Berry Physics and Anomalous Hall Effect

Recall the Bloch’s theorem

Some basic results of the Bloch’s theorem is reproduced here. For more detailed discussion, refer to my earlier post here.

The Bloch function,

where $u_{n\mathbf{k}}(\mathbf{r})$ is periodic with the lattice periodicity. Substituting into the full Schrödinger equation $H\psi_{n\mathbf{k}} = E_{n\mathbf{k}}\psi_{n\mathbf{k}}$ gives an equation for the periodic part alone:

where the $\mathbf{k}$-dependent Bloch Hamiltonian is:

Berry phase

Adiabatic evolution

Consider a quantum system with a Hamiltonian $H(\boldsymbol{\lambda})$ depending on a set of parameters $\boldsymbol{\lambda} = (\lambda_1, \lambda_2, \ldots)$. Suppose the parameters are varied slowly (adiabatically) along some path $\boldsymbol{\lambda}(t)$ in parameter space. By the adiabatic theorem, if the system starts in the $n$-th instantaneous eigenstate $\vert n(\boldsymbol{\lambda}(0))\rangle$, it remains in the $n$-th instantaneous eigenstate throughout the evolution – it never jumps to another eigenstate (provided there is a gap).

:::info

• Here we are talking about the general situation where the Hamiltonian contains some parameters. A little forward-leaping hint for the following discussion is, think about the the Bloch Hamiltonian I put above for each single value of $\mathbf{k}$, isn’t the Hamiltonian just depending on ‘some’ parameter? Yes, the parameter is just $\mathbf{k}$. So we derive things here generally for $\lambda$ and the result will naturally follow, when we replace $\lambda$ with $\mathbf{k}$, as we will see below.
• Explanation about ‘provided there is a gap’ can be found here.

:::

The time-dependent Schrödinger equation is,

Write the ansatz,

The second exponential is the familiar dynamical phase — the time-integrated energy. The first factor $e^{i\gamma_n(t)}$ is what we want to find.

:::info Here, let’s stop by visiting why the ansatz solution is actually proposed. See here. This includes detailed discussion about why we have the second exponential term, i.e., the dynamical phase and why we still need the first term, i.e., the geometric phase. :::

Substituting into the Schrödinger equation and using $H(\boldsymbol{\lambda})\vert n(\boldsymbol{\lambda})\rangle = E_n(\boldsymbol{\lambda})\vert n(\boldsymbol{\lambda})\rangle$, the energy terms cancel and we get,

Projecting onto $\langle n\vert $,

Using the chain rule \(\frac{d}{dt} = \dot{\lambda}_\mu \partial_{\lambda_\mu}\),

:::info $\mu$ here refers to the different component of the $\pmb{\lambda}(t)$ parameter, and the Einstein notation is used in \(\frac{d}{dt} = \dot{\lambda}_\mu \partial_{\lambda_\mu}\) where the repeated index $$\mu$ is summed over. :::

Integrating over the path,

The geometric phase is real

Differentiate the normalization condition $\langle n(\boldsymbol{\lambda}) \vert n(\boldsymbol{\lambda}) \rangle = 1$ with respect to $\lambda_\mu$,

Therefore $\langle n \vert \partial_{\lambda_\mu} n \rangle = -\langle \partial_{\lambda_\mu} n \vert n \rangle = -(\langle n \vert \partial_{\lambda_\mu} n \rangle)^*$, which means $\langle n \vert \partial_{\lambda_\mu} n \rangle$ is purely imaginary. So $i\langle n \vert \partial_{\lambda_\mu} n \rangle$ is real, and $\gamma_n$ is a real number.

Berry connection

Define the Berry connection (also called the Berry vector potential),

Then \(\gamma_n = \int \mathcal{A}_{n,\mu}\, d\lambda_\mu\) is a line integral of \(\mathcal{A}_n\) through parameter space. This is exactly analogous to \(\int \mathbf{A}\cdot d\mathbf{l}\) in electromagnetism, where \(\mathbf{A}\) is the magnetic vector potential.

