Correlation inequalities for noninteracting Bose gases

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by  Andras Suto

Correlation inequalities for noninteracting Bose gases
Andr´as Su¨t˝o
Research Institute for Solid State Physics and Optics
4
0 Hungarian Academy of Sciences
0 P. O. Box 49, H-1525 Budapest
2
Hungary

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E-mail: [email protected]
a
J

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Abstract

2
v ForanoninteractingBosegaswith afixedone-bodyHamiltonian H0 independentofthenumber
3 of particles we derive the inequalities hNiiN < hNiiN+1, hNiNjiN < hNiiNhNjiN for i 6= j,
0 ∂hN0iN/∂β >0 and hNii+N <hNiiN. Here Ni is the occupation numberof the ith eigenstate of
0 H0,β istheinversetemperatureandthesuperscript+referstoaddinganextraleveltothoseof
9 H0. The results follow from theconvexityof the N-particle free energy as a function of N.
0
3
PACS numbers: 05.30.Jp, 02.50.Cw
0
/
h
p Correlation inequalities have been playing a particular role in statistical physics. They have
- always been limited to special models, most of them being valid for classical spins on lattices [1,
h
2].Nevertheless,whentheycouldbeapplied,theyprovedtobeveryusefulinderivingrigorousresults,
t
a for example, on phase transitions or on the existence of limits of correlationfunctions.
m
InthispaperIpresentcorrelationinequalitiesforafamilyofcontinuousorlatticequantumsystems:
: noninteracting Bose gases in boxes, on tori or in some sufficiently fast increasing confining external
v
i potential. Such a system is characterizedby a one-body Hamiltonian H0 having a discrete spectrum
X
ε0 <ε1 ≤ε2 ≤.... In the continuum the number of eigenstates is infinite and we have to impose the
r condition that the sum of the Boltzmann factors x =exp(−βε ) is finite,
a i i
∞ ∞
tre−βH0 = e−βεi ≡ x <∞ . (1)
i
i=0 i=0
X X
The relevant random variables are the occupation numbers N of the eigenstates of H0. In the
i
canonicalensembletheyareweaklydependent,theironlyinter-dependencecomingfromtheconstraint
N =N. Hence, their joint probability distribution is
i
P PN(N0 =n0,N1 =n1,...)=ZN−1δN, ni xnjj , ZN = xnjj (2)
P jY≥0 {nj}:Xnj=NjY≥0
P
Note that Z0 = 1. Let H+0 be a one-body Hamiltonian whose spectrum specH+0 = specH0 ∪{ε}.
Below h·i and h·i+ denote mean values in the N-particle canonical ensembles generated by H0 and
N N
H0, respectively.
+
Theorem. (i) hNiiN <hNiiN+1
(ii) ∂hN i /∂lnx =hN N i −hN i hN i <0 for i6=j
i N j i j N i N j N
(iii) ∂hN0iN/∂β >0
(iv) hN i+ <hN i .
i N i N
1
The content of these inequalities is intuitively obvious but not more obvious than that of the
inequalities of Griffiths, Kelly and Sherman (GKS) which say that in ferromagnetic Ising models the
spinsarepositivelycorrelated[3,4]. HavinginmindthesubtletiesoftheproofoftheGKSinequalities,
one cannot expect a genuinly simple proof for this theorem either.
Inequality (iv) canbe put in a more generalform. Let H1 be a one-particleHamiltonian, suppose
that exp(−βH1) is trace class and specH0 ⊂specH1 where repeated eigenvalues are considered sep-
arately. If ε is a common eigenvalue then the occupation number N of the corresponding eigenstate
i i
(which may be different for H0 and H1) satisfies
(iv’) hN i1 <hN i
i N i N
whereh·i1 denotesthecanonicalexpectationvaluewithrespecttoH1. Oneobtains(iv’)byarepeated
N
applicationof(iv)ordirectlybyasimplemodificationoftheproofof(iv). Onemaywonderaboutthe
relevance of the fourth inequality. One-particle Hamiltonians with a modified spectrum can serve as
auxiliarytools, as for instance in [5, 6]. Actually, the motivation ofthe presentpaper was the need of
thisinequalityinprovingtheoccurrenceofageneralizedBose-Einsteincondensationinaninteracting
trapped Bose gas, when both the interaction and the one-dimensional harmonic trap potential are
scaled as N tends to infinity [6].
