty is avoided,

\[\begin{align} E_1(k)&= \int_k^\infty \frac{dt}{t(t+1)}+\int_0^\infty(e^{-t}-\frac{1}{t+1})\frac{dt}{t}-\int_0^k (e^{-k}-\frac{1}{t+1})\frac{dt}{t} \notag \\ &= -\gamma -log k +k -\frac{k^2}{4}+\cdots \end{align}\]

Laplace Type Integrals

A Laplace Type Integral is to study the asymptotic behavior as $k\rightarrow +\infty$ of the form

\[I(k)=\int_a^b f(t) e^{-k\phi(t)}dt\]

A special case is a Laplace transform, which is why integrals are referred to as Laplace-type integrals.

Integration by parts

Suppose that $\phi(t)$ is monotonic in $[a,b]$, then one needs to analyze the behavior of the integral near the boundaries. Because integration by parts is based on such an analysis, we expect that it is a useful technique for studying Laplace type integral when $\phi(t)$ is monotonic.

We can look at the first example

\[\begin{align} I(k) & = \int_0^\infty \frac{e^{-kt}}{(1+t^2)^2}dt,\quad k\rightarrow+\infty \notag \\ & = \frac{(1+t^2)^{-2}e^{-kt}}{-k}|_0^\infty +\frac{1}{k}\int_0^\infty (-4t)(1+t^2)^{-3}e^{-kt}dt \notag \\ & = \frac{1}{k}+O(\frac{1}{k^3}) \end{align}\]

The rigorous justification of the integration by parts, in general, follows from the following lemma

==Lemma==

Consider the integral
\[I(k)=\int_a ^b f(t)e^{-kt}dt\]
Where the interval $[a,b]$ is a finite segment of the real axis. Let $f(m)(t)$ Denote the $m$th derivative of $f(t)$. Suppose that $f(t)$ has $N+1$ Continuous derivatives while $f^{(N+2)}(t)$ is piecewise continuous on $a\leq t\leq b$. Then
\[I(k)\sim \sum_{n=0}^N \frac{e^{-ka}}{k^{n+1}} f^{(n)},k\rightarrow +\infty\]

The proof is straightforward.

Watson’s Lemma

If $f(t)$ is not sufficiently smooth at $t=a$, then the integration by parts approach for the asymptotic evaluation may not work. Instead, one may resort to Watson’s lemma.

==Waston’s Lemma==

Consider the integral
\[I(k)=\int_0^b f(t)e^{-kt}dt, \quad b>0\]
Suppose that $f(t)$ is integrable in $(0,b)$ and that it has the asymptotic series expansion
\[f(t)\sim t^\alpha \sum_{n=0}^\infty a_n t^{\beta n},\quad t\rightarrow 0^+,\alpha>-1,\beta>0.\]
Then
\[I(k)\sim \sum_{n=0}^\infty a_n\frac{\Gamma(\alpha+\beta n+1)}{k^{\alpha+\beta n+1}},\quad k\rightarrow \infty\]
Proof We break the integral in two parts, $I_1(k)$ and $I_2(k)$,
\[I_1(k) = \int_0^R f(t)e^{-kt }dt ,\quad I_2(k) = \int_R^b f(t)e^{-kt }dt\]
and $R<b$ is a positive constant. The integral is exponentially small as $k\rightarrow \infty$. For finite $b$, because $f(t)$ is bounded for $t>0$, there exists a positive constant $A$, such that $|f|\leq A$ for $t\geq R$. Thus,
\[|I_2(k)|\leq A\int_R^b e^{-kt}dt = \frac{A}{k}(e^{-kR}-e^{-kr})~.\]
Thus $I_2(k)=O(\frac{e^{-kR}}{k})$. Due to the series, we have
\[I_1(k)=\int_0^R [\sum_{n=0}^R a_nt^{\alpha+\beta n}+ O(t^{\alpha +\beta (N+1)}) ]e^{-kt}dt,\quad k\rightarrow \infty~.\]
However,
\[\begin{align} \int_0^R t^{\alpha+\beta n}e^{-kt }dt & = \int_0^\infty t^{\alpha+\beta n}e^{-kt}dt -\int_R^\infty t^{\alpha+\beta n}e^{-kt}dt \notag \\ &=\frac{\Gamma(\alpha+\beta n +1)}{k^{\alpha+\beta n +1}} +O(\frac{e^{-k R}}{k}) \end{align}\]

