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The Pseudospectral Properties of non-self-adjoint Schr\"odinger Operators in the semi-classical limit

We describe the general qualitative behaviour of the resolvent norm for a very wide class of non-self-adjoint Schroedinger operators in the semi-classical regime, as the spectral parameter varies over the complex plane.


Introduction
Several authors [4,9,11] have demonstrated that the spectrum of certain Schrödinger operators with complex potentials is unstable; this is shown by observing that the resolvent norm (H − λ) −1 becomes unbounded as some parameter associated with the operator H varies, even though λ may be far from the spectrum of H. This phenomena is commonly quantified using the concept of the pseudospectral sets for some (and hence all) t > s. It is then well-known that, provided C\Num(T ) is a connected set (T − λ) −1 ≤ 1 dist(λ, Num(T )) (1) for all λ / ∈ Num(T ) [7, p.268]. We will be concerned with the Schrödinger operator with complex-valued V and h > 0, acting in L 2 (Ω) where Ω is some region in R N . We assume Dirichlet boundary conditions throughout. The h-independent set will be of fundamental importance. In Section 3 we show that for any λ ∈ Φ(V ), the resolvent norm tends to infinity in the semi-classical limit, h → 0. The fact that V may take complex values means that conv(Φ(V ))\Φ(V ) (conv denoting the convex hull) is in general non-empty, and in Section 4 we give new results (Theorems 5 and 6) which show that in the semi-classical limit a bound analogous to (1) holds for λ in this set.

Preliminaries
First we discuss the conditions which we shall impose on V and the question of the domain of the operator H h . We assume that there exists a closed set M ⊆ Ω with Lebesgue measure zero, such that the restriction V Ω\M is continuous and bounded. Since the operator H h will not be changed on sets of measure zero, we allow V to be undefined on M. Defining H h,0 to be the self-adjoint operator −h 2 ∆ with domain W 1,2 0 (Ω), it is trivial that V has relative bound zero with respect to H h,0 since V ∈ L ∞ (Ω). Therefore, Dom(H h ) = W 1,2 0 (Ω), and the fact that for all λ > 0, together with an argument similar to the proof of [3,Theorem 1.4.2] shows that H h is maximal on W 1,2 0 (Ω). Moreover, since we may integrate by parts for all f ∈ W 1,2 0 (Ω). Since for all h > 0. .
Proof H h satisfies the conditions for (1) to hold, and then one can apply the lemma.

Spectral instability
In a sense our result in this section generalises that of [5, Section 2]; however, our conclusion is not as strong since we do not show super-polynomial growth of the resolvent norm in the semi-classical limit. Note that the specific form of the phase function x → γ · x below has been chosen to ensure that the second derivative disappears, and to exploit the fact that λ := V (c) + |γ| 2 . We have not attempted to optimise the choice of phase function, but instead have aimed for a clear demonstration of the underlying processes.
Proof Let λ ∈ {Ran(V ) + [0, ∞)} so that λ := V (c) + |γ| 2 , where c ∈ Ω\M, and γ ∈ R N . Let φ ∈ C ∞ c (R N ) be a function whose support is contained within the open ball B(0; 1), and put where p > 0 is a constant to be determined, and Then supp(f λ,h ) ⊆ B(c; h p ) and Therefore, it follows that Taking the limit, where k 1 , k 2 are constants dependent upon our choice of φ and γ. Thus, by taking 0 < p < 1, the continuity of V on Ω\M ensures that Now let w ∈ Φ(V ) and, aiming for a contradiction, suppose that For any δ > 0 there exists λ ∈ {Ran(V ) + [0, ∞)} such that |λ − w| < δ, by the definition of the closure. Hence, using the resolvent identity, Since δ may be taken arbitrarily small, we have as h → 0, contradicting the result just obtained, and completing the proof.

Bounded behaviour of the Resolvent norm
We have seen that when λ ∈ Φ(V ) the resolvent norm becomes infinitely large as h → 0 (Theorem 3). Conversely, when λ / ∈ conv(Φ(V )) the resolvent norm is uniformly bounded in h > 0 (Proposition 2). Therefore, the natural question arises: how does lim h→0 (H h − λ) −1 behave for λ ∈ conv(Φ(V ))\Φ(V )? On the one hand, Proposition 2 cannot in general be extended to λ ∈ conv(Φ(V ))\Φ(V ), as the following counter-example shows: But, we have shown elsewhere [10] that countably many of the eigenvalues {λ h,n } ∞ n=1 of H δ,h are positive real, for every h > 0, in which case On the other hand, for a wide class of potentials, we have the positive result (initially suggested by numerical simulations of the associated discrete problem using Matlab) that for any given λ ∈ conv(Φ(V ))\Φ(V ) the resolvent norm becomes bounded eventually, as h → 0. In the one-dimensional case we have the following result, which we believe to be new.

