Quantum walk with a general coin: exact solution and asymptotic properties

Abstract

In this paper, we present closed-form expressions for the wave function that governs the evolution of the discrete-time quantum walk on the line when the coin operator is arbitrary. The formulas were derived assuming that the walker can either remain put in the place or proceed in a fixed direction but never move backward, although they can be easily modified to describe the case in which the particle can travel in both directions. We use these expressions to explore properties of magnitudes associated to the process, as the probability mass function or the probability current, even though we also consider the asymptotic behavior of the exact solution. Within this approximation, we will estimate upper and lower bounds, examine the origins of an emerging approximate symmetry, and deduce the general form of the stationary probability density of the relative location of the walker.

Appendices

Appendix 1: Exact solution

In this appendix, we provide explicit details on the derivation of the closed-form expressions for the two components of the wave function given in the main text, Eqs. (20) and (21), starting from the recursive formulas (18) and (19). The approach that follows is similar to the one taken in our previous work [30] and differs from the broadest method in the use of the discrete Fourier transform (DFT) instead of the discrete-time Fourier transform [7].

Let $f(n)$ be a given set of $N$ complex numbers, $n~\in ~\{0,\ldots , N-1\}$, and denote its DFT by $\tilde{f}(r)$,

$$\begin{aligned} \tilde{f}(r)\equiv \sum _{n=0}^{N-1} f(n) {\text {e}}^{i 2 \pi r n/N}, \end{aligned}$$

(68)

for $r\in \{0,\ldots , N-1\}$. The $N$ complex quantities $\tilde{f}(r)$ thus defined keep exactly the same information contained in the original series, and therefore, one can recover $f(n)$ from $\tilde{f}(r)$ by means of the so-called inverse DFT formula

$$\begin{aligned} f(n)\equiv \frac{1}{N}\sum _{r=0}^{N-1} \tilde{f}(r) {\text {e}}^{-i 2 \pi r n/N}. \end{aligned}$$

(69)

The DFT is well suited to convert recursive relationships in the position domain, as Eqs. (18) and (19), into a set of algebraic equations in the Fourier domain. In our case, however, the recursive formulas involve not only different locations but different instants of time, so we must carefully choose $f(n)$ and $N$ itself to preserve the overall coherence.

To this end, let us introduce the auxiliary time horizon $T$, $T\ge 0$, set $N\equiv T+1$, and consider the following definition for the DFT of $\psi _{0,1}(n,t)$, valid for any $t$, $t~\in ~\{0,\ldots , T\}$, ⁶

$$\begin{aligned} \tilde{\psi }_{0,1}(r,t;T)\equiv \sum _{n=0}^{N-1} \psi _{0,1}(n,t) {\text {e}}^{i 2 \pi r n/N}, \end{aligned}$$

(70)

where it is implicitly assumed that $\psi _{0,1}(n,t)=0$ for any $n\ge t+1$. This definition entails that $\tilde{\psi }_{0,1}(r,t;T)$ is an explicit function of $T$: that is, for a fixed value of $r$ and a fixed value of $t$, different choices of $T$ lead to different values for $\tilde{\psi }_{0,1}(r,t;T)$. Nonetheless, the final result that one gets after applying the corresponding inversion formula,

$$\begin{aligned} \psi _{0,1}(n,t)\equiv \frac{1}{N}\sum _{r=0}^{N-1} \tilde{\psi }_{0,1}(r,t;T) {\text {e}}^{-i 2 \pi r n/N}, \end{aligned}$$

(71)

does not depend on $T$ for a fixed choice of $n$ and $t$, as long as $0\le n\le t\le N-1$. ⁷ Therefore, the conclusion is that Eq. (71) will be valid for any $N$, $N\ge t+1$.

At this point, we can safely move Eqs. (18) and (19) into the Fourier domain:

$$\begin{aligned} \tilde{\psi }_{0}(r,t;T)&= \cos \theta \, \tilde{\psi }_{0}(r,t-1;T) +{\text {e}}^{-i \varphi }\sin \theta \,\tilde{\psi }_1(r,t-1;T), \end{aligned}$$

(72)

$$\begin{aligned} \tilde{\psi }_{1}(r,t;T)&= {\text {e}}^{i \varphi }\sin \theta \, {\text {e}}^{i 2 \pi r/N}\tilde{\psi }_{0}(r,t-1;T)\nonumber \\ {}&- \cos \theta \, {\text {e}}^{i 2 \pi r/N}\tilde{\psi }_{1}(r,t-1;T). \end{aligned}$$

(73)

The initial values for $\tilde{\psi }_{0,1}(r,t;T)$ are $\tilde{\psi }_{0}(r,0;T)=\cos \eta$, $\tilde{\psi }_{1}(r,0;T)=\sin \eta$, for $r \in \{0,\ldots ,N-1\}$. The resolution of Eqs. (72) and (73) can be tackled through standard matrix techniques, thus resulting in

$$\begin{aligned} \tilde{\psi }_{0}(r,t;T)&= \frac{{\text {e}}^{-i\pi \frac{r}{N}}}{2\cos \omega _{\frac{r}{N}}}\Big \{ (\lambda _+)^t \left[ \left( \cos \theta -\lambda _-\right) \cos \eta + {\text {e}}^{-i \varphi }\sin \theta \sin \eta \right] \nonumber \\&\quad -(\lambda _-)^t \left[ \left( \cos \theta -\lambda _+\right) \cos \eta + {\text {e}}^{-i \varphi }\sin \theta \sin \eta \right] \Big \}, \end{aligned}$$

