Large Gaps Asymptotics for the 1-dimensional Random Schrödinger Operator

J o u r n a l o f P r o b a b i l i t y Electron. Abstract We show that in the Schrödinger point process, Schτ , τ > 0, the probability of having no eigenvalue in a fixed interval of size λ is given by exp − λ 2 4τ + 2 τ − 1 4 λ + o(λ) , as λ → ∞. It is a slightly more precise version of the formula given in a previous work.

We want to compute the asymptotics of E τ (0, λ) as λ → ∞.An asymptotic of this probability was computed in [3] and is given by: as λ → ∞.Our aim in this paper is to compute the asymptotics of E τ (0, λ) to the next order.We will be proving the following theorem: dφ λ (t) = λdt + dB(t) + e −iφ λ dW (t) , φ λ (0) = 0, coupled together for all values of λ ∈ R, where B and W are standard real and complex Brownian motions respectively.By Corollary 4, [3], this SDE has a unique strong solution and for each time t, the function λ → φ λ (t) is strictly increasing and real-analytic with probability one.We define Sch τ as follows: Sch τ := λ : φ λ/τ (τ ) ∈ 2πZ .
We consider the strong solution of the following coupled one-parameter family of SDE where Z is a complex Brownian motion and f (t) = (β/4) exp (−(β/4)t) .For a single λ, (1.6) In [4], it is shown that the quantity lim t→∞ α λ (t) 2π , exists for every λ and is equal to an integer a.s.We define N (λ) as the unique random right-continuous function which agrees with for every λ a.s.For λ 1 < λ 2 , we have N (λ 1 ) < N (λ 2 ) a.s.N (λ) is just the counting function of the Sine β process, giving the number of point of the process in an interval of lenght λ. (1.7) Observe that our SDE (1.5) and (1.7) are very similar, except that in our case, we have Our main tool to prove Theorem 1.1 will be the Cameron-Martin-Girsanov formula, which allows one to compare the measure on paths given by two diffusions.If we knew the conditional distribution of the diffusion X under the event that it does not blow up before time τ /2, then we could use the Cameron-Martin-Girsanov formula to compute E τ (0, λ) explicitely.While we cannot do this, the next best option is to find a new diffusion Y which approximates this conditional distribution.The next section gives the statement of the Girsanov theorem we will be using.

The Cameron-Martin-Girsanov formula
We will use the same version of the Girsanov formula than the one introduced in [5].
Proposition 2.1.Consider the following stochastic differential equations: ) on the interval (0, T ] where B, B are standard Brownian motions.Assume that (8) has a unique solution and assume that: (A) g 2 − h 2 and g − h are bounded when x is bounded above.(Then G s is almost surely well defined when X s is finite.)(B) G s is bounded above by a deterministic constant.
(C) G s → ∞ when s ↑ S if X hits +∞ at time S.In this case, we define G s := −∞ for s ≥ S. Consider the process Ỹ whose density with respect to the distribution of the process Xis given by e G T .Then, Ỹ satisfies the second SDE (9) and never blows-up to +∞ almost surely.Moreover, for any nonnegative function F of the path of X that vanishes when X blows up we have (2.4) The proof of Proposition 2.1 is given in [5].

Construction of the diffusion Y
In this section, we construct a diffusion Y which approximates the conditional distribution of X under the event that it does not explode before time τ /2.We will construct a drift function h(x) for which the diffusion Y , is well defined, a.s, finite for t ≤ τ /2 and with Radon-Nikodym derivative e G τ /2 with G τ /2 as in Proposition 2.1.Lemma 3.1.For the diffusion X λ/τ (which we simply denote X), and T = τ /2, there exists a function h(x) so that conditions (A)-(C) of Proposition 2.1 hold, and G τ /2 has the following form: where ω 1 , ω 2 , Φ are continuous with and there exists with h 4 continuous and such that there exists c > 0, with h 4 ∞ < c/λ.
Proof.In order to construct the function h, we follow [5].We will use Our goal is to find the appropriate drift term h in a way that the diffusion Y will approximate the conditional distribution of X given that is does not blow up in the interval [0, τ /2].We will do this term by term starting with the highest order; we write as it yields the nice cancellation, The contribution to the drift terms h 1 and g 1 to the stochastic integral part of −G s is given by, s 0 We will use Ito's formula to evaluate integrals with respect to dX.Let b be a continuously differentiable function and denote by b its antiderivative.We have, Thus, by applying (3.2), we have λ τ s 0 e X(t) dt.
We choose h 2 so that the integral term in the right-hand side of (3.3) simplifies, that is ).This gives, Thus, EJP 19 (2014), paper 82.
Page 5/12 ejp.ejpecp.org Large gaps asymptotics for the 1-dimensional random Schrödinger operator The coefficient 1/8 in front of s in the first line in the expression above come from the −3/8 in front of s in (3.4) and the constant term in (h 2 + h 3 ) 2 /2.The term η collects the integrand in (3.4), the terms (h 2 + h 3 ) 2 /2 with the constant term removed and −g 2 2 /2.
Explicitly, we have We choose h 4 so that h 1 h 4 = −η, that is .
By Ito's formula, where − h4 is the antiderivative of h 4 .Thus, Using the fact that log cosh(x) − |x| is bounded, one can rewrite the second line of −G s as follows: where ω 1 (x) = 1 2 log (cosh(x)) + log (cosh(x/2)) − |x| and so is bounded.We now plug s = τ /2 to get, We now need to check that that the proposed choice for h satisfies conditions (A)-(C) from Proposition 2.1.First, observe that h 2 , h 3 , h 4 , h4 , ∂ x h 4 are all bounded above by an absolute constant and in particular there is a 1/λ coefficient in front of h 4 , h4 , ∂ x h 4 .This implies that we can write the third line of G s and that there exists Large gaps asymptotics for the 1-dimensional random Schrödinger operator We now need to check that the conditions (A)-(C) of Proposition 2.1 are satisfied.We proceed as in [5].
where |ĝ| ≤ Ce x and | ĥ| ≤ C e x for some positive constants C and C .Thus, g − h and g 2 − h 2 are both bounded if x is bounded above.

