Resonances in the One Dimensional Stark Effect in the Limit of Small Field

Abstract

We discuss the resonances of Hamiltonians with a constant electric field in one dimension in the limit of small field. These resonances occur near the real axis, near zeros of the analytic continuation of a reflection coefficient for potential scattering, and near the line $\arg z = - 2\pi /3$. We calculate their asymptotics. In conclusion we make some remarks about the higher dimensional problem.

Appendices

Appendix 1: The Airy Function $\mathop {{\mathrm{Ai}}}(z)$ in the Sector $\arg z \in (-2\pi /3 + \delta , -\delta )$

We believe the results in this appendix are known but could not find a suitable reference.

Theorem 24

With $\eta =\sqrt{w}$, $\arg \eta \in (-\pi /2,\pi /6)$, $\sqrt{\eta }e^{2\eta ^3/3}\mathop {{\mathrm{Ai}}}(w)$ extends to an analytic function, J, of $\zeta ^{-1}, \zeta = 2\eta ^3/3$ with $\arg \zeta ^{-1} \in (-\pi /2, 3\pi /2)$. The function J and all its derivatives extend continuously to the origin in this sector.

Proof

We start with the integral representation $\mathop {{\mathrm{Ai}}}(w) = \lim _{R \rightarrow \infty } (2\pi )^{-1}\int _{-R}^R e^{i(sw+s^3/3)}ds$ where $w\ge 0$ and deform the contour to $s = i\sqrt{w} + u$ with $u \in \mathbb {R}$ to obtain

$$\mathop {{\mathrm{Ai}}}(w) = \frac{e^{-2\eta ^3/3}}{2\pi } \int _{-\infty }^{\infty } e^{-\eta u^2}e^{iu^3/3}du $$

with $\eta = \sqrt{w}$. We can analytically continue to $\arg \eta \in (-\pi /2, \pi /2)$. Thus using polar coordinates with $\zeta ^{-1} = 3w^{-3/2}/2 = r e^{i\theta }$, we have

$$2\pi \sqrt{\eta }e^{2\eta ^3/3}\mathop {{\mathrm{Ai}}}(w) = e^{-i\theta /6}\int _{-\infty }^{\infty } \exp (-e^{-i\theta /3}t^2)g(t^6r)dt$$

where $g(\psi ) = \cos (\sqrt{2\psi /27})$ is a $C^{\infty }$ function of its argument. With an integration by parts it is easy to verify the polar form of the Cauchy-Riemann equations ($(\partial /\partial r +i r^{-1} \partial /\partial \theta )J = 0$). We take $\theta = \arg \zeta ^{-1} \in (-\pi /2, 3\pi /2)$. Because J satisfies the Cauchy-Riemann equations $J^{(n)}$ is given by $e^{-in\theta }\partial ^n J/\partial r^n$ which is easily shown to have a limit as $r\rightarrow 0$ which is independent of $\theta$.    $\square$

Thus in particular J has an asymptotic expansion

$$J(\zeta ^{-1}) \sim \sum _{n=0}^\infty c_n \zeta ^{-n}$$

which can be differentiated term by term.

Appendix 2: The Airy Function $\mathop {{\mathrm{Ai}}}(z)$ in the Sector $\arg z \in [-\pi , -\pi + \delta ]$

We follow [2] but the latter reference does not go far enough for our purposes. We write the Airy function $\mathop {{\mathrm{Ai}}}(w)$ for w real as

$$\mathop {{\mathrm{Ai}}}(w) = (2\pi )^{-1}\lim _{R \rightarrow \infty }\int _{-R}^R e^{-i(t^3/3 +wt)}dt.$$

We set $w = f^{1/3}(L-z/f) = -(z/f^{2/3})(1-fL/z)$ where we first keep z real and positive. We set $\eta = z(1-fL/z) = z- fL$. With a change of variable we have

$$\begin{aligned} \mathop {{\mathrm{Ai}}}(w) = (2\pi f^{1/3})^{-1}\int _{\mathcal {C}} e^{(-i/f)(s^3/3 - \eta s)}ds. \end{aligned}$$

(39)

