Bose crystal as a standing sound wave

0.38 MB·English

by  Maksim Tomchenko

Bose crystal as a standing sound wave
Maksim Tomchenko
∗
Bogolyubov Institute for Theoretical Physics, 14-b Metrolohichna Street, Kiev 03680, Ukraine
A new class of solutions for Bose crystals with a simple cubic lattice consisting of N
atoms is found. The wave function (WF) of the ground state takes the form Ψ =
0
eSw(l)+Sb {sin(klxxj)sin(klyyj)sin(klzzj)}, where eSb is the ground-state WF of a fluid, Sw(l) =
j
Q q+k=0 N
2 q=0S1(l)(q)ρ−q+ q,k=6 0 S2(l)(q,k)ρkρ−k−q+..., ρk= √1N j=1e−ikrj, andkl =(π/al,π/al,π/al) (al
6 6
1 iPs the lattice constPant). The state with a single longitudPinal acoustic phonon is described by the
0
2 WF Ψk = "ρ−k+ q P2(k,q)N−1/2ρqρ−k−q+ qx Q1(qx)ρ−k−qx +7 permutations#Ψ0, where the
n permutations give tPhe terms with different signsPof components of k. The structure of Ψk is such
a that the excitation corresponds, in fact, to the replacement of k in some triple of sines from Ψ
l 0
J by k. Sucha structureof Ψ0 and Ψk means that the crystal is created by sound: theground state
2 of a cubic crystal is formed by N identical three-dimensional standing waves similar to a longitu-
1 dinal sound. It is also shown that the crystal in the ground state has a condensate of atoms with
k=k. ThenonclassicalinertiamomentobservedincrystalsHe4canberelatedtothesynchronous
l
] tunnelingof condensate atoms.
r
e
h PACSnumbers: 61.50.Ah,67.80.-s,67.80.bd,67.80.de
t
o
.
t I. INTRODUCTION By the modernideas,a crystalis formedsince its energy
a
is less than that of a fluid.
m
The science on crystals is developed for many years, In what follows, we propose the basically new wave
-
d andmostpropertiesofcrystalsaresuccessfullyexplained. solution for the WF of a Bose crystal. In Ref. 33 (cited
n The current interest is focused on the regions of un- below as I), it was shown that the states of a system of
o conventional and yet unguessed properties manifesting N interacting Bose particles positioned in a rectangular
[c themselves, inparticular,at extra-lowtemperatures (see box Lx×Ly×Lz in size include the states with the WF
surveys1–3). For the crystals with a charged lattice and

v1 tthemepBeorasetucrreysstuaples,rcsouncdhucrteigviiotnysaanrde,surepsepresoctliidveplyh,ehniogmh-- Ψ0 =AeSw(l)+S˜b N {sin(klxxj)sin(klyyj)sin(klzzj)},
3
ena. The splash-up of interest arose after the excellent jY=1
2 (2)
experimentsbyE.KimandM.Chan4,5,whereanonclas-
6 where (k ,k ,k ) k = (l π/L ,l π/L ,l π/L ),
2 sicalinertiamoment(NCIM)ofacrystalHe4wasdiscov- lx ly lz ≡ l x x y y z z
l ,l ,l are integers, and the remaining designations
. ered. Later on, a number of new interesting properties x y z
1 are given in Sec. 2. In I, we analyzed Ψ only with
joined by the term “supersolid” were found6–16. There 0
0 k =k =(π/L ,π/L ,π/L )(here,l ,l ,l =1),which
2 are almost no doubts that the “supersolid” phenomenon l 1 x y z x y z
describes the ground state of a gas and a fluid. In this
1 isrelatedtothesuperfluidity ofquantumcrystals,which
case,the sinesin (2) forma standinghalf-wavethat cov-
: was predicted long ago17. However, the physical nature
v ers the whole system and rests on the boundaries. But,
of the superfluidity and the “supersolid” phenomenon is
Xi notclearyet,thoughalotofmodelsweredeveloped18–28. lx,ly andlz mayobviouslytakeinthesolutionanyother
integer values except zero. It is natural to assume that
r The results obtained below indicate that the basic
a ifthe half-waveis equalto the latticeconstant,thenWF
property of crystals, i.e., the nature of crystalline order-
(2) describes a rectangular crystal lattice. Moreover,(2)
ing, is not completely clear as well.
is one of the exact solutions of the Schr¨odinger equation
Commonly accepted is the following structure of the
with zero boundary conditions (BCs). In what follows,
WF of a Bose crystal29–32:
wewillstudythissolutionandthe solutionforalongitu-
dinal acoustic phonon. We will show that the solutions
Ψ =e−iP>jSJ(ri−rj) N ϕ(r R ), (1) agree with observable properties of crystals and predict
0 i i
− a number of specific features, in particular, the conden-
i=1
Y
sates of phonons and atoms in a crystal. Solution (2)
whereN isthetotalnumberofatomsofacrystal,rj and testifies to the wave nature of a crystal. I did not deal
Rj are coordinates of atoms and sites of the lattice, the with crystals earlier and have found the solutions acci-
exponential function is the Bijl-Jastrow function taking dentally, while studying the microstructure of a fluid.
correlations into account, and ϕ(r) is usually written in A shortannouncement of the results will be published
theapproximationofsmalloscillations: ϕ(r)=e α2r2/2. separately34.
−
2
II. GROUND STATE OF A CRYSTAL has the coordinate x=0 by a step
U x 0,
Due to the presence of a product of sines in Ψ0 (2), Uw(x)≈ 0s x>≤0. (7)
the system can be partitioned into lx ly lz identical (cid:20)
× ×
domains separated by plane surfaces, on which Ψ = 0.
0 The potential of the face with the coordinate x = L
x
We may assume that if each domain contains one atom,
is U (L x). Analogously, we can consider four other
w x
and the size of domains is close to the equilibrium in- −
faces. For simplicity, we take U = , which corre-
w
teratomic distance, then the system is stable. A crystal ∞
sponds to zero BCs.
corresponds, obviously, to a system of domains with the
In I, the product of assigning sines
size at which the energy of the system is minimum. If
the number of domains is equal to the number of atoms, N
then lx = Lx/R¯x, ly = Ly/R¯y, lz = Lz/R¯z, where R¯x, Ψbsacre(N)= {sin(klxxj)sin(klyyj)sin(klzzj)} (8)
R¯y andR¯z aretheperiodsofthelatticealongtheappro- jY=1
priateaxes. Foracubiccrystal,they areequaltoR¯. Let
has been factor out from the equations for Ψ and Ψ .
us study the properties of such system and make clear 0 k
In this case, we obtain the cotangent cot(πl x/L ),
whether they correspond to the properties of crystals. x x
which cannot be expanded in a Fourier series, because
We consider only a simple cubic (sc) lattice.