Theorem 5 Let K h be the non-self-adjoint operator
When a or b are finite we impose Dirichlet boundary conditions. We assume that there exists a partition .
Proof We first prove the case when −∞ < a < b < +∞, noting that (iii) is then automatically satisfied. Our proof will involve adding extra points to those in the given partition of the real interval (a, b). We will then estimate the resolvent norm on each of the sub-intervals (x j , x j+1 ). Without loss of generality therefore, we may initially assume that V is twice continuously differentiable and bounded on the interval (−1, 1), since the extension to the general case follows easily. Our method uses the so called WKB approximations [6,8] to the solutions of the differential equation does not cross the negative real axis, condition (ii) also ensures that we may choose a twice continuously differentiable branch of V (x) − λ such that for all x ∈ (−1, 1). Denoting the definite integral where a ∈ (−1, 1) is arbitrary and introduces a constant term which we will omit from our later calculations, it follows from (6) that the function x → Re ξ(x) is increasing on the interval (−1, 1). This fact will be called upon several times in our proof. Property (6) will greatly simplify our application of the WKB approximations, since questions about the Stokes' phenomenon and valid domains do not then arise.
Now defining the operatorK h formally by (4) but with the finite number of boundary conditions f (x j ) = 0 j = 0, . . . , n, K h then effectively acts on By induction, our argument so far shows that the operator is of at most rank n, and we have the norm resolvent convergence Moreover, letting V j denote the potential V restricted to the interval (x j , x j+1 ), condition (i) ensures that one can apply Proposition 2 toK h separately on each interval (x j , x j+1 ). Therefore, taking we have uniformly on h > 0. As the partition of (−1, 1) becomes increasingly fine, it is clear that for every λ ∈ C\Φ(V ) one has Then, by (12) and (13) it follows that lim sup to complete the proof in the finite interval case.
On the interval (a, +∞), the WKB approximations to the classical solutions of (5) take the formỹ (1)) as x → ∞, for all h > 0, provided (iii) holds (see [6, p50]). Comparingỹ 1,2 with y 1,2 in the proof above, and noting thatỹ 2 (x) → 0 exponentially as x → +∞, one can check that the proof just given still carries through. The interval (−∞, b) is dealt with similarly, by a change of signs, completing the proof for the general case.

A Different Approach
In Theorem 5 we relied upon the theory of ODEs to prove the norm resolvent convergence (12). The proof of the next theorem uses a powerful but technically simple construction, the so-called 'Twisting Trick' [3, Section 8.6] or [2]. .
Proof If λ / ∈ conv(Φ(V )) then Proposition 2 applies and we are done. So, assume that λ ∈ conv(Φ(V ))\Φ(V ) is given. For any δ > 0 we define {S j } to comprise N-dimensional cubes of the form {(x 1 , . . . , x N ) : δr i < x i < δ(r i + 1)}, where the r 1 , . . . , r N take integer values. Then by the uniform continuity of V onΩ, there exists a covering ofΩ by disjoint cubes In addition, for any given 0 < α < 1 we can always take δ > 0 small enough so that The proof proceeds by a series of bisections in each of the N dimensions of Ω. Choose a point c = (c 1 , . . . , c N ) such that each c i = δr i for some integer r i . Then the hyperplane {x : x i = c i } splits the cubes into two families; one covering the region Ω 1 := {x : x i − c i > 0}, and the other covering the region Ω 2 := {x : where m is some sufficiently large real number. Then consider the two operators both acting in the Hilbert space H := L 2 (Ω) ⊕ L 2 (Ω). We will show that we define the unitary operator U h : H → H by where γ > 0 is to be determined. Thus To ease notation we denote the following functions, which are to be regarded as multiplication operators on L 2 (Ω), together with the partial differentiation operator D i := ∂/∂x i . Then, one can show using elementary matrix calculations that where P h , Q h and G h are the matrix-valued functions on Ω given by , as h → 0. We therefore take the optimal value γ = 2/3. Now, for our given λ, one may write to H, and U h is uniformly bounded from W 1,2 0 to W 1,2 0 for all h ≤ 1. Thus for some β < ∞ and all h ≤ 1. Therefore, from (16), we obtain where, applying Proposition 2, one has .
One can now bisect each of Ω 1 and Ω 2 in the same manner, and carry out the above process on four copies of L 2 (Ω). Repeating until Ω has been divided into 'strips' of thickness δ, one changes to another coordinate direction and repeats the process until all N dimensions have been decomposed. Then, recalling (14) we have as h → 0, where 0 < α < 1 was arbitrarily chosen. Taking α as close as one likes to 1 will then complete the proof.

Appendix
We give a proof of the following well-known result.
Lemma 7 Let L be the Sturm-Liouville operator acting in L 2 (a, b), together with boundary conditions Here V is a complex-valued continuous function on [a, b], and λ is a complex constant. We may assume that a < 0 < b, and letL denote the operator given formally by (17)  Proof Let u, v be a pair of independent classical solutions to the differential equation (17), with u, v satisfying the boundary conditions at a, b respectively. Multiplying u by the constant v(0)/u(0), we may further assume that u(0) = v(0). Then the Green function is given by For any x ∈ (a, b), the Wronskian is given by Imposing the boundary condition f (0) = 0,L effectively acts on the Hilbert space L 2 (a, 0) ⊕ L 2 (0, b). Then, putting we see that w(0) = 0, and the new Wronskian on L 2 (a, 0): By a similar calculation, the Wronskian on L 2 (0, b) is given by W + = W . Hence we construct the new 'Green' functioñ k(x, y) := −W −1 It is then straightforward to check that putting the 'Green' function of (L − λ) −1 − (L − λ) −1 is given by for all x, y in (a, b). Clearly σ ∈ L 2 (a, b), and for all f ∈ L 2 (a, b) is a compact rank one operator on L 2 (a, b), which completes the proof.