Dirac quantum walks with conserved angular momentum

Abstract

A quantum walk (QW) simulating the flat $(1 + 2)$D Dirac equation on a spatial polar grid is constructed. Because fermions are represented by spinors, which do not constitute a representation of the rotation group ${\mathrm{SO}}(3)$, but rather of its double cover ${\mathrm{SU}}(2)$, the QW can only be defined globally on an extended spacetime where the polar angle extends from 0 to $4 \pi$. The coupling of the QW with arbitrary electromagnetic fields is also presented. Finally, the cylindrical relativistic Landau levels of the Dirac equation are computed explicitly and simulated by the QW.

Appendices

Appendix A: The polar Dirac equation

In $(1+2)$D flat spacetime, the Cartesian Dirac equation (CDE) can be written

$$\begin{aligned} {\mathcal D}^A_B\Psi ^B =0, \end{aligned}$$

(A1)

with the operator $\mathcal D$ defined (in natural units $\hbar =c=1$) by

$$\begin{aligned} {\mathcal D} = \mathrm {i}(\gamma ^0 \partial _t + \gamma ^1 \partial _x + \gamma ^2 \partial _y) - m, \end{aligned}$$

(A2)

where (t, x, y) are Minkowski coordinates and m is the mass of the particle. The indices $(A, B) \in \{ L, R \}^2$ refer to components on a Cartesian, point-independent spin basis which we denote by $\left( b_L, b_R\right)$. We choose a representation where, in this basis, the $\gamma$ operators read

$$\begin{aligned}{}[(\gamma ^0)^A_B] = \sigma _1,\quad [(\gamma ^1)^A_B] = \mathrm {i}\sigma _2,\quad [(\gamma ^2)^A_B] = \mathrm {i}\sigma _3, \end{aligned}$$

(A3)

where the $\sigma$’s are the Pauli matrices and the notation $[(\gamma ^i)^A_B]$ represents the matrix formed by the components of the operator $\gamma ^i$ in the basis $\left( b_L, b_R\right)$.

To obtain the PDE from the CDE, one must first use polar coordinates $(r, \theta )$ instead of Cartesian coordinates, and then also change the spin basis basis, replacing the Cartesian spin basis by the polar spin basis $(b_-, b_+)$ obtained from $(b_L, b_R)$ by performing a rotation of angle $\theta$ in spin-space. According to general spinor theory [36], this rotation reads:

$$\begin{aligned} b_-= & {} \cos \frac{\theta }{2} b_L - \mathrm {i}\sin \frac{\theta }{2} b_R, \end{aligned}$$

(A4)

$$\begin{aligned} b_+= & {} -\mathrm {i}\sin \frac{\theta }{2} b_L + \cos \frac{\theta }{2} b_R. \end{aligned}$$

(A5)

Performing both operations leads to

$$\begin{aligned} {\mathcal D}^a_b \Psi ^b = 0, \end{aligned}$$

(A6)

where

$$\begin{aligned}&{\mathcal D}^a_b= \mathrm {i}\left( \gamma ^1\right) ^a_b\partial _t + \mathrm {i}\left( {\tilde{\gamma }}^2(\theta )\right) ^a_b\partial _r \nonumber \\&\quad + \frac{\mathrm {i}}{r}\left( \left( {\tilde{\gamma }}^3(\theta )\right) ^a_b\partial _\theta +\frac{1}{2}\ \left( {\tilde{\gamma }}^2(\theta )\right) ^a_b\right) -m \end{aligned}$$

(A7)

and

$$\begin{aligned}{}[(\gamma ^0)^a_b] = \sigma _1,\qquad [(\tilde{\gamma }^1)^a_b]=\mathrm {i}\sigma _2,\qquad [(\tilde{\gamma }^2)^a_b]=\mathrm {i}\sigma _3. \end{aligned}$$

(A8)

Note that the operators $\gamma ^i$ and $\tilde{\gamma }^i$ are represented by the same matrices, but in different bases.

