Berry curvature in rotating traps A quantum system in a rotating (harmonic) trap is equivalent to a stationary system in the presence of a vector potential $\mathbf{A}$.
The proof can be found in chapter 5 here, but in short it goes like this:


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*Start with a time independent Hamiltonian for a harmonic 2D trap:
$$ H =\frac{\mathbf{p}^2}{2m} + \frac{1}{2}m(\omega_x^2\, x^2 + \omega_y^2\,y^2) $$

*We rotate it to $H(t) = R H R^\dagger$, where $R = e^{-i \phi L_z}$ is the rotation operator, $L_z$ dimensionless. 

*We solve the TDSE in our intertial frame for a time-dependent $H(t)$: $$ i\hbar \frac{\partial \Psi}{\partial t} = H(t)\Psi$$

*We go to the rotating frame by looking at the rotated wavefunction $\Psi' = R^\dagger \Psi$, and we find the following new TDSE: $$ i\hbar \frac{\partial \Psi'}{\partial t} = H(t)'\Psi', $$ where $$H(t)' = \frac{(\mathbf{p} - m\mathbf{A})^2}{2m} + \frac{1}{2}m(\omega_x^2\, x^2 + \omega_y^2\,y^2 - \Omega r^2), $$
where $\mathbf{A} = \Omega \times r = -\Omega y \boldsymbol{\hat{\imath}} + \Omega x \boldsymbol{\hat{\jmath}}$, and the $-\Omega r^2$ is a centrigular potential.

Questions


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*1) I should have got an expressio for the connection $\mathbf{A}$ as a Berry connection, since all I have done is introduce an (adiabatic?) time dependence to the state $\Psi \rightarrow e^{-i(\Omega t)\,L_z}\Psi$.


If I assume this to be my Berry phase $\gamma$, then the connection should come from:
$$ -\Omega\,t \,L_z = \gamma = \int \mathrm{d}\mathbf{R} \cdot \mathbf{A}, $$ where $\mathbf{R}$ should be my adiabatic parameter which I guess here is $\Omega t$?  I guess I could write $\Omega L_z$ as $\mathbf{\Omega} \cdot \mathbf{L} = \mathbf{\Omega} \cdot (\mathbf{r} \times \mathbf{p})= (\Omega \times r ) \cdot \mathbf{p}$, but I cannot get the desired $\mathbf{A} = \Omega \times r$ !


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*2) Can I get the scalar centrigular potential from the Berry argument too? The same reference on chapter IV gives a formula for the scalar potential:


$$ V(R) = \frac{h^2}{2m} \left ( \frac{d}{dR} |n(R) \rangle|^2 - \langle \frac{d}{dR} n(R)|n(R)\rangle \langle n(R)|\frac{d}{dR} n(R)\rangle \right ), $$
where again $R$ is the adiabatic parameter that in my case I assume is $\Omega t$, and $|n(R)\rangle = e^{-i \Omega t L_z} \Psi$? 
If I am completely off track, what would this $V$ be and how do I get the centrifugal potential from Berry?
 A: In this answer, I'll give an explicit formula of the state $|n\rangle$ and show how to obtain the Berry connection and scalar potential using the usual formula of the Berry connection and Shanker Berry potential formulas.
The situation is as follows: 
Our initial system is a (non-rotating) two dimensional (non-isotropic) harmonic oscillator. We are going to couple it adiabatically to an additional quantum system parameterized by the harmonic oscillator coordinates as slow coordinates (and having possibly other fast coordinates) such that in the adiabatic approximation governed by the additional system's (parametrized) ground state, the composite system will be the rotating harmonic oscillator.
Since we are working in the adiabatic limit, we need only to specify the additional system parametrized ground state eigenvector; we do not need to know the full dynamics. 
In order for the solution not to seem as a wild guess, I'll justify it before going into the actual computations. In the case of spin coupled to a magnetic field, the Berry curvature is the field of a Dirac monopole, which is uniform on the surface of the sphere and radially directed. In our case, the solution is a uniform magnetic field on the plane. So in principle we can choose our system to be a sphere and take the limit of the radius to infinity in order to obtain the plane. This procedure is based on the process known as the Wigner-İnönü contraction.  In this contraction the $SU(2)$ algebra of the sphere contracts to the Heisenberg-Weyl algebra of the plane. Since we know that the ground state of the spin system is a spin coherent state vector. We replace it in the case of the plane by the standard Glauber coherent state vector. The only complication is that in the plane case, the representation of the Heisenberg-Weyl algebra is infinite dimensional. But this is not really a problem because all the sums in the following are absolutely convergent.
Defining
$$z = x+iy$$
(We will need to scale $z$ as: $z\rightarrow \sqrt{\Omega} z$ in order to obtain the right formula. We can do that at the end of the computation for simplicity).
Then, the Glauber coherent state vector is given as follows:
$$|n\rangle = e^{-\frac{z\bar{z}}{2}}\begin{pmatrix}1  \\ \frac{z}{\sqrt{1!}}\\ \frac{z^2}{\sqrt{2!}} \\ .\\.\\.\\\frac{z^i}{\sqrt{i!}} \\ .\\.\\.\end{pmatrix} $$
Please check that the vector $|n\rangle$ is properly normalized:
The Berry potential is given by:
$$A = \langle n| d | n \rangle = \langle n| \frac{\partial}{\partial z} | n \rangle dz  + \langle n| \frac{\partial}{\partial \bar{z}} | n \rangle d \bar{z}$$
Please verify the result:
$$A = \frac{1}{2}(\bar{z} dz – z d\bar{z})$$
Returning to the Cartesian coordinates, we obtain:
$$A = i(x dy – y dx)$$
This after scaling gives the right result.
Similarly a direct application of the formula:
$$V(z, \bar{z}) = \langle n| \frac{\partial}{\partial z} | n \rangle  \langle n| \frac{\partial}{\partial \bar{z}} | n \rangle  - \langle n| \frac{\stackrel{\leftarrow}{\partial}}{\partial \bar{z}} \frac{\stackrel{\rightarrow} {\partial}}{\partial z} | n \rangle$$
Here we obtain:
$$V(z, \bar{z}) = - z \bar{z}$$
This gives the right result after scaling.