The basic ingredient of the proof of the theorem is the convexity of the N-particle free energy.
Lemma. The N-particle free energy is a convex function of N, namely
ZmZn+1 ≤Zm+1Zn for any m<n. (3)
Introducing FN = −lnZN, (3) is equivalent to Fm+1−Fm ≤ Fn+1−Fn which is just convexity.
The usual statement about the convexity of the free energy in homogenous systems is that in the
thermodynamic limit the free energy density is a convex function of the particle number density.
This holds quite generally true, whenever there is asymptotic equivalence between canonical and
grand-canonical ensembles. Indeed, in this case the free energy density is the Legendre transform
of the pressure which is trivially convex as a function of the chemical potential. Inequality (3) is
more model dependent. It perhaps remains true in the so-called diagonal model of a Bose gas [7] if
the pair interaction has a nonnegative Fourier transform, because then there is a repulsion between
the occupationnumbers of different plane-wavestates. Also, in classicallattice models with repulsive
interactions(3)mayholdtruebecausedividingafinitepieceofalatticeintotwoparts,themoreequal
thesizesofthepartsthelargerthenumberofrepulsivelinkstobecut(andtheproductofthepartition
functions of the two parts with it). Note that the subadditive property lnZm+n ≤lnZn+lnZm is a
special case of (3) (Z0 =1).
Proof of the theorem.
We prove the first inequality by using the lemma. The scheme of the proof of the rest will then be
(i)⇒(ii)⇒(iii) and (i)⇒(iv).
(i) Let us introduce the notation Z for the N-particle partition function in a system where the
N,i
level i is missing,
Z = xnj (4)
N,i j
{nj}j6=Xi: nj=NjY≥0
The lemma applies to these partition functions asPwell. For any i
N
Z = xkZ (5)
N i N−k,i
k=0
X
Using (5) and the corresponding identity for ZN+1,
N+1 N
1 1
hNiiN+1−hNiiN = ZN+1 m=1mxmi ZN+1−m,i− ZN m=1mxmi ZN−m,i
X X
2
N N+1 N+1 N
1
= ZNZN+1 (k=0m=1mxki+mZN−k,iZN+1−m,i− k=0 m=1mxki+mZN−m,iZN+1−k,i)
X X X X
N N
1
= ZNZN+1 k=0m=1mxki+m[ZN−k,iZN+1−m,i−ZN−m,iZN+1−k,i]
X X
N N
1
+ (N +1)xk+N+1Z − mxm+N+1Z
ZNZN+1 (k=0 i N−k,i m=1 i N−m,i)
X X
N N
1
= ZNZN+1 k=1m=1mxki+m[ZN−k,iZN+1−m,i−ZN−m,iZN+1−k,i]
X X
N
1
+ZNZN+1 (m=1mxmi [ZN,iZN+1−m,i−ZN−m,iZN+1,i]
X
N
+(N +1)xN+1Z + (N +1−m)xm+N+1Z .
i N,i i N−m,i
)
m=1
X
The expressionin the last line is strictly positive and, due to the lemma, in the penultimate line each
termofthesinglesumisnonnegative. Letusrewritethedoublesum. Sincethediagonaltermsvanish
and the difference in the square bracket changes sign if k and m are interchanged, we obtain
xki+m(m−k)[ZN−k,iZN+1−m,i−ZN−m,iZN+1−k,i].
1≤k<m≤N
X
Because the smallest among the four subscripts is N −m, we find again with the lemma that each
term is nonnegative. This proves inequality (i).
(ii) For i6=j let us introduce
n
1
hN i = kxkZ , (6)
i n,j Z i n−k,i,j
n,j
k=1
X
the mean value of N in the n-particle canonicalensemble of a system in which the level j is missing.
