Derivations

Band Structure

The tight-binding dispersions for the following lattices are given. The $E_\text{nn}$ term is added if self.warp = "on"

\[\begin{split}\begin{equation} \begin{split} \text{Square}: \quad E_\text{n}&=E_0+2t[\cos(k_x a)+\cos(k_y a)]\\ \quad E_\text{nn}&=rt[\cos(2k_x a)+\cos(2k_y a)]\\ \text{Rectangle}: \quad E_\text{n}&=E_0+2t_1\cos(k_x a)+2t_2\cos(k_y b)]\\ \quad E_\text{nn}&=E_0+rt_1\cos(2k_x a)+rt_2\cos(2k_y b)\\ \text{Hexagonal}: \quad E_\text{n}&=E_0+2t[\cos(a/2(k_x+\sqrt{3}ky))+\\ &\quad\quad\cos(a/2(k_x-\sqrt{3}ky))+\cos(k_x a)]\\ \quad E_\text{nn}&=E_0+2rt[\cos(a/2(3k_x+\sqrt{3}ky))+\\ &\quad\quad\cos(a/2(\sqrt{3}ky)-3kx)+\cos(\sqrt{3}k_y a)]\\ \text{Honeycomb}: \quad E_\text{n}&=E_0\pm \\ &\quad\quad t\sqrt{3+2\cos(\sqrt{3}k_x a)+4\cos(\sqrt{3}k_x a/2)\cos(3k_y a/2)}\\ \quad E_\text{nn}&=2rt[\cos(a/2(\sqrt{3}k_x+3k_y))+\\ &\quad\quad\cos(a/2(\sqrt{3}k_x-3k_y))+\cos(a\sqrt{3}k_x)]\\ \end{split} \end{equation}\end{split}\]

Matrix elements

If ME["type"] = "symm": The matrix elements the same symmetry as that of the tight-binding model, in that they are periodic in $k$-space in the same way. The equations are:

\[\begin{split}\begin{equation} \begin{split} \text{Square}: \quad M_n&=t[\cos(k_x a)+\cos(k_y a)]+P(k_x, k_y)\\ \text{Rectangle}: \quad M_n&=t_1\cos(k_x a)+2t_2\cos(k_y b)]+P(k_x, k_y)\\ \text{Hexagonal}: \quad M_n&=t[\cos(a/2(k_x+\sqrt{3}ky))+\\ &\quad\quad\cos(a/2(k_x-\sqrt{3}ky))+\cos(k_x a)]+P(k_x, k_y)\\ \text{Honeycomb}: \quad M_n&=t\sqrt{3+2\cos(\sqrt{3}k_x a)+4\cos(\sqrt{3}k_x a/2)\cos(3k_y a/2)}+P(k_x, k_y)\\ \end{split} \end{equation}\end{split}\]

Here, if Bands.symmetry = "custom" from imported bands, then this mode does not do anything.

If ME["type"] = "rot": The matrix elements give rotational symmetry, which mimics the orbital selectivity of linear and/or circularly polarized light. Here,

\[\begin{equation} \phi = \arctan\left(\frac{k_y}{k_x}\right), \quad M_\text{rot} = \cos(n\phi) \end{equation}\]

If this mode is selected, then $n$ (specified in the field ME["rotN"]) gives the order of rotational symmetry. If ME["rotN"] is not specified, then the order is randomly selected according to the rotational symmetry of the lattice defined in the tight-binding model:

Rectangle: $n = 1, 2$
Square: $n = 1, 2, 4$
Hexagonal/Honeycomb: $n = 1, 2, 3, 6$

If ME["type"] = "rot": The matrix elements are given by a random polynomial. If this mode is selected, then the field ME["polyN"] gives the order of the polynomial. If ME["polyN"] is not specified, then the order is randomly selected.

Self-energy

Fermi-liquid self-energy

Real part: $\Sigma' = 0$

Imaginary part: $\Sigma'' = D \, \omega^2 + \Sigma_0$

Dictionary keys:

SE["val"] = $D$

SE["ImS0"] = $\Sigma_0$

Default values (if not provided):

$\Sigma_0 = 0.01$

$D \in [0.05, 0.25]$.

