Continuous time limit of the DTQW in 2D+1 and plasticity

Abstract

A Plastic quantum walk admits both continuous time and continuous spacetime. The model has been recently proposed by one of the authors in Di Molfetta and Arrighi (Quant Inf Process 19(2): 47, 2020), leading to a general quantum simulation scheme for simulating fermions in the relativistic and non-relativistic regimes. The extension to two physical dimensions is still missing and here, as a novel result, we demonstrate necessary and sufficient conditions concerning which discrete time quantum walks can admit plasticity, showing the resulting Hamiltonians. We consider coin operators as general 4 parameter unitary matrices, with parameters which are functions of the lattice step size $\varepsilon$. This dependence on $\varepsilon$ encapsulates all functions of $\varepsilon$ for which a Taylor series expansion in $\varepsilon$ is well defined, making our results very general.

Appendices

Appendix A W expansion

We wish to expand ${\widehat{W}}^\tau$ to first order in $\varDelta _t$. We begin by expanding $\widehat{S_m}C_m$ up to $O(\varDelta _t^2)$, where $m=x,y$:

$$\begin{aligned} \begin{aligned} \widehat{S_m}C_m&=e^{i\delta _m} [R_z(\zeta '_{m})R_y(\theta _{0m})R_z(\phi _{m})-\frac{i \varDelta _t}{2}(\zeta _{1m} \sigma _z R_z(\zeta '_{m})R_y(\theta _{0m})R_z(\phi _{m})\\&\quad +\theta _{1m} \sigma _y R_z(-2\zeta '_{m})R_z(\zeta '_{m})R_y(\theta _{0m})R_z(\phi _{m})\\&\quad +\phi _{1m} R_z(\zeta '_{m})R_y(\theta _{0m})R_z(\phi _{m}) \sigma _z)+O(\varDelta _t^2)]\\&=e^{i\delta _m}(A_m-\frac{i \varDelta _t}{2} B_m+O(\varDelta _t^2)) \end{aligned} \end{aligned}$$

(78)

Now we can combine the product of $\widehat{S_x}C_x$ and $\widehat{S_y}C_y$ up to $O(\varDelta _t)$:

$$\begin{aligned} \begin{aligned} {\widehat{W}}=\widehat{S_x}C_x\widehat{S_y}C_y&=e^{i\delta _x}\left( A_x-\frac{i \varDelta _t}{2} B_x+O(\varDelta _t^2)\right) e^{i\delta _y}\left( A_y-\frac{i \varDelta _t}{2} B_y+O(\varDelta _t^2)\right) \\&=e^{i(\delta _x+\delta _y)}\left( A_xA_y-\frac{i\varDelta _t}{2}(A_xB_y+B_xA_y)+O(\varDelta _t^2)\right) \\&=e^{i\delta }(A-\frac{i\varDelta _t}{2}B+O(\varDelta _t^2)) \end{aligned} \end{aligned}$$

(79)

And now we expand ${\widehat{W}}^\tau$ in powers of $\varDelta _t$:

$$\begin{aligned} \begin{aligned} {\widehat{W}}^\tau&=\left( \widehat{S_x}C_x\widehat{S_y}C_y\right) ^\tau \\&=e^{i\delta \tau }\left( A-\frac{i \varDelta _t}{2} B+O(\varDelta _t^2)\right) ^\tau \\&=e^{i\delta \tau }\left( A^\tau -\frac{i \varDelta _t}{2}(A^{\tau -1}B+A^{\tau -2}BA+\cdots +ABA^{\tau -2}+BA^{\tau -1})+O(\varDelta _t^2)\right) \\&=e^{i\delta \tau }\left( A^\tau -\frac{i \varDelta _t}{2}\sum _{j=0}^{\tau -1}A^{\tau -1-j}BA^j+O(\varDelta _t^2)\right) \\&=(e^{i\delta }A)^\tau \left( 1-\frac{i \varDelta _t}{2}A^{-1}\sum _{j=0}^{\tau -1}A^{-j}BA^j+O(\varDelta _t^2)\right) \end{aligned} \end{aligned}$$

(80)

Appendix B $\{A,B\}$ expansion

We begin by expanding $\{A,B\}$ in terms of $A_x$, $A_y$, $B_x$, and $B_y$, using $A=A_xA_y$ and $B=A_xB_y+B_xA_y$:

$$\begin{aligned} \begin{aligned} \{A,B\}&=A_xA_yA_xB_y+A_xB_yA_xA_y+A_xA_yB_xA_y+B_xA_yA_xA_y. \end{aligned} \end{aligned}$$

(81)