:::info The Einstein notation is used in \(\gamma_n = \int \mathcal{A}_{n,\mu}\, d\lambda_\mu\) – see the derivation of \(\gamma_n\) above. :::

Gauge freedom

The eigenstates $\vert n(\boldsymbol{\lambda})\rangle$ are only defined up to an arbitrary $\boldsymbol{\lambda}$-dependent phase:

Under this gauge transformation, the Berry connection transforms as:

This is identical to the gauge transformation $\mathbf{A} \to \mathbf{A} - \nabla\phi$ in electromagnetism. So $\mathcal{A}_n$ is gauge-dependent and not physically observable by itself.

:::info Here we just need to plug in the transformed version of $\vert n(\boldsymbol{\lambda})\rangle$ into the definition of the Berry connection $\mathcal{A}_{n,\mu}$ and the exponential term from the left and right will be canceled out, giving us the exact form of the Berry connection presented above, regardless of the extra phase factor. :::

Berry phase

For a closed path $\mathcal{C}$ in parameter space ($\boldsymbol{\lambda}(T) = \boldsymbol{\lambda}(0)$), the Berry phase is,

Under a gauge transformation $\mathcal{A} \to \mathcal{A} - \nabla\phi$, the extra term is $-\oint \nabla\phi \cdot d\boldsymbol{\lambda} = 0$ for a closed loop (assuming $\phi$ is single-valued, i.e., only dependent on $\boldsymbol{\lambda}$). So the Berry phase around a closed loop is gauge-invariant and physically observable.

:::info Applying Stokes’ theorem,

Since the curl of any scalar gradient is zero ($\nabla \times \nabla\phi = \mathbf{0}$), the surface integral evaluates to,

\(-\iint_{\mathcal{S}} \mathbf{0} \cdot d\mathbf{S} = 0\) :::

Berry curvature

By Stokes’ theorem, the line integral around a closed loop equals the flux of the curl of $\mathcal{A}_n$ through any surface $\mathcal{S}$ bounded by $\mathcal{C}$,

where the Berry curvature is,

It is antisymmetric, i.e., $\Omega_{n,\mu\nu} = -\Omega_{n,\nu\mu}$, and it is gauge-invariant – the gauge transformation $\mathcal{A} \to \mathcal{A} - \nabla\phi$ shifts $\mathcal{A}$ by a gradient, whose curl vanishes (see the derivation above). So, $\Omega$ is unaffected by the choice of gauge – it is a physically meaningful, observable quantity.

:::info $d\lambda_\mu \wedge d\lambda_\nu$ in the formulation refers to the wedge product, which is like an infinitesimal integration area with direction, i.e., directional area. The whole derivation involves some heavy exterior calculus which is beyond my capability so I won’t include that here. But anyhow, from the Berry connection $\mathcal{A}_n$ to the Berry curvature, it is just the application of the Stokes’ theorem, turning a line integral to an area integral. :::

Berry curvature in Bloch bands

Now we have the general results of Berry physics with the general parameter $\boldsymbol{\lambda}$, and we can apply it to the Bloch band where the parameter space is the Brillouin zone, i.e., $\boldsymbol{\lambda} = \mathbf{k}$, and the states are Bloch states $\vert u_{n\mathbf{k}}\rangle$ (see the note below).

:::info The full Bloch states are $\psi_{n\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} u_{n\mathbf{k}}(\mathbf{r})$ as presented earlier. However, regarding the quantities involved in the Berry physics, the plane wave modulation part will be cancelled out during the inner product evaluation and therefore we can stay with the periodic part of the Bloch states, i.e., $\vert u_{n\mathbf{k}}\rangle$ :::

The Berry connection is:

The Berry curvature is:

Expanding this:

:::info Derivation of the Berry curvature

Start from the Berry connection,

and differentiate it with the product rule,

Swapping $\alpha \leftrightarrow \beta$ gives

Now,

The last term vanishes because partial derivatives commute, $\partial_{k_\alpha}\partial_{k_\beta} = \partial_{k_\beta}\partial_{k_\alpha}$ (assuming $\vert u_n(\mathbf{k})\rangle$ is smooth enough in $\mathbf{k}$), and we are left with,

\(\Omega_{n,\alpha\beta} = i\left(\langle \partial_{k_\alpha}u_n\vert \partial_{k_\beta}u_n\rangle - \langle \partial_{k_\beta}u_n\vert \partial_{k_\alpha}u_n\rangle\right)\) :::

Insert $\sum_{m} \vert u_m\rangle\langle u_m\vert = \mathbf{1}$ between the derivatives,

The $m = n$ term vanishes. Since,

and we know that,

For a complex number to satisfy $a = -a^*$, $a$ has to be purely imaginary. So, we know that both $\langle \partial_{k_\alpha} u_n \vert u_n \rangle$ and $\langle u_n \vert \partial_{k_\beta} u_n \rangle$ are purely imaginary. Therefore,

is real, and hence the $m = n$ term in the expression for $\Omega_{n,\alpha\beta}$ is canceled out.

