Splitting

Reference

Background

Once we've created a quantum degenerate gas of atoms, we have the ability to dynamically deform the potential these atoms see. By applying resonant radio-frequency (rf) fields, we couple different internal magnetic substates of the atom, and the new potential includes contributions from the static magnetic and the oscillating magnetic (rf) fields. We call this new potential a "dressed" potential. The exact shape of this new potential depends on the frequency of the rf field, such that we can sweep the frequency to deform the trap.

In the series of images below, we represent a one-dimensional dressed potential seen by the atoms as the frequency is increase left to right. At low frequencies, the potential is only slightly deformed. As we increase the frequency, a barrier is raised in the centre of the trap and we go from a single to a double well.

doubleup6.jpg
doubleup1.jpg
doubleup4.jpg
doubleup2.jpg
doubleup5.jpg

Once we've split the cloud in two, we're interested in learning how the atoms decided to go either to the right side or the left side. For bosons in a BEC, this decision is dominated by the effects of interactions. In 87Rb, where there exists a repulsive interaction between the bosons in the BEC, the system wishes to minimise its total energy. Since it costs energy to put bosons together on one side of the well, the overall energy is minimised when the exact same number of atoms go to each side of the well. In contrast, if there were no interactions between the atoms, the choice to go right or left would be random, and therefore governed by binomial statistics.

We are interested in looking at the fluctuations in the numbers of atoms that go right or left over a large number of experimental realisations. We take a BEC, split it in two, then count the number that went to the right and the number that went to the left. We then repeat this many times. Then, we perform statistics on these results and look at the fluctuations in the numbers - for instance: what is the spread in the fraction of atoms that chose to go right?

If this were a random process, we expect that this spread is proportional to $\sqrt{N}$, where N is the total number of atoms in the system. If the interactions in the BEC play a role, we expect this number to be less than $\sqrt{N}$. This reduction in fluctuations would indicate that the system of atoms somehow acts collectively to reduce its overall energy.


Experiment

To actually count the numbers of atoms that went into the right and left wells, we use absorption imaging. Resonant light is shone through the atoms at a CCD camera, and the shadow image is recorded on our computer. The atoms are then transferred into a state out of resonance with this light, and a second pulse is shone on the CCD camera, for reference. After dividing the two images, we can determine the number of atoms that were in each well from the darkness of the shadow. Below, we see one such image.

splitclouds2.jpg

Another tool we have is to look at the phase difference between the wavefunctions of the atoms on the right and the left. To do this, we let the two clouds fall under gravity. The clouds expand into each other, and if we take another absorption image, we see interference fringes develop between the two clouds. The phase of the fringes (relative locations of the peaks and valleys with respect to the centre of the cloud) tells us about the phase of the wavefunction before release. This information is useful in determining how well we do our splitting (is it coherent?) and in determining the point at which the two condensates stop communicating with each other (how much tunnelling through the barrier?)

fringes1.jpg

Theory

References

A bosonic Josephson junction R. Gati & M.K. Oberthaler, J. Phys. B: At. Opt. Mol. Phys. 40, R61 - R89 (2007).

Two-mode model

To understand the behaviour of a BEC in a double well potential, we must consider the interactions between atoms and the tunnelling between the two wells. We assume (for now) that the temperature is very low and that the condensate resides in the ground state of the potential. We restrict ourselves to a "two-mode" model, where we consider only the ground and (nearly-degenerate) first excited state of the double well potential.

Two-mode Hamiltonian

In constructing a hamiltonian, we include a term describing the kinetic and potential energy (regular Schroedinger equation) and a term that accounts for interactions.

(1)
\begin{align} \hat{H} = \hat{H_0} + \hat{H_{int}} = \int d{\bf r} \left[\frac{-\hbar^2}{2m} \hat{\Psi}^{\dagger} \nabla^2 \hat{\Psi}+\hat{\Psi}^{\dagger} V_{dw} \hat{\Psi}\right] + \frac{g}{2}\int d{\bf r}~\hat{\Psi}^{\dagger}\hat{\Psi}^{\dagger} \hat{\Psi}\hat{\Psi} \end{align}

where $V_{dw}$ is the double well potential. One can show (Gati & Oberthaler, 2007) that if you rewrite the field operators as a sum of contributions from the ground ($\phi_g$) and first excited $\phi_e$ states,

(2)
\begin{align} \hat{\Psi} = \hat{c}_g \phi_g + \hat{c}_e \phi_e \end{align}

and then change to the right-left basis, where

(3)
\begin{align} \hat{c}_r = \frac{1}{\sqrt{2}}(\hat{c}_g + \hat{c}_e);~~~ \hat{c}_\ell = \frac{1}{\sqrt{2}}(\hat{c}_g - \hat{c}_e) \end{align}

you can then rewrite the two-mode hamiltonian as

(4)
\begin{align} \hat{H}_{2M} = \frac{E_C}{8} (\hat{c}_r^{\dagger}\hat{c}_r - \hat{c}_{\ell}^{\dagger}\hat{c}_{\ell})^2 - \frac{E_J}{N} (\hat{c}_r^{\dagger}\hat{c}_{\ell} + \hat{c}_{\ell}^{\dagger}\hat{c}_{r}) + \frac{\delta E}{4}(\hat{c}_r^{\dagger}\hat{c}_{\ell} + \hat{c}_{\ell}^{\dagger}\hat{c}_{r})^2 \end{align}

