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B + n i f N n F j f ] and ; . Q (5) (4) is µ 2 exchange of x’s while antisymmetric under the exchange prove that the ground state of HG contains no or only of y’s. We find (up to a normalization constant) one fermion. Let us assume the ground state con- tains more than one fermion, and has the wave function Qψ(x , , y F ) = S 1 N 1 ψ(x , , y F ),(7) 1 N x ,···,x B ,y 1 N ψ(x , , xNB ; y , , yNF ). Due to the time reversal ··· ··· 1 ··· 1 ··· † symmetry of HG, ψ can be chosen to be real. Now con- Q ψ(x1, , yNF ) = Ax1,y1,···,yNF ψ(x1, , yNF ),(8) ··· ··· struct a trial state: where Sx1 xN y1 is the symmetrization operator for co- ,···, B , ˜ ψ(x , , x B ; y , , y F )= ψ(x , , y F ) . (13) ordinates x , , xNB , y , and Ax1,y1,···,yN is the anti- 1 N 1 N 1 N 1 ··· 1 F ··· ··· | ··· | symmetrization operator for coordinates x , y , , yNF . 1 1 Obviously, ψ˜ is non-negative and different from ψ for Physically, S turns a fermion into a boson by symmetriz-··· ing one fermion coordinate with respect to all boson co- NF > 1, as the latter changes sign under the exchange of y’s. In fact ψ˜ is a two-component or “spin”-1/2 bo- ordinates, and A turns a boson into a fermion by anti- son wave function as it is also symmetric under the ex- symmetrizing one boson coordinate with respect to all y ˜ fermion coordinates. change of ’s. Since ψ and ψ differ in phase only, we find the potential/interaction energy does not change: If the bosons and fermions have the same disper- ψ Vˆ ψ = ψ˜ Vˆ ψ˜ , because Vˆ depends only on sion and interaction strength, namely tB = tF , and 1st 1st 1st ij ij hdensity| | buti noth | the phase| i of wave function. The situa- U BB = U BF = U F F , it is easy to show that the canoni- ij ij ij tion is different for Tˆ , which is sensitive to the wave cal ensemble Hamiltonian H is supersymmetric, i.e., 1st function phase. Because ψ˜ and ψ are the same for certain [Q,H] = [Q†,H]=0. (9) configurations (when ψ 0) and differ by a sign for others (when ψ < 0), we≥ find the expectation− value of ˜ For completeness we also present a generic example of every term in Tˆ1st is either the same for ψ and ψ , or h i | i | i supersymmetric H in the continuum, in 1st quantization: differ by a sign. Because ψ˜ is non-negative, and all t’s in Tˆ are non-negative− (meaning Tˆ only has negative ma- ′ 2 2 H = [ + V (xi)] + [ + V (yi)] trix elements), we find every single term in ψ˜ Tˆ st ψ˜ is 1st −∇xi −∇yi h | 1 | i i≤XNB i≤XNF non-positive. We thus have ψ Tˆ st ψ ψ˜ Tˆ st ψ˜ , and h | 1 | i ≥ h | 1 | i as a result ψ H1st ψ ψ˜ H1st ψ˜ . Since in general ψ˜ + U(xi xj )+ U(yi yj ) h | | i ≥ h | | i | i − − is not an eigen state of H1st, and the ground state of such i
H1st = Tˆ1st + Vˆ1st, (11) NF can only be 0 or 1[12]. We now show that the ground states have a double where the form of Tˆ st and Vˆ st can be deduced from Eqs. degeneracy. Assume ψ is the ground state with NF = 1 1 | 0i (3) and (4). If ψ(x , , yNF ) is an eigen wave function of 0. We can then construct a different state with NF = 1 1 ··· the Hamiltonian H1st with NB bosons and NF fermions, and one fewer boson then Qψ(x , , yNF ) is an eigen wave functionof H st 1 ··· 1 † with NB + 1 bosons and NF 1 fermions with exactly ψG = Q ψ0 , (14) − † | i | i the same eigen energy. Similarly Q ψ(x , , y F ) is 1 N which is also an exact eigen state of H with exactly an eigen wave function of H with N 1··· bosons and G 1st B the same energy E , as guaranteed by supersymmetry N +1 fermions, also with exactly the same− eigen energy. 