2. Electron-boson kink self-energy The self-energy is phenomenological:

\[\begin{split}\Sigma'(\omega) &= R \left[ \frac{\gamma}{(\omega + \omega_0)^2 + \gamma^2} - \frac{\gamma}{(\omega - \omega_0)^2 + \gamma^2} \right] \\ \Sigma''(\omega) &= \Sigma_1 - \frac{\Sigma_1 - \Sigma_0}{1 + e^{-(\omega + \omega_0)\gamma}}\end{split}\]

Dictionary keys and initialized random values (if not provided):

SE["Amp"] = $R \in [0, 0.01]$

SE["Ekink"] = $\omega_0 \in [0.05, 0.3]$

SE["gamma"] = $\gamma \in [0, 50]$

SE["ImS0"] = $\Sigma_0 \in [0.015, 0.025]$

SE["ImS1"] = $\Sigma_1 \in [0.115, 0.125]$

Photoemission intensity

The spectral function and matrix elements are calculated at once using S.Make_specfun(m). Depending on the dimension of the calculation:

\[\text{cube:}\quad I(k_x, k_y, \omega) = \sum_m M_m(\omega)\frac{\Sigma_m''(\omega)}{(\omega-\epsilon_\mathbf{k}-\Sigma'_m(\omega))^2+\Sigma''_m(\omega)^2}\]

\[\text{sliceEk:}\quad I(k_x, k_{y_0}, \omega) = \sum_m M_m(\omega)\frac{\Sigma_m''(\omega)}{(\omega-\epsilon_\mathbf{k}-\Sigma'_m(\omega))^2+\Sigma''_m(\omega)^2}\]

\[\text{slicekk:}\quad I(k_x, k_y) = \sum_m \int_{\Delta\omega} M_m(\omega)\frac{\Sigma_m''(\omega)}{(\omega-\epsilon_\mathbf{k}-\Sigma'_m(\omega))^2+\Sigma''_m(\omega)^2} d\omega\]

where $\Delta\omega = k_B T + \Delta E$

Then, S.Make_specmod(m) adds the Fermi-Dirac distribution and resolution broadening.

The Fermi-Dirac is added by multiplication:

\[\begin{split}I_\text{FD}(\mathbf{k}, \omega) &= f_\text{FD}(\omega, T) I(k_x, k_y, \omega), \\ f_\text{FD}(\omega, T) &= \frac{1}{1 + e^{\omega - E_F / k_B T}}\end{split}\]

The resolution effects are added by convolving $I_\text{FD}(k_x, k_y, \omega)$ in energy and momenta with Gaussian functions whose full-width-half-max are defined by $\Delta E$ and $\Delta k$, respectively:

\[I_\text{res}(\mathbf{k}, \omega) = G(\Delta k, k_x, k_y) \ast \big(G(\Delta E, \omega) \ast I_\text{FD}(\mathbf{k}, \omega)\big)\]

The probe photon energy

For real values of theta and alpha (i.e., for the photoelectron to be emitted), the condition

\[E_k = h \nu - \Phi - E_b \geq \frac{\hbar^2 (k_x^2 + k_y^2)}{2 m_e}\]

must be satisfied for the deepest binding energies and largest momenta. This gives:

\[h \nu \geq \frac{\hbar^2 k_\text{lim}^2}{m_e} + \min(\omega) + \Phi\]

To ideally collect angles within ~45°, we add a factor of 2 to the RHS and a random number r in [0,5] eV:

\[h \nu = \frac{2 \hbar^2 k_\text{lim}^2}{m_e} + \min(\omega) + \Phi + r\]

The momentum-to-angle conversion

We define the sample’s out-of-plane direction as $\hat{z}$. The axis parallel (perpendicular) to the slit of the hemispherical analyzer is denoted $\hat{y}$ ($\hat{x}$).

The polar angle $\theta$ is a rotation about the $\hat{y}$ axis.
The tilt angle $\phi$ is a rotation about the $\hat{x}$ axis.
The azimuth angle $\alpha$ is a rotation about the $\hat{z}$ axis.