ConsiderWF(2). Init, S˜b isgivenbytheformula(see Lx cot(πlxx/Lx)dx = . To overcome this prob-
| | ∞
I) 0
lRem, we will use the following. It is easy to see that,
under the change sinkx sinkx, the derivative
(π)
a (k ) → | |
S˜ = 2 1 ρ ρ (d sinkx/dx)(1/ sinkx) gives again cot(kx). There-
b 2! k1 −k1 for|e, we r|eplace Ψ|bare (8|) by the WF
kX16=0 sc
(π)k1+k26=0a (k ,k ) N
+ 33!√1N2 ρk1ρk2ρ−k1−k2 +..., (3) Ψbsacre(N)= {|sin(klxxj)|×|sin(klyyj)|×|sin(klzzj)|}
k1X,k26=0 jY=1
(9)
where and pass to
1 N N
ρk = √N e−ikrj (k6=0), (4) Ψbsacre(N) = {(|sin(klxxj)|+δ) (10)
Xj=1 jY=1
sin(k y ) +δ ( sin(k z ) +δ) ,
and the summation is carried on over the wave vectors × | ly j | × | lz j |
k=π jx , jy , jz (5) wtiohnere δ >0,δ →(cid:0) 0. For such Ψbs(cid:1)acre, we obtain the fu(cid:9)nc-
L L L
(cid:18) x y z(cid:19)
f˜(x)=(df/dx)(1/f), f = sin(k x ) +δ (11)
(jx,jy,jz are integers), which are multiple to π/L. This | lx j |
is denoted by the symbol (π) above the sums (under instead of cot(k x). The former can be expanded in a
cyclic BCs, k are multiple to 2π/L, and the solutions lx
Fourier series. For the singular point x , the series gives
0
arestrongly changed). The function Sw(l) in (2) takesthe the arithmetic mean value of those at the points x0 0
−
form of an infinite series (see I) and x +0. In the final formulas, we transit to the limit
0
δ 0,whichreturnsustofunctions(9)and(8). Bearing
Sw(l) = S1(l)(q)ρ−q+q+k16=0S2(l)√(qN,k1)ρk1ρ−k1−q+ tehx→ipsafnascitoinnsmatinδd=,w0e:willimmediatelyuseinformulasthe
qX=6 0 q,Xk16=0
+ q+k1+k26=0S3(l)(qN,k1,k2)ρk1ρk2ρ−k1−k2−q+...(6) cot(klxx)≡cot(πlxx/Lx)=Xjx Clx(qx)eiqxx. (12)
q,kX1,k26=0 Here, qx =2πjx/Lx, jx runs all integers, and
where k run values (5), and q take values
j i for j =l ,2l ,3l ,...
x x x x
2π jx, jy, jz . In(3), the firstsumis the Bijl-Jastrow C (q ) C (j )= −i for j = l , 2l , 3l ,...
Lx Ly Lz lx x ≡ lx x  x − x − x − x
fun(cid:16)ction writte(cid:17)n in the variables ρk. 0 for the rest jx.
Let the faces of the crystalbe ideally plane and paral-  (13)
lel to atomic planes of the lattice. Every face creates a Theproofofformulas(12)and(13)isgiveninAppendix.
potentialbarrierfor atoms of the crystal. We model this We now have all the required in order to write the
barrier for the face that is perpendicular to the X and WFs of a Bose crystal. The WF of the ground state
3
is set by formula (2), where S˜ and S(l) are given by can be obtained analogously from the equations for S(1)
b w j
formulas (3) and (6). The equations for the functions (see I):
a and the ground-state energy of an atom (E ) follow
l 0
fCro1(mq)e→quCatli(oqn),sSinj(1)I→wiSthj(l)t:he changes k1 → kl ≡ kc, = S−1(il√)(Nq)2ǫk0(cqxq)x=a2(−qx)Clx(qx)δq,qx +(x→y,z)
E =E˜b+A , (14) 1 (π)
0 0 1 + 2(q2+q q)S(l)(q,q )+6q2S(l)(q,q , q )
N 1 1 2 1 1 3 1 − 1
qX16=0n
A = ~2 k2 1 (2π)q2S(l)(q)S(l)( q) (15) + √N4q12S1(l)(q1)S2(l)(q−q1,q1)
1 2m c − N 1 1 − + √N(q2 q q)S(l)(q )S(l)(q q )
qX=6 0 1 − 1 1 1 1 − 1
 o
(2π) (2π)
i 2k q C (q )S(l)( q )+(x y,z) , + 2ik C (p ) (q p )S(l)(q p )
− √N cx x lx x 1 − x →   cx lx x x− x 1 − x
qXx6=0 pXx6=0 h
 − 2pxS2(l)(q−px,−q) +(x→y,z) , (22)
N 1 1 (π) 1 (π) ~2k2 i o
E˜b = − nν (0) nν (k) a (k),
0 2d+1N 3 −2d+1N 3 −N 2m 2 S(l)(q,q )[ǫ (q )+ǫ (q+q )]
kX6=0 kX6=0 2 1 0 1 0 1
(16) + 2S(l)(q)a ( q )qq q2S(l)(q)a (q,q )
1 2 − 1 1− 1 3 1
nν3(k)m +a (k)k2 a2(k)k2 =A (k) (17) = δq,qx√NikcxClx(qx){2q1xa2(q1)−qxa3(q1,qx)}
2d~2 2 − 2 2 (2π)
(π) (π) + ik C (p ) 4(q +q p )S(l)(q p ,q )
+ N1 a3(k,q)(q2+kq)+ 21N q2a4(q,−q,k), pXx6=0 cx lx x h x 1x− x 2 − x 1
qX6=0 qX=6 0 − 6pxS3(l)(q−px,q1,−q−q1) +(x→y,z)
(π) i
a3(k,q)≈A3(k,q)−2ǫ (k)+ǫR((qk),+q)ǫ (k+q), (18) + √1N √2N(q22−q1q2)S3(l)(q,q1−q2,q2)
0 0 0 q =0(cid:26)
X26
4
+ q (q +q +q)S(l)(q,q ,q )
R(k,q) = kqa (k)a (q) k(k+q)a (k)a (k+q) √N 2 1 2 3 1 2
2 2 2 2
−
q(k+q)a (q)a (k+q), (19) 4q (q q +q)S(l)(q )S(l)(q q ,q )
− 2 2 − 2 1− 2 1 2 2 − 2 1
+ 6q2S(l)(q )S(l)(q q ,q , q q )
2 1 2 3 − 2 1 − − 1
(2π) + 4(q +q )2S(l)(q ,q )S(l)(q q ,q +q ) . (23)
A (k) = 1 ik C ( q ) 4(q +k )S(l)(q ,k) 1 2 2 2 1 2 − 2 1 2
2 √N cx lx − x x x 2 x o
qXx6=0 h Equations (14)–(23) are a complicated system of non-
+ 6q S(l)(q , q ,k) +(x y,z) linear integral equations, whose solutions determine the
x 3 x − x → properties of the ground state of the crystal with a rect-
i
(2π) angular lattice. Let us analyze these equations. Of a
1
+ 4(q2+qk)S(l)( q)S(l)(q,k)+ paramountinterest are the value of E and the distribu-
N 1 − 2 0
qX=6 0h tion of atoms in the crystal.
