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I recently read about BEC loaded into the optical lattice p.200

Looking at a condensate released from a lattice after a time of flight typically on the order of a few milliseconds amounts to observing its momentum distribution. A harmonically trapped condensate has a Gaussian momentum distribution in the limit of small interactions, whereas in the Thomas-Fermi limit in which the interactions dominate over the kinetic energy contribution it has a parabolic density profile and expands self-similarly after being released. By contrast, a condensate in a periodic potential contains higher momentum contributions in multiples of 2kL, their relative weights depending on the depth of the lattice. In fact, in the tight-binding limit see Sec. IV we can consider the condensate to be split up into an array of local wave functions that expand independently after the lattice has been switched off. Eventually they all overlap and form an interference pattern that in the absence of interactions is the Fourier transform of the initial condensate.

When there is no lattice potential all particles occupy the same state, but when the lattice appears is it still a single mode condensate or multi-mode one due to appearance of additional interference peaks (particle condense in more than one state)?

These were obtained after suddenly releasing the atoms from an optical lattice potential with different potential depths V0 after a time of flight of 15 ms. Values of V0 were: a, 0 Er; b, 3 Er; c, 7 Er; d, 10 Er; e, 13 Er; f, 14 Er; g, 16 Er; and h, 20 Er. source

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In a lattice, forgetting about the interactions, for now, the eigenstate of the Hamiltonian are the Bloch functions $|u_{n,q}\rangle$, which have the same periodicity as the lattice (I will do everything in 1D to alleviate the notations, but it is easily generalized to other lattices). This implies that it has a Fourier series decomposition $$ u_{n,q}(x)=\sum_ {m\in \mathbb {Z}} \tilde u_{n,q}(m) e^{i m G x}, $$ with $G$ the primitive vector of the reciprocal lattice.

Assume that, as is usually the case, the lowest energy state is the state in the band $n=0$ with quasimomentum $q=0$. The BEC then forms in this state, which is macroscopically occupied. When the trap is released to perform a time-of-flight experiment, the atoms start in the state $|u_{0,0}\rangle$, but evolve with the free particle Hamiltonian. It can then be shown that under some hypothesis (no collisions, long enough times, etc.) that the density of atoms is proportional to the Fourier transform of the initial wave-function, with the wave-vector replaced by $\frac{m x}{\hbar t}$.

In the present case, this means that the density $n(x,t)$ measured after a time-of-flight of duration $t$ will be given by $$ n(x,t)\propto \sum_ {m\in \mathbb {Z}} \tilde u_{n,q}(m) \delta\left(\frac{m x}{\hbar t}-m G\right), $$ where the $\delta$ function comes from the Fourier transform of the exponentials. Thus, the density measurement gives a sum of picks corresponding to the reciprocal lattice (picks of finite width in practice due to the finite duration of the flight), with weight proportional to the Fourier coefficients of the Bloch wavefunction.

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TL;DR: The "BEC in a lattice" does not really make sense. It is a BEC before being loaded into the lattice, but once it's in you have to write it in the lattice eigenbasis. It is still a single condensate according to the pre-lattice basis (momentum). EDIT: It makes sense to talk about a BEC in a lattice when it's non-interacting, so that you can still use single-particle states. Then a BEC means a macroscopic occupation of a single one of such states. But when you have interactions, the true ground state in a many-body ground state. And now stuff like $U$ in your question can destroy the off-diagonal long-range order associated with a BEC. This is what I mean by "does not really make sense". Once in a lattice, other forces are at play, which will eventually dominate the physics.

The key thing is that they are not doing a BEC in the optical lattice.

They are loading an existing BEC (obtained in a harmonic trap) into said optical lattice.
Once the lattice is on, the lattice energy eigenstates are the ones that determine the physical evolution of the system, so forget about the BEC.

(You can't really obtain a BEC directly in a lattice, as the atoms are spread across minima of the potential which reduces the number density, crucial for increasing phase space density and getting a BEC).


If you read the paper where that plot is from, they slowly relax the harmonic trap while ramping up the optical lattice so as to ensure the condensate always remains in the many-body ground state the potential. But in the lattice, the state time evolution is given by the lattice energy eigenstates (Bloch waves) which, by Bloch's theorem, have the same periodicity of the lattice. Which is why you are seeing square sets of peaks (because of the cubic lattice). But if you ramped the lattice strength back down, slowly, you'd get the BEC back. The single peak from before.