Differentiating $H(\mathbf{k})\vert u_{n\mathbf{k}}\rangle = E_{n\mathbf{k}}\vert u_{n\mathbf{k}}\rangle$ with respect to $k_\alpha$ and projecting onto $\langle u_m\vert$ with $m \neq n$,

Therefore:

:::info The Hellmann-Feynman Trick

Start from

and differentiate both sides with respect to $k_\alpha$,

Project onto $\langle u_m\vert $ with $m \neq n$,

To simplify, we realize,

• Since $\vert u_m\rangle$ and $\vert u_n\rangle$ are eigenstates of the Hermitian operator $H$ with $m\neq n$, they’re orthogonal, giving $\langle u_m\vert u_n\rangle = 0$.

• $H$ is Hermitian, so, $\langle u_m\vert H = E_m\langle u_m\vert $, and $\langle u_m\vert H\vert \partial_{k_\alpha}u_n\rangle = E_m\langle u_m\vert \partial_{k_\alpha}u_n\rangle$.

Accordingly, the result simplifies to,

Rearranging,

\(\langle u_m\vert \partial_{k_\alpha}H\vert u_n\rangle = (E_n - E_m)\langle u_m\vert \partial_{k_\alpha}u_n\rangle,\) :::

:::info Some notes about Hermitian operators

First, the definition of the Hermitian conjugate follows,

or,

If $A$ is Hermitian, that means $A^\dagger = A$, and accordingly,

Since the Hamiltonian operator is Hermitian, we have,

which is applied in the derivation in the Helmann-Feynman trick above. Several basic properties of the Hermitian conjugate are applied here – the Hermian conjugate carries out different actions when performing on different objects,

• $\vert \rangle \Rightarrow \langle\vert$, i.e., bra $\Rightarrow$ ket

• $H \Rightarrow H^\dagger$, i.e., operator becomes its Hermitian conjugate

• $(AB)^\dagger = B^\dagger A^\dagger$ :::

:::info Comparison to the standard Hellmann-Feynman theorem

The expression above,

is very similar to the Hellmann-Feynman theorem where we only care about the diagonal term, where $m = n$. :::

Here, performing the Hellman-Feynman trick to write down,

is very helpful. It lets us rewrite the Berry connection/curvature – which, in its original form, needs $\vert \partial_{k_\alpha}u_n\rangle$, a quantity sensitive to the arbitrary phase choice of $\vert u_n\rangle$ – in terms of matrix elements of $\partial_{k_\alpha}H$ between different bands, which are gauge-invariant and computable directly from $k\cdot p$ perturbation theory.

Wait, what is $k\cdot p$ perturbation theory?

Go back to the Bloch Hamiltonian given at the very top,

Expanding the square,

The first term $H_0$ is the Hamiltonian without perturbation, corresponding to the stationary Schrödinger equation at exactly $\mathbf{k} = 0$. The third and forth terms together are all about the perturbation and both the eigen states and eigen values can be given following the perturbation theory (see Wiki page here). In the perturbation term of the Hamiltonian, we see $\mathbf{k}\cdot\mathbf{p}$ and that is exactly what we mean by $k\cdot p$ perturbation theory.

Now take $\partial_{k_\alpha}$ derivative over the Hamiltonian,

Defining \(\hat{v}_\alpha \equiv \dfrac{\hat{p}_\alpha+\hbar k_\alpha}{m}\) (\(\hat{p}_\alpha+\hbar k_\alpha \Rightarrow\) free electron momentum plus the crystal momentum), we have,

Substituting this back into the expression for the Berry curvature given above,

:::info For the $\langle \partial_{k_\alpha} u_n \vert u_m \rangle$ term in the Berry curvature expression given earlier, it can be given as,

The ‘$-$’ sign originates from the multiplication of $i$ as the prefactor in the Berry curvature $\Omega_{n,\alpha\beta}(\mathbf{k})$ and the pure imaginary number within the bracket of the $\Omega_{n,\alpha\beta}(\mathbf{k})$ expression (‘$\dots - c.c.$’). :::

The antisymmetry $\Omega_{n,\alpha\beta} = -\Omega_{n,\beta\alpha}$ is preserved – swapping $\alpha \leftrightarrow \beta$ exchanges the two matrix elements in the numerator, taking the product to its complex conjugate, and since $\mathrm{Im}(z^*) = -\mathrm{Im}(z)$, we get a sign flip.

