Superfluid transport dynamics in a capacitive atomtronic circuit
Aijun Li, Stephen Eckel, Benjamin Eller, Kayla E. Warren, Charles W. Clark, Mark Edwards
aa r X i v : . [ c ond - m a t . qu a n t - g a s ] J un Superfluid transport dynamics in a capacitive atomtronic circuit
Aijun Li,
1, 2
Stephen Eckel, Benjamin Eller, Kayla E. Warren, Charles W. Clark, and Mark Edwards
1, 3 Department of Physics, Georgia Southern University, Statesboro, GA 30460–8031 USA Key Laboratory of Coherent Light, Atomic and Molecular Spectroscopy, College of Physics, Jilin University Joint Quantum Institute, National Institute of Standards and Technologyand the University of Maryland, Gaithersburg, MD 20899, USA (Dated: October 2, 2018)We simulate transport in an atomtronic circuit of a Bose-Einstein condensate that flows from asource region into a drain through a gate channel. The time-dependent Gross–Pitaevskii equation(GPE) solution matches the data of a recent experiment. The atomtronic circuit is found to besimilar to a variable–resistance RLC circuit, which is critically damped at early times and showsLC oscillations later. The GPE also predicts atom loss from the drain. Studies of the dependenceof condensate transport upon gate parameters suggest the utility of the GPE for investigation ofatomtronic circuits.
PACS numbers: 03.75.Gg,67.85.Hj,03.67.Dg
The “lumped abstraction” model provides an interfacebetween the physics of electromagnetism and the engi-neering of electronic circuits. By attributing ideal inde-pendent macroscopic properties (e.g. resistance, induc-tance, and capacitance) to the individual components acircuit, the lumped abstraction model makes possible thedesign of highly complex functional circuits. [1, 2]. A ma-jor challenge in the emerging field of atomtronics is theestablishment of a comparable interface for the design ofatom circuits. Lee et al. [3] have found equivalents of elec-tronic resistance, capacitance and inductance in a simpleatomtronic circuit. We believe that the time–dependentGross–Pitaevskii equation can be useful in both deter-mining the validity of an atomtronic lumped abstractionmodel and, if valid, determining the values of circuit pa-rameters. Here we test this hypothesis by applying it toa recent atomtronic experiment [4].A typical example of an atom circuit consists of a Bose–Einstein condensate (BEC) harmonically confined to ahorizontal plane by a red–detuned laser and arbitrarilyconfined within the plane by a combination of red– andblue–detuned lasers [5–9]. Atomtronic devices analogousto batteries, diodes, transistors, and fundamental logicgates have been proposed [10–16]. Atom circuits in theform of a BEC confined in a ring geometry have beenstudied as potential rotation sensors [17–25].Recently, a series of experiments was conducted inwhich a gas of thermal atoms [3] and Bose–Einstein–condensed atoms [4] were confined in a quasi–2D po-tential consisting of two wells connected by a channel.The atoms were initially confined the source well andthen released to flow down the channel into the drain well. The difference between the number of atoms in thesource, N S ( t ), and the number in the drain, N D ( t ), nor-malized by their sum, the number imbalance ∆ N ( t ) =( N S − N D ) / ( N S + N D ), as a function of time was in-ferred from image data. Similar experiments in Fermigases have been recently reported [26, 27]. -60 -30 0 30 60 x ( µ m ) -60 -30 0 30 60 y ( µ m ) V ho r i z ( x , y ) / k ( n K ) r r (x ,y )(x ,y ) FIG. 1. (color online) The potential in the horizontal ( xy )plane consists of the mask potential and the xy part of the(harmonic) sheet potential. The mask potential is zero insidehard–walled circular wells (with centers ( x k , y k ) and radii r k where k = 1 ,
2) and is equal to V d outside; inside the channelthe mask potential is harmonic along the y direction plus aconstant step. In this paper we present simulations of the experimentof Ref. [4] using the time–dependent Gross–Pitaevskii(GP) equation and show that it quantitatively capturesthe physics of the evolution of the number imbalance. Wethen describe the features of the GP–predicted transportdynamics in terms of a variable–resistance RLC circuit.The dynamics includes atom loss from the drain and weprovide a possible mechanism for this loss. Finally wesummarize our study of the transport dynamics depen-dence on the length and effective width of the channel.The time–dependent GP equation [28, 29] has the form i ¯ h ∂ Ψ ∂t = (cid:20) − ¯ h M ∇ + V trap ( r , t ) + gN | Ψ | (cid:21) Ψ( r , t ) , (1)where M is the mass of a condensate atom, g =4 π ¯ h a s /M measures the strength of binary atom scat-tering where a s is the s –wave scattering length, N is thenumber of condensate atoms, and V trap ( r , t ) is the trappotential in which the condensate atoms move. In simu-lating the experiment in Ref. [4] we found that obtainingagreement with the data strongly depended on carefulmodeling of the trap potential.The optical dipole trap present in Ref. [4] was producedby the superposition of a horizontal, planar red–detunedlight sheet and a vertical, blue–detuned, Gaussian laserbeam (the “mask beam) partially blocked by a dumbbell–shaped mask. We modeled this as a 3D harmonic “sheet”potential plus a 2D dumbbell–shaped ”mask” potentialplus a “gate” potential, a high step function that blocksthe channel only during condensate formation. The maskpotential was, in turn, modeled as the superposition of“well” and “channel” potentials. The full model potentialthus had the form V trap ( r , t ) = 12 M (cid:0) ω , x x + ω , y y + ω , z z (cid:1) V well ( x, y ) + V channel ( x, y )+ ǫ ( t ) V gate ( x ) . (2)For the gate potential, ǫ = 1 during condensate formationand is zero otherwise. The z axis is vertical, the x axislies along the line joining the two well centers, and the y axis is perpendicular to the channel (see Fig. 1).The well and channel potentials had the form V well ( x, y ) = V d X k =1 , (cid:20) (cid:18) ρ k ( x, y ) − r k b (cid:19)(cid:21) ,ρ k ( x, y ) ≡ p ( x − x k ) + ( y − y k ) , k = 1 , ,V channel ( x, y ) = V step + 12 M ω y y . (3)The well potential was zero inside two wells with cen-ters at ( x k , y k ) and radii r k ( k = 1 ,