+ 4(k+q)2S(l)(q,k)S(l)( q, k) It is seen from Eq. (22) that the nonzero value of
2 2 − − S(l)(q)isdeterminedbythefirst“one-dimensional”term
+ 6q2S1(l)(−q)S3(l)(q,k,−k) , (20) ∼1Clx(qx)δq,qx on the right-hand side. This yields the
i one-dimensional solutions S(l)(q )δ , S(l)(q )δ ,
twhheesryesǫt0e(mk,)a=ndk2(1−2a2(k)), d =3 is the dimension of andS1(l)(qz)δq,qz. However,th1etexrmsq,∼qxS1(l)(1q1)Sy2(1)q(,qq−y
q ,q ) and some other ones on the right-hand side gen-
1 1
erate also not one-dimensional solutions of the form
ν3(k)= U3(r)e−ikrdr (21) S1(l)(qx + qy)δq,qx+qy and S1(l)(qx + qy + qz) of the
Z same order ( √N) as one-dimensional solutions. Since
∼
istheFouriertransformoftheinteractionpotentialU (r) C (j ) are nonzero only at the “resonance” points j =
3 lx x x
oftwoBoseparticles. TheequationsforthefunctionsS(l) l , 2l , 3l ,..., the function S(l)(q) is nonzero only
j ± x ± x ± x 1
4
for the “resonance” wave vectors where j = 1, 2, 3,... and k = π/R¯. Despite the
x cx
± ± ±
approximate character of the formulas, we may expect
r l r l r l
qres(r)=2π x x, y y, z z , (24) that they give the generalformof a probabilitydistribu-
(cid:18) Lx Ly Lz (cid:19) tion in a cell.
Since S(l)(q ) = 0 only at the resonance points and
rwehgearredrfo=r t(hrexs,eryre,rlazt)i,onrxs,,rwye,rczan=w±r1it,e±t2h,e±s3o,lu..t.i.onWith S1(l)(−qx)1=S1(xl)(q6 x), we have
S(l)(q )
S(l)(q=q +q +q )=S(l)(q) δ +δ s (x) 1 x 2cos(q x), (31)
1 x y z 1 q,qrxes q,qryes 1 ≃ √N x
+ δ +δ +δ n+δ rx=X1,2,...
q,qres q,qres+qres q,qres+qres q,qres+qres
z x y x z z y
+ δq,qres (25) where qx = rx2πlx/Lx = rx2π/R¯. We will determine
} S(l)(q ) from Eq. (26), by using the zero approximation
1 x
and,analogously,S(l) (q,q ,...q ).AsisseenfromEq. for a (q),
j 2 1 j 1 2
(22), the values of S≥(l)(q ) depe−nd significantly on the
1 x
sumSs(l)wSit(lh).S2(Hl)o,wSe3(vl)era,nwdewriethstrtihcet toeurrmseslv∼esSt1(ol)Sth1(le),zaenrdo a2(q)= 21 −s14 + nν83~(2qq)2m, (32)
∼ 1 2
approximation
which follows from (17) if all sums are neglected. Note
that we took the solution with the sign “minus” before
S(l)(q ) i√N2kcxClx(qx)a2(−qx)δqx,qrxes, (26) the root (see I).
1 x ≈− q 2q a (q )
x− x 2 x We choose the interatomic interaction potential for
atoms of the crystal, as in I:
S(l)(q ,q ) δ i√NkcxClx(qx) Ub r a
2 x 1 ≈ qx,qrxes 1−2a2(qx) U3(r)≈ Ubd a≤≤r ≤b (33)
2q a (q ) q a (q ,q ) 0 r >b.