Now we expand further using $B_m=\zeta _{1m}\sigma _z A_m+\theta _{1m} \sigma _y R_z(-2\zeta '_{m})A_m+\phi _{1m} A_m \sigma _z$ (where $m=x$ or y):

$$\begin{aligned} \begin{aligned} \{A,B\}=&\zeta _{1y}A_xA_yA_x\sigma _zA_y+\theta _{1y}A_xA_yA_x\sigma _yR_z(-2\zeta _{0y}')A_y+\phi _{1y}A_xA_yA_xA_y\sigma _z\\&+\zeta _{1y}A_x\sigma _zA_yA_xA_y+\theta _{1y}A_x\sigma _yR_z(-2\zeta _{0y}')A_yA_xA_y+\phi _{1y}A_xA_y\sigma _zA_xA_y\\&+\zeta _{1x}A_xA_y\sigma _zA_xA_y+\theta _{1x}A_xA_y\sigma _yR_z(-2\zeta _{0x}')A_xA_y+\phi _{1x}A_xA_yA_x\sigma _zA_y\\&+\zeta _{1x}\sigma _zA_xA_yA_xA_y+\theta _{1x}\sigma _yR_z(-2\zeta _{0x}')A_xA_yA_xA_y+\phi _{1x}A_x\sigma _zA_yA_xA_y. \end{aligned} \end{aligned}$$

(82)

Using $\sigma _zA_x=(-1)^\nu A_x\sigma _z$ and $\sigma _zA_y=(-1)^{\nu +1} A_y\sigma _z$, it can be shown that the first and third columns of Eq. (82) cancel. Now we expand the remaining terms using $A_m=R_z(\zeta '_{m})R_y(\theta _{0m})R_z(\phi _{m})$:

$$\begin{aligned} \begin{aligned} \theta _{1y}A_xA_yA_x\sigma _yR_z(-2\zeta '_{0y})A_y&=-\theta _{1y}R_z(-2\phi _{0y})\sigma _y\\ \theta _{1y}A_x\sigma _yR_z(-2\zeta '_{0y})A_yA_xA_y&=-\theta _{1y}R_z(2\zeta '_{0x}+2\phi _{0x}(-1)^\nu +2\zeta '_{0y}(-1)^\nu )\sigma _y\\ \theta _{1x}A_xA_y\sigma _yR_z(-2\zeta '_{0x})A_xA_y&=-\theta _{1x}R_z(2\zeta '_{0y}(-1)^\nu -2\phi _{0y}+2\phi _{0x}(-1)^\nu )\sigma _y\\ \theta _{1x}\sigma _yR_z(-2\zeta '_{0x})A_xA_yA_xA_y&=-\theta _{1x}R_z(2\zeta '_{0x})\sigma _y. \end{aligned} \end{aligned}$$

(83)

Putting it all together, we have the following for $\{A,B\}$:

$$\begin{aligned} \begin{aligned} \{A,B\}=&-\theta _{1y}(R_z(-2\phi _{0y})+R_z(2\zeta '_{0x}+2\phi _{0x}(-1)^\nu +2\zeta '_{0y}(-1)^\nu ))\sigma _y\\&-\theta _{1x}(R_z(2\zeta '_{0x})+R_z(2\zeta '_{0y}(-1)^\nu -2\phi _{0y}+2\phi _{0x}(-1)^\nu ))\sigma _y \end{aligned} \end{aligned}$$

(84)

Appendix C Cross term constraint

In this section, we will deduce for which constraints from Sect. 4 do the resulting continuum limit PDEs include a cross derivative term. We will find that the only constraints which will have cross terms will be the pair $\cos (\frac{\phi _{x}+\phi _{y}+\zeta _{x}+\zeta _{y}}{2})=\cos (\frac{a_1+a_2}{2})=0\rightarrow a_1+a_2=2\pi m +\pi$ and $\cos (\frac{a_1-a_2}{2})=0\rightarrow a_1-a_2=2\pi t +\pi$. We first reiterate the definitions of ${\hat{\varGamma }}_{l_xl_yn_xn_y}$ and $\nu '_{l_xl_yn_xn_y}$:

$$\begin{aligned} \begin{aligned} {\hat{\varGamma }}_{l_xl_yn_xn_y}&=R_z(\zeta _{x})\sigma _z^{l_x}\sigma _y^{n_x}R_y(\theta _{0x})R_z(\phi _{x})R_z(\zeta _{y})\sigma _z^{l_y}\sigma _y^{n_y}R_y(\theta _{0y})R_z(\phi _{y})\\ \nu '_{l_xl_yn_xn_y}&=\frac{(\partial _x)^{l_x}(\partial _y)^{l_y}(-\frac{i\theta _{1x}}{2})^{n_x}(-\frac{i\theta _{1y}}{2})^{n_y}}{l_x!l_y!n_x!n_y!} \end{aligned} \end{aligned}$$

(85)