2) and equal tothe well depth, V d , outside. The hardness of the welledges was controlled by the value of b . The channelpotential was modeled as a step plus a harmonic os-cillator along y . The channel length was defined by L c = ( x − x ) − ( r + r ) and the mask potential was setequal to min( V well , V channel ) when x ≤ x ≤ x and equalto V well otherwise.The sheet–potential frequencies were determined to be ω sh , x / π = 10 Hz, ω sh , y / π ≈ ω sh , z / π = 529Hz. The well radii were r = r = 24 µ m, the welldepth was V d /k = 83 nK, and the hardness parameterwas b = 0 . µ m. The well centers were separated by x − x = 74 µ m. In the channel V step /k ≈
20 nK and ( N S - N D ) / ( N S + N D ) time (seconds) ω y /2 π = 112 Hz ω y /2 π = 121 Hz ω y /2 π = 129 Hz FIG. 2. (color online) Atom number imbalance (with errorbars) between source and drain wells versus time from Ref. [4]for different channel transverse trapping frequencies, ω y / π :112 Hz (gray x’s, bottom curve); 121 Hz (green circles, middlecurve); and 129 Hz (blue triangles, top curve). For all casesthe channel length was L c = 26 µ m and the well depths where V /k = 83 nK where k is the Boltzmann constant. The redsolid curves show the GP simulation. The number of conden-sate atoms was taken to be N = 480,000. The middle and topcurves have been vertically offset by 0.5 and 1.0, respectively,for clarity. ω y / π varied between 110 and 130 Hz (see Fig. 2). Moredetails are given in the Supplementary Materials.The solution of the 3D time–dependent GP equationwas approximated using the hybrid Lagrangian Varia-tional Method (HLVM) [30]. This approach is valid whenthere is strong confinement to a planar region. The solu-tion is assumed to be a product of an unrestricted func-tion in the plane and a Gaussian in the strongly con-fined direction. The HLVM equations were solved us-ing the split–step, Crank–Nicolson algorithm [31] in a150 µ m × µ m box with a space step of ∆ x = ∆ y =0 . µ m. The initial, variationally stable condensatewas determined as described in Ref. [30] and evolved fora total of 0.5 seconds using 700,000 time steps. Conden-sate transport dynamics was modeled by integrating theGP equation on the same with ǫ = 0 in Eq. (2).As described below, the evolution of the system exhib-ited atom loss from the dumbbell area. This significantlydegraded the GP simulation because atoms leaving thedumbbell region were reflected from the grid boundaryback into the dumbbell. Thus, at each time step, the GPsolution was multiplied by a windowing function havingunit value inside an area surrounding the dumbbell andthat sloped gradually to zero outside this region. Thisabsorbing boundary condition enabled the determinationthe atom loss as a function of time.Figure 2 compares the GP results with the data ofRef. [4]. The number imbalance, ∆ N ( t ), is plotted versustime for three progressively wider (top to bottom, curves -0.2 0 0.2 0.4 0.6 0.8 1 0 50 100 150 200 250 300 ( N S - N D ) / ( N S + N D ) time (ms)
15 ms0 ms33 ms89 ms op ti ca l d e n s it y ( a r b . un it s ) t = t t = t (a) (c) atom loss = 35 degreesstraight−through atomswall−bounce atoms θ (b) θ FIG. 3. (color online) (a) Number imbalance, ∆ N ( t ), for a dumbbell potential with channel length L c = 20 µ m, Thomas–Fermiwidth of w TF ≈ µ m (or transverse frequency ω y / π = 63 Hz), well depth of V d /k = 97 nK, and N = 436 ,
000 atoms.The bottom (red) curve is the GP solution. The top (blue) curve shows the fraction of initial number of atoms remainingin the dumbbell area versus time. (b) Dumbbell optical density snapshots at four times during filling of the drain well. (c)Enlargement of the dumbbell optical density at t = 15 ms. Atoms that flow straight along the dumbbell axis can collide withatoms that bounce off the drain wall at angle θ . vertically offset for clarity) channels. As an aid to intu-ition, we introduce a Thomas–Fermi (TF) approximatechannel width w TF = ((64 /π ) gn ch ω z /ω y ) / , (where n ch is the number of atoms per unit length in the channel,see Supplementary Materials for details). The channelwidths range between 12 and 14 µ m. The agreementbetween theory (shown in red), with no adjustable pa-rameters, and experiment is good suggesting that theGP equation can accurately portray the behavior of av-erage quantities such as ∆ N ( t ). It is therefore interestingto understand the GP–predicted transport dynamics andhow it depends on channel shape.The behavior of the number imbalance can be conve-niently described in terms of a model RLC circuit witha time–dependent resistance, an initially charged capac-itor and a switch that is closed at t = 0. The capacitorcharge ratio, q ( t ) /q (0), in this circuit is equivalent to thenumber imbalance, ∆ N ( t ), [3] and the chemical capaci-tance, in analogy with the electronic capacitance, can bedetermined from the potential shape and the number ofcondensate atoms [3] (See Supplemental Material for afuller description).Figure 3(a) displays typical behavior of ∆ N ( t ) in whichthree distinct regimes of behavior are readily apparent.For 0 ≤ t ≤ t it drops rapidly from unity as the conden-sate flows down the channel and