1x 2 1 − x 3 1 x . (27)
× ǫ (q )+ǫ (q +q ) 
0 1 0 x 1 Below, we use, unless otherwise indicated, a = 2˚A,
The analogous relations are true for S1(l)(qy), S1(l)(qz) bto=H4e˚A4,aUtobmd s=. −T9hiKs,pwotheinchtiaclorisressupffioncdiesnatlpyprcoruxidme,atbeulyt
and S(l)(q ,q ), S(l)(q ,q ). The omitted corrections
2 y 1 2 z 1 it is qualitatively proper and has the analytical Fourier
can renormalize S(l)(q ), by changing its value by sev- transform
1 x
eral times. In this case, the higher corrections are
4π
damped by the decrease of a (q) with increase in q. ν (k)= ν˜ (k), (34)
2 3 k3 3
For the not one-dimensional solutions S(l)(q +q ) and
1 x y
S(l)(q +q +q ),eventhezeroapproximationisacom-
1 x y z ν˜ (k) = [U U ][sin(ak) akcos(ak)]
plicatedsum(overq )oftheterms S(l)(q )S(l)(q q ) 3 b− bd −
and S(l)(q )S(l)(q1 q ,q ). B∼y o1ur es1tim1ates−, th1e + Ubd[sin(bk)−bkcos(bk)]. (35)
∼ 1 1 2 − 1 1
not one-dimensional solutions are significantly less than By formulas (29)–(35), we determine ψ(x) (29). It
the one-dimensional ones. Due to the complexity of the hasaperiodicshape,incorrespondencewiththe domain
equations, we neglect the not one-dimensional solutions: structure. Thedistribution ψ(x)2 inoneofthedomains
is shown in Fig. 1. The valu|e of R|¯ is chosen like that for
S(l)(q) S(l)(q) δ +δ +δ . (28) He II: R¯ = 3.6˚A, which is close to R¯ of the crystalline
1 ≃ 1 q,qrxes q,qryes q,qrzes phases of He4. As is seen from Fig. 1, the distribution
n o
Let us study the distribution of atoms along the x- ψ(x)2 at the height of the potential barrier Ub =300K
|is sim|ilar to the bare one (sin2(k x)), but is more flat-
axis (coinciding with one ofthe axes of a crystal)for the cx
lattice of He4 atoms at R¯x = R¯y = R¯z = R¯ = 3.6˚A. tened. At Ub > 1000K, we see the appearance of two
maximalocated∼symmetricallyrelativeto the center ofa
According to I, the probability for an atom to be at a
point x is approximately determined by the formulas cell. They increase with Ub. So, the probability density
is the highest not at the center of a cell, as is commonly
ψ(x) sin(k x)es1(x), (29) accepted,butatthesemaxima. Inthethree-dimensional
cx
≃ case, the maxima indicate the presence, inside a cell, of
anorbitwithcubicshape. Theorbitdependsonthe val-
(2π) S(l)(q ) ues of a and b: at a=1˚A, b=3˚A and Ub =3000K,the
s (x) 1 x eiqxx, (30) maxima disappear, and ψ(x)2 is similar to the curve of
1
≃ √N trianglesinFig.1. Buts|uchs|malladoesnotcorrespond
qx=2Xπjx/Lx
5

1.0
30
0.8 25
20
0.6

15
0.4
10
0.2
5
0.0 0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
x (A) x (A)
Fig. 1: Probability density |ψ(x)|2 (29), (31) versus the co- Fig.2: Function|ψ(x)|2forasccrystalofHe4athighbarriers
ordinate x of a cell of the sc crystal at the potential with U . Potential(33)witha=2˚A,b=4˚A,U =−9Kandthe
b bd
a = 2˚A, b = 4˚A, U = −9K and the barriers U = 300K barriers U = 9000K (crosses), U = 105K (squares), U =
bd b b b b
(triangles), U = 3000K (circles), and U = 3000K with re- 106K (circles) and U = 107K (triangles). Stars mark the
b b b
gardforonlythefirstterminsum(31)(squares). Starsstand curveforacrystalofkryptonatomswiththesameparameters
for the bare function sin2(k x). The x-axis coincides with except the mass (m = 21m ) and potential (U = 3000K,
cx 4 b
thex-axisof a cell of thecrystal, values of x are given in ˚A. U =−140K).
bd
to the He4-He4 potential. As a increases by 1˚A, there product of sines in (2)) sets the lattice of a crystal and
the motion of atoms in the cells.
appears a clear orbit, like the curve of circles in Fig. 1.
Can an orbit be discovered in experiments? The scat-
The orbit size is approximately equal to a half of the
teringoflightorneutronsinacrystalwillshowordinary
cell size, because the main contribution to the maxima
Bragg–Wolf peaks first of all, since the time average of
is given by the first term (with q = 2k ) in sum (31).
x cx the positions of an atom is the center of a cell. But the
The terms with q = 4k ,6k ,8k are small, and the
x cx cx cx orbitcanberevealedinsomespecificfeaturesofthescat-
peaksforq =4k ariseonlyatU >106K(seeFig.2).
x cx b tering.
Tocalculate ψ(x)2,itissufficientto∼taketwofirstterms
in sum (31)|(due|to the fast decrease of a (q), as q in- Letusestimatetheground-stateenergyE0ofasccrys-
2 tal, using formulas (14)–(17). With regard for solution
creases).
(28) for S(l)(q) with S(l)(q ) (26) and solution (13) for
InFig.2,wepresentthedependenceoftheorbitonthe 1 1 x
C (j), we obtain
barrier height U . As is seen, the orbit becomes narrow l
b
at large U . Moreover, at U = 106K and 107K, the
second orbbit appears at a disbtance of 0.5˚A from the E =E˜b+A , A = ~2kc2 1+8 a2(qx)−3a22(qx) ,
wall of a cell. However, the neglected h≃igher correlative 0 0 1 1 2m " (2a2(qx) 1)2 #
r −
corrections become large at large U , which can lead to X (36)
b
the widening of orbits. where r =1,2,3,..., q =2πr/R¯, k2 =3π2/R¯2, and E˜b
x c 0
In Fig. 2, stars show ψ(x)2 for the lattice of krypton (16)coincideswithE ofaBosefluid. Consideracrystal
0
atoms. The orbit is nar|row a|t the barrier U = 3000K, of He4 atoms with R¯ = 3.6˚A, ~2k2/2m 13.85K, and
whereas the orbit for helium atoms is wide bat such U . the potential (33) with a = 2˚A, bc= 4˚A≈, U = 9K.
b bd
−
However, it is known1,35 that the ground state of the Using the zero approximation(32) for a (q ) and taking
2 x
crystal of heavy inert elements is well described under the barrier U = 1000K, we obtain A 12K, E
b 1 0
≈ − ≈
the assumptionofsmalloscillationsofatomsnearpoints 430K.ForU =200K,wehaveA 4K,E 36K,
b 1 0
− ≈ ≈−
of the lattice. Apparently, there is no contradiction in whereas A 9K, E 1K for U = 81K. The last
1 0 b
thiscase,sincethefunctionr2 ψ(r)2 atsmalloscillations value of E ≈is close to≈th−e experimental one1. For He
0
ischaracterizedbyaspherical|orbit|1 withapproximately II, E corresponds to experimental data also for U
0 b
thesamesize. Nevertheless,theorbitinFig.2hasshape 100K (see I). For the realistic value U 103 104K∼,
b
of the surface of a cube (with edge R¯/2), rather than a the experimental value E 1K can be∼obtain−ed with
0
≃−
sphere, even for the function r2 ψ(r)2. In other words, regard for correlative corrections.
| |
the motion of an atom is oscillatory only approximately. The most essential and unexpected is the conclusion
More exactly, this motion has a wave character. This thatthelatticeiscreatedbyastandingwaveintheprob-
fact is unusual and means that the resonance wave (the ability field. As we see in Sec. 3, this wave is similar to
6
a sound one. In other words, the crystals have a wave satisfy the equations
nature.