Here we have used the fact the $\eta$ is real to move the contour into the lower half plane so that $\mathcal {C}$ is just $s(t) = t - i\alpha$ with $t\in \mathbb {R}$ and $\alpha > 0$. We then see that the Airy function is analytic in $\eta$ for $\eta$ in a neighborhood of a real point. We distort the contour further. We use steepest descents near the critical points of the exponential where $(s^3/3 - \eta s)' = s^2 - \eta = 0$, namely the points $s = \pm \sqrt{\eta }$. Thus near $+\sqrt{\eta }$ we write $s = \zeta + \sqrt{\eta }$. Note that at this critical point we have $s^3/3 - \eta s = -2\eta ^{3/2}/3$ and $-i(s^3/3 - \eta s + 2\eta ^{3/2}/3) = -i(\zeta ^3/3 +\sqrt{\eta }\zeta ^2)$. The steepest descents curve near $\zeta = 0$ will come from setting $\text {Re}(\zeta ^3/3 + \sqrt{\eta }\zeta ^2) = 0$. We will use part of the following contour for s near $\sqrt{\eta }$: We solve the equation $\text {Re}(\zeta ^3/3 + \sqrt{\eta }\zeta ^2) = 0$ and find with $\zeta = x + iy$,

$$y = \frac{-x(\gamma + x/3)}{\nu + \sqrt{\nu ^2 +(\gamma + x)(\gamma + x/3)}}$$

We have

$$-i(\sqrt{\eta } \zeta ^2 + \zeta ^3/3) = -b^2x^2 + \sum _{n = 3}^\infty B_n x^n $$

where $b^2 = 2\gamma ^{-2} |\eta |(\sqrt{|\eta |} -\nu )$. The series converges for $|x| < \gamma$ (see [2]). We have the following estimates in the indicated regions:

$$\begin{aligned} -i(\zeta ^3/3 + \sqrt{\eta }\zeta ^2)&\le -(\gamma /3)^2x^2/\sqrt{\nu ^2 + (\gamma +x)(\gamma +x/3)}; 0 \ge x \ge - \gamma , \nu \ge 0 \\&\le -(\gamma ^2/18|\eta |)x^2; 0 \le x \le 4\gamma , \nu \ge 0 \\&\le -\gamma x^2 ; 0 \le x \le 10\gamma , 0 \ge \nu \ge -\gamma /10 \\&\le -\gamma x^2; - \gamma \le x \le 0, 0 \ge \nu . \end{aligned}$$

For s near $-\sqrt{\eta }$ we write $s=\zeta - \sqrt{\eta }$ which gives $-i(s^3/3 -\eta s - 2\eta ^{3/2}/3) = -i(\zeta ^3/3 - \sqrt{\eta }\zeta ^2)$. Setting $\text {Re}(\zeta ^3/3 - \sqrt{\eta }\zeta ^2) = 0$ and $\zeta = x+ iy$ we find

$$y = x(\nu + \sqrt{\nu ^2 + (\gamma -x)(\gamma -x/3)})(\gamma -x)^{-1}.$$

We find

$$-i(\zeta ^3/3 - \sqrt{\eta }\zeta ^2) = -\tilde{b}^2 x^2 + \sum _{n=3}^{\infty }\tilde{B}_nx^n$$

where $\tilde{b}^2 = 2\gamma ^{-2} |\eta |(|\eta |^{1/2} + \nu )$. The series converges for $|x| < \gamma$.

We have the following estimates in the indicated regions:

$$\begin{aligned} -i(\zeta ^3/3 - \sqrt{\eta }\zeta ^2)&\le -\gamma x^2 \sqrt{\frac{\gamma - x/3}{\gamma -x}}; 0 \le x< \gamma , \nu \ge 0 \\&\le -\gamma x^2 (y/x) \le - \gamma x^2 \frac{\gamma - x/3}{2\sqrt{\nu ^2 + (\gamma -x)(\gamma -x/3)}}; 0 \le x< \gamma , \nu< 0 \\&\le -\gamma x^2/5; -10\gamma \le x \le 0, 0 \le \nu < \gamma \\&\le -(2\gamma /5)x^2; - 8\gamma \le x \le 0, - \gamma /3 \le \nu \le 0. \end{aligned}$$

We write

$$\mathop {{\mathrm{Ai}}}(w) = (2\pi f^{1/3})^{-1}(\int _{\mathcal {C_-}} e^{(-i/f)(s^3/3 - \eta s)}ds + \int _{\mathcal {C_+}} e^{(-i/f)(s^3/3 - \eta s)}ds).$$

We first consider the integral over the part of the contour $\mathcal {C_+}$ near $\sqrt{\eta }$:

$$\int _{C_+'} e^{(-i/f)(s^3/3 - \eta s)}ds = (2\pi f^{1/3})^{-1}e^{(2i\eta ^{3/2}/3f)}\int _{\mathcal {B}_+} e^{(-i/f)(\zeta ^3/3 + \sqrt{\eta } \zeta ^2) }d\zeta .$$

