Interaction between subbands in a quasi-one-dimensional superconductor

Abstract

In the framework of the Bogoliubov–de Gennes equation, we study the spinless $p$-wave superconductor in an infinite strip in the presence of some impurity. We analytically determine the wave functions of stable bound states with energies close to edge points of the energy gap. We prove that for a small impurity potential, the contribution of the nearest subbands to the wave functions in the case of energy values close to edge points is very small, and these energy levels are significantly closer to the gap edge than in the one-dimensional case. We also study the bound states with nearly zero energy values; in contrast to the one-dimensional case, they do not have the “particle–hole” symmetry. In the cases under study, in addition to the bound states, there also exit resonance states related to them.

Appendix

To determine the Green’s function of the Hamiltonian $H^{(n)}$, we solve the equation

$$ (H^{(n)}-E)\Psi^{(n)}=\Phi^{(n)}$$

(A.1)

for $\Psi^{(n)}$. Using (1), we rewrite Eq. (A.1) as

$$ \begin{pmatrix} -\partial_x^2+\biggl(\dfrac{2\pi n}{l}\biggr)^2-\mu-E & \Delta\biggl(\dfrac{-\partial_x+2\pi n}{l}\biggr)\\ \Delta\biggl(\partial_x+\dfrac{2\pi n}{l}\biggr) & \partial_x^2-\biggl(\dfrac{2\pi n}{l}\biggr)^2+\mu-E \end{pmatrix} \begin{pmatrix} \psi_\mathrm{e}^{(n)}(x)\\ \psi_\mathrm{h}^{(n)}(x) \end{pmatrix}= \begin{pmatrix} \varphi_\mathrm{e}^{(n)}(x)\\ \varphi_\mathrm{h}^{(n)}(x) \end{pmatrix},$$

(A.2)

$n=0,\pm 1,\dots$. After the Fourier transformation with respect to $x$, Eq. (A.2) becomes

$$ \begin{pmatrix} p^2+\biggl(\dfrac{2\pi n}{l}\biggr)^2-\mu-E & \Delta\biggl(-ip+\dfrac{2\pi n}{l}\biggr)\\ \Delta\biggl(ip+\dfrac{2\pi n}{l}\biggr) & -p^2-\biggl(\dfrac{2\pi n}{l}\biggr)^2+\mu-E \end{pmatrix} \begin{pmatrix} \widetilde{\psi}_\mathrm{e}^{(n)}(p)\\ \widetilde{\psi}_\mathrm{h}^{(n)}(p) \end{pmatrix}= \begin{pmatrix} \widetilde{\varphi}_\mathrm{e}^{(n)}(p)\\ \widetilde{\varphi}_\mathrm{h}^{(n)}(p) \end{pmatrix},$$

(A.3)

$n=0,\pm 1,\dots$. The determinant of matrix (A.3) is

$$ d_n=E^2-\biggl(p^2+\biggl(\frac{2\pi n}{l}\biggr)^{\!2}-\mu \biggr)^{\!2}-\Delta^2 \biggl(p^2+\biggl(\frac{2\pi n}{l} \biggr)^{\!2}\,\biggr).$$

(A.4)

By (A.4), the dispersion law $d_n=0$ is

$$ \biggl(p^2+\biggl(\frac{2\pi n}{l}\biggr)^{\!2}+\frac{\Delta^2}{2}-\mu \biggr)^{\!2}-\frac{\Delta^4}{4}+\Delta^2\mu-E^2=0.$$

(A.5)

Using the assumption $|\mu|\ll \min\{|\Delta|,\Delta^2\}$ in (A.5), we obtain that the spectrum of the operator $H^{(n)}$, which coincides with matrix (A.2), is determined by the inequality

$$E^2\geqslant \biggl(\frac{2\pi n}{l}\biggr)^{\!4}+2\biggl(\frac{\Delta^2}{2}-\mu\biggr) \biggl(\frac{2\pi n}{l}\biggr)^{\!2}+\mu^2 \geqslant \mu^2$$

and decreases as $|n|$ increases. Therefore, the spectrum of the Hamiltonian $H$, which coincides with the union of the spectra of $H^{(n)}$, $n=0,\pm 1,\dots$ , is described by the inequality $|E|\geqslant |\mu|$.

