7

How do you get the Hamiltonian formula for transition metal dichalcogenides? How to derivate the expression above (equation (1))? You can use the $ k \cdot p$ method to derive this effective Hamiltonian. Please take a look at this paper for the details. $k \cdot p$ theory for two-dimensional transition metal dichalcogenide semiconductors. What does the ...


6

There are a few issues with the derivation: (not necessarily a mistake, but) please note that if you mean the Coulomb Hamiltonian, the RHS of (1) should involve the reciprocals of $|r_i|$ etc., not $r_i^2$ etc. The Coulomb interaction enters the Hamiltonian in the form of potential energy, rather than in the form of force. Please ignore this comment if you ...


6

Finally, I figure out this question on my own. Put Eq.$(2)$ and $(3)$ into Eq.$(1)$: \begin{equation} -e \Re\{\mathcal{E}_a e^{i\omega t}\} (\partial_a f_0+\partial_a f_1)+(\partial_tf_1+\partial_tf_2)=\dfrac{-f_1-f_2}{\tau} \tag{7} \end{equation} in which $\Re\{\mathcal{E}_a e^{i\omega t}\}f_2\propto |\mathcal{E}|^3$ is droped and $\partial_t f_0=0$. ...


6

They start with the time-independent Schrödinger equation: $$\tag{1} H|\psi^\alpha\rangle = E^\alpha |\psi^\alpha\rangle. $$ Then they define $|\psi_0^\alpha\rangle$ to be what they call a "model state", which is an approximation to the true state $|\psi^\alpha\rangle$ but more easily accessible, and is related to the true state by a Møller ...


4

As @Jack suggested, try the $\vec{k} \cdot \vec{p}$ method. The hopping integral $t_{ij}$ is "borrowed" from the Hubbard model. It is the kinetic term in the Hamiltonian that explains electrons being able to "hop" from one atomic site to another. In real space, it is an integral of the orbital overlap between the two neighboring atomic ...


3

The TD-Schrodinger equation is $$H(\lambda)|\psi(t)\rangle=i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle$$ So, we just need to plug in \eqref{4} and remove all terms that aren't first-order in $\dot\lambda$. Let's start with the left-hand side, since its a little simpler. $$\begin{align}H(\lambda)|\psi(t)\rangle&=e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\...


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