i
Here Z is the (n−k)-particle partition function in a system with missing levels i,j. For any
n−k,i,j
N ≥n and m=N −n, hN i agrees with the conditional expectation value of N in the N-particle
i n,j i
system provided that N =m. The inequality (i) is valid also in this case. One can write
j
N
hN i = p (x )hN i (7)
i N m j i N−m,j
m=0
X
where
xmZ
j N−m,j
p (x )= (8)
m j
Z
N
N N
are probabilities, p (x )=1 (and mp (x )=hN i ). Now
m=0 m j m=1 m j j N
P ∂pm(xj) xmj ZN−mP,j
= (m−hN i )=p (x )(m−hN i ) (9)
j N m j j N
∂lnx Z
j N
changes sign when m passes through hN i . Therefore, by applying the inequality (i) to hN i
j N i N−m,j
we find
N
∂hN i ∂p (x )
i N m j
= hN N i −hN i hN i = hN i
i j N i N j N i N−m,j
∂lnx ∂lnx
j m=0 j
X
3
∂p (x ) ∂p (x )
m j m j
= hN i + hN i
i N−m,j i N−m,j
∂lnx ∂lnx
j j
0≤mX<hNjiN hNjiNX<m≤N
∂p (x ) ∂p (x )
m j m j
< hN i +hN i
i N−⌊hNjiN⌋,j ∂lnx i N−⌈hNjiN⌉,j ∂lnx
j j
0≤mX<hNjiN hNjiNX<m≤N
N
∂
≤ hN i p (x )=0 (10)
i N−⌊hNjiN⌋,j∂lnx m j
j m=0
X
which proves the second inequality. The identity
N
hN N i = mp (x )hN i (11)
i j N m j i N−m,j
m=1
X
can be read off from the first line of (10).
(iii)
∞
∂hN0iN
=− N0 Njεj +hN0iN Njεj =− εj[hN0NjiN −hN0iNhNjiN]
∂β N N
D X E DX E Xj=0
∞
=−ε0[hN02iN −hN0i2N)]− εj[hN0NjiN −hN0iNhNjiN]
j=1
X
∞
=− (εj −ε0)[hN0NjiN −hN0iNhNjiN]>0 (12)
j=1
X
where we used
∞
[hN0NjiN −hN0iNhNjiN]=N[hN0iN −hN0iN]=0 (13)
j=0
X
in the last equality and (ii) in the inequality.
(iv)Letx=exp(−βε)andletZ+ denotetheN-particlecanonicalpartitionfunctionofthesystem
N
with the additional level. In analogy with (7) and by applying (i)
N xmZ N
hN i+ = N−mhN i ≡ p+(x)hN i <hN i (14)
i N Z+ i N−m m i N−m i N
m=0 N m=0
X X
because p+(x)=Z /Z+ <1.
0 N N
This finishes the proof of the theorem.
Proof of the lemma.
l
Let p1 ≤ p2 ≤ ··· ≤ pl be a partition of N, that is, p1 ≥ 1 and 1pi = N. We denote this by
p = (p1,p2,...,pl) ⊢ N. For each m ∈ [0,N], ZmZN−m is a sum over the same set of monomials
of x1,x2,..., namely all those jxnjj with jnj = N. For a givenPm the coefficient of a monomial
depends only on the set of nonzero n and not on their order or on the set of the corresponding
j
Q P
subscripts j. Therefore
Z Z = a(p|m) xnj (15)
m N−m j
pX⊢N {nXj}∼pY
where {n } ∼ p means that, after rearranging in increasing order, the vector of nonzero n agrees
j j
with p. We shall prove that for any 1≤p1 ≤···≤pl
1
a(p1,p2,...,pl|m)≤a(p1,p2,...,pl|m+1) if m< (p1+···+pl) (16)
2
4
which implies (3) through a term-by-term inequality. The proof is obtained by induction over l.
a(p1,...,pl|m) is the number of different mth degree monomials within xp11···xpll or, in more
popular terms, the number of possibilities to distribute m Euros among l people in such a way that
they can obtain at most p1,...,pl, respectively. Thus, a(p|m)=1 for any p≥1 and any 0≤m≤p,
so that for l=1 (16) holds with equality. The induction hypothesis is that for an l>1
1
a(p1,...,pl−1|m)≤a(p1,...,pl−1|m+1) if m< (p1+···+pl−1). (17)
2
Because of the identity
a(p|m)=a(p| p −m) (18)
i
the hypothesis (17) is equivalent to X
a(p1,...,pl−1|m)≤a(p1,...,pl−1|n) if m≤min{n, p1+···+pl−1−n}. (19)
Let us introduce the notations
Q(p1,...,pl)={q∈Zl :0≤qi ≤pi, 1≤i≤l} , (20)
l
C(p1,...,pl|m)={q∈Q(p1,...,pl): qi =m} (21)
1
X
and
l−1
PC(p1,...,pl|m)={q∈Q(p1,...,pl−1):m−pl ≤ qi ≤m}. (22)
1
X
We have
a(p1,...,pl|m)=|C(p1,...,pl|m)|=|PC(p1,...,pl|m)| (23)
|A|meaningthenumberofelementsofthesetA. Thefirstequalityisobvious. Thesecondcomesfrom
the fact that PC(p|m) is the orthogonal projection of C(p|m) onto the l−1 dimensional subspace
l
perpendicular to the lth unit vector and, because of the constraint q = m, the projection is a
1 i
bijection. Comparing (22) with
P
l−1
PC(p1,...,pl|m+1)={q∈Q(p1,...,pl−1):m+1−pl ≤ qi ≤m+1} (24)
1
X
we find that
l−1
|PC(p|m+1)\PC(p|m)| = |{q∈Q(p1,...,pl−1): qi =m+1}|
1
X
= a(p1,...,pl−1|m+1) (25)
while
l−1
|PC(p|m)\PC(p|m+1)| = |{q∈Q(p1,...,pl−1): qi =m−pl}|
1
X
= a(p1,...,pl−1|m−pl). (26)
Note that either of (25) and (26) can vanish if m+1 > p1+···+pl−1 or m−pl < 0, respectively.