begins to fill the drain.We find that this behavior accurately matches a criti-cally damped RLC circuit characterized by a capacitivedischarge time, τ . The optical density at several timesduring this period is shown in Fig. 3(b). At t ≈ t , ∆ N exhibits a “kink”, that is an abrupt increase in the (neg-ative) slope, after which the imbalance shows an approx-imately linear decrease between t < t < t . At t = t ,the resistance in the analog circuit abruptly increases andthen begins a linear decrease to zero at t = t . After this,∆ N exhibits oscillations at frequency ω about a positive, but small, average value equivalent to resistanceless LCoscillations in the analog circuit. The Supplemental Ma-terial contains details on this model circuit and how thecapacitance and time–dependent resistance were calcu-lated.The presence of the kink can be understood from thetop (blue) curve appearing in Fig. 3(a). This curve showsthe ratio of the number of atoms located in the dumbbellarea at time t relative to the initial number. It is clearfrom this curve that this ratio drops sharply at time t ≈ t indicating atom loss from the dumbbell area. Thiscoincides with the appearance of the kink. From thesimulation images it can be seen that atoms leaving thedumbbell region area do so chiefly in the drain well. Asudden reduction of N D , while keeping N S fixed, causesan increase in ∆ N or, in this case, a slowing of its rateof decrease.The question arises as to why the atoms are leavingthe drain. Atoms in the initial condensate (shown in thetop picture in Fig. 3(b)) obviously do not have enoughenergy (which is chiefly due to interactions) to jump outof the source well. When the early atoms arrive at thedrain their energy is almost entirely kinetic but their totalenergy hasn’t changed and they will still be unable toleave the drain.One possible mechanism for atoms to gain the neededkinetic energy to escape the drain is through energy–redistributing collisions. This is illustrated in Fig. 3(c).Atoms fan out as they exit the channel. Some atomsflow straight along the dumbbell axis (straight–throughatoms) while others diffract, bounce off the drain wall,and come back to the drain center (wall–bounce atoms).Straight–through atoms colliding elastically with wall–bounce atoms can redistribute kinetic energy.In a classical collision of two identical particles of mass m , speed v , and colliding at an angle θ , kinetic energy istransferred from one particle to the other after the colli-sion. The maximum possible kinetic energy of the moreenergetic particle is given by (see Supplemental Materialfor more details): K f , max = K i (1 + sin θ ) . (4)For the case illustrated in Fig. 3, the angle measuredfrom the simulation images is θ ≈ ◦ . Assuming that theinitial energy of colliding atoms was equal to the chemicalpotential of the initial condensate, for this case we have K i /k ≈
65 nK and so K f /k ≈
102 nK which is greaterthan the 97 nK well depth for this case.Thus even one collision can transfer enough kineticenergy to an atom to enable it to escape the drain.Atom loss from the condensate is an additional dissipa-tion mechanism, increasing the resistance already devel-oped by the creation of vortices and solitons [4]. Atomsleaving the condensate presumably would enter the ther-mal cloud present in the experiment but not accountedfor in the GP model. In the experiment, condensate andthermal atoms cannot be distinguished making this lossprocess hard to detect.Finally we studied the dependence of the transportdynamics on the channel length and width by perform-ing 252 simulations. Each simulation was characterizedby a value of ω y , chosen from 21 values in the range60 Hz < ∼ ω y / (2 π ) < ∼
130 Hz, and a value of L c , chosenfrom 12 values in the range 2 µ m ≤ L c ≤ µ m. We cal-culated the number of atoms in the source, channel, anddrain regions of the dumbbell potential as a function oftime for a total simulation time of 0.5 seconds. For eachcase we fit the number imbalance to find the values ofthe capacitive discharge time, τ , and the LC oscillationfrequency, ω .The dependence of the exponential decay time, τ ( L c , w TF ), on the channel length and width is shownin the density plot in Fig. 4(a). It is clear from this plotthat τ = RC is independent of L c . Since we have fixedthe number of condensate atoms and the shape of thewells, the capacitance, C , is also fixed. The resistanceis therefore independent of channel length suggesting a“contact” resistance in which dissapative processes giv-ing rise to a resistance occur in the wells rather than inthe channel. Examples of such processes include atomloss and the formation of vortices and solitons.The plot also shows that the decay time does dependon the channel width. We averaged τ ( L c , w TF ) over L c keeping w TF fixed and found that the resulting averagedecay time, τ avg ( w TF ), decayed according to a power law w − . This differs from the Feynman model [32] wheresuperfluid flow above a critical velocity from a channelinto an infinite reservoir dissipates energy via the forma-tion of a line of vortices. The Feynman resistance variesinversely with the square of the channel width. This dif-ference may be due to the presence of other dissipationsources such as atom loss and the geometry of the well. (a)
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Supplemental Material: Superfluid transport dynamics in a capacitive atomtroniccircuit