Condition (B):
The first and third line in −G τ /2 are both bounded as well as ω 1 in the second line.Thus, we only need to show that is bounded from below by a constant only depending on λ and τ , which is obvious.Thus G s is bounded from above by a constant (depending only on λ and τ ).

Condition (C):
As s ↑ S, the time when X reaches ∞, we have that goes to ∞ while the other terms in G s are bounded.Thus, G s → −∞.

Proof of Theorem 1.1
Recall that We consider the diffusion Y given by the SDE (3.1) with a drift function h(x) as in Lemma 3.1.That is Y satisfies the following SDE: where there exists c > 0 such that h 4 ∞ ≤ c/λ.We apply Proposition 2.1 with F (X) = 1 (X finite on [0, τ /2]) .Then, where Our aim is to evaluate E [exp (Ψ(Y ))] as λ → ∞.We will prove the following lemma.
Lemma 4.1.As λ → ∞, Large gaps asymptotics for the 1-dimensional random Schrödinger operator This lemma, combined with (4.1) proves Theorem 1.1.In order to prove the lemma we will bound Y above and below in order to get upper and lower bounds on E [exp (Ψ(Y ))] .
For the upper bound, we will bound Y above by the stationary solution of an SDE whose drift is very close to the one of Y.This will give a very precise upper bound which we think is actually the full asymptotics (up to the constant term) for P (Sch τ [0, λ] = 0).
Unfortunately, this method cannot be applied for the lower bound.For the lower bound we will bound Y below by the (random) solution of an ODE.This will give a less precise asymptotics but good enough to get the exp(2τ /λ) that we need.Observe that we could have used a similar method for the upper bound but we chose to proceed differently as the bound obtained is more precise and likely to be the full asymptotics up to the constant term.Proof of Lemma 4.1: If we drive Y 1 and Y with the same Brownian motion and consider Y 1 started from its stationary distribution, then, since Y (0) = −∞ < Y 1 (0) and the drift of Y 1 is greater than the drift of Y we have that Now, let us compute the stationary distribution for Y 1 .The adjoint of the infinitesimal generator of Y 1 is given by The stationary distribution of Y 1 satisfies ∂/∂t (p λ (t, y)) = 0 and A * p λ (t, y) = 0. So, We would like to asymptotically evaluate K(λ).Now, we will use Lemma 5.1 from Appendix A. We can neglect the term Cy/λ in the exponential.We call I(λ) the integral above (after the term cy/λ has been removed).We set g(y) = (2/τ ) cosh(y) and h(y) = exp (−2 log (cosh(y/2)) − 2y) .The minimum of g is attained at 0 and g(0) = 2/τ, g (0) = 0, g (0) = 2/τ > 0 and h(0) = 1 = 0. Thus, by Lemma 5.1 that is, Using (4.2) and the fact that x → 2x + + (λ/τ )e x + ω 1 (y) is strictly increasing, we have that Now, since 2y + − |y| − 2y = −y.Now, we can neglect the term in cy/λ in the exponential above and so we write We make the following change of variables: u = exp(−y).Then, We set g(u) = −u/τ and h(u) = √ u 2 + 1/(u+1).We have g (u) = −1/τ = 0.The maximum is attained in 0 with g(0) = 0, h(0) = 1.Thus, by Lemma 5.2, Thus, Lower Bound For the lower bound, we are going to bound Y stochastically below by the (random) solution of an ODE.We know that h 4 ∞ ≤ c/λ for some c > 0. Thus, −1 − (1/2) tanh(y/2) + h 4 (y) ≥ −(3/2) − c/λ.Since, eventually we will let λ → ∞, one can choose λ large enough so that −(3/2) − c/λ ≥ −2.Then Z 2 ≤ Y where where Thus, for all t ∈ [0, τ /2], we have (after Taylor expansion for large λ) .