The inequality (16) is satisfied if
a(p1,...,pl−1|m−pl)≤a(p1,...,pl−1|m+1) , (27)
5
and this follows from the induction hypothesis if
m−pl ≤min{m+1, p1+···+pl−1−(m+1)}. (28)
Ifm+1≤(p1+···+pl−1)/2thenthisistheminimumontheright-handsideof(28)whichistherefore
fulfilled. If m+1>(p1+···+pl−1)/2, the minimum is p1+···+pl−1−(m+1). But then
m−pl ≤p1+···+pl−1−(m+1)
is equivalent to
1 1
m+ ≤ (p1+···+pl)
2 2
which holds true because m<(p1+···+pl)/2. By this we have finished the proof.
Concerning the proof of the lemma, two remarks are in order. First, it is tempting to present
a more elegant argument which refers to the convexity of the hyper-rectangle Q(p). If we had Rl
instead of Zl in the definition of Q(p), C(p|m) would be the intersection of the rectangle with the
hyperplane q = m. Because of the convexity of Q(p) the l−1 dimensional Lebesgue measure of
i
C(p|m) cannothave a minimum in the interval 0<m< p . By symmetry, it has a not necessarily
i
P
strict maximum at m = p /2, implying that it is nondecreasing for m < p /2. However, we
i i
P
do not need this result for the Lebesgue measure of the cut but for the number of points of integer
P P
coordinates on it, which makes the convexity argument somewhat shaky.
Second, the number in question can be given by a formula. It reads
p1 min{pj,m− ij=−11qi} min{pl−1,m− il−=21qi}
a(p|m)= ··· P ··· P 1 . (29)
qX1=0 qj=max{0,m− Xij=−11qi− li=j+1pi} ql−1=max{0,Xm− il−=21qi−pl}
P P P
Ifm≤p ,inallthesummationsthelowerboundsequalzeroandthusform<p ,a(p|m)≤a(p|m+1)
l l
can explicitly be seen. For m≤p by simple combinatorial considerations we find
i
p1 min{pi−1,m− ji−=21qj} m− i−1q +l−i
a(p|m)= ··· P j=1 j . (30)
l−i
qX1=0 qiX−1=0 (cid:18) P (cid:19)
If m>p the general formula is less transparent. This is why we have opted for the inductive proof.
l
Acknowledgment This work has benefited from the support of the Hungarian Scientific Research
Fund through Grant T 42914.
Noteadded. Afterhavingsubmittedthepaper,IlearnedfromValentinZagrebnovthattheconvexity
of the free energy was earlier shown by Lewis, Zagrebnov and Pul´e [8]. Their proof is somewhat
shorter, the present one provides somewhat more information due to the established connection with
combinatorics and geometry. Also, I was informed about a recent work by Pul´e and Zagrebnov[9] in
which inequality (i) is derived and used.
References
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C. Domb and M. S. Green) 1 7-109 (Academic Press 1972)
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Dissertation, Universit´e Catholique de Louvain 1977
[3] Griffiths R B: Correlations in Ising ferromagnets.I. J. Math. Phys. 8 478-483(1967)
Griffiths R B: Correlations in Ising ferromagnets.II. External magnetic fields. J. Math. Phys. 8
484-489(1967)
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[4] Kelly D G, Sherman S: General Griffiths’ inequalities on correlations in Ising ferromagnets. J.
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