The angles $\theta$ and $\alpha$ are calculated from conservation of momentum:

\[\theta = \text{sign}(k_x)\arcsin\left(\frac{\hbar\sqrt{k_x^2+k_y^2}}{\sqrt{2m_eE_k}}\right), \quad\quad \alpha = \arcsin\left(\frac{\hbar k_y}{\sqrt{2m_eE_k}\sin\theta}\right)\]

The vector defining the direction of photoelectron propagation is:

\[\begin{split}\begin{equation} \begin{split} d'&=R(\alpha, \hat{z})R(\theta, \hat{y})d\\ &=\begin{pmatrix} \cos\alpha\cos\theta & -\sin\alpha & \cos\alpha\sin\theta \\ \sin\alpha\cos\theta & \cos\alpha & \sin\alpha\sin\theta\\ -\sin\theta & 0 & \cos\theta\\ \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} =\begin{pmatrix} \cos\alpha\sin\theta \\ \sin\alpha\sin\theta \\ \cos\theta \\ \end{pmatrix}\\ \end{split} \end{equation}\end{split}\]

To get the emission angles with respect to the analyzer, we find the angles $\theta_m$ and $\phi_m$ needed to map $d'$ onto $d$. We rotate first along the analyzer slit ($\theta_m$), then perpendicular to it ($\phi_m$):

\[\begin{split}\begin{equation} \begin{split} d&=R(\phi_m, \hat{x})R(\theta_m, \hat{y})d'\\ \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} &=\begin{pmatrix} \cos\theta_m & 0 & \sin\theta_m\\ \sin\phi_m\sin\theta_m & \cos\phi_m & -\sin\phi_m\cos\theta_m \\ -\cos\phi_m\sin\theta_m & \sin\phi_m & \cos\phi_m\cos\theta_m \\ \end{pmatrix} \begin{pmatrix} d'_x \\ d'_y \\ d'_z \\ \end{pmatrix}\\ \end{split} \end{equation}\end{split}\]

Adding offset angles is straightforward: we apply the rotation matrices $R(\theta_0,\hat{y})$ and $R(\phi_0,\hat{x})$ to $d$ before calculating the photoemission angles:

\[d' = R(\alpha, \hat{z})R(\theta, \hat{y})R(\phi_0,\hat{x})R(\theta_0,\hat{y})d\]

This leads to the system of equations:

\[\theta_m = -\arctan\left(\frac{d'_x}{d'_z}\right), \qquad \phi_m = \arcsin(d'_y)\]

Alternatively, if we rotate perpendicular to the slit first, then along the slit, we reverse the order of the matrices:

\[d = R(\theta_m, \hat{y})R(\phi_m, \hat{x})d'\]

In this case:

\[\theta_m = -\arcsin(d'_x), \qquad \phi_m = \arctan\left(\frac{d'_y}{d'_z}\right)\]

Having obtained this conversion for all $k_x$ and $k_y$, we define a square mesh of angles $\theta_M$ and $\phi_M$ and interpolate onto it using $(\theta_m, \phi_m, I_\text{res}(\mathbf{k},\omega))$.

The resulting values $I(\theta, \phi, E_k)$ are stored in arpes.intensity.

When simulating a sample with flake-like domains at different offset angles, the conversion is done for each domain. In this case, the function returns the intensity $I(\theta, \phi, E_k, \theta_0, \phi_0)$ directly, so intensity from all flakes can be combined externally.

The angle-to-momentum check

To convert from angle to momentum, we simply define a transformation matrix $T$ consisting of all the rotational matrices:

\[\begin{split}\begin{equation} \begin{split} T&=R_\text{motors}(\theta_m, \phi_m)R_\text{offset}(\theta_0, \phi_0, \alpha_0)\\ &=R(\theta_m, \hat{y})R(\phi_m, \hat{x})R(\theta_0,\hat{y})R(\phi_0,\hat{x})R(\alpha_0,\hat{z}) \end{split} \end{equation}\end{split}\]

Then the normalized vector for photoelectron $k_\text{norm}$ is given by solving $Tk_\text{norm}=d$

\[\begin{equation} k= |k|T/d, \end{equation}\]

where $d=(0,0,1)$, and $|k|=\sqrt{2m_e E_k}\hbar$.