Let's forget the trap for a moment and assume (for the sake of simplicity) that you have a BEC in free space. There, it is the macroscopic occupation of zero momentum $|\mathbf{k}=0\rangle$. Then you load the BEC into the lattice. Momentum is not a good label in a lattice any longer, because of quasimomentum and what not. But the atoms are still "condensed" in one state $|\mathbf{k}=0\rangle$, it's just that you have to write that state in the lattice energy eigenbasis now. So that's where the peaks come in.

So it doesn't really make sense to talk about "the BEC in the lattice". However, a BEC may be considered (incorrectly, but whatever) a "boring" superfluid. I.e. a superfluid with zero critical velocity (with no interactions. Here they use $^{87}$Rb at zero external field which therefore has interactions but hey, longer story). Now a superfluid is something that is defined in condensed matter and in lattices (see later), and it requires off diagonal long range order. The BEC also has off diagonal long range order. So loading a BEC into a lattice is a convenient way of achieving a superfluid in a lattice.


I can add maths, theory, and diagrams if needed. But in short:

To actually understand the phyics, I refer to the paper I linked before. The key thing is that the Hamiltonian has two competing terms, the tunnelling $J$ which is minimised by each atom being delocalised over the whole lattice (superfluid phase), and a repulsive interaction $U$ which is minimised by having each atom localised onto each lattice site (Mott insulating phase). There is going to be a transition at some critical $J/U$.

Experimentally, they are observing this by varying the lattice depth $V_0$. With increasing $V_0$, $U$ increases moderately but the tunnelling $J$ decreases exponentially. The atomic wavefunctions in a certain lattice minimum cannot tunnel to nearby ones, essential for delocalisation and superfluid phase, thereby killing phase coherence.

What you are seeing, then, in the diffraction pattern. Instead of light diffracting off slits, you are diffracting atoms off light (the optical lattice).

In the localised phase, phase coherence has been killed off and hence you don't get any interference. In the superfluid phase, there is still tunnelling, so there is phase coherence and you have the "two-slit" like pattern:

enter image description here

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  • $\begingroup$ I find this answer quite confusing. It is mixing the non-interacting and interacting cases (going back and force between them). "The BEC is defined in the harmonic trap, where it is the macroscopic occupation of zero (close to it anyway) momentum |𝐤=0⟩." does not really make sense, since in the trap, the macroscopically occupated state is the n=0 state of the trap. You can very well talk about a BEC in a lattice (even though it is created from a BEC in a trap). $\endgroup$ – Adam Jun 30 at 7:19
  • $\begingroup$ I changed the harmonic trap / zero momentum bit, I just want to make the point that momentum is not quantity in the lattice. How would you talk about the BEC in a lattice? I tried to address OP's questions on whether the BEC is now multi-mode because of additional peaks when in the lattice. $\endgroup$ – SuperCiocia Jun 30 at 7:25
  • $\begingroup$ I would say the BEC in the harmonic trap is adiabatically connected to the ground state of the lattice. Is that enough to call the latter a BEC as well? $\endgroup$ – SuperCiocia Jun 30 at 7:28
  • $\begingroup$ See my answer, I think that addresses your questions. $\endgroup$ – Adam Jun 30 at 13:09
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There is a single condensate only. The additional peaks appear due to the small occupation of higher-energy states, as the increasing well depth encourages localization of particles.

See also a similar question here.

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  • $\begingroup$ This is exactly what I am not sure about. You start with the condensate - according to definition all particles occupy the same state - and then as you increase lattice depth you are seeing additional peaks which following your interpretation corresponds to small occupation of higher-energy states. Do all particles at this stage occupy the same state (even with admixture of higher-energy states) or some of them jump to the other energy level? If there is a single condensate why do I see the Mott insulator which is definitely not a condensate according to definition? $\endgroup$ – WoofDoggy Nov 8 '16 at 22:28
  • $\begingroup$ It seems that I have been wrong. It turns out that the change of the Wannier functions is important for the emergence of the extra peaks. I.e., as the lattice depth is increased, the Wannier functions get squeezed. I will update my answer once I feel I have a complete understanding of the problem. $\endgroup$ – ffc Nov 12 '16 at 13:43

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