To obtain the part of the contour in the s variable near $\sqrt{\eta }$ we take $s = \zeta + \sqrt{\eta }$. This gives us the contour $\mathcal {B}_+$ with $\zeta = x + iy$ where $x \in (\beta , \infty ) , y = - \alpha$ , where as mentioned above $\alpha > \nu$. We thus see that this integral is analytic in $\sqrt{\eta }$ as long as $f>0$. We will see that after multiplying by $f^{-1/2}$ we can take the limit as $f\rightarrow 0$ uniformly for $\eta$ near a point $k_0 >0$. We deform the contour to obtain a new contour $\Gamma _+$ where

$$\zeta = x+iy, x \in (-\gamma /2, \gamma /2), y = \frac{-x(\gamma + x/3)}{\nu + \sqrt{\nu ^2 +(\gamma + x)(\gamma + x/3)}}.$$

Thus we have

$$I_+ = (2\pi f^{1/3})^{-1}\int _{\Gamma _+}e^{-i(\zeta ^3/3 +\sqrt{\eta }\zeta ^2)/f} d\zeta = \frac{f^{1/6}}{2\pi } J_+ $$

$$J_+ = f^{-1/2}\int _{-\gamma /2}^{\gamma /2} e^{f^{-1}(-b^2x^2 + \sum _{n=3}^{\infty }B_nx^n)}(1+ idy/dx)dx$$

We expand the integrand in a convergent power series keeping the $e^{-b^2x^2}$ to obtain

$$J_+ = \int _{-\gamma f^{-1/2}}^{\gamma f^{-1/2}} e^{-b^2x^2}(1 + \sum _{k=1}^\infty (\sum _{n=3}^\infty B_nf^{-1 + n/2}x^n )^k/k!)(1+i\sum _{m=0}^\infty a_m f^{m/2} x^m)dx$$

The odd powers of x do not contribute and thus we have

$$J_+ = 2(1+ia_0)\int _0^{\gamma f^{-1/2}}e^{-b^2x^2}dx + \sum _{m + |n|/3 \ge 1, l \ge 1 } c_{n,m}f^{(|n| + m -2l)/2} \int _0^{\gamma f^{-1/2}}e^{-b^2x^2} x^{|n|+ m}dx$$

Here n is a multi-index $n = (n_1,n_2, \ldots , n_l)$ and $|n| = n_1 + \cdots + n_l$. Each $n_j \ge 3$ and thus the power of f, $(|n| + m)/2 - l$ is positive and since $|n| + m$ is even this power is a positive integer $\ge (l+m)/2$ . Note that $\int _{0}^ {\gamma f^{-1/2}}e^{-b^2x^2} x^{|n| + m}dx$ is $C^{\infty }$ in f near 0 as long as we define $f^{-n} e^{-b^2\gamma ^2/f}$ to be 0 at $f=0$. Thus $J_+$ is $C^\infty$ for $f\ge 0$ in the sense that the derivatives have limits as $f \downarrow 0$. In the limit $f\downarrow 0$ the first term can be calculated (with some effort - see also [2]) to be $e^{-i\pi /4}\sqrt{\pi /\eta ^{1/2}}$.

We now connect up this curve to infinity as follows: We take $x+iy = \gamma /2 + t -i y(\gamma /2)$ with $t \in [0, \infty )$. For simplicity take $|\nu | < \gamma /10$. It is then easy to see that the real part of $-i(\zeta ^3/3 +\sqrt{\eta }\zeta ^2)\le - C < 0$ independent of $t\ge 0$ and independent of $\eta$ for $\eta$ near a real point $k_0 > 0$. If the extended curve is called $\tilde{\Gamma }_+$ This is easily seen to imply that

$$\tilde{I}_+ = (2\pi f^{1/3})^{-1}\int _{\tilde{\Gamma }_+}e^{-i(\zeta ^3/3 +\sqrt{\eta }\zeta ^2)/f} d\zeta = \frac{f^{1/6}}{2\pi } \tilde{J}_+ $$

with $\tilde{J}_+$ a $C^{\infty }$ function in f for $f_0 >f\ge 0$ and analytic in $\sqrt{\eta }$ for $\sqrt{\eta }$ near a point $k_0 > 0$.

The curve near the critical point $-\sqrt{\eta }$ can be handled similarly. We set $s= \zeta - \sqrt{\eta }$ giving $-i(s^3/3 - \eta s) = -i(\zeta ^3/3 - \sqrt{\eta }\zeta ^2) -2i\eta ^{3/2}/3$ = and $\zeta = x+ iy$. We demand that for $x\in (-\gamma /2, \gamma /2)$ we have $\text {Re}[-i(\zeta ^3/3 - \sqrt{\eta }\zeta ^2)] = 0$. This results in