(π)
1
ǫ(k) =ǫ (k) b (k ,k)2k (k +k)
0 2 2 2 2
− N
kX26=0
(π)
1
6k2b (k , k ,k)
III. STATE WITH A LONGITUDINAL − N 2 3 2 − 2
ACOUSTIC PHONON kX26=0
(2π)
2
Q (q,k)(k+q)
1
− √N
Consider a crystalwith sc lattice. An optical phonons q=0
X6
are absent for it. We consider only longitudinal acoustic qS(l)( q)+2(k+q)S(l)( q, k)
phonons. The WF of a crystal with a single phonon can × 1 − 2 − −
be obtained from the solution for a one-phonon state of h (2π) i
4
the Bose fluid (see I) by the following changes: k k , q2Q ( q,q,k)S(l)(q) (41)
C (q) C (q), and S(1) S(l). We obtain 1 → c − √N q=0 2 − 1
1 → l j → j X6
(2π)
2k (k +q )iC ( q )Q (q ,k)
− cx x x lx − x 1 x
Ψk(r1,...,rN)=ψkΨ0, (37) qXx6=0
(2π)
+ 4k q iC (q )Q ( q ,q ,k)+(x y,z),
cx x lx x 2 − x x →
qXx6=0
ψ =ψ0 +7permutations, (38)
k k
Q (q,k)[ǫ(k) ǫ (k+q)]=
1 0
−
= 2ik C (q )δ [ k +2q b (q,k)]
cx lx x q,qx − x x 2
2S(l)(q) 4k2
+ 1 [qk 2q2b (q,k)] S(l)(q, q k)
(2π) √N − 2 − √N 2 − −
ψk0 =ψkb +b0(k)+ Q1(q,k)ρ q k 1 (π)
− −
q=6X0,−k − N {2q1(q1+q+k)
q+q +k=0 q =0
1 6 Q (q,q ,k) X16
+ q+qq,Xq1+6=q0+k=20 √N1 ρq1ρ−q−q1−k (39) ×+ [6Qq122Q(q3,(qq1,,qk1),−+qQ1,1k(q)+q1,k)√NS1(l)(−q1)]
+ 1 2 6 Q3(q,qN1,q2,k)ρq1ρq2ρ−q−q1−q2−k+..., + 4q12Q2(q−q1,q1,k)√NS1(l)(q1)
q,qX1,q26=0 + 4(q1+q+k)2Q1(q+q1,k)√NS2(l)(−q1,−q−k)
(2π) o
+ 2k iC (p )[( k q +p )Q (q p ,k)
cx lx x − x− x x 1 − x
pXx6=0
ψb =ρ + (π) b2(k2,k)ρ ρ (40) + 2pxQ2(q−px,px,k)]+(x→y,z), (42)
k −k √N k2 −k2−k
+ (π)k2+k3+k=6 0k2b6=X3(0k,−2kN,k3,k)ρk2ρk3ρ−k2−k3−k+..., − 2b0k(ckx)kǫx(ikC)lx=(−−k2xk)2√SN1(l)δ(k−,kkex)δk,ke
k2X,k36=0 2 (π)
q2Q2( k,q,k)δk,ke
− √N −
q=0
where the permutation means ψ0 with the different sign X6
k (2π)
oqfiosnqeuoarntsiezveedralilkceo2mπpjo/nLe,natnsdofthtehevevcetcotrosrkkj,,tqhje,vaencdtokr − 2(k+q)2Q1(q,k)S1(l)(−k−q)δk,ke
are quantized like πj/L. q6=X0,−k
(2π)
waSvoeludteicoanyi(n3g8)indtoesecirgihbtescoaunthterreet-rdaivmeleinnsgiownaavlesst.anding + 2√NkcxqxiClx(qx)Q1(−k−qx,k)δk,ke
qXx6=0
The energyofa phononE(k)andthe functions b , Q + (x y,z), (43)
j j
→
7
b (k ,k)[ǫ(k) ǫ (k ) ǫ (k+k )]= Equations (41)–(44) are very complicated. By setting
2 2 0 2 0 2
− −
= kk2a2(k2) k(k+k2)a2(k+k2) k2a3(k,k2) Ql =0inthem,weobtaintheequationsforaBosefluid,
− − namely, for b and the dispersion curve E(k). Thus, the
(π) 2
2 3k (k +k +k) equations for a crystal are the equations for a fluid plus
3 2 3 b (k ,k ,k)
− √N √N 3 2 3 some additional anisotropic corrections.
kX36=0(cid:26) Letusconsiderdispersioncurves. First,thevalueofb0
+ Q (k ,k)(k +k) 2(k k )S(l)( k ,k ) does not influence E(k). In the simplest approximation
1 3 3 3− 2 2 − 3 2 with regard for the anisotropy, the dispersion curves are
+ 3(k +k)S(l)( kh,k , k k) determined by the formula
3 3 − 3 2 − 2−
+ 2Q2(−k3,k2+k3,k)(k2+k3) i 2 (2π)
× k3S1(l)(k3)+2(k2+k3)S2(l)(k3,k2) ǫ(k) =ǫ0(k)− √N q=0Q1(q,k)(k+q)qS1(l)(−q)
+ h3k2S(l)(k )Q ( k ,k , k k,k)i (2π) X6
3 1 3 3 − 3 2 − 2−
2k (k +q )iC ( q )Q (q ,k)
(2π) o − cx x x lx − x 1 x
+ 2k iC (p )[3p Q ( p ,k , k k,k) qXx6=0
cx lx x x 3 − x 2 − 2− + (x y,z). (47)
pXx6=0 →
+ 2(k2x+px)Q2(−px,k2+px,k)], With regardfor solutions (13), (26), (28), (45), and (46)
+ higher corrections+(x y,z), (44) in the approximationb2 =0, we obtain
→
where ǫ(k) = 2mE(k)/~2, ke is the wave vector with ǫ(k)=ǫ0(k)+ǫcxr(k)+ǫcyr(k)+ǫczr(k), (48)
all even components (i.e., they are multiple to 2π/L),
q = q i , and p = p i . By (x y,z), we denote
x x x x x x → 4k2 k (k +q )
the same terms as one with the separated x-component, ǫcr(k)= cx x x x ,
bofutEqwsi.th(t4h1e)–c(h4a4)n,geesxcxep→t foyr athned fixrs→t onze., Acolnlttaeirnmas x −qxX=qxres (1−2a2(qx))2ǫ(k)−ǫ0(k+qx)(49)
product of two wave vectors (for example, q2, kcxkx, or where the sum is taken over the resonance values of
(kS3o+mke)(okf3t−heks2e))e.quTahteiosenswcaovnetvaeinctothrsemsuumsts,bewnhoernezetrhoe. qxres =±2kcx,±4kcx,±6kcx,.... For a sc crystal,
firstargumentofthefunctionsQl orSj(l) hascomponents E(k)= ~2ǫ0(k) +j~2ǫcxr(k), (50)
multiple to π/L. In this case, we set Q = 0, S(l) = 0, 2m 2m
l j
because, by the definition of these functions, the compo-
where j=1, 2, and 3 for directions (1,0,0), (1,1,0), and
nents of the first argument are multiple to 2π/L. (1,1,1). Since the values of ǫcr(k) and ǫ (k) are of the
x 0
Equations (41)–(44) are written in the approximation
same order, the dispersion curves are different for differ-
of “two sums in the wave vector”, at which the series
ent directions. In other words, the spectrum of longitu-
contain the functions a2, a3, b2, b3, Sj(1)3, and Ql 3 and dinalacousticphononsisanisotropic,whatwasobserved
donotincludethecorrectionsa ,b ≤,S(1) ,and≤Q . in experiments.