From (A.4) and (A.5), we have

$$ \frac{1}{d_n}=-\frac{1}{2a}\biggl(\frac{1}{p^2-p_+^2}-\frac{1}{p^2-p_-^2}\biggr),$$

(A.6)

where

$$ a=\sqrt{\frac{\Delta^4}{4}-\Delta^2\mu+E^2}, \qquad p_\pm=\sqrt{\pm a+\mu-\frac{\Delta^2}{2}-\biggl(\frac{2\pi n}{l}\biggr)^{\!2}}.$$

(A.7)

In what follows, we only consider the case $n=\pm 1$ and use the assumption that $\mu$ and $O(1/l)$ are of the same order of magnitude. We obtain the Green’s function of the operators $H^{(\pm 1)}$ near the edge points of the superconducting gap $(-|\mu|,|\mu|)$. By (A.7), we approximately have

$$ a=\frac{\Delta^2}{2}-\mu-\frac{\mu^2-E^2}{\Delta^2}, \qquad p_+=\frac{2\pi i n}{l}, \qquad p_-=i|\Delta|.$$

(A.8)

We note that

$$ p_+=\frac{i\sqrt{\mu ^2-E^2}}{|\Delta|}, \qquad n=0.$$

(A.9)

For definiteness, we consider the case of a topological phase $\mu >0$. For the energies $E=\mu-\varepsilon$, where $0<\varepsilon\ll \mu$, we have $a=\Delta^2/2-\mu-{2\mu \varepsilon}/(\Delta^2)$. From (A.3), (A.6), and (A.8), we obtain

$$ \begin{aligned} \, \widetilde{\psi}_\mathrm{e}^{(\pm 1)}(p)={}&\frac{1}{\Delta^2}\biggl( \biggl(p^2+\biggl(\frac{2\pi}{l}\biggr)^{\!2}\,\biggr)\widetilde{\varphi}_\mathrm{e}^{(\pm 1)}(p) -\Delta\biggl(ip\mp \frac{2\pi}{l}\biggr)\widetilde{\varphi}_\mathrm{h}^{(\pm 1)}(p)\biggr)\times{}\\ &\times \biggl(\frac{1}{p^2+(2\pi/l)^2}-\frac{1}{p^2+\Delta^2}\biggr),\\ \widetilde{\psi}_\mathrm{h}^{(\pm 1)}(p)={}&\frac{1}{\Delta^2}\biggl(\Delta\biggl(ip\pm \frac{2\pi}{l}\biggr)\widetilde{\varphi}_\mathrm{e}^{(\pm 1)}(p) -\biggl(p^2+\biggl(\frac{2\pi}{l}\biggr)^{\!2}-2\mu\biggr)\widetilde{\varphi}_\mathrm{h}^{(\pm 1)}(p)\biggr)\times{}\\ &\times \biggl(\frac{1}{p^2+(2\pi/l)^2}-\frac{1}{p^2+\Delta^2}\biggr). \end{aligned}$$

(A.10)

