pdf file
pdf file
pdf file
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
BOGOLIUBOV SPECTRUM OF INTERACTING BOSE GASES 37<br />
then there exists a subsequence {x (L)<br />
nk }k≥1 which converges, in the norm induced<br />
by the quadratic form of A, to an eigenvector of A with the eigenvalue<br />
λL(A).<br />
Theproofof Lemma25is elementary, usingthesameargument ofproving<br />
the convergence of ground states, and an induction process. The proof of<br />
Theorem 2 is finished. �<br />
7.2. Proof of Theorem 3.<br />
Step 1. Convergence of the free energy. We need to show that<br />
We can rewrite<br />
lim<br />
N→∞ (Fβ(N)−NeH) = −β −1 logTrF+ [e−βH ].<br />
Fβ(N)−NeH = inf<br />
Γ≥0,Tr F ≤N<br />
+<br />
where � HN := UN(HN −NeH)U ∗ N , and<br />
(Γ)=1 {Tr[� HNΓ]−β −1 S(Γ)}<br />
−β −1 logTrF+ [e−βH ] = Tr[HΓ]−β −1 S(Γ)<br />
where Γ := Z −1 e −βH with Z = Tr � e −βH� .<br />
Upper bound. Let us write Γ = �∞ i=1ti �<br />
�Φ (i) 〉〈Φ (i)�� where {Φ (i) } ∞ i=1 is an<br />
orthonormal family in F+ and t1 ≥ t2 ≥ ... ≥ 0, � ti = 1. Then<br />
TrF+<br />
� HΓ � −β −1 S(Γ) =<br />
∞�<br />
i=1<br />
�<br />
ti〈H〉 Φ (i) +β −1 �<br />
tilogti . (61)<br />
Fix L ∈ N. By using Lemma 23 and the fact that H is bounded from<br />
below, we can findfor every M ≥ 1afamily of normalized states {Φ (i)<br />
M }L i=1 ⊂<br />
F ≤M<br />
+ such that limM→∞〈Φ (i)<br />
M ,Φ(j)<br />
M 〉 = δij and<br />
limsup〈H〉<br />
(i) ≤ 〈H〉<br />
Φ Φ (i)<br />
M→∞ M<br />
for all i,j ∈ {1,2,...,L}. Denote θL = � L<br />
i=1 ti and<br />
L�<br />
�<br />
�<br />
ti<br />
ΓL,M := �Φ<br />
θL<br />
i=1<br />
(i)<br />
M<br />
��<br />
Φ (i)<br />
�<br />
�<br />
�.<br />
M<br />
(62)<br />
Then it is easy to see that ΓL,M ≥ 0 and Tr[ΓL,M] = 1. Moreover, because<br />
limM→∞〈Φ (i)<br />
M ,Φ(j)<br />
M 〉 = δij we get<br />
lim<br />
M→∞ S(ΓL,M) = −<br />
L�<br />
ti<br />
θL<br />
i=1<br />
� �<br />
ti<br />
log . (63)<br />
θL<br />
Choosing M = N 1/3 and applying Proposition 15 we obtain<br />
Tr ≤N[<br />
F+ � HNΓL,M]−TrF+ [HΓL,M] → 0 (64)