Example

The example we present here is

\[I(k) = \int_0^5 \frac{e^{-kt}}{\sqrt{t^2+2t}}dt,\quad k\rightarrow \infty\]

A singularity exists at $t=0$, which forbids the partial integral method. Intuitively, the singularity at $t=0$ makes the dominant contribution. Thus it would be desirable to expand $(t^2+2t)^{-1/2}$ in the neighborhood of the origin. We separate the integral into two parts,

\[I(k) =I_1(k) +I_2(k)= (\int_0^R +\int_R^5 )\frac{e^{-kt}}{\sqrt{t^2+2t}}dt\]

where $R$ is sufficiently small but finite to allow expansion of $\sqrt{t^2+2t}$ around $t=0$,

\[(2t+t^2)^{-1/2}\sim (2t)^{-1/2}-\frac{(2t)^{1/2}}{8}\]

we find

\[I(k) \sim \int _0^R e^{-kt}(2t)^{-1/2}dt -\frac{1}{8}\int_0^R e^{-kt}(2t)^{1/2}dt\]

We intuitively relax the integrating region by replacing $R$ with $\infty$ with an exponentially small error,

\[I(k) \sim \frac{\Gamma(1/2)}{(2k)^{1/2}}-\frac{\Gamma(3/2)}{2(2k)^{3/2}}.\]

The error can be estimated $R_e= \int_R^\infty e^{-kt}t ^\alpha dt = \frac{e^{-kR} R^\alpha}{-k} + \int ^\infty_R\frac{\alpha e^{-kt}t^{\alpha-1}}{-k}dt = O(\frac{e^{-kR}}{k}).$

Laplace’s method

The above discussion is focused on the case with $\phi(t)$ being monotonic in $[a,b]$. Suppose, the local minimum occurs at an interior point $c$ $a<c<b$, and $\phi^\prime(c)=0,\phi^{\prime\prime}(c)>0$. Further, we assume that $\phi^\prime (t)\neq 0$ in $[a,b]$ except at $t=c$ and $f,\phi$ are sufficiently smooth which will be explained below.

==Laplace’s Method==

Consider the integral
\[I(k) = \int _a^b f(t)e^{-k \phi(t)}dt,\quad k\rightarrow +\infty\]
and assume that $\phi^\prime(c)=0$, $\phi^{\prime\prime}(c)>0$ for some point $c$ in the interval $[a,b]$ except at $t=c$, $\phi\in C^4[a,b]$ and $f\in C^2[a,b]$. Then if $c$ is an interior point, we have the leading term as
\[I(k)\sim e^{-k\phi(c)}f(c)\sqrt{\frac{2\pi}{k\phi^{\prime\prime}(c)}},\quad k\rightarrow \infty\]

Fermi liquid theory

2022-08-23T00:00:00+00:00

Introduction to Fermi liquid and Fermi liquid theory
Charge and Current carried by quasiparticles
Bosonization description of Fermi liquid theory
Examples of Applications
- Thermodynamics and specific heat
- Density-Density response

Introduction to Fermi liquid and Fermi liquid theory

Fermi liquid theory is a general theoretical framework (an effective field theory) describing the behavior of interacting fermion systems at low temperature. The random-phase approximation (RPA) and the corresponding excitation spectra are consistent with Fermi liquid behavior. In RPA, the electronic excitation spectrum consists of two pairs: collective and single particle-hole pair excitations. The particle-hole pair excitations spectrum is basically the same as the corresponding spectrum for non-interacting particles.

Fermi liquid theory is a theoretical framework that justifies this qualitative behavior, at least when the excitation energy is low. The basic assumption of Fermi liquid theory is ==adiabaticity==, meaning that there is a one-to-one correspondence between the ground state and low-energy excitation spectrum of fermions in the absence and in the presence of (repulsive) interactions. (Attractive interaction is excluded to avoid Cooper instability).

As a consequence of the assumption, the low-energy particle-hole excitation spectrum can basically be labeled by the same quantum number as for non-interacting fermions and the natures of the excitations are qualitatively similar. To capture this physics, Landau introduced the concept of quasiparticles.

