Following our previous post and Ref. [1], we want to use the equation for the interband coherence
\begin{eqnarray}
\left(i\hbar\frac{d}{dt}-\Delta\right) \rho_{v\mathbf k',c\mathbf k}(t)
&&
+ \sum_{i\neq 0} V_{\mathbf i} \rho_{v\mathbf k'- \mathbf i,c\mathbf k- \mathbf i}(t)
\\
&&=
\langle c\mathbf k | [-(Q/m) \mathbf A(\mathbf r, t) \cdot \mathbf p] | v\mathbf k' \rangle
~~~~~~~~~~ (1)
\end{eqnarray}
(see Below-bandgap excitation. Part 1) with a twisted-light vector potential. Then, the matrix elements in Cartesian coordinates are
$$
\langle c\mathbf k | [-(Q/m) \mathbf A(\mathbf r, t) \cdot \mathbf p] | v\mathbf k' \rangle = \xi(t) \frac{e^{i\theta\ell}}{L q_r}
$$
where $\mathbf k' = \mathbf k - \mathbf q_0$ and $q_0 = q_r \cos(\theta) \hat{x} + q_r \sin(\theta) \hat{y} + q_z \hat{z}$ with variable $\theta$ [2]. Following the procedure for plane waves, we find
\begin{eqnarray}
&& \left[
\hbar \omega -10
\left(E_g + \frac{\hbar^2\,\mathbf q_0^2}{2|m_v^*|}\right) +
i\, \frac{\hbar^2\,\mathbf q_0}{|m_v^*|} \cdot \nabla +
\frac{\hbar^2}{2\, \mu} \nabla^2 +
V(\mathbf{r})
\right] \rho_{\mathbf q_0 } (\omega, \mathbf{r})
\nonumber \\
&& \hspace{30mm}
= L^3
\, \xi(\omega)\,
\frac{e^{i\theta\ell}}{L q_r}
\delta(\mathbf{r})
\hspace{30mm} (2)
\,.
\end{eqnarray}
Notice the new factor in the RHS.
Once again we can solve by completing squares and using the unitary transformation $U(\mathbf r) = exp[i(\mu/|m_v^*|) \mathbf q_0 \cdot \mathbf r]$:
\begin{eqnarray}\label{Eq_Coherence_TS_final}
\rho_{v \mathbf{k} - \mathbf q_0 ,c \mathbf{k}}(\omega)
&=& L^3\, \xi(\omega)\,\frac{e^{i\theta\ell}}{L q_r}
\times \nonumber \\
&& \hspace{-20mm}
\sum_\nu \,
\frac{\psi^*_\nu(\mathbf{r}=0)}
{\hbar\,\omega - E_g - \frac{\hbar^2 \mathbf q_0^2}{2 M} - E_\nu}
\psi_\nu \left(\mathbf{k} - \frac{\mu}{|m_v^*|} \mathbf q_0 \right)
\hspace{20mm} (3)
\,.
\end{eqnarray}
Notice the new phase factor.
Let us focus first on the center-of-mass (COM) motion of the exciton-like particle, signaled by $\hbar^2 \mathbf q_0^2/(2 M)$ in the denominator of Eq. (3). Thanks to the square, all interband coherences (differing in their $\mathbf q_0$ or more specifically in the angle $\theta$) share the same COM momentum given by $\{q_r, q_z\}$. For a plane wave, $\hbar q_z$ is clearly the momentum of the photon traveling along $z$; for twisted light, $q_z$ is the component of the wave-vector in the $z$ direction. If we calculated the linear momentum of the field $\mathbf P = \epsilon_0 \mathbf E \times \mathbf B$ in the paraxial approximation $q_r<q_z$ we would observe that the radial and angular components of $\mathbf P$ are proportional to $q_r$, and so $q_r$ is the transverse wave-vector. The excitons are then absorbing the momentum of the twisted light beam, which contributes the kinetic energy $\hbar^2 \mathbf q_0^2/(2 M)$ that adds to $E_g$. To extend our understanding, recall how we expanded the twisted-light vector potential in terms of plane-waves (end of Absorption in Bulk. Part 1): we could then say that each plane-wave with a particular linear momentum transfers it to the COM of an exciton (exciton-like).
The second interesting feature is the argument of $\psi_\nu$ in Eq. (3). We know that a pure exciton is represented by $\psi_\nu(\mathbf k)$; them this tells that twisted light creates state that are not exactly excitons, or in other words, the relative-motion wave particle is perturbed.
Finally, and also in connection to the finding in Absorption in Bulk. Part 1 related to the expansion of $\mathbf A(\mathbf r,t)$ we see that Eq. (3) contains a particular phase factor for each value of $\theta$.
Macroscopic quantities
Equation (3) is the building-block of several macroscopic quantities. I show as an example the polarization [1]. Because the twisted light field has spatial variation, one must consider a local polarization
\begin{eqnarray}\label{Eq_Pol}
{\mathbf P}({\mathbf R}, t)
\hspace{0mm}
&=&
\hspace{0mm}
2
\left(\frac{L_{\cal R}}{L}\right)^{\!3}
\! \sum_{\mathbf k, \, \theta }
\Re\left\{e^{i \mathbf q_0 \cdot \mathbf R}
\,\mathbf d_{vc} \,
\rho_{v \mathbf k - \mathbf q_0, c \mathbf k }(t)
\right\}
\!, \hspace{4mm}\,
\end{eqnarray}
corresponding to a macroscopic cell located at $\mathbf R = (R,\Phi,Z)$ with linear size $L_{\cal R}$; $\Re$ is the real part. Now it is evident that the response of the system to the twisted-light beam takes into account a superposition of states differing in the value of $\theta$ -for plane waves, the polarization still retains the sum over $\mathbf k$. From the polarization one can derive the susceptibility which clearly shows the absorption lines, in this case shifted by the COM term.
References
[1] I recall that, in problems of twisted light-matter interaction, one should not neglect of the photon's momentum. We therefore have abandoned the "vertical transition" approximation.
[2] Notice the connection between the vector $\mathbf q_0$ of electrons and the vector $\mathbf q(\eta)$ of the light entering the superposition of planes waves that form a twisted-light vector potential - see the end of the previous post Absorption in Bulk. Part 1.
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