Perturbation determinant and Levinson’s formula for Schrödinger operators with 1-D general point interaction

Abstract

We consider the one-dimensional Schrödinger operator with properly connecting general point interaction at the origin. We derive a trace formula for trace of difference of resolvents of perturbed and unperturbed Schrödinger operators in terms of a Wronskian which results in an explicit expression for perturbation determinant. Using the estimate for large-time real argument on the trace norm of the resolvent difference of the perturbed and unperturbed Schrödinger operators we express the spectral shift function in terms of perturbation determinant. Under certain integrability conditions on the potential function, we calculate low-energy asymptotics for the perturbation determinant and prove an analog of Levinson’s formula

Appendix A: Estimate on trace norm of resolvent difference

Here we’ll provide an explicit derivation of the estimate

$$\begin{aligned} ||R_{V}^\mathcal {A}(-t)-R_{0}^\mathcal {A}(-t)||_1\le C_\epsilon t^{-\frac{3}{2}+\epsilon } \end{aligned}$$

for all $\epsilon >0$ and large t. We will need the following lemma.

Lemma A.1

[6], Lemma 3.4] Let $\mathcal {H}$ be a Hilbert space and $f, g \in \mathcal {H}$. Assume that $\mathcal {R}=(\cdot ,\,f)f-(\cdot ,\,g)g$ is an operator of rank two on $\mathcal {H}$. Then, the trace norm of $\mathcal {R}$ is given by

$$\begin{aligned} ||\mathcal {R}||_1=\sqrt{(||f||^2+||g||^2)^2-4|(f,g)|^2}. \end{aligned}$$

(A.1)

Moreover, if we let $g=f+h$ then,

$$\begin{aligned} (||f||^2+||g||^2)^2-4|(f,g)|^2\le 6||h||^2||f||^2+3||h||^4. \end{aligned}$$

(A.2)

Lemma A.2

Assume that the potential V satisfies condition (1.2). Then the resolvents $R_V^\mathcal {A}$ of $H_V^\mathcal {A}$ and $R_0^\mathcal {A}$ of $H_0^\mathcal {A}$ satisfy, for all $\epsilon >0$ and for large t, 

$$\begin{aligned} ||R_{V}^\mathcal {A}(-t)-R_{0}^\mathcal {A}(-t)||_1\le C_\epsilon t^{-\frac{3}{2}+\epsilon }. \end{aligned}$$

Proof

Let $\mathcal {R}_1=R^{\mathcal {A}}_{V}(-t)-R^D_{V}(-t)+R^D_{0}(-t)-R^{\mathcal {A}}_{0}(-t)$ and $\mathcal {R}_2=R^D_{V}(-t)-R^D_{0}(-t)$ then,

$$\begin{aligned} R^{\mathcal {A}}_{V}(-t)-R^{\mathcal {A}}_{0}(-t)=\mathcal {R}_1+\mathcal {R}_2. \end{aligned}$$

Hence

$$\begin{aligned} ||R^{\mathcal {A}}_{V}(-t)-R^{\mathcal {A}}_{0}(-t)||_1\le ||\mathcal {R}_1||_1+||\mathcal {R}_2||_1. \end{aligned}$$

(A.3)

Trace norm on $\mathcal {R}_2$ can be estimated as $||\mathcal {R}_2||_1\le ct^{-\frac{3}{2}+\epsilon }$ (c.f. Lemma 4.5.6. [24]). We only need to find an estimate on the trace norm of rank two operator $\mathcal {R}_1$. By adding and subtracting the quantity

$$\begin{aligned} \dfrac{\theta _j^2(x_j,i\sqrt{t})}{L(i\sqrt{t})\theta _j^2(0,i\sqrt{t})}+\dfrac{e^{-2 x_j\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}\hspace{1cm}j=1,2 \end{aligned}$$

in the diagonal of $\mathcal {R}_1$ we can re-write it as

$$\begin{aligned} \mathcal {R}_1=\mathcal {S}_1+\mathcal {S}_2, \end{aligned}$$

where

$$\begin{aligned} \mathcal {S}_1=\begin{bmatrix} \frac{(a-\sqrt{t} b+1)e^{-2 x_1\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left( a+b\frac{\theta _2'(0,i\sqrt{t})}{\theta _2(0,i\sqrt{t})}+1\right) \theta _1^2(x_1,i\sqrt{t})}{L(i\sqrt{t})\theta _1^2(0,i\sqrt{t})} &{}0\\ 0 &{}\frac{(1-\sqrt{t} b-d)e^{-2 x_2\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left( b\frac{\theta _1'(0,i\sqrt{t})}{\theta _1(0,i\sqrt{t})}-d+1\right) \theta _2^2(x_2,i\sqrt{t})}{L(i\sqrt{t})\theta _2^2(0,i\sqrt{t})} \end{bmatrix} \end{aligned}$$

