One-dimensional quantum walks subject to next-nearest-neighbour hopping decoherence

Abstract

It is believed that the decoherence will lead to a crossover from quantum to classical for the diffusive behaviour of discrete quantum walk in the long time limit. However, a few systems with sub-ballistic diffusive behaviours have been found in some non-unitary quantum walks. In this paper, we study the one-dimensional discrete quantum walks subject to a short-range hopping decoherence, i.e. with a probability the walker could hop to next-nearest-neighbour lattices unilaterally and/or bilaterally in one time step. We find that, when the decoherence effects only come from the bilateral hopping operation, the diffusive behaviours of quantum walks are sub-ballistic and the distributions of position exhibit three peaks. These results are quite different from those of the previous non-unitary quantum walks. Our results could be used to improve the algorithmic properties of quantum walk due to its faster diffusive speed and more uniform spreading.

Appendix: the eigenvalues of superoperator \({\mathcal {L}}_{k}\)

It is easy to see that $\lambda =1$ is an eigenvalue of ${\mathcal {L}}_{k}$, and the other eigenvalues are eigenvalues of the three by three submatrix

$$\begin{aligned} M_{k}=\left( \begin{array}{cccc} 0 &{} a &{} b \\ 0 &{} -b &{} a \\ 1 &{} 0 &{} 0\end{array} \right) . \end{aligned}$$

(35)

Let $\lambda _{1}$, $\lambda _{2}$ and $\lambda _{3}$ are eigenvalues of $M_{k}$, then we have

$$\begin{aligned} det\left| I-M_{k}\right| =(1-\lambda _{1})(1-\lambda _{2})(1-\lambda _{3}). \end{aligned}$$

(36)

If $\lambda =1$ is an eigenvalue of $M_{k}$, then $det\left| I-M_{k}\right| \equiv 0$. When $0<p\le 1$ and $0\le q<1$, we can find that $det\left| I-M_{k}\right| =1-a^{2}-b^{2}\ne 0$ for all $k$. Therefore, $\lambda =1$ is not an eigenvalue of $M_{k}$. Similarly,

$$\begin{aligned} det\left| I+M_{k}\right| =(1+\lambda _{1})(1+\lambda _{2})(1+\lambda _{3})=a^{2}+(1-b)^{2}\ne 0. \end{aligned}$$

(37)

Hence, $\lambda =-1$ is also not an eigenvalue of $M_{k}$.

Now suppose $\lambda _{1}=e^{i\theta }$ ($\theta$ is a real number) is a complex eigenvalue of $M_{k}$, then the conjugate $\lambda _{2}=e^{-i\theta }$ must also be an eigenvalue of $M_{k}$. The third eigenvalue is $\lambda _{3}=a^{2}+b^{2}$ due to $det\left| M_{k}\right| =\lambda _{1}\lambda _{2}\lambda _{3}$. Therefore, the characteristic polynomial of $M_{k}$ is given by

$$\begin{aligned} det\left| \lambda I_{3}-M_{k}\right|= & {} (\lambda -\lambda _{1}) (\lambda -\lambda _{2}) (\lambda -\lambda _{3}) \nonumber \\= & {} \lambda ^{3}- \left( a^{2}+b^{2}+2\cos \theta \right) \lambda ^{2} \nonumber \\&+\,\left[ 1+2(a^{2}+b^{2})\cos \theta \right] \lambda -a^{2}-b^{2}. \end{aligned}$$

(38)

On the other hand, we can obtain the characteristic polynomial of $M_{k}$ directly from Eq. (35), i.e.

$$\begin{aligned} det\left| \lambda I_{3}-M_{k}\right| =\lambda ^{3}+b\lambda ^{2}-b\lambda -a^{2}-b^{2}. \end{aligned}$$

(39)

Comparing corresponding coefficients of Eqs. (38) and (39), we obtain the following equations:

$$\begin{aligned} b= & {} -(a^{2}+b^{2}+2\cos \theta ), \end{aligned}$$

(40)

$$\begin{aligned} -b= & {} 1+2(a^{2}+b^{2})\cos \theta ). \end{aligned}$$

(41)

From the two equations, we find $(1-a^{2}-b^{2})(1-2\cos \theta )=0$. Then, $\cos \theta$ must be $1/2$ due to $1-a^{2}-b^{2}\ne 0$. Substituting $\cos \theta =1/2$ back into one of equations, we infer that $1+a^{2}+b^{2}=-b$. This result is impossible since $1+a^{2}+b^{2}>1>-b$ when $0<p\le 1$ and $0\le q<1$. This contradiction implies that $M_{k}$ has not complex eigenvalue of unit modulus. In conclusion, when $0<p\le 1$ and $0\le q<1$, $\lambda =1$ is only unit modulus eigenvalue of ${\mathcal {L}}_{k}$, the other eigenvalues $|\lambda |<1$.

On the other hand, when $0<p<1$ and $q=1$, we find that $a=0$ and $b=1$ for $k=0,\pm \pi$ and the superoperator ${\mathcal {L}}_{k}$ has four unit modulus eigenvalues.