ItfollowsfromEq. (42)thatjt≥h4e sjo≥lu4tionj≥f4or the ful≥nc4- Due to a large value of ǫcxr, the method of iterations
does not converge, and the more exact numerical meth-
tion Q (q,k) has a “resonance” form analogous to (25).
1
ods require much time for the analysis. To demonstrate
We restrict ourselves to the one-dimensional approxima-
the influence of the correction ǫcr, we decrease it by two
tion: x
ordersofmagnitudesothatthe methodofiterationscan
Q(l)(q,k) Q(l)(q,k) δ +δ +δ . be applied. In Fig. 3, we present the dispersion curves
1 ≃ 1 q,qrxes q,qryes q,qrzes calculated by formula (50). The curves are different for
n o(45) differentdirections andaresimilar to the observedones2
In the zero approximation without regard for the sums
for a bcc crystal of He4.
in Eq. (42), we have
The qualitative behavior of the dispersion curves can
be understood, by using the simple formula
Q (q ,k) = 2ik C (q ) 2q S(l)(q )/√N
1 x cx lx x − x 1 x ~2ǫ (k) ~2k2
h k +2q b (q ,k) i E(k) 0 = (1 2a (k)). (51)
− x x 2 x (46) ≈ 2m 2m − 2
× ǫ(k) ǫ (k+q )
− 0 x
i2kcxClx(qx) −kx+2qxb2(qx,k). Wmuitlahbreecgoamrdesfor approximation (32) for a2(k), this for-
≈ 1 2a (q ) × ǫ(k) ǫ (k+q )
− 2 x − 0 x
The analogous relations can be given for Q1(qy,k), and E(k) ~2k2 2+ nν3(k)~2k2. (52)
Q1(qz,k). ≈s(cid:18) 2m (cid:19) 8 m
8
In Fig. 3, the dispersion curves are broken at kmax
equal to a half of the minimum vector g0 of the recip-
60 rocal lattice (for directions (1,0,0), (1,1,0), and (1,1,1)
of a sc crystal, g = π(1/R¯,0,0),π(1/R¯,1/R¯,0) and
0
50 π(1/R¯,1/R¯,1/R¯), whereas g is twice larger for the bcc
0
lattice), since the dispersion curves for crystals are peri-
40 odic with period g:
)
K
( 30
E E(k+g)=E(k). (53)
20
The proofofthis fact for one-particle WFs37 canbe eas-
10 ily generalized to our case of N-particle WFs. However,
it is true for the cyclic BCs. For a three-dimensional
0 crystal, the BCs are quite different and are close to zero
0.0 0.5 1.0 1.5 2.0 2.5 3.0 ones. However, the majority of quasiparticles are local-
-1 ized wave packets (this is supported by the fact that the
k (A )
theory of transfer was constructed for wave packets and
Fig.3: DispersioncurvesE(k)(50)withǫcr decreasedbytwo
x agrees with experiment). For such packets, the trans-
orders of magnitude for directions (1,0,0) (triangles), (1,1,0)
lational invariance with the period of a lattice holds,
(circles), and (1,1,1) (squares) for a sc crystal of He4 atoms
with R¯ = 3.6˚A and potential (33) with a = 2˚A, b = 4˚A, and the conclusions of the theorem are valid. As for
the standing waves (37)–(40), they are the sum of eight
U = −9K and U = 3000K. Stars show the zero approxi-
bd b
mation E(k)=~2ǫ (k)/2m. For the comparison with exper- travelingwaves,fromwhichthewavepacketscanbecon-
0
iments, the dispersion curves are drawn to kmax of the bcc structed.
lattice. Wenotethatthedensityofphononstatesatthequan-
tizationofkbylaw(5)isthesameasthatforcyclicBCs.
In the literature1,38,39, the dispersion curves of crys-
It is the well-known Bogolyubov formula with the addi-
tals are usually calculated in the approximationof small
tional multiplier 1/8 arising due to the boundaries (see
oscillations (1).
I). Indeed, for each of the directions (1,0,0), (1,1,0), and
(1,1,1), we need to continue curve (52) to kmax for a
givendirection. We obtainednoanisotropy,butthe gen-
eral shape of the curve is proper: the phonon curve has
IV. THE CONDENSATE
no minimum for direction (1,0,0), a slightly pronounced
minimum for (1,1,0), and a clear minimum for (1,1,1).
This corresponds to experiments2 for the bcc lattice of In Ref. 40 and recently41, it was shown that the ideal
He4. The minimum arise due to a displacement of kmax crystalinthegroundstatecontainsnocondensateatoms
to the side of large k and to the presence of a minimum with k = 0. However, the condensate is possible in the
on the liquid-like curve E(k)=~2ǫ (k)/2m. To describe presence of vacancies and other defects17,20–22,24,26. The
0
the curve more exactly and to obtain the anisotropy, we condensate can be a key factor for the explanation of
need to consider the next corrections to the equations. NCIM.Buttheexperimentdoesnotconfirm42 aconden-
It is of interest that the simple formula (52) describes sate of atoms with k=0.
the curves qualitatively correctly. This formula is uni- It follows from Ψ (2) that the ideal sc crystal does
0
versalandis validfor a gas,a liquid, anda crystal,since not possess off-diagonal long-range order, but it have a
thesethreestatesofsubstancearedescribedbyWFswith condensate of atoms with k = k. It is easy to verify
l
the same structure (2), (37)–(40) (the difference is only if we switch-off the interatomic interaction: in this case,
in the values of kl). the exponents in Ψ0 (2) become equal to 1, and we have
The theoretical dispersion curves for a crystal and He theproductofsineswithk=k,i.e.,allatomsareinthe
l
II (see I) correspond to the experiment at Ub ∼ 103K. condensate with k = kl. If the interaction is switched-
For E0, the agreement with experiment for a liquid and on,the condensateis exhausted. Forafluid, the conden-
a crystal holds at Ub 102K (see Sec. 2). However, it sateisalsodeterminedbyafactorbeforetheexponential
∼
is assumed that the best potential for He4 atoms is the function: under cyclic BCs,this factor( eiprj p=0 =1)
Azizpotential36withmuchmorelargervalueofUb,about j |
2 106K. This disagreement is related to the neglect of generates the condensate of atoms withQp = 0. For the
hi×ghercorrectionsor(moreprobably)tothefactthatthe fluid in a vessel, the factor is the product of sines with
potential at small distances has only an effective mean- k = k1 (see I), and the condensate is on the levels with
ing. Therefore, the values of Ub can be very different for k=k1,3k1,5k1,...43.