0 F (12). ψ can be viewed as a zero momentum, fermionic If we further have the same chemical potential for the G zero mode| i of the ground state, which is known as the bosons and fermions: µ = µ , the grand canonical F B Goldstino mode in the high-energy literature. Hamiltonian is also supersymmetric: As a simple example illustrating the results presented † [Q,HG] = [Q ,HG]=0. (12) above, consider the special case of non-interacting par- ticles with all U = 0. For non-interacting bosons we Given the great tunability of parameters in cold atom always have µB = 0 (we measure energy from the bot- systems, we expect such conditions can be reached in a tom of single particle dispersion) at zero temperature, variety of systems. In the following we discuss conse- regardless of boson number. Supersymmetry of HG re- quences of such supersymmetry when present. quires µF = µB = 0, as a result we can have either no We start by considering the case where HG is su- fermion, or a single fermion occupying the k = 0 state persymmetric: [Q,HG] = 0. In this case we can (with zero energy) in the ground state of HG. 3
It should be clear from the discussion above that not From (17) we can immediately conclude the following: only the ground states, but all eigen states of HG (ex- (i) Q ψ0 is an exact excited state with excitation energy | i † cept for vacuum with N = 0) come in degenerate pairs ∆µ; and (ii) Q ψ0 = 0, because if it were not null, it that have the same total number of particles but differ would be a state| withi negative excitation energy ∆µ. in fermion number by one due to (12). The vacuum is We thus find even though in this case supersymmetry− special in that it is the only state that is supersymmetric, is explicitly broken by ∆µ, we still have a sharp zero as it is annihilated by both Q and Q†. momentum fermionic collective mode generated by Q, We can also construct Goldstino at finite momentum: which is now gapped. The situation is somewhat similar to what happens to a ferromagnet in an external mag- † ψG(q) = R ψ , (15) netic field: the field breaks rotation symmetry and opens | i q| 0i a spin wave gap, but the spin wave remains a sharp col- by boosting Q† to finite momentum: lective mode. This is a “hole-like” Goldstino mode since it is created by Q instead of Q†, which creates a hole † −iq·xi † † Rq = e aifi = akfk+q. (16) in the occupied Goldstino Fermi sea. The analogous (to Xi Xk Eq. (15)) finite momentum states are Rq ψ0 , which are expected to have downward quadratic dispersion| i of the † 2 Since the statistics of a single fermion created by Rq has form Eq E ∆µ α q . − 0 ≈ − | | no physical consequence, ψG(q) is just like ferromag- Again due to (17), all eigenstates of H except for | i G netic spin wave states for SU(2) bosons[11]; they have vacuum come in pairs whose energies differ by ∆µ and exactly the same dispersion and differ only in statistics. fermion numbers differ by one. This is because the canon- While ψG(q) is not an exact eigen state of HG for finite ical Hamiltonian remains supersymmetric. | i q, it approaches one in the long-wave length limit q 0, We now turn the discussion to possible experimental 2 → and has quadratic dispersion: Eq E0 q [11]. detection of the Goldstino modes. Normally one would − ∝ | | It is appropriate at this point to discuss the relation expect that these modes can only be detected in processes between the sharp fermionic collective mode we call Gold- in which a boson is turned to a fermion or vice versa, as stino here, and the Goldstino in high-energy context. In that is what Q† or Q does. There is no such process that high-energy context, Goldstino refers to the Goldstone can be easily engineered in cold atom systems that we fermion arising from spontaneous breaking of the global are aware of, except for possible co-tunneling processes supersymmetry; it is a Weyl spinor with spin-1/2. In in which a fermion leaves the system and a boson enters our non-relativistic model, the gapless Goldstone fermion the system simultaneosuly, or vice versa. In the follow- mode is also the result of the spontaneous breaking of the ing we show that in the presence of a Bose condensate, global supersymmetry. In this sense, they are quite sim- the Goldstino mode contributes a finite spectral weight ilar. In fact, if we had a