In this Supplemental Material section we provide extra material on the following topics: (1) a description of how thepotential parameters used to model the NIST experiment were determined, (2) an explanation of the fitting procedureused in the systematic study of transport behavior for different channel shapes, (3) the details of the variable–resistanceRLC circuit model, (4) a derivation of the model capacitance of the dumbbell circuit, (5) a derivation of the Thomas–Fermi channel width, w TF , and (6) a derivation of the formula for maximum kinetic energy transfer in a classical,elastic collision. DETERMINATION OF DUMBBELL POTENTIAL PARAMETERS
We reiterate the potential here for convenience. The full model dumbbell potential consists of sheet, well, andchannel potentials: V trap ( r , t ) = 12 M (cid:0) ω , x x + ω , y y + ω , z z (cid:1) + V well ( x, y ) + V channel ( x, y ) (1)The z axis is vertical, the x axis lies along the line joining the two well centers, and the y axis is perpendicular to thechannel. There is also a gate potential but it was modeled as a high step located along the y axis and was presentonly during formation of the initial condensate we neglect it here.The model well and channel potentials have the form V well ( x, y ) = V d X k =1 , (cid:20) (cid:18) ρ k ( x, y ) − r k b (cid:19)(cid:21) ,ρ k ( x, y ) ≡ p ( x − x k ) + ( y − y k ) , k = 1 , ,V channel ( x, y ) = V step + 12 M ω y y . (2)The channel potential has two circular wells having centers ( x k , y k ) and radii r k where k = 1 , V d . Thehardness of the well edges was varied by adjusting the value of b which is the range over which the step rises fromzero to one. The channel potential is harmonic along the y direction (due to instrument resolution [4]) plus a step, V step .The full mask potential is equal to min( V well , V channel ) between the wells ( − r ≤ x ≤ r ) and equal to V well outsidethis region. The gate potential is a high step function parallel to the y axis and located in the center of the channel.The sheet–potential frequencies were determined to be ω sh , x / π = 10 Hz, ω sh , y / π ≈ ω sh , z / π = 529 Hz.The well radii were r = r = 24 µ m, the well depth was V d = 83 nK, and the hardness parameter was b = 0 . µ m.The well centers were separated by 74 µ m. In the channel V step ≈
20 nK and ω y / π varied between 110 and 130 Hz.Because of imaging aberrations, the exact channel potential is unknown and cannot be determined a priori . Wetherefore chose to model the channel potential in the simple way described above, using the combination of a harmonictrapping potential ω y ∝ √ V d and a V step that is independent of V d . With two free parameters – the proportionalityconstant between ω y and √ V d and V step – we found we could accurately predict the measured 1D equilibrium densitiesin the channel. Other observables like the Thomas-Fermi width or the 2-D density of atoms are compromised by theaberrations; as such, the 1-D density is the only reliable measure by which we can model the channel potential. Thereservoir potentials, which are much larger than the channel, are less affected by the imaging aberrations and thus wefound parameters ( ω sh ,x and ω sh ,y , b , and V d ) that best reproduced the measured 2-D density. SHAPE–STUDY FITTING PROCEDURE
The study of the transport dynamics of condensate released into the dumbbell potential across a range of differentchannel lengths and widths described in the main text consisted of two phases. The first phase was the simulation ofthe condensate behavior using the Gross–Pitaevskii (GP) equation for each of the 252 cases of fixed channel lengthand width. For each case, the number of atoms in the source well, channel, and drain well as a function of elapsed timefollowing condensate was calculated and saved. In the second phase, for each channel shape, the number imbalance,∆ N ( t ), was calculated and a fit was performed to extract the capacitive decay constant, τ , and the LC oscillationfrequency, ω .The simulations performed in the first phase were done by numerical solution of the GP equation. The procedurefor this was the same as described in the main text for the experimental simulations. The hybrid Lagrangian varia-tional equations (HLVM) of motion for the 3D GP were solved using the split–step, Crank–Nicolson algorithm underconditions of space and time step size the same as for the experiments. Each simulation produced a time–tagged wavefunction, stored on a 2D space grid, at 400 equally spaced times during the interval 0 ≤ t ≤ . x, y, z, t ) = (cid:18) √ πw ( t ) (cid:19) / exp (cid:26) − z w ( t ) (cid:27) ψ ( x, y, t ) . (3)The number of atoms in the source region at time t is given by N S ( t ) = Z D sourceregion d r | Ψ( x, y, z, t ) | = (cid:18) √ πw ( t ) (cid:19) Z + ∞−∞ dz exp (cid:26) − z w ( t ) (cid:27) Z Z D sourceregion dx dy | ψ ( x, y, t ) | N S ( t ) = Z Z D sourceregion dx dy | ψ ( x, y, t ) | , (4)and similarly for the channel and drain regions. The source, channel, and drain populations were computed numericallyat each time during a given simulation and the result was stored. This was done for all 252 channel–shape cases.In phase two of the shape study, the time–dependent number imbalance, ∆ N ( t ) = ( N S ( t ) − N D ( t )) / ( N S ( t ) + N D ),was fitted with a function that enabled the estimation of the decay constant, τ , and the LC oscillation frequency, ω .The fitting function had the following form:∆ N ( t ) =
11 + exp n t − t c t w o e − t/τ +
11 + exp n − t − t c t w o (cid:18) b cos (cid:18) πtT + φ (cid:19) + c (cid:19) . (5)This function only assumes capacitive discharge, e − t/t F , and LC–oscillation behavior, b cos (cid:0) πtT + φ (cid:1) + c , and consistsof these functions multiplied by turn–off and a turn–on functions, respectively. These turn–off/turn–on functions areset off in square brackets in the above equation. The fit parameters are t c , t w , τ , b , T , φ , and c . The decay constant, τ , is already one of the fit parameters and the LC oscillation frequency can be calculated by ω = 2 π/T .The number imbalance associated with each channel–shape simulation was fit with this function to get τ and ω .These results were then, themselves, analyzed for dependence on channel length and width as described in the maintext. VARIABLE RESISTANCE RLC CIRCUIT MODEL
