# Metadata
**Title:** Spin Polarization of Photoelectrons from Topological Insulators
**Authors:** Park, Cheol-Hwan; Louie, Steven G.
**Year:** 2012
**Tags:** #TI #ARPES
[URL](http://dx.doi.org/10.1103/physrevlett.109.097601)|[Bookends](bookends://sonnysoftware.com/18744)
---
# Maximum spin Polarization and Photocurrent
The measured [[Physics/Spin|spin]]-polarization in spin-[[Angle-Resolved Photoemission Spectroscopy|ARPES]] experiments is defined by:
$
P_{\text{max}} = \max_{\hat{t}} \frac{I_{\hat{t}} - I_{-\hat{t}}}{I_{\hat{t}} + I_{-\hat{t}}}
$
Where $I_\hat{t}$ and $I_{-\hat{t}}$ are the intensities for the electron spin aligned and anti-aligned with $\hat{t}$ respectively. Here, the unit [[vector]] $\hat{t} = \hat{t}_{\max}$ that maximizes the equation above defines the direction of the electron spin-polarization.
The photocurrent from a detector (i.e. an apparatus that selects a specific 3D wavevector and hence an energy) with the spin quantization axis aligned with $\hat{t}$ is:
$
I_\hat{t} \propto \left| \sum_{\{f|E_f = E_i + h\nu\}} \langle \hat{t}, \pmb{R}_D | f \rangle \langle f | H^{\text{int}} | i\rangle \right|^2
$
Here $|i\rangle$ is the initial TSS state, $|f\rangle$ is the final photoexcited state, $E_i$ and $E_f$ are their energies. $h\nu$ is the photon energy, and $H^\text{int}$ is the light-matter interaction [[Hamiltonian]] connecting the two states. Finally, $|\hat{t}, \pmb{R}_D\rangle$ is the detected state. Its spin part is the [[eigenstate]] of $\pmb{s}.\hat{t}$ with [[eigenvalue]] +1 and its spatial part is localized at $\pmb{R}_D$, the location of the detector far away from the sample. Here we use $\pmb{s}$ to denote the [[vector]] of [[Pauli matrices]] for [[Spin-Half|spin-1/2]] systems.
## Deriving $I_\hat{t}$
Consider an electronic system described by a single particle [[Hamiltonian]] $H_0$ and a time dependent pertubation of the form $H_{\text{int}}(t) = H^{\text{int}}(\pmb{A}) e^{-2 \pi i \nu t}$, where $\pmb{A}$ is the [[Vector Potential]] and $h\nu$ is the [[photon]] [[energy]]. We are interested in the evolution of an electron which is at [[eigenstate|state]] $|i\rangle$ before turning on the time-dependent perturbation into a final state $|f\rangle$. Thus, the total [[Hamiltonian]] is:
$
H(t) = H_0 + H^{\text{int}}(\pmb{A})e^{-2\pi i \nu t}
$
Assuming the perturbation is turned on at $t = 0$, the normalized [[wavefunction]] of the electron at time $t$ can be written, to first order, as
$
|\psi(t)\rangle = e^{-i E_i t/\hbar} |i\rangle + \sum_f c_f(t) e^{-i E_f t/\hbar}|f\rangle
$
where $|f\rangle$ are the possible final states and $c_f(t)$ is the probability amplitude of finding the electron in the state $|f\rangle$ at time $t$. Clearly, $c_f(0) = 0 \, \forall \, f$.
Solving the time-dependent [[Schrödinger equation]], we get:
$
c_f(t) = -\frac{i}{\hbar}e^{it \frac{\Delta E_f}{2\hbar}} \frac{\sin \frac{\Delta E_f}{2\hbar}}{\frac{\Delta E_f}{2\hbar}} \langle f | H^\text{int} | i \rangle
$
where $\Delta E_f \equiv (E_f - E_i) - h\nu$.
The transition rate is given by [[Fermi's Golden Rule]] as follows:
$
\Gamma_{f \leftarrow i} = \lim_{t\rightarrow \infty} P_f(t)/t = \frac{2 \pi}{\hbar} \delta(E_f - E_i - h\nu) |\langle f | H^{\text{int}}|i\rangle|^2
$
Now, in a spin-resolved [[photoemission]] experiment, the photocurrent is given by
$
I_\hat{t} \propto \lim_{t \rightarrow \infty} | \langle \hat{t}, \pmb{R}_D | \psi(t) \rangle|^2/t
$
Now, we can write the [[inner product]]
$
\begin{align}
\langle \hat{t}, \pmb{R}_D | \psi(t) \rangle &= \sum_E \sum_{\{f | E_f = E\}} -\frac{i}{\hbar}e^{it \frac{\Delta E_f}{2\hbar}} \frac{\sin \frac{\Delta E_f}{2\hbar}}{\frac{\Delta E_f}{2\hbar}} \langle f | H^\text{int} | i \rangle e^{-i E_f t/\hbar} \langle \hat{t}, \pmb{R}_D|f\rangle,\\
&= \sum_E -\frac{i}{\hbar}e^{it \frac{\Delta E - 2E}{2\hbar}} \frac{\sin \frac{\Delta E}{2\hbar}}{\frac{\Delta E}{2\hbar}}
\sum_{\{f | E_f = E\}} \langle f | H^\text{int} | i \rangle\langle \hat{t}, \pmb{R}_D|f\rangle
\end{align}
$
where $\Delta E = E - E_i - h\nu$. Here we have used the fact that $\langle \hat{t}, \pmb{R}_D | i\rangle = 0$ since the initial state is localized at the crystal and far away from the detector. ==Obviously==, a product of the terms in the equation above corresponding to different $E$ values does not contribute to the photocurrent calculated from the equation for $I_\hat{t}$
Plugging back into $I_\hat{t}$, and using the equation:
$
\lim_{a\rightarrow \infty} \frac{1}{\pi} \frac{\sin^2(ax)}{ax^2} = \delta(x)
$
We end up with:
$
\begin{align}
I_\hat{t} &\propto \sum_E \delta(E - E_i - h\nu) \sum_{\{f | E_f = E\}} |\langle \hat{t}, \pmb{R}_D|f\rangle\langle f | H^\text{int} | i \rangle|^2 \\
&\propto \sum_{\{f | E_f = E_i + h\nu\}} |\langle \hat{t}, \pmb{R}_D|f\rangle\langle f | H^\text{int} | i \rangle|^2
\end{align}
$
## Interpretation of $I_\hat{t}$
The detector reads the spin character of the photoexcited state at $\pmb{R}_D$, far away from the surface sample, thus the near-surface part of the [[wavefunction]] only affects the measured spin indirectly through the [[Matrix Element]] $\langle f | H^{\text{int}}| i \rangle$. Moreover, a summation of the transition amplitude over [[degeneracy|degenerate]] photoexcited states is necessary because we cannot tell, even in principle, which state $|f\rangle$ is involved in detection.
We can write
$
|f'\rangle = \sum_{\{f|E_f = E_i + h\nu\}} |f\rangle\langle f|H^\text{int}|i\rangle
$
Then
$
I_\hat{t} = |\langle \hat{t}, \pmb{R}_D|f'\rangle|^2
$
Since at $\pmb{R}_D$, where the measurement is performed, the $|f\rangle