Since cross terms have all $n_{vm}$s equal to zero and two $l_{vm}$s equal to one, we write the proportionality expression for $\nu '_{l_{1x}l_{1y}00}\nu '_{l_{2x}l_{2y}00}{\hat{\varGamma }}_{l_{1x}l_{1y}00}{\hat{\varGamma }}_{l_{2x}l_{2y}00}$ (where $a_1=\phi _{x}+\zeta _{y}$ and $a_2=\phi _{y}+\zeta _{x}$):

$$\begin{aligned} \begin{aligned}&\nu '_{l_{1x}l_{1y}00}\nu '_{l_{2x}l_{2y}00}{\hat{\varGamma }}_{l_{1x}l_{1y}00}{\hat{\varGamma }}_{l_{2x}l_{2y}00}\\&\quad \propto \partial _{x}^{l_{1x}+l_{2x}}\partial _{y}^{l_{1y}+l_{2y}}\\&\qquad \times R_z(\zeta _{x})\sigma _z^{l_{1x}}R_y(\theta _{0x})R_z(\phi _{x})R_z(\zeta _{y})\sigma _z^{l_{2x}}R_y(\theta _{0y})R_z(\phi _{y})\\&\qquad \times R_z(\zeta _{x})\sigma _z^{l_{1y}}R_y(\theta _{0x})R_z(\phi _{x})R_z(\zeta _{y})\sigma _z^{l_{2y}}R_y(\theta _{0y})R_z(\phi _{y})\\&\quad =\partial _{1x}^{l_{1x}+l_{2x}}\partial _{1y}^{l_{1y}+l_{2y}}\\&\qquad \times R_z(\zeta _{x})\sigma _z^{l_{1x}}R_y(\theta _{0x})R_z(a_1)\sigma _z^{l_{2x}}R_y(\theta _{0y})R_z(a_2)\\&\qquad \times \sigma _z^{l_{1y}}R_y(\theta _{0x})R_z(a_1)\sigma _z^{l_{2y}}R_y(\theta _{0y})R_z(\phi _{y})\\&\quad =\partial _{1x}^{l_{1x}+l_{2x}}\partial _{1y}^{l_{1y}+l_{2y}}\sigma _z^{l_{1x}+l_{1y}+l_{2x}+l_{1y}} R_z(\zeta _{x})\\&\qquad \times R_y((-1)^{l_{1y}+l_{2x}+l_{1y}}\theta _{0x})R_z(a_1)R_y((-1)^{l_{2x}+l_{1y}}\theta _{0y})\\&\qquad \times R_z(a_2)R_y((-1)^{l_{2y}}\theta _{0x})R_z(a_1)R_y(\theta _{0y})R_z(\phi _{y}) \end{aligned} \end{aligned}$$

(86)

Since each cross term will be proportional to $\partial _x\partial _y$, we define the following matrix to contain the relevant parts of the above equation to our analysis:

$$\begin{aligned} \begin{aligned} {\hat{J}}_{l_{1x}l_{1y}l_{2x}l_{2y}}&=R_y((-1)^{l_{1y}+l_{2x}+l_{2y}}\theta _{0x})R_z(a_1)R_y((-1)^{l_{2x}+l_{2y}}\theta _{0y})R_z(a_2)R_y((-1)^{l_{2y}}\theta _{0x}) \end{aligned} \end{aligned}$$

(87)

For the cross derivative terms in Eq. (53) to cancel, the following must be true:

$$\begin{aligned} \begin{aligned}&{\hat{J}}_{1100}+{\hat{J}}_{1001}+{\hat{J}}_{0110}+{\hat{J}}_{0011}=0\\&\quad \rightarrow R_y(-\theta _{0x})R_z(a_1)R_y(\theta _{0y})R_z(a_2)R_y(\theta _{0x})\\&\qquad +R_y(-\theta _{0x})R_z(a_1)R_y(-\theta _{0y})R_z(a_2)R_y(-\theta _{0x})\\&\qquad +R_y(\theta _{0x})R_z(a_1)R_y(-\theta _{0y})R_z(a_2)R_y(\theta _{0x})\\&\qquad +R_y(\theta _{0x})R_z(a_1)R_y(\theta _{0y})R_z(a_2)R_y(-\theta _{0x})=0 \end{aligned} \end{aligned}$$

(88)

We see that either the first two terms can cancel when the $\partial _y$ non-divergent constraint in Eq. (62) is imposed, or the second and fourth term can cancel when the $\partial _x$ non-divergent constraint in Eq. (62) is imposed. When the $\partial _y$ constraint is imposed again, the above equation reduces to the following:

$$\begin{aligned} R_y(-2\theta _y)-R_y(2\theta _y)=0\rightarrow \theta _y=n\pi \text { for} n=1, 2, 3, \ldots . \end{aligned}$$

(89)

Similarly, when the $\partial _x$ constraint is imposed again, we recover the following:

$$\begin{aligned} R_y(-2\theta _x)-R_y(2\theta _x)=0\rightarrow \theta _x=m\pi \text { for} m=1, 2, 3, \ldots . \end{aligned}$$

(90)

Thus, we see that these cross derivative terms will cancel with either $\theta _x$ or $\theta _y$ equal to an integer multiple of $\pi$. Therefore, most of the constraints will contain no cross terms. The only set of constraints which will have cross terms will be the pair $\cos (\frac{\phi _{x}+\phi _{y}+\zeta _{x}+\zeta _{y}}{2})=\cos (\frac{a_1+a_2}{2})=0\rightarrow a_1+a_2=2\pi m +\pi$ and $\cos (\frac{a_1-a_2}{2})=0\rightarrow a_1-a_2=2\pi t +\pi$, as there is no constraints on $\theta _x$ or $\theta _y$ equalling an integer multiple of $\pi$.