Kubo formula

Let’s summarize what we have achieved briefly. We derived the expression for the Berry curvature, showing that it is fundamentally determined by the velocity operator. So far, for the Berry curvature derivation, there is no external field involved. Once the external field is introduced, the field-involved (and gauge invariant) Hamiltonian can be constructed. With the field effect being treated as perturbation, the induced current can be derived given the applied electric field. Accordingly the conductivity can be deduced – here the dissipationless (thus corresponding to the anomalous Hall effect) conductivity part of the Kubo formulation is presented,

:::info The derivation for the Kubo formula is much too involved for me and here I am referring to Refs. [1-4] for derivation and some relevant results. :::

where $f_{n\mathbf{k}}$ is the equilibrium Fermi-Dirac occupation of the Bloch state $\vert u_{n\mathbf{k}}\rangle$,

$\mu$ is the chemical potential, which equals $E_F$ at $T=0$. $f_{m\mathbf{k}}$ is the same function evaluated at $E_{m\mathbf{k}}$. We can write,

Then we can swap \(n \leftrightarrow m\). Because \(\hat v\) is Hermitian, \(\langle u_m\vert \hat v_\alpha\vert u_n\rangle = \langle u_n\vert \hat v_\alpha\vert u_m\rangle^*\). So the product of matrix elements becomes its complex conjugate. The denominator is unchanged, and \(\mathrm{Im}(z^*) = -\mathrm{Im}(z)\), so,

Then we can split the summation as,

For the second term, we can rename the dummy indices $n \leftrightarrow m$ and use $S_{mn} = -S_{nm}$,

Then we have,

Looking at the inner summation, we can realize that it is exactly the Berry curvature from earlier derivation,

Putting this back into the Kubo formula, we have,

Since $\Omega_{n,\alpha\beta} = -\Omega_{n,\beta\alpha}$, it follows immediately that,

The conductivity tensor is antisymmetric.

Hall vector

The $3\times 3$ antisymmetric Hall conductivity can be given as,

Define the Hall vector $\mathbf{G}$ with components,

:::info Einstein notation is used here, i.e., repeated indices are summed. :::

where $\epsilon_{\gamma\alpha\beta}$ is the Levi-Civita symbol. Explicitly,

For example,

The Hall current is,

So,

The cross product $\mathbf{G} \times \mathbf{E}$ is always perpendicular to both $\mathbf{G}$ and $\mathbf{E}$, and therefore the Hall current is always perpendicular to the driving field.

Symmetry

The Berry curvature is a rank-2 tensor and transforms as,

In the matrix-form, it can be written as,

When the matrix $M$ is diagonal, the transformation becomes simplified,

For example, for the mirror reflection with respect to the $xz$ plane, the matrix $M$ is given as,

and accordingly, the Berry curvature transforms as,

So, if $M$ is a symmetry of the system, under the $M$ operation, we are not expecting anything to change, including the Berry curvature for sure. Therefore, given the situation above, $\Omega_{xy}$ and $\Omega_{yz}$ are forced by the symmetry to be zero. Only $\Omega_{xz}$ is allowed to exist. The same logic applies to the Hall conductivity tensor $\sigma_{\alpha\beta}$ as well.

Sometimes, we could also see in the literature that the Berry curvature is given in its vector form,

This is actually the same thing as for the tensor form. To see why, let’s take the following example,

N.B. This is just the definition of the Berry curvature. We can do the relabeling, just calling it as,

Exactly the same expression as $\Omega_{n,xy}$ above. Generally, we have,

which is exactly the Hall vector as defined in earlier discussion. Why can we do such a relabeling to turn a rank-2 tensor that has 9 components into a rank-1 tensor that has 3 components? The thing is, we know that the Berry curvature tensor is antisymmetric and therefore it only has 3 independent components – all the diagonal components are forced to be $0$ and the off-diagonal components are antisymmetric (thus giving 3 independent components out of the 6 components).