different processes. In some models, the agreement with Letuscalculatethe condensateforthegroundstateof
experiment was attained with the Aziz potential. a sc crystal. Under zero BCs, the condensate is deter-
9
mined by the formula43 vectors k=((1+2j )k ,(1+2j )k ,(1+2j )k ) with
x lx y ly z lz
j ,j ,j =0,1,2,3,...; i.e., to the vector k =k and to
x y z l c
Lx,Ly,Lz larger vectors with odd multiple components.
N 8
kn = dr dr ρ(r ,r )sin(k x )sin(k x ) The distribution of atoms over levels depends on the
N V2 1 2 1 2 nx 1 nx 2
Z0 barrier height Ub, here we have the interesting pic-
ture. The numerical calculation for He4 atoms with
sin(k y )sin(k y )sin(k z )sin(k z ), (54)
× ny 1 ny 2 nz 1 nz 2 R¯ = 3.6˚A and potential (33) with a = 2˚A, b = 4˚A,
U = 9K, and realistic values 500K < U < 5000K
bd b
−
gives g 1, g 0.1, and the sub∼sequen∼t g are
2 4 j
ρ(r ,r )=V d´rΨ (r ,´r)Ψ (r ,´r), (55) ∼ ∼ −
a b ∗0 a 0 b small. With such gj, we obtain Nk=kc ≈ (1 −g2/2+
Z g2/4)6ρc N/8. For the higher levels, we have: N
ymwihealerdrkses naxc,onlyle,cntzionareofinvteecgteorrss, rn2,=...(,nrNx,.nyR,nelza)t,ioannd(2´r) (Nwg2ik2t=/h27kk−c∞=g≈4/k2c)(,6g3ρ2kgc∞c4N/a4/n)86d,ρc∞NthNke=/5i8nk,tce≈remtc(e.gd22i/aH8te+enocgne4e,/s2t)wh6iρket=c∞h3lNekkvc/e≈=8ls,
(k ,3k ,k ), (k ,3k ,3k )(andwithpermutations)
cx cy cz cx cy cz
ρ(ra,rb) = sin(klxxa)sin(klyya)sin(klzza) are filled. The rest levels are almost empty. Thus, we
× sin(klxxb)sin(klyyb)sin(klzzb) ohnaveea8chlevoeflsthweimth,Ninkt∼he(gs2u/m2)6Nρc∞N/ρ8c∼Nρ/c∞64N. /F8o3ractoomms-
c
A2V d´r exp (S(l)(r ,´r)) (56) parison, N ρ N/2 for a liquid∼43.∞It is easy to verify
× | | Z h n w a ∗o that k1 ≈ ∞
exp S(l)(r ,´r)+S˜ (r ,´r)+S˜ (r ,´r)
× w b b a b b Lx
dx
N n o ρc ρl /P3, P = [fx(x)]2. (62)
sin2(k x )sin2(k y )sin2(k z ) . ∞ ∼ ∞ Z Lx
× { lx j ly j lz j } 0
j=2
Y Withtheobtainedg ,wehaveP (1 g +g2)/2 1/2,
 j ≃ − 2 2 ≃
In Sw(l) (6), we consider only the one-particle part S1(l). im.eu.,laρc∞for∼a8flρul∞id..ThLeetvaulusecoonfsρidl∞eristhgeive“ncrbyystatlh”e cfoorr--
Then
rection A (k) (20) in Eq. (17) for a (k). Then, for
2 2
ρ(r ,r ) = ρc sin(k x )sin(k y )sin(k z ) Ub 102-104K, the values of a2(k) decrease twice (in
a b ∞ lx a ly a lz a mod∼ulus) on the average in the significant interval of k
sin(k x )sin(k y )sin(k z )
× lx b ly b lz b (where a2 is not small). In the zero approximation44,
ρl = exp[ 1 a22(k) ]. We know from experiment
× exp Rρ(qx)+(x→y,z).(57) ∞ −N k=6 01−2a2(k)
qxX=qrxes  tfohratHρe∞4 ≈ato0m.0s7Pρfolr He I1I/.3Aatntdwiρcce less8a/23(k. ),Hwenecoeb,ttahine
  condensate levels∞of≃a sc solid He∞4 c∼ontain N 0.04N
c
1 ∼
Rρ(qx)= √N (S1(l)(qx))∗e−iqxxa +S1(l)(qx)eiqxxb . atoAmts.UTh>is1is04o-n1l0y5aKr,owugehheasvteimgate. 1, g takes val-
b 2 4
(cid:16) ((cid:17)58) ues from∼-2 down to -10, g ∼0; the levels with
j 6
We use approximation (28), (26). Then S1(l) is real, and k = 3kc,5kc,7kc and the int≥erm≈ediate ones are filled.
we can replace e−iqxx cosqxx and can sum only over At Ub > 106K, the values of g6, g8, and g10 be-
→
q >0. We obtain come lar∼ge, which generates the condensates with k =
x
9k ,11k ,13k .
c c c
ρ(ra,rb)=ρc fx(xa)fx(xb)fy(ya)fy(yb)fz(za)fz(zb), Asisseen,thecondensatestructureinacrystalissim-
∞ (59) ilar to that of a fluid in a vessel43, but with the vector
k instead of k and with a different distribution over
c 1
levels. Within our method, it is impossible to calculate
2
fx(x)=sin(klxx)exp√N S1(l)(qrxes)cos(qxresx). the condensate with high accuracy, because the signifi-
 qxrXes>0  cantcorrectionswereomitted inalmostallequations. In
(60) this case, the Monte-Carlo method can be efficient41.
