# 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:

• 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

• 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 $\mathbf{A} = \Omega \times r$ !

• 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?

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.

• Thanks, but I have time $t$ and $\phi = \Omega t$. How do I relate this to $x$, $y$ and your harmonic oscillator $r = |z|$? – SuperCiocia Feb 26 '18 at 20:16
• Sorry so does this mean the argument is only valid for coherent states? What about a generic state, eigenstate of the Hamiltonian? – SuperCiocia Feb 27 '18 at 1:18
• In the following I'll call the system given in the question: "Our system" and the system that we couple to our system the "additional system". Our system, i.e., the system given in the question is a harmonic oscillator. It has two natural frequencies $\omega_x$ and $\omega_y$. What I explained in my answer is how to obtain the dynamics of our system (i.e., the harmonic oscillator) in a rotating frame of reference, not by means of a coordinate transformation as done in the article, but instead by adiabatically coupling it to an additional system. – David Bar Moshe Feb 27 '18 at 8:21
• Cont. This is gives meaning to the vector $|n\rangle$ as the ground state of the additional system. The additional system is not necessarily a Harmonic oscillator. In fact I was not specific about its Hamiltonian. The only requirement was the form of its ground state (and its dependence on the parameter space which is the configuration space in our case). This state can be a ground state of many Hamiltonians, but we do not need to know which. – David Bar Moshe Feb 27 '18 at 8:23
• Cont. Our system which is eventually a $2D$ Harmonic oscillator in a rotating system does not have to be in a coherent state. It can assume any state compatible with its dynamics. It is the additional system coupled to our system which has to be in a coherent state, because only this specific coherent state gives rise to a Berry curvature of the form of a uniform magnetic field in the $z-$ direction. – David Bar Moshe Feb 27 '18 at 8:23