In this section we give more details about the variable–resistance RLC circuit model described in the main paper.This circuit model consists of an initially charged capacitor of capacitance C , an inductor with inductance L , aninitially open switch, and a resistor with time–dependent resistance, R ( t ). This circuit is shown in Fig. 1(a).The goal of this modeling exercise is to find out if such a model circuit with some variable resistance, R ( t ), iscapable of reproducing the behavior of the number imbalance, ∆ N ( t ), of the dumbbell circuit as produced by theGP equation. Thus we need to find the functional form of R ( t ). The method for determining it will be to equate thenormalized charge on the capacitor, ¯ q ( t ) ≡ q ( t ) /q (0) , (6)where q ( t ) is the capacitor charge in the model circuit, to the number imbalance. Next we find a formula for R ( t ) interms of ¯ q ( t ) using Kirchhoff’s circuit rules. Finally we find a fit to ¯ q ( t ) and use this fitting function to get R ( t ). ( N S - N D ) / ( N S + N D ) time (ms) GPEFit 0 0.02 0.04 0.06 0.08 0.1 0 50 100 150 200 250 300 m od e l r e s i s t a n ce ( a r b . un it s ) time (ms) C L switch
R(t) (a) (b) (c)
FIG. 1. (color online) (a) Model RLC circuit with variable resistance. (b) Number imbalance, ∆ N ( t ), from the Gross–Pitaevskii(GP) equation and fitted result (see text) for a dumbbell potential with channel length L c = 20 µ m and and Thomas–Fermiwidth of w TF ≈ µ m. (c) The time–dependent resistance, R ( t ), derived from the fitted number imbalance. We can apply Kirchhoff’s rule to the circuit in Fig. 1(a) where we assume an instantaneous current i ( t ) flowing inthe clockwise direction and an instantaneous charge q ( t ) on the capacitor. Then we have the following two equations: L didt + R ( t ) i ( t ) + 1 C q ( t ) = 0 dqdt = i ( t ) (7)and we can combine these yielding one equation for q ( t ): L d qdt + R ( t ) dqdt + 1 C q ( t ) = 0 R ( t ) = − (cid:18) ¨¯ q ( t ) + ω ¯ q ( t ) Cω ˙¯ q (cid:19) (8)where ω ≡ / √ LC is the frequency of LC oscillations that the circuit would have if R = 0 and we have written theresult in terms of ¯ q . The values of q ( t ) and ω will be found by fitting. The value of the capacitance can be calculatedgiven the dumbbell potential and the number of condensate atoms and so is assumed to be known here.The number imbalance has distinct behavior during Interval (a): 0 ≤ t ≤ t , Interval (b): t ≤ t ≤ t and Interval(c): t > t as explained in the main text. Thus we chose a different function for ∆ N ( t ) on each interval. Our fittingfunction was the following: ∆ N ( t ) = q ( t ) q (0) ≡ ¯ q ( t ) = (1 + ω t ) e − ω t ≤ t ≤ t q c − I c ( t − t ) t ≤ t ≤ t b cos( ω t + φ ) + c t < t . (9)We fit the frequencies appearing in Intervals (a) and (c) separately. However, we found that these frequencies werenearly the same in most cases. The fitting parameters were ω , q c , I c , b , ω , φ , c , t , and t . Thus, for the final valueof the fitting function, we set ω = ω = ( ω + ω ) / L c = 20 µ m and TF width w TF = 22 µ m (red curve) along with theresult of the fit (green curve). The fit is very good as would be expected given the number of fitting parameters.Finally, this fitting function can be used to find the time–dependent resistance, R ( t ), as already described. Thisquantity is plotted in Fig. 1(c). The plot shows that R is essentially constant during 0 ≤ t ≤ t . At t = t , where∆ N displays a kink, the resistance abruptly increases after which is decreases linearly to zero at t = t and remainszero thereafter. A summary of this behavior appears in the main text. CALCULATION OF THE CAPACITANCE OF THE DUMBBELL–POTENTIAL BECDefinition of chemical capacitance
Here we derive the capacitance of the BEC in the dumbbell potential. The chemical capacitance is defined inanalogy with the definition for an electronic capacitance. For the electronic capacitor, if an external agent moves a(positive) charge δq from one initially neutral plate (drain plate) to the other plate (source plate), then the sourceplate has a net charge of + δq while the drain plate has a net charge − δq . The source reservoir gets its name fromthe fact that moving “charges” (either real positive charges or atoms) will initially flow from source to drain toregain equilibrium. The charge difference (source charge - drain charge) is +2 δq however the charge on the electroniccapacitor is regarded as being + δq . Once there is a charge on the capacitor, it causes a voltage difference, δV , todevelop and the electronic capacitance is defined as C elec ≡ δqδV . (10)This motivates the definition of the chemical capacitance.The chemical capacitance is defined analogously. If the total number of atoms is N and there are equal numbersof atoms ( N/
2) in each well (this is equilibrium) and the external agent moves δN atoms from the drain well to thesource well, then the number of atoms in each well is N S = 12 N + δN and N D = 12 N − δN. (11)Analogous to the electronic capacitor the difference in the number of atoms between the wells is +2 δN . However,we shall regard the charge on the chemical capacitor as being δN atoms. Once there is a charge on the chemicalcapacitor, it causes a chemical potential difference, δµ , to develop and the chemical capacitance is defined as C chem ≡ δNδµ . (12)We can find an approximate expression for this based on the TF approximation for the chemical potential of a BECin a cylindrical well potential. The chemical potential for a BEC in a cylindrical well