$$y = \frac{x(\gamma - x/3)}{-\nu + \sqrt{\nu ^2 + (\gamma -x)(\gamma -x/3)}}.$$

We connect this curve to infinity and to the imaginary axis just as in the case where we extended the curve $\Gamma _+$. The result is the same. It remains to connect the two curves on the imaginary axis. The curve on the left ends on the imaginary axis at $s = i(2\nu + \sqrt{\nu ^2 + 5\gamma ^2/12})$ or $\zeta = \gamma + i(\nu + \sqrt{\nu ^2 + 5\gamma ^2/12})$. The curve on the right ends on the imaginary axis at $s = i(-2\nu + \sqrt{\nu ^2 + 5\gamma ^2/12})$. Thus we extend the curve on the left as $\zeta = \gamma + i(u+ \nu + \sqrt{\nu ^2 + 5\gamma ^2/12})$ or $s = \gamma + i( u + 2\nu + \sqrt{\nu ^2 + 5\gamma ^2/12})$ with u going from 0 to -$4\nu$. It is not hard to see that this piece is $C^\infty$ in f for $f\ge 0$ with the function and all derivatives equal to zero at $f=0$. (Of course the derivative at zero is the right hand derivative.) It is convenient in the estimates to restrict $|\nu | < \gamma /10$. Even though the contours involve $\sqrt{\eta }$ in a non-analytic way, it is not hard to see that they can be distorted in such a way that given $\sqrt{\eta _0}$ the contour can be chosen to depend on this quantity but then $\sqrt{\eta }$ can be varied in a small disk around this point and the result is analytic in the disk with estimates similar to what we have derived.

Thus what we have shown is that

$$\mathop {{\mathrm{Ai}}}(w) = f^{1/6} e^{-i\pi /4}\frac{1}{2 \sqrt{\pi \eta ^{1/2}}}(e^{2i\eta ^{3/2}/3f} a_1(k,f) + ie^{-2i\eta ^{3/2}/3f} a_2(k,f))$$

where $a_j(k,f)$ is $C^{\infty }$ in f for $f_0 > f \ge 0$ and analytic in k for k near a point on the positive real axis. Since the limits involved in taking derivatives in f converge uniformly in k, these derivatives are also analytic in k in a neighborhood of a positive real point. We have $a_j(k,0) = 1$.

We also need the derivative of $\mathop {{\mathrm{Ai}}}(w)$ with respect to the spatial variable which we have called L in this appendix. Note that in (39) L occurs only in $\eta = z - fL$ so that differentiation with respect to L brings down $-is$ in this integral. In the integral near $+\sqrt{\eta }$ the change of variable is $s=\zeta + \sqrt{\eta }$. $-i\zeta$ contributes something of at most order f while the term $-i\sqrt{\eta }$ gives the main contribution. A similar analysis near the critical point $-\sqrt{\eta }$ shows that the main contribution is a factor of $+ i \sqrt{\eta }$. Thus we obtain

$$(d/dL)\mathop {{\mathrm{Ai}}}(w) = f^{1/6} e^{-i\pi /4}{\sqrt{\frac{\eta ^{1/2}}{4\pi }}}(-ie^{2i\eta ^{3/2}/3f} \tilde{a}_1(k,f) -e^{-2i\eta ^{3/2}/3f} \tilde{a}_2(k,f))$$

where $\tilde{a}_j(k,f)$ is $C^{\infty }$ in f for $f_0 > f \ge 0$ and analytic in k for k near a point on the positive real axis. Since the limits involved in taking derivatives in f converge uniformly in k, these derivatives are also analytic in k in a neighborhood of a positive real point. We have $\tilde{a}_j(k,0) = 1$.

We now must do similar estimates with $A_1(x)$ for z near the line $\arg = -2\pi /3$. Here the argument of the Airy function $\mathop {{\mathrm{Ai}}}$ is $w = f^{1/3}(x-z/f)$. We set $z_1 = e^{2\pi /3} z$ which is near a positive real point and $\eta = z_1(1-fx/z)$ which is also near a point on the positive real axis when f is small. Thus after a change of variable we can write

$$\begin{aligned} \mathop {{\mathrm{Ai}}}(w) = (2\pi f^{1/3})^{-1}\int _{\mathcal {C}} e^{(-i/f)(s^3/3 - \eta s)}ds. \end{aligned}$$

(40)

where the contour $\mathcal {C}$ is the same as above. Thus the same analysis as above works in this situation. Note that when we differentiate with respect to the spatial variable the $\eta$ dependence on x gives a contribution to the exponential from $-i\eta s/f$ of $ixe^{2\pi i /3}$. Thus differentiation with respect to x brings down a factor of $ie^{2\pi i/3}s$, and thus a main contribution of $\sqrt{\eta }ie^{2\pi i/3}$ for the integral near the critical point $\sqrt{\eta }$. Similarly there is a main contribution of $-\sqrt{\eta }ie^{2\pi i/3}$ for the integral near the critical point $-\sqrt{\eta }$.