and

$$\begin{aligned} (\mathcal {S}_2)_{j,k}=\begin{bmatrix} \frac{\theta _j(x_j,i\sqrt{t})\theta _k(x_k,i\sqrt{t})}{L(i\sqrt{t})\theta _j(0,i\sqrt{t})\theta _k(0,i\sqrt{t})}-\frac{e^{-(x_j+x_k)\sqrt{t}}}{(d-a)\sqrt{t}+bt-c} \end{bmatrix}\hspace{1cm}j,k=1,2. \end{aligned}$$

Now, as $\mathcal {S}_1$ is a diagonal matrix and to apply the definition of trace norm, we can let $\mathcal {S}_1=I\mathcal {S}_1I$, where I is the second order identity matrix. This implies,

$$\begin{aligned} ||\mathcal {S}_1||_1=|\sigma _1|+|\sigma _2| \end{aligned}$$

where

$$\begin{aligned} |\sigma _1|=\int _0^{\infty }\left| \frac{(a-\sqrt{t} b+1)e^{-2 x_1\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left( a+b\frac{\theta _2'(0,i\sqrt{t})}{\theta _2(0,i\sqrt{t})}+1\right) \theta _1^2(x_1,i\sqrt{t})}{L(i\sqrt{t})\theta _1^2(0,i\sqrt{t})}\right| dx_1 \end{aligned}$$

and

$$\begin{aligned} |\sigma _2|=\int _0^{\infty }\left| \frac{(1-\sqrt{t} b-d)e^{-2 x_2\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left( b\frac{\theta _1'(0,i\sqrt{t})}{\theta _1(0,i\sqrt{t})}-d+1\right) \theta _2^2(x_2,i\sqrt{t})}{L(i\sqrt{t})\theta _2^2(0,i\sqrt{t})}\right| dx_2. \end{aligned}$$

For t large enough, these expressions can be simplified to

$$\begin{aligned} |\sigma _1|&\approx \left| \frac{(a-\sqrt{t} b+1)}{(d-a)\sqrt{t}+bt-c}\right| \int _0^{\infty }\left| e^{-2 x_1\sqrt{t}}-\theta _1^2(x_1,i\sqrt{t})\right| dx_1\\&=\left| \frac{(a-\sqrt{t} b+1)}{(d-a)\sqrt{t}+bt-c}\right| \int _0^{\infty }\left| \theta _1(x_1,i\sqrt{t})-e^{-x_1\sqrt{t}}\right| \left| \theta _1(x_1,i\sqrt{t})+e^{- x_1\sqrt{t}}\right| dx_1 \end{aligned}$$

and

$$\begin{aligned} |\sigma _2|\approx \left| \frac{(1-\sqrt{t} b-d)}{(d-a)\sqrt{t}+bt-c}\right| \int _0^{\infty }\left| \theta _2(x_2,i\sqrt{t})-e^{-x_2\sqrt{t}}\right| \left| \theta _2(x_2,i\sqrt{t})+e^{- x_2\sqrt{t}}\right| dx_2. \end{aligned}$$

To find estimates for $|\sigma _1|$ and $|\sigma _2|$ we use the following estimates for large t

$$\begin{aligned} \left| \theta _k(x_k,i\sqrt{t})-e^{-\sqrt{t}x_k}\right| \le \dfrac{m_1}{\sqrt{t}}e^{-\sqrt{t}x_k} \end{aligned}$$

(A.4)

and

$$\begin{aligned} \left| \theta _k(x_k,i\sqrt{t})+e^{-\sqrt{t}x_k})\right| \le \left( \frac{m_2}{\sqrt{t}}+2\right) e^{-\sqrt{t}x_k} \end{aligned}$$

(A.5)

This yields the following bounds

$$\begin{aligned} |\sigma _1|\le \left| \frac{(a-\sqrt{t} b+1)}{(d-a)\sqrt{t}+bt-c}\right| \frac{m_1(m_2+2\sqrt{t})}{2t^{3/2}}=|-m_1t^{-3/2}+O(t^{-2})|,\\ |\sigma _2|\le \left| \frac{(1-\sqrt{t} b-d)}{(d-a)\sqrt{t}+bt-c}\right| \frac{m_1(m_2+2\sqrt{t})}{2t^{3/2}}=|-m_1t^{-3/2}+O(t^{-2})|. \end{aligned}$$