Consider a BEC having N atoms confined in a cylindrical well potential. To get an expression for the chemicalpotential we will assume that the Thomas–Fermi approximation applies and that the BEC is confined radially in ahard–walled cylindrical well of radius R and axially in a harmonic potential. We assume that the confining potentialis therefore (using cylindrical coordinates ( r, θ, z ) V trap ( r, θ, z ) = 12 M ω z z + V well ( r ) (13)where V well ( r ) = (cid:26) ≤ r ≤ R ∞ R < r < ∞ (cid:27) . (14)In the potential actually present in the experiment, there is also a weak harmonic confinement along the x directionwhose zero is offset from the origin (assumed here to be at the center of the source well). The effect of this extraharmonic potential is to produce a small but noticeable gradient along the x direction in the initial condensatedensity profile. Since this extra potential is so weak ( ω sh , x / π ≈ N .The condensate wave function, ψ ( r ), satisfies the time–independent Gross–Pitaevskii (GP) equation: − ¯ h M ∇ ψ ( r ) + V trap ( r, θ, z ) ψ ( r ) + gN | ψ ( r ) | ψ ( r ) = µ ψ ( r ) . (15)where M is the mass of a condensate atom, N is the number of condensate atoms, g = 4 π ¯ h a s /M is the strength ofthe binary scattering of condensate atoms, a s is the s –wave scattering length, and µ is the chemical potential of thecondensate. The chemical potential is the energy required to add another atom to the condensate.We will derive an approximate expression for the chemical potential by using the Thomas–Fermi (TF) approxima-tion. The TF approximation is valid whenever the interaction energy is much larger than the kinetic energy. Whenthis is the case we have ψ ( r ) ≈ ψ (TF)0 ( r ), µ ≈ µ (TF)0 , and these TF–approximate quantities satisfy the GP equationwhere the kinetic–energy term is neglected: V trap ( r, θ, z ) ψ (TF)0 ( r ) + gN | ψ (TF)0 ( r ) | ψ (TF)0 ( r ) = µ (TF)0 ψ (TF)0 ( r ) . (16)The formal solution of this equation is gN | ψ (TF)0 ( r ) | = (cid:26) µ (TF)0 − V trap ( r, θ, z ) µ (TF)0 − V trap ( r, θ, z ) ≥
00 otherwise (cid:27) (17)We can now insert the particular form of the potential into the above to get gN | ψ (TF)0 ( r ) | = (cid:26) M ω z z − M ω z z ≤ r ≤ R, and | z | ≤ z TF (cid:27) (18)Where we have defined the TF z radius as µ (TF)0 ≡ M ω z z . (19)The values of z TF and thus µ (TF)0 are found by requiring that the TF wave function be normalized to unity or, moreconveniently gN = gN Z d r | ψ (TF)0 ( r ) | . (20)Thus we have gN = Z π dθ Z + z TF − z TF dz Z R rdr M ω z (cid:0) z − z (cid:1) = (cid:18) M ω z (cid:19) (2 π ) (cid:18) R (cid:19) (cid:18) z (cid:19) = (cid:18) πR (cid:19) (cid:16) µ (TF)0 (cid:17) / (cid:0) M ω z (cid:1) / . (21)So the expression for the TF chemical potential is µ (TF)0 = ( gN ) (cid:0) M ω z (cid:1) / (cid:0) πR (cid:1) ! / ≡ αN / (22) The chemical capacitance in the Thomas–Fermi approximation
Now consider the chemical potential difference between the source well having N S atoms and the drain well with N D atoms. Using Eq. (22) we have δµ = µ ( N S ) − µ ( N D ) = αN / S − αN / D = α (cid:18) N + δN (cid:19) / − α (cid:18) N − δN (cid:19) / = α (cid:18) N (cid:19) / "(cid:18) δN N (cid:19) / − (cid:18) − δN N (cid:19) / ≈ α (cid:18) N (cid:19) / (cid:20)(cid:18) δN N (cid:19) − (cid:18) − δN N (cid:19)(cid:21) δµ = α (cid:18) N (cid:19) / δN N = 4 αδN (cid:0) N (cid:1) / (23)where we assumed that δN ≪ N in the above derivation. This linear approximation is good to 10% over a numberimbalance up to δN/ ( N ) = ± .
93 [4].Finally we can get an approximate expression for the chemical capacitance: C chem = δNδµ = 3 (cid:0) N (cid:1) / α . (24)This expression for C chem only depends on the shape of the potential and the total number of condensate atoms. THE THOMAS–FERMI CHANNEL WIDTH
In this section we provide a derivation of the Thomas–Fermi condensate width in the dumbbell potential. Thiswidth enables us to define the channel shape in a more intuitive way. The channel is assumed to have a length L c along the line joining the two wells of the dumbbell ( x axis) with hard walls located at x = ± L c /
2. In between thesewalls, the potential is assumed to be harmonic in y and z plus a constant step: V ch ( x, y, z ) = (cid:26) V step + M ω y y + M ω z z | x | ≤ L c ∞ | x | > L c (cid:27) (25)The solution of the time–independent Gross–Pitaevskii (GP) equation, ψ , can be approximated by the Thomas–Fermi solution ψ TF where the kinetic–energy term in the GP is neglected: − ¯ h M ∇ ψ ( r ) + V ch ( r ) ψ ( r ) + gN ch | ψ | ψ ( r ) = µψ ( r ) V ch ( r ) ψ TF ( r ) + gN ch | ψ TF ( r ) | ψ TF ( r ) = µ TF ψ TF ( r ) (26)where we have assumed that there are N ch atoms in the channel. The Thomas–Fermi–approximate solution can bewritten as gN ch | ψ TF ( r ) | = (cid:26) µ TF − V ch ( r ) µ TF ≥ V ch ( r )0 otherwise (cid:27) = (cid:26) µ TF − (cid:0) V step + M ω y y + M ω z z (cid:1) µ TF ≥ V ch ( r )0 otherwise (cid:27) (27)The value of µ TF is determined by normalization.Thus we require that Z d r | ψ TF ( r ) | = 1 or , equivalently , Z d r gN ch | ψ TF ( r ) | = gN ch . (28)Using the explicit TF solution we have gN ch = Z d r gN ch | ψ TF ( r ) | = Z + L c / − Lc/ dx Z + y TF − y TF dy Z + z TF ( y ) − z TF ( y ) dz (cid:20) ( µ TF − V step ) − M ω y y − M ω z z (cid:21) (29)where z TF ( y ) is satisfies 12 M ω z z ( y ) = ( µ TF − V step ) − M ω y y . (30)Performing the integral over z gives us gN ch = 43 (cid:18) M ω z (cid:19) Z + L c / − Lc/ dx Z + y TF − y TF dy ( µ TF − V step ) − M ω y y M ω z ! / (31)Now y TF is the edge where the y integrand goes to zero and satisfies the following condition:12 M ω y y = µ TF − V step (32)This expression enables us to define the Thomas–Fermi width, w TF , as follows: w TF = 2 y TF = 2 µ TF − V step12 M ω y ! / (33)So now we need to find µ TF − V step by evaluating the rest of the integral in Eq. (31). The result is gN ch = (cid:16) π (cid:17) y (cid:0) M ω y (cid:1) / (cid:0) M ω z (cid:1) / L c , (34)where y = µ TF − V step12 M ω y ! . (35)Inserting this into the equation for gN ch gives us an expression from which we can evaluate µ TF − V step : gN ch = (cid:16) π (cid:17) µ TF − V step12 M ω y ! (cid:0) M ω y (cid:1) / (cid:0) M ω z (cid:1) / L c = πL c (cid:0) M ω y (cid:1) / (cid:0) M ω z (cid:1) / ( µ TF − V step ) (36)We can use this equation to solve for µ TF − V step : µ TF − V step = ( gN ch ) (cid:0) M ω y (cid:1) / (cid:0) M ω z (cid:1) / πL c ! / . (37)With this we can now get the final result for w TF : w TF = 2 y TF = 2 "(cid:18) π (cid:19) ( gN ch ) (cid:0) M ω z (cid:1) / L c (cid:0) M ω y (cid:1) / / (38)The channel width can be obtained from this expression for each of the simulations in the channel–shape study byusing the simulation value of N ch . We compared the predictions of this formula with the channel widths as determinedby inspection of the images at the end of each simulation and found good agreement between them. This formulaenables us to present the channel–shape study results in terms of channel length and width. This is more intuitive interms of the channel length and transverse harmonic frequency, ω y . KINETIC ENERGY TRANSFER IN A CLASSICAL COLLISION
In this section we derive a formula for the final kinetic energies of two identical particles that collide at an angle θ .This result enables us to estimate the amount of kinetic energy that can be transferred in a collision of wall–bounceatoms in the drain well with atoms that flow directly into the well. This estimate can be compared with the depth ofthe well to determine if this kinetic–energy transfer is a viable mechanism for atom loss from the dumbbell region. y θ1 2 xvv FIG. 2. Classical model for a collision of atoms that bounce off the wall with those that come straight through. Two particleseach having mass m and speed v collide at an angle θ . We assume first that the colliding atoms have the same initial kinetic energy and that this kinetic energy comesentirely from the interaction energy of the initially confined condensate in the source well. In the source well, almostall of the particle energy is interaction energy since the potential energy and kinetic energies are (nearly) zero. Afterrelease, particles that flow into the channel have all of their initial interaction energy converted to kinetic energy.To develop a classical model we imagine two identical particles of mass M and initial speed v colliding elastically atan angle θ as shown in Fig. 2. The central question is: how much kinetic energy can be transferred from one particleto the other in this perfectly elastic, momentum–conserving collision? We will impose these conservation laws on thecollision in the center–of–mass (CM) frame. These lab and CM frames are illustrated in the following figure: cm y θ y lab x lab x cm θ particle 1 particle 2A BOv v FIG. 3. This figure shows the positions of the two particles as they move toward one another before the collision.
This figure shows the tracks of the particles in the “lab” frame whose origin is chosen to be at the point where theparticles meet and the x lab axis is chosen along the direction of particle 1. The track of particle 2 runs through theorigin and makes an angle θ with the x lab axis. In the figure the particle positions are shown at six different timesbefore the collision. The pair of particle locations corresponding to a given time are connected by a line. These linesdenote the x cm axis of the CM frame of reference. The line perpendicular to these lines is the y cm axis in this frame.The x cm axis is inclined from the x lab axis by angle θ/
2. This can be seen by considering the triangle AOB in thefigure which runs from the open–circle particle furthest from the origin (point A) to the filled–circle particle furthestfrom the origin (point B) to the origin (point O) and back to point A. Since the particles are equidistant to the origin,this is an isosceles triangle. Hence angles BAO and ABO are equal and the sum of these angles is supplemental toangle AOB. But, since angle AOB is also supplemental to θ it follows that the sum of the two equal angles BAO andABO is equal to θ so both angles BAO and ABO equal θ/ i lab and ˆ j lab that point along the x lab and y lab axes respectively.Additionally we can define unit vectors that point along the x cm and y cm and denote them as ˆ i cm and ˆ j cm , respectively.Although the CM frame is moving with respect to the lab frame, the relative orientation of the two pairs of axesremains fixed with the CM axes rotated with respect to the lab axes by the angle θ/
2. Thus the CM unit vectors canbe expressed in terms of the lab unit vectors as follows:ˆ i cm = ˆ i lab cos( θ/
2) + ˆ j lab sin( θ/ j cm = − ˆ i lab sin( θ/
