If and V are defined as above, we show that if has the eigenvalue less than zero, then the smallest eigenvalue of is simple. We use Theorems XIII.44, XIII.45 [4] to prove it.
Proposition 3.1 Suppose has an eigenvalue at the bottom of its spectrum. Then this eigenvalue is simple and the corresponding eigenfunction can be chosen to be a positive function.
Proof Let be a smooth nonincreasing function such that if and if . Let . Set , , , . From the proof of Theorem XIII.47 [4], we know that is positivity preserving and acts irreducibly on . Hence, by Theorem XIII.45 [4], if converges to H and converges to in the strong resolvent sense, then is positivity preserving and acts irreducibly on . By Theorems XIII.43 and XIII.44 [4], we can get the result. Since is the core for all and P, and for any , in , then we have the necessary strong resolvent convergence by Theorem VIII.25(a) [5]. This ends the proof. □
Proposition 3.2 Assume that . in for any , .
Proof If , then . For any test function , we have . If , then we have in for . It follows because ϕ and belong to . Hence, in . □
Proposition 3.3 Assume that . is a compact operator for . And converges to in operator norm sense.
Proof For , . Since is a bounded operator from to , and V is a compact operator from to , then is a compact operator on . Using a similar method to that in Proposition 2.2, we can show that and have the same non-zero eigenvalues, and for the same eigenvalue , the multiplicity of as the eigenvalue of and the multiplicity of as the eigenvalue of are the same. Hence, is a compact operator. Because
and if , then in . Hence, in operator norm sense as . This means that is a compact operator. □
Lemma 3.4 Suppose , are two bounded self-adjoint operators on a Hilbert space H. Set
then .
Proof By the definition of , one has
This ends the proof. □
Lemma 3.5 Suppose is a family of compact self-adjoint operators on a separable Hilbert space H, and for α near zero. Set
Then:
-
(a)
is an eigenvalue of , and converges when . Moreover, if , then is an eigenvalue of .
-
(b)
Suppose that is an eigenvalue of of the multiplicity of m. Then there are m eigenvalues (counting multiplicity), (), of near . Moreover, we can choose such that (), is the eigenvector of corresponding to (), and converges as . If converges to , then is the eigenvector of corresponding to .
Proof (a) By the min-max principle, we know that is an eigenvalue of . By Lemma 3.4, one has
It follows that converges to the eigenvalue of .
(b) Because is a compact operator and is an eigenvalue of , then is a discrete spectrum of . Then there exists a constant small enough such that has only one eigenvalue in (). For α small enough, has exactly m eigenvalues (counting multiplicity) in because the eigenvalues of converge to the eigenvalues of by part (a) of lemma. Suppose the m eigenvalues, near , of are , and the corresponding eigenvectors are such that . Let
Then . Let , then . For α near zero, one has
Let . Let be an element in A such that acquires the minimum value. Then we have
and
In the last equality, we use the fact
and
It follows that . Let . Then for because if , and . This ends the proof. □
Let and
Here, or 1, , (), are compact operators, and for β near zero. Set
Then, by the min-max principle, is an eigenvalue of . Moreover, if , then is a discrete eigenvalue of because is a compact operator. If is an eigenvalue of of multiplicity m, without loss, we can suppose that . Then there exist exactly m eigenvalues (counting multiplicity), , of near . By Lemma 3.5, we know that there exists a family of normalized eigenvectors of such that , (), and () converge as . Suppose that converge to for all j such that . Then . can be extended to a standard orthogonal basis. Set
Then we have the following.
Lemma 3.6 , are given as before. Then the eigenvalue of , () has the following form:
Here
Proof If , then is the discrete eigenvalue of . Suppose that the multiplicity of is m, and suppose that as before. Hence, we can choose small enough such that there is only one eigenvalue in . We know that () converge to . It follows that if δ is small enough, there are exactly m eigenvalues (counting multiplicity) of in for β small. Set
Then
Since
then
Then
In particular,
In the last step, we use that
Similarly, we can get
and
□
First, we study the asymptotic expansion of the smallest eigenvalue of . By Proposition 3.1, we know that is a simple eigenvalue of , and the corresponding eigenfunction can be chosen to be positive. We suppose is a positive eigenfunction corresponding to . Then . Without loss of generality, we can suppose that . Then we can get the following result.
Lemma 3.7 Assume that . Set . is defined as above. Then converges in when . If and converges to ϕ, then ϕ is the eigenfunction of , and .
Proof By the assumption of , one has . One can check that converges in as by Lemma 3.5. And also, by Lemma 3.5, we know that ϕ is the normalized eigenfunction of corresponding to . ϕ is a positive function since is a positive function. Let , then and u is a positive function because . Then
In the last equality, we use the fact that
with , and . This ends the proof. □
Theorem 3.8 Assume that . ϕ is defined in Lemma 3.7. If , one of three exclusive situations holds:
-
(a)
If , then
with .
-
(b)
If , then
with .
-
(c)
If , then
with .
Proof
-
(a)
By Theorem 2.1, one has
Then if , we can get in . Because is the simple eigenvalue of , then is the simple eigenvalue of . Since , one has that is monotonous with respect to λ and so is the . Hence, and the eigenvalues of are monotonous with respect to λ. Therefore, we have that is the biggest eigenvalue of . If not, suppose that is an eigenvalue of , then by the continuity and monotony of the eigenvalue of with respect to λ, we know that there exists a constant such that . It follows that is an eigenvalue of . This is contradictory to that is the smallest eigenvalue. By Lemma 3.7, we know the normalized eigenfunction of converges to ϕ. It follows with . Then
Here is the eigenvalue of corresponding to the eigenfunction . By Lemma 3.6, we should compute . Let . From the definition of ϕ, one has . Hence,
In the last equality, we use the fact , which can be obtained by Proposition 3.2. Since , we have by Theorem 3.1 [1]. So, ψ is the ground state of . We also have
Hence,
It follows
So, with . By the Proposition 2.2, one has . It follows
Since , we can get the leading term of is with .
-
(b)
If , then
By Lemma 3.7, one has
Then we have
with . As in (a), using and
one has . To get the leading term of , we can suppose that . Then, by comparing the leading term, we can get . It follows
with .
-
(c)
If , one has
By Lemma 3.7, we know that . Using the same argument as before, we can conclude
with . As above, we can get that
with . □