As usual, the coupling of the Dirac fermion with an electromagnetic field with 3-potential $(A_\mu )=(A_t,A_r,A_\theta )$ is achieved by adding $+\mathrm {i}e A_\mu$ to $\partial _\mu$ for $\mu = 0, 1, 2$. We choose to set the charge e to $-1$ and introduce ${\mathcal D}_\mu = \partial _\mu - i A_\mu$, so the Dirac equation for $\Psi$ in polar coordinates and polar spin basis cab be abbreviated into

$$\begin{aligned} \Bigg (\mathrm {i}\sigma _1{\mathcal D}_t -\sigma _2{\mathcal D}_r -\frac{1}{r}\left( \sigma _3{\mathcal D}_\theta +\frac{1}{2}\sigma _2\right) -m\Bigg ) \Psi = 0, \end{aligned}$$

(A9)

which we call the polar Dirac equation (PDE). We use this compact form in the article when no confusion with the CDE seems possible. Note that polar coordinates are not defined at $r = 0$; therefore, the PDE is non-singular over the definition domain of polar coordinates.

Let us conclude this section by pointing out a very important property of the PDE. The second polar coordinate $\theta$ is an angle. Thus, the components $\Psi ^L$ and $\Psi ^R$ of $\Psi$ in the Cartesian spin basis, when written as functions of r and $\theta$, are $2 \pi$-periodic functions of $\theta$. So are the time component $A_t$, the Cartesian components $A_x$, $A_y$ and the polar components $A_r$ and $A_\theta$ of the potential. The components $\Psi ^-$ and $\Psi ^+$ of $\Psi$ in the polar spin basis are linear combinations of $\Psi ^L$ and $\Psi ^R$ with coefficients $\cos (\theta /2)$ and $\sin (\theta /2)$. These two coefficients are $2 \pi$-anti-periodic in $\theta$ i.e. they obey $f (\theta + 2 \pi ) = - f(\theta )$ for all $\theta \in [0, 2 \pi [$. It follows that the polar components $\Psi ^-$ and $\Psi ^+$ are also $2 \pi$-anti-periodic in $\theta$. This expresses the fact that spinors belong to representations of the double cover of the rotation group $\text{ SO }(2, {\mathbb R})$ and, thus get an extra minus sign after a rotation by $2 \pi$. Thus, the PDE is defined over $\{(r, \theta ), r \in {\mathbb R}^*_+, \theta \in [0, 4 \pi [ \}$ and should only be used with initial conditions which are $2 \pi$-anti-periodic in $\theta$. By construction, the PDE conserves this anti-periodicity over time. Finally, only half integer modes $k = s + 1/2$, $s \in \mathbb Z$ enter the decomposition of the polar spinor components $\Psi ^\pm$ in terms of Fourier modes $\exp (\mathrm {i}k \theta )$.

Appendix B: Angular momentum from the polar Dirac equation

The angular Fourier transform $\tilde{f}$ of an arbitrary function f of the variables $(r, \theta )$ is defined by

$$\begin{aligned} {\tilde{f}} (r, \kappa ) = \int _{\theta = 0}^{4 \pi } {f} (r, \theta ) \exp (i \kappa \theta ) d\theta . \end{aligned}$$

(B1)

Since $\theta$ is bounded by $4 \pi$, its conjugate variable $\kappa$ is discrete, with step $\Delta \kappa = (2 \pi )/(4 \pi ) = 1/2$. Since $\theta$ is continuous, $\kappa$ is unbounded. A simple and common choice for the range of $\kappa$ is therefore $\left\{ 0, \pm 1/2, \pm 1, ...\right\}$. As explained above, the potential components have integer angular Fourier modes while the polar spinor components only have half-integer angular Fourier modes.

If the potential components $A_t$, $A_r$ and $A_\theta$ do not depend on $\theta$, the Fourier transform of the polar spinor components obeys

$$\begin{aligned} \partial _t \left( {\tilde{\Psi }}(r, \kappa )\right) = ({L}^D {\tilde{\Psi }})(r, \kappa ) \end{aligned}$$

(B2)

with

$$\begin{aligned} ({ L}^D {\tilde{\Psi }}) (r, \kappa )= & {} \left( \sigma _3 (\partial _r - i A_r) + \frac{1}{r} \left( i \sigma _2 (\kappa + A_\theta ) + \frac{\sigma _3}{2} \right) \right. \nonumber \\&\left. - i m \sigma _1 + i A_t \right) {\tilde{\Psi }}(r, \kappa ). \end{aligned}$$

(B3)

Thus, each angular Fourier component evolves independently of the others. Moreover, each Fourier component evolves in a unitary way i.e. the operator ${ L}^D$ conserves the norm of each angular Fourier component.