Landau imagined that the ground state of an interacting fermion liquid can be labeled as filled Fermi gas of non-interacting fermions. In the presence of interaction, these fermions are not the same as the original fermions in the problem. __ Landau imagined them as __the original free fermions dressed by the interaction and called them quasi-particles. The energy of the system can be written in terms of the occupation number of quasi-particles $n_\mathbf k$. In particular, the ground state energy as a functional of the density distribution is,

\[E_G = E[\{n_\mathbf k=\theta(k_F-|k|\}],\]

which corresponds to a filled Fermi sea.

Low-energy excitations can be labeled by small changes in the occupation number of quasi-particles, $n_\mathbf k=\theta(k_F - |k|)+\delta n_\mathbf k$. Assuming that $\delta n_\mathbf k$ is small, Landau expanded the energy functional to second order,

\[E[n_\mathbf k] \sim E_\mathrm G + \sum_\mathbf k \xi_{\mathbf k}^*\delta n_\mathbf k+\frac{1}{2V}\sum_{\mathbf k\mathbf k ^\prime}f_{\mathbf k\mathbf k^\prime}\delta n_\mathbf k \delta n_{\mathbf k^\prime}\]

where

\[\xi_\mathbf k^*\sim \frac{\hbar k_F}{m^*}(|k|-k_F).\]

Here $m^*$ is called the effectiveness and $f_{\mathbf k\mathbf k^\prime }$ are called the Landau parameters. The form of $\xi_\mathbf k^*$ ensures that the ground states (filled Fermi sea) is stable. It recedes to a free fermi gas with $m^*=m$ and $f_{\mathbf k\mathbf k^\prime}=0$.

Landau next derived the (low-energy) dynamics of the quasi-particles. This is quite impossible from a microscopic point of view since the nature of the quasi-particles is not specified. Landau realized that he could bypass this problem if he interpreted the above energy functional as a local Hamiltonian function for classical particles, with

\[E[n_\mathbf k] \rightarrow E_\mathrm G+ \sum_{\mathbf k}\xi_{\mathbf k}^* \int \delta n_\mathbf k(\mathbf r ) d^d \mathbf r +\frac{1}{2}\sum_{\mathbf k \mathbf k^\prime} \int \delta n_\mathbf k(\mathbf r ) \delta n_{\mathbf k^\prime}(\mathbf r) d^d \mathbf r,\]

where the dynamics of the distribution function $\delta n_\mathbf k(\mathbf r )$ is determined by the classical equation of motion that followed from the Hamiltonian (essentially a Boltzmann equation). This assumption is justifiable in the $\mathbf q, \omega\rightarrow 0$ limit, where the system should become ‘classical’ like (correspondence principle). Recall that the uncertainty principle forbids simultaneous identification of the position and momentum of particles. Therefore, for this description to be valid, the distribution function $\delta n_\mathbf k(\mathbf r)$ should not be changing too rapidly in space so that the uncertainty principle is not violated (ie. $\Delta r\Delta k \geq 1$).

As we shall see later, we shall be interested in states $\mathbf k ,\mathbf k^\prime$ in the vicinity of the Fermi surface. In this regime, we may replace $f_{\mathbf k \mathbf k^\prime}$ by $f_{\mathbf k_F \mathbf k_F^\prime}$, where $\mathbf k_F$ are wave vectors on the Fermi surface. For a rotationally symmetric system, we also expect that $f_{\mathbf k_F \mathbf k_F^\prime}\rightarrow f(\theta)$, where $\theta$ is the angle between the directions $\mathbf k_F$ and $\mathbf k_F^\prime$. This form of Landau interaction is assumed in most studies of Fermi liquid theory and will be assumed in the following.

If a particle with momentum $\mathbf k$ is located at $\mathbf r$ at $t$ then, at time $t+dt$, it is located at $\mathbf r +\delta r=\mathbf r +\mathbf v(\mathbf r) dt$, with momentum $\mathbf k+ \delta\mathbf k =\mathbf k + \mathbf F(t)dt$,

\[\begin{align} & n(\mathbf r +\delta \mathbf r,\mathbf k +\delta \mathbf k ,t+\delta t)=n(\mathbf r ,\mathbf k,t) \\ \Rightarrow &\frac{\partial n}{\partial t}+\mathbf v(\mathbf k)\cdot \nabla_\mathbf r n+\mathbf F \cdot \nabla_\mathbf kn=\frac{dn}{dt}=0 \end{align}\]