2) + ˆ j lab cos( θ/ . (39)With these two sets of unit vectors, we can now analyze the collision.We will analyze the collision by requiring that the total momentum and total kinetic energy be conserved. Ourstrategy will be to conserve these quantities in the CM frame. Thus we will write down the initial velocities of particles1 and 2 in the lab frame, transform these to the CM frame, find the final velocities there, transform the velocities backto the lab frame, and then compute the final kinetic energies in that frame. In this way we will be able to calculatethe difference in the final kinetic energies to learn how much kinetic energy can be transferred due to the collision.Before we implement this procedure, we need to know how to transform between the two frames.Consider two particles of masses m and m with position vectors r ( lab )1 and r ( lab )2 . In what follows the superscriptin a variable name will denote the reference frame to which the quantity is referred. Thus, r ( lab )1 is the vector thatstretches from the origin of the lab frame to the location of mass m while r ( cm )1 would be the vector that stretchesfrom the origin of the CM frame to m . The position and velocity of the center of mass of the two–particle system,referenced to the lab frame, is given by r ( lab ) cm = 12 r ( lab )1 + 12 r ( lab )2 v ( lab ) cm = 12 v ( lab )1 + 12 v ( lab )2 (40)where we have assumed m = m . This gives the velocity of the CM frame as measured by an observer in the labframe. We can transform velocities between the lab and CM frames using v ( lab ) = v ( lab ) cm + v ( cm ) . (41)We will use this equation to express the initial velocities of the colliding particle with respect to the CM frame.From the picture of the collision in Fig. 2 we can write the velocities of the colliding particles in the lab frame.These initial velocities are v ( lab )1 i = v ˆ i lab v ( lab )2 i = − v cos( θ )ˆ i lab − v sin( θ )ˆ j lab . (42)These velocities can now be used to compute the velocity of the center of mass in the lab frame: v ( lab ) cm = 12 v ( lab )1 i + 12 v ( lab )2 i = 12 v (1 − cos( θ )) ˆ i lab − v sin( θ )ˆ j lab = (cid:0) − v sin( θ ) (cid:1) ˆ j cm . (43)Now we can express the initial velocities with respect to the CM frame.0The initial velocities of particles 1 and 2 relative to the CM frame can be calculated as follows: v ( cm )1 i = v ( lab )1 i − v ( lab ) cm = v ˆ i lab − (cid:18) v (1 − cos( θ )) ˆ i lab − v sin( θ )ˆ j lab (cid:19) = v cos( θ )ˆ i cm , v ( cm )2 i = v ( lab )2 i − v ( lab ) cm = − v cos( θ )ˆ i lab − v sin( θ )ˆ j lab − (cid:18) v (1 − cos( θ )) ˆ i lab − v sin( θ )ˆ j lab (cid:19) = − v cos( θ )ˆ i cm . (44)In the CM frame, the colliding particles have equal and opposite velocities along the x cm axis as expected.In the CM frame the total momentum of the system is zero before and after collision. Also, since kinetic energy andtotal momentum are conserved in the lab frame and because the CM frame moves at a constant velocity relative tothe lab frame, the total momentum and kinetic energy, as measured in the CM frame, will also be conserved. Figure4 shows before and after pictures of the collision as observed in the CM frame. θ f θ f x cm y cm y cm x cm v v vvbefore collision after collisionCM frame FIG. 4. Collision as observed in the CM frame. The particle speeds before and after the collision are the same because thecollision is perfectly elastic.
The particles have oppositely directly velocities before and after the collision with equal speeds guaranteeing thatthe total momentum of the system is always zero. The directions of their final velocities don’t have to be the same asthe initial velocities. In Fig. 4 the angle between the velocities before and after the collision is denoted as θ f whichwe will refer to as the “CM scattering angle”. This angle can vary between 0 ◦ and 180 ◦ . The CM–frame speeds ofthe particles before the collision are v ′ ≡ v cos( θ ). If the speed of the particles after the collision is denoted as v ′′ ,then conservation of kinetic energy requires that12 M ( v ′ ) + 12 M ( v ′ ) = 12 M ( v ′′ ) + 12 M ( v ′′ ) (45)Thus we have v ′′ = v ′ = v cos( θ ).From the figure it is easy to write down the CM–frame final velocities for the two particles. These are v ( cm )1 f = v ′ cos( θ f )ˆ i cm + v ′ sin( θ f )ˆ j cm v ( cm )2 f = − v ′ cos( θ f )ˆ i cm − v ′ sin( θ f )ˆ j cm (46)In expressing these velocities back in the lab frame and it will be convenient to express them in terms of the CM–frameunit vectors: v ( lab )1 f = v ( lab ) cm + v ( cm )1 f = v cos( θ ) cos( θ f )ˆ i cm + (cid:0) v cos( θ ) sin( θ f ) − v sin( θ ) (cid:1) ˆ j cm v ( lab )2 f = v ( lab ) cm + v ( cm )2 f = − v cos( θ ) cos( θ f )ˆ i cm − (cid:0) v cos( θ ) sin( θ f ) + v sin( θ ) (cid:1) ˆ j cm (47)1The last step will be to compute the difference in lab–frame final kinetic energies.The difference in the lab–frame kinetic energies of the two particles after the collision can be written as δK ( lab ) ≡ M (cid:16) v ( lab )2 f (cid:17) − M (cid:16) v ( lab )1 f (cid:17) . (48)We can calculate the squares of the vectors v ( lab )1 f and v ( lab )12 f using their components in the CM–frame coordinatesystem since vector lengths are independent of the coordinate system. Thus we have( v ( lab )1 f ) = v − v sin( θ ) sin( θ f )( v ( lab )2 f ) = v + v sin( θ ) sin( θ f ) (49)Substituting these squared velocities into the expression for the kinetic–energy difference the result is δK ( lab ) = 12 M (cid:0) v + v sin( θ ) sin( θ f ) (cid:1) − M (cid:0) v − v sin( θ ) sin( θ f ) (cid:1) = 12 M v (2 sin( θ ) sin( θ f )) . (50)The final kinetic energies in the lab frame can be written as K ( lab )1 f = 12 M v (1 − sin( θ ) sin( θ f )) (51)and K ( lab )2 f = 12 M v (1 + sin( θ ) sin( θ f )) . (52)We can now find the maximum possible transfer of kinetic energy in a collision of the type considered here. Since θ f can take any value from 0 ◦ to 180 ◦ , the maximum and minimum kinetic energies for fixed θ occur when θ f = 90 ◦ : K ( lab )1 f,min = 12 M v (1 − sin( θ )) = K ( lab )1 i (1 − sin( θ )) K ( lab )2 f,max = 12 M v (1 + sin( θ )) = K ( lab )2 i (1 + sin( θθ