From this, by using the well-known relations

$$\begin{aligned} \, &\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{e^{ipx}\widetilde{\varphi}(p)\,dp}{p^2-p_0^2}= -\frac{1}{2ip_0}\int_{-\infty}^{\infty} e^{ip_0|x-x'|}\varphi(x')\,dx',\\ &\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{p e^{ipx}\widetilde{\varphi}(p)\,dp}{p^2-p_0^2}= -\frac{1}{2i}\int_{-\infty}^{\infty} e^{ip_0|x-x'|}\operatorname{sgn}(x-x')\varphi(x')\,dx' \end{aligned}$$

we obtain $\Psi^{(\pm 1)}=(H^{(\pm 1)}-E)^{-1}\Phi^{(\pm 1)}$ (i.e., in fact, the resolvent):

$$ \begin{aligned} \, \psi_\mathrm{e}^{(\pm 1)}(x)={}&\frac{1}{2\Delta}\biggl( \operatorname{sgn}(\Delta)\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{e}^{(\pm1)}(x')\,dx'+{}\\ &+\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\pm{} \\ &\pm \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\,dx'-{}\\ &-\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx' \biggr),\\ \psi_\mathrm{h}^{(\pm 1)}(x)={}&\frac{1}{2\Delta}\biggl( \frac{\mu}{\Delta(\pi/l)} \int_{-\infty}^{\infty} e^{-|2\pi/l||x-x'|}\varphi_\mathrm{h}^{(\pm1)}(x')\,dx'-{}\\ &-\operatorname{sgn}(\Delta)\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{h}^{(\pm1)}(x')\,dx'-{}\\ &-\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\pm{}\\ &\pm \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\,dx'+{}\\ &+\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx' \biggr). \end{aligned}$$

(A.11)

In the case $E=-\mu+\varepsilon$, $\varepsilon>0$, we similarly obtain

$$ \begin{aligned} \, \psi_\mathrm{e}^{(\pm 1)}(x)={}&\frac{1}{2\Delta}\biggl( -\frac{\mu}{\Delta(\pi/l)} \int_{-\infty}^{\infty} e^{-|2\pi/l||x-x'|}\varphi_\mathrm{e}^{(\pm1)}(x')\,dx'+{}\\ &+\operatorname{sgn}(\Delta)\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{e}^{(\pm1)}(x')\,dx'+{}\\ &+\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\pm{} \\ &\pm \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\,dx'-{}\\ &-\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx' \biggr),\\ \psi_\mathrm{h}^{(\pm 1)}(x)={}&\frac{1}{2\Delta}\biggl( -\operatorname{sgn}(\Delta)\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{h}^{(\pm1)}(x')\,dx'-{}\\ &-\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\pm{}\\ &\pm \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\,dx'+{}\\ &+\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx' \biggr). \end{aligned}$$

(A.12)

We now assume that $|E|\ll \mu$. As above, we obtain $\Psi^{(\pm 1)}=(H^{(\pm 1)}-E)^{-1}\Phi^{(\pm 1)}$, and in this case, we have

$$ \begin{aligned} \, \psi_\mathrm{e}^{(\pm 1)}(x)={}&\frac{1}{2\Delta}\biggl( -\frac{\mu}{2\Delta (\pi/l)}\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm1)}(x')\,dx'+{}\\ &+\operatorname{sgn}(\Delta)\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{e}^{(\pm1)}(x')\,dx'+{}\\ &+\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\pm{} \\ &\pm \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\,dx'-{}\\ &-\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{h}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx' \biggr),\\ \psi_\mathrm{h}^{(\pm 1)}(x)={}&\frac{1}{2\Delta}\biggl( \frac{\mu}{2\Delta(\pi/l)} \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{h}^{(\pm1)}(x')\,dx'-{} \\ &-\operatorname{sgn}(\Delta)\int_{-\infty}^{\infty} -e^{-|\Delta||x-x'|}\varphi_\mathrm{h}^{(\pm1)}(x')\,dx'-{}\\ &-\int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\pm{}\\ &\pm \int_{-\infty}^{\infty} e^{-(2\pi/l)|x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\,dx'+{}\\ &+\int_{-\infty}^{\infty} e^{-|\Delta||x-x'|}\varphi_\mathrm{e}^{(\pm 1)}(x')\operatorname{sgn}(x-x')\,dx'\biggr). \end{aligned}$$

(A.13)