One-qubit Aharonov–Bhom phase gate constructed by Majorana zero-energy states

Abstract

A well-known scheme for generating Majorana zero-energy modes is a heterojunction of s-wave superconductor and topological insulator that induces an equivalent two-dimensional p-wave superconductor at the interface. In this work, we construct a cylindrical system based on the heterojunction with the external magnetic field. We investigate the Aharonov–Bhom phase (A–B phase) of Majorana zero-energy state, find that the A–B phase between two Majorana zero-energy states located at the inner and outer boundaries of the ring system will interfere with each other. The intensity of interference fringes may change with A–B phase, which indicates that we can store the quantum information through the A–B phase and construct a single qubit A–B phase gate. We further explore the splitting of zero-bias conductance peak caused by the A–B phase and adjust the splitting energy by the A–B phase.

Graphic Abstract

Appendix

According to the BdG equation $H \psi = E \psi =0$,

$$\begin{aligned} \begin{aligned} \left[ \frac{\left( -i \hbar \nabla \right) ^{2}}{2 m}-\mu \right] u +\frac{i \hbar }{p_{F}}\left\{ \Delta _0 e^{\frac{i e^{*} n \phi _0}{2 \pi \hbar }\theta }, \partial _{z}\right\} v=0, \end{aligned} \end{aligned}$$

(A1)

where $u=v^{*}$. We first consider the second term of the equation:

$$\begin{aligned}{} & {} \frac{i \hbar }{p_{F}}\left\{ \Delta _0 e^{\frac{i 2e n \phi _0}{2 \pi \hbar }\theta }, \partial _{z}\right\} \nonumber \\{} & {} \quad = \frac{i \hbar }{p_{F}}\left[ \Delta _0 e^{i(1+\frac{ 2e n \phi _0}{2 \pi \hbar })\theta }\left( \frac{\partial }{\partial r}+\frac{i}{r}\frac{\partial }{\partial \theta }\right) +e^{i \theta }\right. \nonumber \\{} & {} \qquad \left. \left( \frac{\partial }{\partial r}+\frac{i}{r}\frac{\partial }{\partial \theta }\right) \Delta _0 e^{i\frac{ 2e n \phi _0}{2 \pi \hbar }\theta }\right] . \end{aligned}$$

(A2)

Since the wave function needs to satisfy the particle–hole symmetry, we can write the angular part of u as $e^{i(\frac{1}{2} + \frac{ e n \phi _0}{2 \pi \hbar })\theta }$, and its conjugate of v is $e^{-i(\frac{1}{2} + \frac{ e n \phi _0}{2 \pi \hbar })\theta }$. Thus, the second term in Eq. (A1) can be reduced to

$$\begin{aligned}&\frac{i \hbar }{p_{F}}\left\{ \Delta _0 e^{\frac{i 2e n \phi _0}{2 \pi \hbar }\theta }, \partial _{z}\right\} v \nonumber \\&\quad =\frac{i \hbar }{p_{F}}\left[ \Delta _0 e^{i(1+\frac{ 2e n \phi _0}{2 \pi \hbar })\theta }\left( 2\frac{\partial }{\partial r}+\frac{1}{r}\right) \right] e^{-i(\frac{1}{2} + \frac{ e n \phi _0}{2 \pi \hbar })\theta } v(r)\nonumber \\&\quad =\frac{i \hbar }{p_{F}}\left[ \Delta _0 e^{i(\frac{1}{2}+\frac{ e n \phi _0}{2 \pi \hbar })\theta }\left( 2\frac{\partial }{\partial r}+\frac{1}{r}\right) \right] v(r). \end{aligned}$$

(A3)