Therefore

$$\begin{aligned} |\sigma _1|=|\sigma _2|=O(t^{-3/2}) \end{aligned}$$

and hence

$$\begin{aligned} ||\mathcal {S}_1||_1=O(t^{-3/2}). \end{aligned}$$

To find an estimate on the norm of $\mathcal {S}_2$, we use Lemma (A.1). Clearly, $\mathcal {S}_2=(\cdot ,f)f-(\cdot ,g)g$ is an operator of rank two, where

$$\begin{aligned}f_k(x_k)=\dfrac{-e^{-x_k\sqrt{t}}}{\sqrt{(d-a)\sqrt{t}+bt-c}}\quad \text { and }\quad g_k(x_k)=\dfrac{-\theta _k(x_k,i\sqrt{t})}{\sqrt{L(i\sqrt{t})}\theta _k(0,i\sqrt{t})}.\end{aligned}$$

Let $h=g-f$, then

$$\begin{aligned} h_k(x_k)&=\dfrac{e^{-x_k\sqrt{t}}}{\sqrt{(d-a)\sqrt{t}+bt-c}}-\dfrac{\theta _k(x_k,i\sqrt{t})}{\sqrt{L(i\sqrt{t})}\theta _k(0,i\sqrt{t})}\\&=-\left( \dfrac{\theta _k(x_k,\sqrt{t}i)-e^{-\sqrt{t}x_k}}{\sqrt{L(i\sqrt{t})}\theta _k(0,i\sqrt{t})}+\left( \dfrac{1}{\sqrt{L(i\sqrt{t})}\theta _k(0,i\sqrt{t})}-\dfrac{1}{\sqrt{(d-a)\sqrt{t}+bt-c}}\right) e^{-\sqrt{t}x_k}\right) . \end{aligned}$$

To compute an estimate on $||h||^2=\sum _{k=1}^{2}\int _0^{\infty }\left| h_k(x_k)\right| ^2dx_k$, we use (2.13) and (A.4) and after some simplifications we obtain

$$\begin{aligned} ||h||^2\le \dfrac{m_1^2}{ \left( (d-a)\sqrt{t}+bt-c \right) {t}^{3/2}}. \end{aligned}$$

Similarly,

$$\begin{aligned} ||f||^2=\sum \limits _{k=1}^{2}\int \limits _0^{\infty }\left| \dfrac{-e^{-\sqrt{t}x_k}}{\sqrt{(d-a)\sqrt{t}+bt-c}} \right| ^2dx_k=\dfrac{1}{ \left( (d-a)\sqrt{t}+bt-c \right) \sqrt{t}}. \end{aligned}$$

The estimate on the right-hand side of inequality (A.2) is now given by

$$\begin{aligned} 6||h||^2||f||^2+3||h||^4&\le \frac{6m_1^2}{ \left( (d-a)\sqrt{t}+bt-c \right) ^2t^2}+\frac{3m_1^4}{ \left( (d-a)\sqrt{t}+bt-c \right) ^2t^3} \\&=\frac{3m_1^4+6m_1^2t}{\left( (d-a)\sqrt{t}+bt-c \right) ^2t^3}\\&=\frac{6m_1^2}{b^2t^4}+\frac{12m_1^2(a-d)}{b^3t^{\frac{9}{2}}}+O(t^{-5})\\&=O(t^{-4}). \end{aligned}$$

This implies

$$\begin{aligned} ||\mathcal {S}_2||_1=O(t^{-2}) \end{aligned}$$

and hence

$$\begin{aligned} ||\mathcal {R}_1||_1=O(t^{-\frac{3}{2}}). \end{aligned}$$

$\square$

The current estimates have been improved compared to those in [22] and [24], as they impose a condition $(\alpha >1/4)$ leading to $||R-R_0 ||\le Ct^{(-2.5)}$ when $\alpha$ approaches 1/4. In contrast, our results involve the parameter $>0$, and as $\epsilon$ approaches 0, we have $||R-R_0 ||\le Ct^{(-1.5)}$. It is also worth noting that while Demiral in [6] proved a similar estimate for the Kirchhoff case, our case involves general 1D point interactions with additional parameters, making it challenging to derive estimates. In dealing with $R_1=S_1+S_2$, where $S_1$ doesn’t follow Lemma A1 of [6] entirely but only a part $(S_2)$, we observed that $S_1$ is a diagonal matrix. To find an estimate for $S_1$, we employed singular value decomposition ($\sigma _1$ and $\sigma _2$) to compute its trace norm in the above proof.