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  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 , we know that is positivity preserving and acts irreducibly on . Hence, by Theorem XIII.45 , 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 , 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) . 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
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
is an eigenvalue of , and converges when . Moreover, if , then is an eigenvalue of .
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
In the last equality, we use the fact
It follows that . Let . Then for because if , and . This ends the proof. □
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:
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
In the last step, we use that
Similarly, we can get
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:
If , then
If , then
If , then
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 . So, ψ is the ground state of . We also have
So, with . By the Proposition 2.2, one has . It follows
Since , we can get the leading term of is with .
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
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 . □