Since qres =2k ,4k ,6k ,..., we have  ThestructureofΨ (2)helpsustoimaginetheconden-
x lx lx lx 0
sate: Ψ contains N identical standing waves (the prod-
0
fx(x)=sin(klxx)e{g2cos(2klxx)+g4cos(4klxx)+...}, (61) uct of sines) resting on the walls by their wings. These
wavesformasingleresonanceclassicalwave,whichmod-
where g = 2S(l)(jk i )/√N 1. Function (61) co- ulates the motion of atoms. Therefore, many atoms are
j 1 lx x ∼
incides with (29), (31). Substituting (59) and (61) in characterized by the wave vector k of this wave. It will
l
(54) and expanding (61) in a series, we obtain that the be discussed in Sec. 8 that this wave can be considered
condensate levels with N N correspond to the wave asaparticularkindoflongitudinalsound. Therefore,the
k
∼
10
condensateofN “zero-phonons”ispresentintheground Itisofimportancetoconfirmthe existenceofthe con-
state. Thus,theresonancewaveformsacrystalandsup- densate in experiments. A neutron or a photon are scat-
portsthecondensateofatoms. Inafluid,thecondensate teredwiththecreationofaquasiparticleorelastically. In
is also formed by a wave, but with k = k (see I and the second case, the wave vector of the scattering atom
1
Ref. 43). eitherisnotchangedorchangesonlythedirection(tothe
Wave function (2) describes a simple rectangular lat- opposite one), and the state of the crystal is invariable
tice. The more complicated lattices can be constructed, (except for the recoil). If the condensate is present, then
by setting arelevantbareWF Ψbare insteadofthe prod- the scattering with the momentum transference 2k (or
c
uct of sines Ψbare (8). The important point is whether with a change in one-two components of the vector 2k )
sc c
Ψbare reflects the structure of a Wigner–Seitz cell, i.e., gives the intense peak. At Ψbare = 0 and at Ψbare = 0,
cell cell 6
whether it equals zero on its surface. This is true for thesechangesinthemomentumareequaltothevectorg
the sc lattice, but is not necessarily for other lattices. If of the reciprocal lattice, and the scattering corresponds
Ψbare =0onthecellsurface(wedenoteΨbare =0),then to the Bragg–Wolf (BW) peaks. But if Ψbare = 0, then
it is ea6 sy to guess the form of Ψbare for thceellb6cc and fcc the number of condensate peaks is less thacnellthat of BW
lattices: peaks. At Ψbare =0, the intensity ofall BW peaks must
cell 6
increaseatsmallT (inthepresenceofacondensate)pro-
N
portionally to the amount of the condensate. For bare
Ψbare = Ψbare(N/2) sin(k (x a /2))
bcc sc { lx j − l WFs(8),(63)–(65),theBWpeaks(noncondensateones)
j=1Y+N/2 canbe, apparently,interpretedas a result ofthe scatter-
sin(k (y a /2))sin(k (z a /2)) ing by zero-phonons with k=k .
× ly j − l lz j − l c
+ permutations. (63) For the bcc, fcc, and hcp lattices, the Wigner–Seitz
(cid:9)
cell is complicated. Moreover, at Ψbare = 0, its cre-
cell
ation requires a wave with complicated structure. But,
Ψbare = Ψbare(N/4)Ψxy(N/4)Ψzx(N/2)Ψyz(3N/4) atΨbare =0,thelatticecanbeformedfromwaveswitha
fcc sc z y x cell 6
+ permutations. (64) simplerstructureofthe type(63)-(66). Since the Nature
uses simple structures, we suppose that Ψbare =0.
cell 6
We note that k=k +δk for the higher condensates,
c
N0+N/4 and the components δk are multiple to the components
Ψxzy(N0) = sin(√2klxx´j)sin(√2klyy´j) 2kc. Therefore, all condensates must be revealed in ex-
j=YN0+1n perimentsasapartofthe“base”condensatewithk=kc.
sin(k [z +a /2]) , (65)
× lz j l }
V. COMPARISON WITH THE TRADITIONAL
x´j =(xj al/2)cos(π/4)+yjsin(π/4), (66) APPROACH
−
y´ = (x a /2)sin(π/4)+y cos(π/4). (67) Let us compare the wave solution (2) with the tradi-
j j l j
− − tional one (1). In the last case, it is assumed that the
Here, the permutations symmetrize the WF to the Bose atoms carry on small oscillations near lattice points.
form, a is the period of a lattice, and four functions in Thetraditionalapproachinvolvesseveralassumptions.
l
(64) onthe right-handside depend onthe coordinatesof 1) The lattice is set “by hands”, it is not obtained from
atoms with the numbers 1,...,N/4; N/4+1,...,N/2; theSchr¨odingerequation. 2)Itisassumedthattheprob-
N/2 + 1,...,3N/4 and 3N/4 + 1,...,N. These WFs abilitydensitymaximuminacellislocatedatthelattice
implythat,forthebcclattice,k =π(i +i +i )/a ,and point, near which the atom carries on random oscilla-
c x y z l
the condensate k are the same as that for the sc lattice. tions (atom is fixed likewise on rubber string, which role
Forthe fcc lattice, therearefour compositecondensates: is played by the function ϕ(r R)). Here, one more
−
with kc = π(ix+iy +iz)/al and k′c = π(√2ix´+√2iy´+ courageousassumptionishidden: thatthemechanismof
i )/a with permutations of x,y,z. appearance of a lattice does not influence the motion of
z l
At Ψbare = 0, k are different. For the sc lattice, the atoms in it. 3) It is assumed also that the boundaries
cell c
basis vectors of the reciprocal lattice are b = 2πi /a , have no effect on the solutions. Therefore, the realistic
1 x l
b = 2πi /a , and b = 2πi /a, and the relation k = BCs close to zero ones are replaced by the cyclic condi-
2 y l 3 z l c
g3D/2 (b +b +b )/2 holds. For the bcc lattice, tions.
0 ≡ 1 2 3
we have b = 2π(i +i )/a , b = 2π(i +i )/a , and Solution (2) has no above-mentioned drawbacks. It is
1 y z l 2 x z l
b = 2π(i + i )/a , and the relation k = g3D/2 = found with the use of natural zero BCs. The lattice and
3 x y l c 0
2π(i +i +i )/a . the probability distribution in a cell do not postulated,
x y z l
We note that the solution for a one-particle WF with buttheyfollowfromthesolutionoftheSchr¨odingerequa-
spherical orbit was studied in Ref. 25. There, it was tion. Itis seenfromthe probabilitydistributionthatthe
assumed that the atoms of the condensate have k motion of atoms is strongly affected by the mechanism
2.1˚A−1. This is close to our results. ∼ (wave one) of formation of a crystal. In addition, the