Appendix-A

$\vert n\rangle$ is the eigenstate of the Hamiltonian at $t = 0$. Then as the Hamiltonian changes with the parameter $\pmb{\lambda}(t)$, generally we will have the original state changing accordingly. Suppose originally we have two eigenstates $\vert m\rangle$ and $\vert n\rangle$ and they should be orthogonal, if $m \neq n$. If we have the eigenstate $\vert n\rangle$ changing as $\pmb{\lambda}(t)$ with the Hamiltonian, the orthogonality $\langle m\vert n \rangle$ will no longer be guaranteed. Accordingly, we can use the quantity $\langle m\vert \dot{n} \rangle$ to characterize how fast the eigenstate $\vert n\rangle$ is changed towards $\vert m\rangle$.

so,

and therefore,

The time scale associated with such an eigenstate variation rate is therefore,

The time scale accosicated with the actual transition between the two energy levels is determined by the gap, and therefore,

If we want to consider the situation where the eigenstate variation is slow, we are basically aiming at $\tau_{\text{variation}} \gg \tau_{\text{gap}}$. Accordingly,

This means if we have the changing rate of the Hamiltonian is small, as compared to the gap between the original eigenstate and any other eigenstates, the variation of the eigenstate (with $\pmb{\lambda}(t)$) can be ignored, i.e., the original eigenstate stays as the eigenstate of the Hamiltonian, ‘on-the-fly’. Physically, this is understandable – when the Hamiltonian changes very slow, the original eigenstate (of the original Hamiltonian) has enough time to relax to the new eigenstate (of the new Hamiltonian) and therefore if the system starts from the original eigenstate $\vert n[\pmb{\lambda}(0)]\rangle$, it will always stay as the eigenstate of $H[(\pmb{\lambda}(t)]$, evolving as $\vert n[\pmb{\lambda}(t)]\rangle$.

Appendix-B

The ansatz for the time-dependent wave function under adiabatic evolution is explicitly constructed to satisfy the time-dependent Schrödinger equation (TDSE) while enforcing the physical constraints of the adiabatic theorem.

The Adiabatic Assumption The adiabatic theorem dictates that if a system starts in a non-degenerate eigenstate $\vert{}n(\lambda(0))\rangle$ and the Hamiltonian changes infinitely slowly, the system will not transition to other energy levels. Therefore, the state $\vert{}\Psi(t)\rangle$ is allowed to be different from the eigenstate $\vert{}n(\lambda(t))\rangle$ but only by a phase factor,

Dynamical Phase For a simple time-independent Hamiltonian, a stationary state evolves with the standard phase factor $e^{-iE_n t/\hbar}$ $^\dagger$. When the external parameter $\lambda(t)$ makes the Hamiltonian time-dependent, the instantaneous energy $E_n(\lambda(t))$ fluctuates. The natural generalization of this standard phase is the time integral of the instantaneous energy$^\shortparallel$,

Factoring this dynamical phase out of the total phase $\theta(t)$ separates the expected kinematic evolution of the state from any phase changes caused by the changing parameters.

:::info $^\dagger$ Stationary state$^\ddagger$ evolving with time The standard phase factor arises directly from solving the time-dependent Schrödinger equation (TDSE) for an energy eigenstate.

When a quantum system is governed by a time-independent Hamiltonian $H$, its evolution is dictated by,

If the system is prepared in a stationary state (an eigenstate $\vert{}n\rangle$ of the Hamiltonian), it satisfies the time-independent eigenvalue equation,

To find how this state evolves over time, we assume the solution takes the form of the original state multiplied by some time-dependent scalar function $f(t)$,

Substitute this ansatz into the TDSE,

Because $\vert{}n\rangle$ has no explicit time dependence, the time derivative only acts on $f(t)$. Because $f(t)$ is a scalar, it commutes with the linear operator $H$,

Apply the eigenvalue relation $H \vert{}n\rangle = E_n \vert{}n\rangle$ to the right side,

Since this must hold for the non-zero vector $\vert{}n\rangle$, we can drop the ket to get a simple first-order ordinary differential equation for $f(t)$,

Integrating this equation with the initial condition $f(0) = 1$ (so that $\vert{}\Psi(0)\rangle = \vert{}n\rangle$) yields,