To complete the equation, we have to determine the velocities $\mathbf v(\mathbf r )$ and forces $\mathbf F(t)$ from the Landau Fermi liquid theory. They are given in classical mechanism by

\[\mathbf v(\mathbf k )=\frac{\partial \mathcal E(\mathbf k )}{\partial k_i},\mathbf F_i(t)= - \frac{\partial\mathcal E(\mathbf k)}{\partial r_i},\]

where $\mathcal E(\mathbf k ) = \xi_\mathbf k^* +\sum_{\mathbf k^\prime}f_{\mathbf k\mathbf k^\prime}\delta n_{\mathbf k^\prime}(\mathbf r)$. Therefore, to leading order where $f_{\mathbf k \mathbf k^\prime} \rightarrow f_{\mathbf k_F \mathbf k_F^\prime}$,

\[\mathbf v(\mathbf r) = \frac{\partial \xi_\mathbf k^*}{\partial \mathbf k } =\frac{\hbar \mathbf k_F }{m^*}, \quad \mathbf F(t) = - \sum_{\mathbf k^\prime} f_{\mathbf k_F \mathbf k_F^\prime}\nabla_\mathbf r n_{\mathbf k^\prime }(\mathbf r)\]

Writing $n_\mathbf k (\mathbf r, t )=n_F(\mathbf k)+\delta n_\mathbf k(\mathbf r ,t)$ where $n_F$ is the equilibrium Fermi distribution, the Boltzmann equation becomes,

\[\frac{\partial \delta n_\mathbf k(\mathbf k,t)}{\partial t}+\mathbf v_\mathbf k\cdot \nabla _\mathbf r \delta n_\mathbf k(\mathbf r ,t)+\delta(\xi_\mathbf k^*)\sum_{\mathbf k^\prime} f_{\mathbf k\mathbf k^\prime}\mathbf v_{\mathbf k^\prime}\cdot \nabla_\mathbf r\delta n_{\mathbf k^\prime}(\mathbf r,t)=0 \label{coneq}\]

To linear order in $\delta n$, where we consider $T\rightarrow 0$ so that $n_F(\mathbf k)\sim \theta(-\xi_\mathbf k^*)$. Note that in this limit, __only states in the vicinity of the Fermi surface are involved in the transport equation. __

Remark: for the last term in Eq. $\eqref{coneq}$, we use
\[\nabla_k n_{\mathbf k}(\mathbf r) =\nabla_\mathbf k n_F(\mathbf k)+\nabla_\mathbf k\delta n_\mathbf k(\mathbf r)\\ \simeq \nabla_\mathbf k \theta (-\xi_\mathbf k^*) =-\mathbf v_\mathbf k(\mathbf r)\delta(\xi_\mathbf k^*)\]
Where a $\delta$ function emerges because in the limit $T\rightarrow 0$ only electrons at Fermi surface can be excited.

In this way, Landau derived results that cannot be obtained directly from the quantum energy functional. Note that interactions between quasiparticles represented by the Landau interaction $f_{\mathbf k\mathbf k^\prime}$ exist in the transport equation. We shall see later that the interactions are not small and cannot be neglected in general. Therefore, Landau’s theory predicts that the low-energy effective theory of Fermi liquids is not a theory of free fermions. Interactions between quasiparticles persist down to zero temperature. The validity and meaning of Landau’s approach were not fully understood until the 1990s when the mathematical techniques of renormalization groups and bosonization mature.

Charge and Current carried by quasiparticles

Now we study the charge and current carried by quasi-particles. We expect that the charge carried by a quasiparticle is exactly equal to the charge carried by the original particle because of the basic assumption that there is a one-to-one correspondence between bare particles and quasiparticles and that total charge is conceived. It turns out that the same conclusion cannot be drawn so readily for the current. To see why this is so, we examine the transport equation. Summing the equation over $\mathbf k$, we obtain the continuity equation,

\[\frac{\partial \rho(\mathbf r, t)}{\partial t}+\nabla\cdot \mathbf j(\mathbf r,t)=0,\]

where

\[\rho(\mathbf r,t)=\sum_\mathbf k \delta n_\mathbf k(\mathbf r,t).\]