The first term in Eq. (A1) can also be written as an equation that only relates to the radial direction:

$$\begin{aligned} \begin{aligned}&\left[ \frac{\left( i \hbar \nabla \right) ^{2}}{2 m}-\mu \right] u \\&\quad = \left[ -\frac{\hbar ^2}{2m}(\frac{1}{r} \frac{\partial }{\partial r}+\frac{\partial ^2}{\partial r^2}+\frac{1}{r^2}\frac{\partial ^2}{\partial \theta ^2})- \mu \right] e^{i(\frac{1}{2} + \frac{ e n \phi _0}{2 \pi \hbar })\theta }u(r)\\&\quad = \left[ -\frac{\hbar ^{2}}{2 m}\frac{\partial ^2}{\partial r^2}-\frac{\hbar ^{2}}{2 m r}\frac{\partial }{\partial r}+\frac{e^{2} B^{2} r^{2}}{8 m}+\frac{\hbar ^{2}}{8 m r^{2}} -\frac{\hbar e B}{4 m} - \mu \right] \\&\qquad e^{i(\frac{1}{2} + \frac{ e n \phi _0}{2 \pi \hbar })\theta }u(r), \end{aligned} \end{aligned}$$

(A4)

and we can thus obtain Eq. (3):

$$\begin{aligned} \begin{aligned}&\left( \frac{\hbar ^{2}}{2 m}\frac{\partial ^2}{\partial r^2}+\frac{\hbar ^{2}}{2 m r}\frac{\partial }{\partial r}-\frac{e^{2} B^{2} r^{2}}{8 m}-\frac{\hbar ^{2}}{8 m r^{2}} +\frac{\hbar e B}{4 m}+ \mu \right) u(r) \\&\qquad -\left( \frac{2 i \hbar \Delta _{0}}{m v_{F}}\frac{\partial }{\partial r}+\frac{i \hbar \Delta _{0}}{m v_{F} r}\right) v(r)=0. \end{aligned} \end{aligned}$$

(A5)

The formula contains both real and imaginary coefficients and due to particle–hole symmetry, we add the constant coefficient $e^{i\frac{\pi }{4}}$ and $e^{-i\frac{\pi }{4}}$ in the wave function:

$$\begin{aligned} \begin{aligned}&\left( \frac{\hbar ^{2}}{2 m}\frac{\partial ^2}{\partial r^2}+\frac{\hbar ^{2}}{2 m r}\frac{\partial }{\partial r}-\frac{e^{2} B^{2} r^{2}}{8 m}-\frac{\hbar ^{2}}{8 m r^{2}} +\frac{\hbar e B}{4 m}+ \mu \right) \\&\quad e^{-i\frac{\pi }{4}} u(r) -\left( \frac{2 i \hbar \Delta _{0}}{m v_{F}}\frac{\partial }{\partial r}+\frac{i \hbar \Delta _{0}}{m v_{F} r}\right) e^{i\frac{\pi }{4}} u(r)\\&\quad =\left( \frac{\hbar ^{2}}{2 m}\frac{\partial ^2}{\partial r^2}+\frac{\hbar ^{2}}{2 m r}\frac{\partial }{\partial r}-\frac{e^{2} B^{2} r^{2}}{8 m}-\frac{\hbar ^{2}}{8 m r^{2}} +\frac{\hbar e B}{4 m}+ \mu \right) u(r)\\&\qquad +\left( \frac{2 \hbar \Delta _{0}}{m v_{F}}\frac{\partial }{\partial r}+\frac{ \hbar \Delta _{0}}{m v_{F} r}\right) u(r)=0. \end{aligned} \end{aligned}$$

(A6)

Because the wave function of MZM is localized at the edge of the superconductor around the radius $R_{\text {in}}$ and $R_{\text {out}}$, we approximate the variable r in the coefficient as a constant R [1]:

$$\begin{aligned} \begin{aligned}&\left( \frac{\hbar ^{2}}{2 m}\frac{\partial ^2}{\partial r^2}+\frac{\hbar ^{2}}{2 m R}\frac{\partial }{\partial r}-\frac{e^{2} B^{2} R^{2}}{8 m}-\frac{\hbar ^{2}}{8 m R^{2}} +\frac{\hbar e B}{4 m}+ \mu \right) u(r)\\&\quad +\left( \frac{2 \hbar \Delta _{0}}{m v_{F}}\frac{\partial }{\partial r}+\frac{ \hbar \Delta _{0}}{m v_{F} R}\right) u(r)=0, \end{aligned} \end{aligned}$$

(A7)

the wave function of Majorana Eq. (4) can be obtained by solving Eq. (A7).