two-component boson system to the spectral function of single fermion Green’s func- instead of a Bose-Fermi mixture, we would have a pseu- tion; as a result it can be detected through processes that dospin ferromagnet that breaks SU(2) symmetry, with a involve a single fermion. Physically this is possible be- branch of gapless spin-wave mode; the spin-wave mode is cause in the presence of a Bose condensate, the boson the Goldstone boson associated with breaking of SU(2) number is not fixed in the ground state; as a result a sin- symmetry. In the Bose-Fermi mixture, the second com- gle fermion (hole) can grab a boson from the condensate ponent is fermionic, and the SU(2) symmetry becomes and propagate as the Goldstino mode. To demonstrate supersymmetry that is generated by the fermionic oper- this we calculate the overlap between the fermionic single ator Q. The difference here is that the Goldstino is a hole state fq ψ with the normalized Goldstino mode =0| 0i spinless fermion with quadratic (instead of linear) dis- (1/√N)Q ψ0 : persion, due to absence of Lorentz symmetry. | i 1 † 1 † We now turn to the more interesting case in which ψ Q fq ψ = ψ f akfq ψ √ h 0| =0| 0i √ h 0| k =0| 0i there is a finite density of fermions in the ground state N N Xk ψ . In order to sustain this we must have a higher | 0i chemical potential for the fermions, thus ∆µ = µF 1 † † = ψ0 f f0a0 ψ0 + ψ0 f akfq=0 ψ0 µ > 0. ∆µ can be viewed as a chemical potential for− √ h | 0 | i h | k | i B N kX=06 Goldstino; in its presence the fermionic Goldstino modes 0 are “filled up” to some Fermi wavevector kF . NB f 1 † = nq=0 + ψ0 fkakfq=0 ψ0 , (18) In the presence of ∆µ> 0, HG is no longer supersym- r N rN h | | i kX=06 metric; it instead has a nonzero commutation relation 0 with the supersymmetry generators: where NB is the number of bosons in the condensate f † < 0 and nq=0 = ψ0 fq=0fq=0 ψ0 1. When NB is macro- [Q,HG]= [Q,µF NF + µBNB]= ∆µQ; h | | i ∼ − − scopic (as is the case in the presence of a condensate), † † † 0 [Q ,HG]= [Q ,µF NF + µB NB] = ∆µQ . (17) a ψ = N ψ , and the first term in (18) dominates − 0| 0i B| 0i p 4 the second. We thus find the zero momentum fermion Klein-Gordon type. As a result the anti-commutation Green’s function has finite weight on the Goldstino, and relation between the super-charge and its hermitian con- 0 the weight is approximately NB/N, proportional to the jugate is the Hamiltonian in that case, instead of the condensate density. As a result the fermion spectral func- total particle number in the present work. tion A(q =0,ω) has a sharp δ-function peak at ω = ∆µ. This is highly unusual as in electronic or other fermionic We thank Lee Chang, Xiaosong Chen, Xi Dai, Miao systems, one normally expects sharp (coherent) spectral Li, Hualing Shi, Zhan Xu, Zhengyu Wen, Li You, Lu peak for fermions with momentum near kF and corre- Yu, Guangming Zhang, and Zhongyuan Zhu for useful sponding energy near zero in a Fermi liquid, correspond- discussions, and Kavli Institute for Theoretical Physics ing to Landau quasiparticles which are well defined only (supported in part by the National Science Foundation near the Fermi surface. The sharp, coherent fermion peak under Grant No. PHY05-51164), and Kavli Institute for at zero momentum and finite energy we find here is a re- Theoretical Physics China for hospitality. This work was markable consequence of the combination of supersym- supported by Chinese National Natural Science Founda- metry and Bose condensation, which is unique to such su- tion, a fund from CAS, and MOST grant 2006CB921300 persymmetric Bose-Fermi mixtures. Another remarkable (YY), and by National Science Foundation grants No. property is that this spectral peak remains sharp at finite DMR-0225698 and No. DMR-0704133 (KY). temperature T as long as T is below the Bose condensa- tion temperature Tc. This is because (17) guarantees that the Goldstino mode is sharp at any T , while for T