Is the total charge fluctuation and

\[\mathbf j(\mathbf r,t)= \sum_\mathbf k \mathbf v_\mathbf k\delta n_\mathbf k(\mathbf r,t)+\sum_{\mathbf k\mathbf k^\prime}\delta(\xi_{\mathbf k^\prime}^*)f_{\mathbf k^\prime\mathbf k }\mathbf v_{\mathbf k^\prime}\delta n _\mathbf k(\mathbf r,t )=\sum_\mathbf k\mathbf j_\mathbf k\delta n_\mathbf k(\mathbf r,t)\]

where

\[\mathbf j_\mathbf k=\mathbf v_\mathbf k+\sum_{\mathbf k\mathbf k^\prime}\delta (\xi_{\mathbf k ^\prime}^*)f_{\mathbf k^\prime\mathbf k}\mathbf v_{\mathbf k^\prime}\]

is by definition the current carried by a quasi-particle. This is quite different from the naive expectation, $\mathbf j_\mathbf k =\mathbf v_\mathbf k$ for single quasi-particle. Note that this result is a consequence of having interaction terms in the transport equation, and is rather independent of quantum mechanics. This reason why $\mathbf j_\mathbf k \neq \mathbf v_\mathbf k$ is the quasiparticles are never really independent from one another when $f_{\mathbf k\mathbf k^\prime}\neq 0$. It is impossible to excite just one quasiparticle without creating a cloud of other quasiparticles surrounding it. You can imagine the situation of a man trying to push his way through a crowd of people, When he pushes through the crowd, people in his way will all be affected and have to move away a little bit. This phenomenon is called backflow in the study of fluids. In mathematical terms, one has to distinguish a quasiparticle from an elementary excitation, which is what is physically excited in the system.

An elementary excitation = a quasi-particle + crowd of other quasi-particles around it.

The crowd also contributes to the total current carried by the elementary excitation and is responsible for the $\sum_{\mathbf k \mathbf k^\prime}\delta(\xi_{\mathbf k^\prime}^*)f_{\mathbf k^\prime \mathbf k}\mathbf v_{\mathbf k^\prime}$ factor.

For a translationally invariant system, another interesting result occurs. First, we recall that quasiparticles and bare particles carry the same charge, and the quasiparticle density $\rho(\mathbf r ,t)$ is equal to the density of bare particles. Therefore, the current $\mathbf j(\mathbf r, t)$ is equal to the current of the bare particles. For the fermion system with Hamiltonian, it is easy to show that the expression for the current is

\[\mathbf j(\mathbf r ,t )=\sum_\mathbf k \mathbf v_{\mathbf{k}B}^0(\mathbf r, t)=\sum_\mathbf k \mathbf v_\mathbf k^0\delta n_\mathbf k(\mathbf r ,t)\]

Where $\mathbf v_\mathbf k^0 =\frac{\hbar \mathbf k}{m}$ with $m$ being the bare mass of particles, and $\delta_{\mathbf k B}(\mathbf r, t)$ is the density of bare particles. Thus, we have

\[\mathbf v_\mathbf k^0 = \mathbf v_\mathbf k+\sum_{\mathbf k\mathbf k^\prime}f_{\mathbf k\mathbf k^\prime}\mathbf v_{\mathbf k^\prime}.\]

Note that in this case $f_{\mathbf k_F\mathbf k_F^\prime}\rightarrow f(\theta)$, and by symmetry

\[\sum_{\mathbf k \mathbf k^\prime}\delta(\xi_{\mathbf k^\prime}^*)f_{\mathbf k^\prime \mathbf k}\mathbf v_{\mathbf k^\prime}=\frac{1}{(2\pi)^3}\int kdk\sin\theta d\theta d\phi\delta(\xi_\mathbf k^*)f(\theta)\frac{\mathbf k_F\cos\theta}{m^*}.\]

The significant result is often written as a relation between the effective mass $m^*$ and the Landau parameter $F_{1s}$

\[\frac{1}{3}F_{1s} =\frac{1}{(2\pi)^3}\int kdk\sin\theta d\theta d\phi\delta(\xi_\mathbf k^*)f(\theta)\mathbf \cos\theta\]

where

\[\frac{1}{m}=\frac{1}{m^*}(1+\frac{F_{1s}}{3})\]