Thus, the full time-dependent state is simply the initial eigenstate rotating by a complex phase at a constant frequency determined by its energy,

\(\vert{}\Psi(t)\rangle = e^{-i E_n t / \hbar} \vert{}n\rangle\) :::

:::info $^\ddagger$ Stationary state A stationary state is a quantum state for which we can cleanly separate out the spatial part and the time part. Several key characteristics of a stationary state are listed below,

• Static Probability Density. The probability of finding a particle at a given location is determined by the absolute square of the wavefunction, $\vert{}\Psi(\mathbf{r}, t)\vert{}^2$. If we can manage to separate out the time evolution part purely, squaring the wavefunction, the time evolution part will be like this, $\vert{}e^{-iE_n t/\hbar}\vert{}^2 = 1$. The spatial probability distribution never changes.

• Constant Observables. The expected measurement outcome (expectation value) for any time-independent observable operator $\hat{A}$ remains completely fixed. The time-evolution phase applied to the bra $\langle\Psi(t)\vert{}$ mathematically cancels the phase applied to the ket $\vert{}\Psi(t)\rangle$.

By contrast, if a system is prepared in a superposition of multiple energy eigenstates (e.g., $\vert{}\Psi\rangle = \frac{1}{\sqrt{2}}\vert{}E_1\rangle + \frac{1}{\sqrt{2}}\vert{}E_2\rangle$), it is a non-stationary state. The different energy components carry different time evolution parts. When squared to calculate probabilities, those cross-terms cause the probability distribution and the expectation values of observables to change over time. :::

:::info $^\shortparallel$ Phase factor evolution with time

For a time-independent Hamiltonian, the energy $E_n$ is a constant value. The rate at which the quantum phase evolves is also a constant (refer to the stationary state wavefunction form presented above),

Accordingly, the total accumulated phase over a period $t$ is simply the constant rate multiplied by the time duration: $\theta(t) = -\frac{E_n t}{\hbar}$.

When the Hamiltonian is time-dependent, its instantaneous energy eigenvalue $E_n(\lambda(t))$ (following the adiabatic assumption) fluctuates as the system evolves. Accordingly we have,

and,

\(\theta_{dyn}(t) = \int_0^t d\theta = -\frac{1}{\hbar} \int_0^t E_n(\lambda(t')) dt'\) :::

Geometric Phase If the eigenstate vector $\vert{}n(\lambda)\rangle$ were perfectly rigid in Hilbert space, the dynamical phase alone would be the complete solution. However, as the external parameter $\lambda(t)$ changes, the eigenstate vector itself varies. When the TDSE operator $i\hbar \frac{d}{dt}$ operates on the state, the product rule infers that the derivative must act on both the phase exponential and the ket vector $\vert{}n(\lambda(t))\rangle$. This causes some problems if we only have the dynamical phase presented above in the wavefunction. To see why, let’s put down the ansatz wavefunction form as (here below I am using $\vert{}n(t)\rangle$ in place of $\vert{}n(\lambda(t))\rangle$ for the sake of simplicity),

Then plug this into the TDSE,

The left-hand side (LHS) gives,

through the application of the production rule of the derivative and the following bit of result,

Now, the right-hand side (RHS),

Equating the LHS and RHS, the $E_n(t) \vert{}\Phi(t)\rangle$ term cancels out, and we are left with,

inferring,

This is the problem – with only the dynamical phase factor, we conclude that the eigen state is not varying with time at all, which apparently does not make sense since we know the state will be evolving with time. To solve the issue, we can introduce an extra phase factor $e^{i\gamma_n(t)}$ in the ansatz so that the application of the production rule while evaluating the LHS will generate an additional term $i\hbar \dot{\gamma}_n(t)\vert{}n(t)\rangle$. This then avoids problematic situation of $\vert{}n(t)\rangle$ not evolving with time. As for how to find out $\gamma_n(t)$, the discussion will be presented in the main story.

References

[1] https://davidtong.org/pdfs/teaching/quantum-hall-effect/qhe.pdf

[2] Henrik Bruus and Karsten Flensberg, Introduction to many-body quantum theory in condensed matter physics.

[3] J. Sinova, et al., Phys. Rev. B. 67, 235203, 2003.

[4] Y. Yao, et al., Phys. Rev. Lett. 92, 037204, 2004.