where

(5)
\begin{align} E_C = 8 \kappa_{g,e} \end{align}
(6)
\begin{align} E_J = \frac{N}{2}(\mu_e-\mu_g) - \frac{N(N+1)}{2}(\kappa_{e,e} - \kappa_{g,g}) \end{align}
(7)
\begin{align} \delta E = \frac{\kappa_{e,e} + \kappa_{g,g} - 2\kappa_{g,e}}{4} \end{align}
(8)
\begin{align} \kappa_{i,j} = \frac{g}{2} \int d{\bf r} |\phi_i|^2 |\phi_j|^2 \end{align}
(9)
\begin{align} \mu_{e,g} = \int d{\bf r} \frac{-\hbar^2}{2m} \phi_{e,g}^{*} \nabla^2 \phi_{e,g}+\phi_{e,g}^{*}( V_{dw} + gN|\phi_{e,g}|^2 ) \phi_{e,g} \end{align}

Discussion of terms

Question: What do each of these expressions mean and what do these quantities depend on?

  • $E_C$ is called the charging energy. It belongs to the term that multiplies the number difference, so it tells us how much energy it costs to have a different number of atoms in the right and the left wells. It depends on the probability we exist in both of the two modes, and is maximised when $\phi_e = \phi_g$, or there is an equal superposition of ground and excited states, which is saying that all of the atoms are in the left or all are in the right mode. The energy is minimized when the right and left modes are equally populated. This energy term also depends on the size and sign of the interactions between atoms, and disappears for a non-interacting system.
  • $E_J$ is the tunnelling energy. It belongs to the tunnelling term, which tells us what happens when we move a particle from the right to the left well, of vice-versa. There are two parts to this expression. The first, which depends on the chemical potentials of the ground and first excited states, tells us about the energy cost to move a particle from one well to another - the energy it takes to move an atom from right to left. Examining this term, we see that (overall) it is positive if the ground state is more populated (the wells are balanced) and zero when the symmetric and antisymmetric modes are equally populated (tendency for atoms to be all in left or all in right). Thus, this drives the system to want to tunnel all to one side or all to the other. The second term in the tunnelling energy depends explicitly on interactions, in addition to the interactions included in the chemical potential term. If we are in a state where the wavefunction is primarily of a symmetric character ($\phi_g > \phi_e$, balanced wells) then this energy term is overall negative. If the wavefunction is an equal superposition of symmetric and antisymmetric ($\phi_g \approx \phi_e$), all on right or all on left), then this part of the tunnelling term is zero. This drives the system to tunnel such that the overall interaction energy is reduced by balancing the splitting. The two terms are in competition with each other, such that the tunnelling behaviour is not straightforward.
  • $\delta E$ is a small term, and I'm going to assume it doesn't matter for us like it doesn't matter for the Oberthaler group.

Simplified Hamiltonian

We can define the number difference $\hat{n}$ and tunnelling operators $\hat{\alpha}$ from the left and right mode operators

(10)
\begin{align} \hat{n} = \frac{\hat{c}_r^{\dagger}\hat{c}_r - \hat{c}_{\ell}^{\dagger}\hat{c}_{\ell}}{2}; ~~~ \hat{\alpha} = \frac{\hat{c}_r^{\dagger}\hat{c}_{\ell} + \hat{c}_{\ell}^{\dagger}\hat{c}_{r}}{N}, \end{align}

and simplify the Hamiltonian

(11)
\begin{align} \hat{H}_{2M} = \frac{E_C}{2} \hat{n}^{2} - E_J \hat{\alpha}. \end{align}

This is usually called the Bose-Hubbard Hamiltonian.

Mean-field, Gross-Pitaevskii

Assuming a mean-field picture of a BEC, where the temperature is low and interactions are local, we assume we can write the operators for each side of the well as complex numbers describing the properties of the wavefunction in each half of the well, ie, $\hat{c_R} = \sqrt{N_R} exp(-i\phi_R)$ where $N_R$ is the number of atoms on the right side, and $\phi_R$ is the phase of the wavefunction on the right side. In so doing, we can rewrite the Hamiltonian as

(12)
\begin{align} {H}_{2M, GP} = \frac{E_C}{2} {n}^{2} - E_J \sqrt{1 - \frac{4 n^2}{N^2}}\cos\phi, \end{align}

where $n = (N_L - N_R) / 2$ is the number difference between the wells and $\phi = \phi_R - \phi_L$ is the phase difference. We can expand this Hamiltonian to second order in [[ $ \phi $]] and [[ $ n$ ]]to get something that looks much like the canonical harmonic oscillator Hamiltonian

(13)
\begin{align} {H}_{simple} = \left(E_C + \frac{4 E_J}{N^2}\right) \frac{{n}^{2}}{2} + E_J \frac{\phi^2}{2}, \end{align}

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