Bosonization description of Fermi liquid theory

Landau’s Fermi liquid theory is an effective low-energy theory for fermion systems. It is natural to ask whether the theory can be put into the framework of effective quantum field theory. For example, one may imagine that the wave function of $N$ particle fermion system in Fermi liquid theory is written as

\[\Psi(\mathbf r_1,\mathbf r_2,\cdots,\mathbf r_N) = \int D\mathbf x\Phi(\mathbf x_1,\cdots,\mathbf x_M)\Psi_0([\mathbf r]:\mathbf x_1,\cdots,\mathbf x_M)\]

where $(\mathbf r_1,\mathbf r_2,\cdots, \mathbf r_N)$ are the coordinates of the bare fermions and $(\mathbf x_1,\cdots,\mathbf x_M)$ are the coordinates of special points in the wave function corresponding to positions of the quasiparticles and there is a one-to-one correspondence between $(\mathbf x_1,\cdots,\mathbf x_M)$ and $(\mathbf r_1,\mathbf r_2,\cdots, \mathbf r_N)$ (With $N=M$) in Fermi liquid theory. We may imagine that the low-energy properties of the system are characterized by the wave function $\Phi(\mathbf x_1,\cdots, \mathbf x_M)$ such that when we apply the principle of least action to the action

\[S=\int _a^b dt \langle \Psi(t)|[i\hbar \frac{\partial}{\partial t}-H]|\Psi(t)\rangle .\]

We obtain the Landau transport equation.

It turns out that the trial wave function approach is not a convenient starting point to understand Fermi liquid theory. The correct starting point is to realize that the ground state of the system is a filled Fermi sea and low-energy excitations in the system can be viewed as local deformations in the shape of the Fermi surface. The shape of the Fermi surface can be labeled by a wave vector $\mathbf k(\mathbf x)=(k_F+\delta u(\hat n,\mathbf x))\hat n$ where $\delta u(\hat n ,\mathbf x)$ labels local small changes in the shape of the Fermi surface in the direction $\hat n$. $\delta u(\hat n, \mathbf x)$ can be identified with $\delta n_\mathbf k (\mathbf x)$ in Fermi liquid theory with $\mathbf k=k_F\hat n$. Note that $\delta u(\hat n, \mathbf x)\hat n$ is a real number field and a quantum field theory can be constructed by a canonical quantization scheme if the equation of motion for $\delta u(\hat n, \mathbf x)$ is known. This is the spirit of bosinzation theory. The Landau energy functional $E[n]$ Represents the energy cost for a small distortion in the shape of Fermi surface in the theory and the Boltzmann equation becomes the quantum equation of motion for the operator $\delta u(\hat n ,\mathbf x)\hat n$. The excitations are bosons representing harmonic oscillatory modes of the Fermi surface. The most non-trivial result is perhaps that there is one-to-one mapping between the excitation spectrum obtained in the bosonization theory and the fermionic excitation spectrum.

Examples of Applications

Thermodynamics and specific heat

We first have to evaluate the system’s free energy starting from the energy function.

It is easier to start with the microcanonical ensemble of the system. We consider the system with fixed energy

\[E= E[n_\mathbf k]\sim E_G +\sum_\mathbf k \xi_\mathbf k^*\delta n_\mathbf k+\frac{1}{2}\sum_{\mathbf k\mathbf k^\prime}f_{\mathbf k\mathbf k^\prime}\delta n_\mathbf k \delta n_{\mathbf k^\prime}\]

and fixed particle number $N$. The particles are distributed with equal probability to each available state $\mathbf k$ provided that the above constraint is satisfied. To derive the free energy, we first note that for $N_j$ particles distributed in $G_j$ states, the total number of possibilities is $\Gamma_j\frac{G_j!}{N_j!(G_j-N_j)!}$ and the corresponding entropy is

\[\ln G_j \sim G_j \ln G_j - N_j\ln N_j -(G_j-N_j)\ln (G_j-N_j)\\ = -G_j(n_j\ln n_j+(1-n_j)\ln(1-n_j))\]

Applying this to Fermi liquid theory, we identify $j$ as a group of states with momentum $p$ between $\mathbf k$ and $\mathbf k+\Delta \mathbf k$ with $\Delta \mathbf k$ small enough so that $n_\mathbf p\sim n_\mathbf k$ and $G_j\rightarrow \frac{V}{(2\pi)^d} (\Delta k)^d$. The entropy per unit volume is thus

\[S\sim -k \int \frac{d^d k }{(2\pi)^d}(n_\mathbf k \ln n_\mathbf k +(1-n_\mathbf k)\ln (1-n_\mathbf k))\]

where $n_\mathbf k$ can be determined by maximizing the entropy with the energy constraint and $\sum n_\mathbf k=N$. The constraints can be handled by introducing Lagrange multipliers $\alpha,\beta$ where we maximize $S/k-\alpha N -\beta E$ with respect to $n_\mathbf k$. It is easy to obtain

\[n_\mathbf k =\frac{1}{1+\exp(\frac{\mathcal E(\mathbf k -\mu)}{kT})}\\ \delta n_\mathbf k = n_\mathbf k-\theta (-\xi_\mathbf k^*)\]

