Tuesday, 19 March 2019

quantum mechanics - How to include Berry connection in Hamiltonian?


When we calculate Berry connection, $A(R)=i<\psi(x,y)|\frac{d}{dR}|\psi(x,y)>\hat{R}$ corresponding to the Berry phase of any system, the gauge potential is related to the $R$ of the parameter space. It is not dependent on $x$ like $\psi(x)$ because in the inner product the spatial part is integrated over.


We also know that in presence of a vector potential, the Hamiltonian can be written as $\mathcal{H}=\dfrac{(p-A)^2}{2m}+V$. Here $p=p_x \hat{i}+p_y \hat{j}$, (in 2D) and $A$ is also supposed to be $A_x \hat{i}+A_y\hat{j}$.


But in case of Berry connection, $A=A(R) \hat{R}$ corresponding to parameter space. So, how to include this Berry gauge potential in Hamiltonian? I cannot just subtract a vector in parameter coordinates from a vector in spatial coordinate, right?



Answer



The Berry connection lives in the parameter space, thus it appears not in the microscopic Hamiltonian given in the question but in the effective Hamiltonian equation of motion in the parameter space. The aim, in the following, is to show the details in the variational approximation.



To be precise, the bra-ket notation that I'll be using is explained in the following two equations: $$\mathbf{A} = \langle\psi(R) |\mathbf{\nabla_R} | \psi(R) \rangle \triangleq \int d^3x \ \psi(x,R)^{\dagger} \mathbf{\nabla_R}( \psi(x,R)) $$


Where $\psi(x,R)$ are the scalar wave functions corresponding to the state vectors $| \psi(R) \rangle$


In components the above equation takes the form:


$$A_i = \langle\psi(R) |\frac{\partial} {\partial R^i} | \psi(R) \rangle = \int d^3x \ \psi(x,R)^{\dagger}\frac{\partial} {\partial R^i}\psi(x,R)) $$


Given a microspcopic Hamiltonian (which can also have an explicit dependence on the parameter space$H(x, R)$, (For a fixed $R$, $H(x, R)$ can be a Schroedinger operator in the real space), the effective Hamiltonian on the parameter space is defined by:


$$\mathcal{H}(R) = \langle\psi(R) |H | \psi(R) \rangle = \int d^3x \ \psi(x,R)^{\dagger}H\psi(x,R)) $$


$H$ can have the form given in the question with $A$ being an additional external field not related to the Berry connection.


The exact time dependent shroedinger equation:


$$ \frac{\partial | \psi \rangle }{\partial t} = H | \psi\rangle $$


can be derived from by the variation of the Lagrangian:



$$ L = \langle\psi(R) |(\frac{\partial}{\partial t}- H )| \psi(R) \rangle$$


According to the variational approximation, we seek a solution by varying the state vectors not on the whole Hilbert space but only within the parameter space. The meaning of this approximation that we are not allowing the state vectors to vary in the directions of the real space, thus we are close to the lowest excitation state of the hamiltoniam for a fixed $R$.


Thus, we vary the Lagrangian only with respect to the parameter space and find a solution to the Lagrange equation of motion:


$$\frac{d}{dt} (\frac{\partial L}{\partial \dot{R}^i}) - \frac{\partial L}{\partial R^i} = 0$$


Using the above notation we have:


$$ L = A_i \dot{R}^i - \mathcal{H}(R) $$


The Lagrange equations of motion:


$$\frac{dA_i }{dt} - \frac{\partial A_j}{\partial R^i} \dot{R}^j - \frac{\partial \mathcal{H}}{\partial R^i} = 0$$


Using:


$$\frac{dA_i }{dt} = \frac{\partial A_i}{\partial R^j} \dot{R}^j$$



We obtain:


$$( \frac{\partial A_i}{\partial R^j} - \frac{\partial A_j}{\partial R^i} )\dot{R}^j = \frac{\partial \mathcal{H}}{\partial R^i} $$


Recognizing the expression of the Berry curvature:


$$ F_{ij}(R) = \frac{\partial A_i}{\partial R^j} - \frac{\partial A_j}{\partial R^i}$$


We obtain:


$$ F_{ij}(R)\dot{R}^j = \frac{\partial \mathcal{H}}{\partial R^i} $$


If the Berry curvature is invertible, i.e., there is a matrix $\Omega^{ij}(R)$ such that:


$$\Omega^{ij}(R) F_{jk}(R) = \delta^i_k,$$


we get the equation of motion on the parameter space:


$$ \dot{R}^i = \Omega^{ij}\frac{\partial \mathcal{H}}{\partial R^j} $$



These are classical Hamilton equation of motion, with the symplectic structure equal to the Berry curvature. Thus the quantum evolution can be approximated by a classical evolution in the parameter space.


Update: Effect on the system spectrum


In many cases, the parameter space is compact integrable system. In this case, its quantization splits each spectral line of the system's spectrum into a finite number of energy levels. In this case there will be a finite number of periodic solutions of the equations of motion


$$R(t) = R(t+T),$$


living on fixed energy hypersurfaces $\mathcal{H} = E = const.$. We should fix our system parameters such that the Bohr-Sommerfeld quantization condition:


$$\int_{0}^{T} A_i(R(t) )\dot{R}^i(t) dt = 2\pi n$$


Where $n$ is a fixed integer. The above conditions are implicit equations which fix the energy levels $E_k$ and the corresponding periods $T_k$.


The corresponding state is given by:


$$| \psi_k\rangle = \frac{1}{T_k}\int_0^{T_k} dt \ e^{i\int_0^{t} A_i(R(\tau) )\dot{R}^i(\tau) d\tau} | \psi(R(t))\rangle $$


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