The angular momentum operator $\hat{J}$ is defined by

$$\begin{aligned} ({\hat{J}} {\tilde{\Psi }}) (r, \kappa ) = \kappa {\tilde{\Psi }}(r, \kappa ). \end{aligned}$$

(B4)

This definition is equivalent to

$$\begin{aligned} ({\hat{J}} {\Psi }) (r, \theta ) = - \mathrm {i}\left( \partial _\theta {\Psi }\right) _{\mid r, \kappa }. \end{aligned}$$

(B5)

The expectation value of $\hat{J}$ is conserved by $L^D$ if all potential components are independent of $\theta$.

It is instructive to rewrite the expectation value of the operator $\hat{J}$ in terms of the Cartesian spin components. One finds

$$\begin{aligned} \left\langle \hat{J}\right\rangle= & {} - \mathrm {i}\int \lambda _{AB} \Psi ^{A *} \left( (x \partial _y - y \partial _x)\delta ^B_C \right. \nonumber \\&\left. + \frac{\mathrm {i}}{2} (\sigma _1)^B_C \right) \Psi ^C \mathrm{d}x \, \mathrm{d}y, \end{aligned}$$

(B6)

where the metric $\lambda$ in spin space is defined by $\lambda _{AB} = 1$ if $A = B$ and 0 otherwise. Equation (B6) shows that the total angular momentum of the spinor is the sum of the kinetic angular momentum and of the spin.

Appendix C: Relativistic Landau levels

The eigenstates common to the Dirac Hamiltonian and the angular momentum operator are of the form

$$\begin{aligned} \Phi _{E, \kappa }(t, r, \theta ) = \exp (- i E t)\Xi _{E, \kappa } (r) \exp (- i \kappa \theta ), \end{aligned}$$

(C1)

with

$$\begin{aligned} \Xi =\xi ^- b_- + \xi ^+ b_+ \, . \end{aligned}$$

(C2)

The computation of these eigenstates is best carried out by replacing the components $\xi ^-$ and $\xi ^+$ of $\Xi$ by the new unknown functions

$$\begin{aligned} u^-= & {} \frac{\mathrm {i}}{\sqrt{2}} \exp \left( \frac{\mathrm {i}\pi }{4}\right) ( \xi ^- + \xi ^+), \nonumber \\ u^+= & {} \frac{1}{\sqrt{2}} \exp \left( \frac{\mathrm {i}\pi }{4}\right) (- \xi ^- + \xi ^+). \end{aligned}$$

(C3)

The eigenfunctions $u^\pm _{E,\kappa }$ associated with energy E and angular momentum $\kappa$ obey

$$\begin{aligned} \pm \partial _ru^\pm _{E,\kappa }(r)+\left( \frac{\kappa }{r}-\frac{1}{2}Br\right) u_{E,\kappa }^\pm (r)-(E\mp m)u_{E,\kappa }^\mp (r)=0. \end{aligned}$$

(C4)

The equations (C4) are solved by changing, first the unknown functions, then the variable. One first introduces $v^\pm =r^{\pm \kappa }e^{\mp \frac{1}{4}B r^2}u^\pm$ and then changes variable to $x = \mid B \mid r^2/2$. These substitutions transform (C4) into standard equations solved by Laguerre polynomials. For example, introducing $n=\frac{m^2-E^2}{2B}$ and $\alpha =-\kappa -\frac{1}{2}$, one finds that $v^+$ obeys

$$\begin{aligned} x\frac{\hbox {d}^{2}v^{+}}{\hbox {d}x^{2}}+(\alpha +1-x)\frac{\hbox {d}v^{+}}{\hbox {d}x}+nv^+=0. \end{aligned}$$

(C5)

which is solved by

$$\begin{aligned} v^+=CL_n^\alpha (x)=CL_n^\alpha \left( -\frac{1}{2}Br^2\right) , \end{aligned}$$

(C6)

where C is a constant and $L_n^\alpha$ is an associated Laguerre polynomial. The final, normalized expression for $u^\pm$ is

$$\begin{aligned} u(r)=\left( \begin{array}{cc}\frac{B}{E-m}Cr^{1-\kappa }\mathrm {e}^{\frac{1}{4}Br^2}L_{n-1}^{\alpha +1}\left( -\frac{1}{2}Br^2\right) \\ Cr^{-\kappa }\mathrm {e}^{\frac{1}{4}Br^2}L^{\alpha }_n\left( -\frac{1}{2}Br^2\right) \end{array}\right) . \end{aligned}$$

(C7)

with

$$\begin{aligned} \vert {C}\vert ^2=\frac{(m-E)^2(-B)^{\alpha +1}n!}{\pi 2^{\alpha +1}(n+\alpha )!(-2Bn+(m-E)^2)}. \end{aligned}$$

(C8)