Where $\mathcal E(\mathbf k )=\frac{\delta E[n]}{\delta n_\mathbf k}=\xi_\mathbf k^*+\sum_{\mathbf k^\prime}f_{\mathbf k\mathbf k^\prime}\delta n_{\mathbf k^\prime}$. In principle, this equation has to be solved self-consistently. At low temperatures, one can perform a Sommerfeld-type expansion of the free energy. To the lowest order in temperature, the specific heat is $C=\gamma T$ where

\[\gamma =\frac{k_B^2}{3}\frac{m^*k_F}{\hbar^2}=\frac{m^*}{m}{\gamma_0}\]

where $\gamma_0T$ is the corresponding specific heat for the non-interacting Fermi gas. Note that to the lowest order in temperature only effect of interaction on the specific heat is to renormalize the electron mass in Fermi liquid theory.

Density-Density response

To determine the density-density response function to an external scalar field, we solve the Boltzmann equation

Where we have added an external force term in the equation. For an external scalar potential $\mathbf F^\mathrm{ext}=e\mathbf E^\mathrm{ext} =-e\nabla\Phi^\mathrm{ext}$ and $F^\mathrm{ext}\cdot \nabla_\mathbf k n_\mathbf k ^\mathrm{(0)}\rightarrow e\mathbf v_\mathbf k\cdot \nabla\Phi^\mathrm{ext}\delta(\xi_\mathbf k^*)$ at $T\rightarrow 0$.

Defining $\delta n_\mathbf k(\mathbf q,\omega)=\int e^{i(\mathbf q\cdot \mathbf r -\omega t )}\delta n_\mathbf k(\mathbf r, t)d^dr dt$, we obtain

\[(\mathbf q \cdot \mathbf v_\mathbf k-\omega)\delta n_\mathbf k+\mathbf q \cdot \mathbf v_\mathbf k\delta(\xi_\mathbf k^*)\sum_{\mathbf k^\prime}f_{\mathbf k\mathbf k^\prime}\delta n_{\mathbf k^\prime} = -e \mathbf v_\mathbf k \cdot \mathbf q \Phi^\mathrm{ext}\delta(\xi_\mathbf k^*)\]

For rotationally symmetric systems, we can expand the solutions in terms of spherical harmonics. Here we shall consider a very simple situation $f_{\mathbf k\mathbf k^\prime}\sim f_0$ . In this case, it is easy to see that the solution for $\delta \rho = \sum_\mathbf k \delta n_\mathbf k$ is

\[\delta\rho(\mathbf q, \omega)=\frac{\chi_{0F}(\mathbf q,\omega)}{1-f_0 \chi_{0F}(\mathbf q,\omega)}e\Phi^\mathrm{ext}(\mathbf q,\omega)\]

Where

\[\chi_{0F} (\mathbf q,\omega) =\sum_\mathbf k \frac{\mathbf v_\mathbf k \cdot \mathbf q}{\omega-\mathbf v_\mathbf k\cdot \mathbf q}\delta(\xi_\mathbf k^*)\]

Note that this is of the same form as the density-density response function deduced from RPA. More interestingly, it is easy to see that in the small $q$ Limit the Lindblad $\chi_0(\mathbf q,\omega)$ has the same expression as $\chi_{0F}(\mathbf q ,\omega)$ except that the bare electron mass $m$ is replaced by the effective mass $m^*$, indicating that RPA is in fact an example of Fermi liquid theory with un-renormalized mass $m=m^*$ and interaction $f_0=V(q\rightarrow 0)$.

It suggests that besides quasi-particle-hole pair excitations, collective excitations can also be produced in Landau’s Fermi liquid theory as in RPA. The limitation of Fermi liquid theory is that it is valid only in the long-wavelength $\mathbf q\rightarrow 0$ Limit, where a semi-classical description of the system is valid.

Ref： Introduction to Classical and quantum field theory by T. K. Ng