Graduate How to calculate the interband velocity matrix in a graphene system

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In graphene systems, the velocity operator can be expressed as v = ∂H/ħ∂p, with matrix elements calculated as <ψ|v|ψ>, yielding v_x = v_F cos(θ) and v_y = v_F sin(θ} for intraband velocity. A new method for calculating the velocity matrix, as presented by Mikhailov, involves deriving the velocity operator from the equations of motion rather than directly from the Hamiltonian. The two approaches yield the same results but differ in their mathematical methods; the first is a direct calculation from the Hamiltonian, while the second is an indirect calculation based on motion equations. Understanding these differences is crucial for applying the appropriate method in research. Both methods are valid and widely recognized in condensed matter physics.
Sunny Huang
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In graphene system, the velocity operator sometimes is v= ∂H/ħ∂p, and its matrix element is calculated as <ψ|v|ψ>, i.e., v_x = v_F cos(θ) and v_y = v_F sin(θ) [the results are the same with Eq. 25] for intraband velocity. Recently, I see a new way to calculate the velocity matrix (Mikhailov, Sergey A. "Quantum theory of the third-order nonlinear electrodynamic effects of graphene." Physical Review B 93.8 (2016): 085403.). But I cannot understand the Eq. 20 (see the following picture). Additionally, what the difference between the two ways to define the velocity operator?
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The two ways of calculating the velocity operator are essentially the same; they just use different mathematical methods to calculate the same values. The first way is to use the Hamiltonian to derive the velocity operator as v= ∂H/ħ∂p, where H is the Hamiltonian and p is the momentum. This method has been used for a long time and is well known in condensed matter physics.The second way, as described in Mikhailov's paper, is to calculate the velocity operator by using the equations of motion. In this method, the equations of motion for electrons in graphene are written as \frac{\partial \psi (r)}{\partial t} = \frac{1}{i\hbar} \left[ H_{0} + \sum_{i=1}^{4} H_{i} , \psi (r) \right]where H_0 is the kinetic energy and H_i are the interaction terms. Then, the velocity operator is derived from these equations of motion asv_{x,y} = \frac{1}{i\hbar} \left[ H_{0} + \sum_{i=1}^{4} H_{i} , x_{x,y} \right]The main difference between the two approaches is that the first approach is a direct calculation of the velocity operator from the Hamiltonian, while the second approach is an indirect calculation of the velocity operator from the equations of motion.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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