(74)

and

$$\begin{aligned} \tilde{\psi }_{1}(r,t;T)&= \frac{\left( \lambda _+ -\cos \theta \right) \left( \cos \theta -\lambda _-\right) {\text {e}}^{-i\pi \frac{r}{N}}}{2\cos \omega _{\frac{r}{N}}}\nonumber \\&\quad \times \Bigg \{ (\lambda _+)^t \left[ \frac{\cos \eta \, {\text {e}}^{i \varphi }}{\sin \theta } +\frac{\sin \eta }{\cos \theta -\lambda _-} \right] \nonumber \\&\quad -\,(\lambda _-)^t \left[ \frac{\cos \eta \, {\text {e}}^{i \varphi }}{\sin \theta } -\frac{\sin \eta }{\lambda _+ -\cos \theta } \right] \Bigg \}, \end{aligned}$$

(75)

with $\lambda _{\pm }$ functions of $r/N$,

$$\begin{aligned} \lambda _+&\equiv {\text {e}}^{-i(\omega _{\frac{r}{N}}-\pi \frac{r}{N})},\end{aligned}$$

(76)

$$\begin{aligned} \lambda _-&\equiv -{\text {e}}^{i(\omega _{\frac{r}{N}}+\pi \frac{r}{N})}, \end{aligned}$$

(77)

and where $\omega _{\frac{r}{N}}$ is an angle that, given $r$ and $N$, satisfies

$$\begin{aligned} \sin \omega _{\frac{r}{N}}=\cos \theta \,\sin \frac{\pi r}{N}. \end{aligned}$$

(78)

Note that, since $r\in \{0,\ldots ,N-1\}$, we have

$$\begin{aligned} 0\le \sin \omega _{\frac{r}{N}}\le \cos \theta \le 1, \end{aligned}$$

so, to prevent any uncertainty, we consider that $\omega _{\frac{r}{N}}$ is the only solution that Eq. (78) has in $[0,\frac{\pi }{2}]$.

Now, we can simply introduce the value of $\tilde{\psi }_{0,1}(r,t;T)$ given in (74) and (75) into Eq. (71) and recover $\psi _{0,1}(n,t)$ after the computation of a finite sum. Note that, at this point, our procedure has yielded a closed-form expression for the wave function; thus, we could simply stop here. However, we are going to simplify the final formulas as much as possible: In this way, we can get a more detailed view of the properties of the solution.

To manage the complexity of this endeavor we analyze the particular case $\eta =0$ first. Let us begin with $\psi _{0}(n,t)$:

$$\begin{aligned} \psi _{0}(n,t)&= \frac{1}{N}\sum _{r=0}^{N-1} \frac{{\text {e}}^{-i\pi (2n +1) \frac{r}{N}}}{2\cos \omega _{\frac{r}{N}}}\left\{ (\lambda _+)^t\left( \cos \theta -\lambda _-\right) \right. \nonumber \\&\quad -\left. (\lambda _-)^t \left( \cos \theta -\lambda _+\right) \right\} . \end{aligned}$$

(79)

If we use Eq. (78) in conjunction with the definition of $\lambda _{\pm }$, Eqs. (76) and (77), we can show that

$$\begin{aligned} {\text {e}}^{-i\pi r /N}\left( \cos \theta - \lambda _{\pm }\right) =\cos \theta \cos \frac{\pi r}{N} \mp \cos \omega _{\frac{r}{N}} \end{aligned}$$

(80)

holds, and that consequently

$$\begin{aligned} \psi _{0}(n,t)&= \frac{1}{2N}\left[ \left( \cos \theta +1\right) - (-1)^t\left( \cos \theta -1 \right) \right] \nonumber \\&\quad +\frac{1}{2N}\sum _{r=1}^{N-1} \frac{\cos \theta \cos \frac{\pi r}{N}}{\cos \omega _{\frac{r}{N}}} {\text {e}}^{-i\left[ \pi (2n-t)\frac{r}{N}+ \omega _{\frac{r}{N}}t\right] }\nonumber \\&\quad +\frac{1}{2N}\sum _{r=1}^{N-1} {\text {e}}^{-i\left[ \pi (2n-t)\frac{r}{N}+ \omega _{\frac{r}{N}}t\right] }\nonumber \\&\quad -\frac{(-1)^t}{2N}\sum _{r=1}^{N-1} \frac{\cos \theta \cos \frac{\pi r}{N}}{\cos \omega _{\frac{r}{N}}} {\text {e}}^{-i\left[ \pi (2n-t)\frac{r}{N}- \omega _{\frac{r}{N}}t\right] }\nonumber \\&+\quad \frac{(-1)^t}{2N}\sum _{r=1}^{N-1} {\text {e}}^{-i\left[ \pi (2n-t)\frac{r}{N}- \omega _{\frac{r}{N}}t\right] }. \end{aligned}$$

(81)

Here, we have isolated in the first term all the contribution coming from the $r=0$ case. This step is necessary to rearrange the two last summations by introducing a new variable $s$, $s\equiv N-r$,

$$\begin{aligned} \psi _{0}(n,t)&= \frac{1}{2N}\left[ \left( \cos \theta +1\right) - (-1)^t\left( \cos \theta -1 \right) \right] \nonumber \\&\quad +\,\frac{1}{2N}\sum _{r=1}^{N-1} \frac{\cos \theta \cos \frac{\pi r}{N}}{\cos \omega _{\frac{r}{N}}} {\text {e}}^{-i\left[ \pi (2n-t)\frac{r}{N}+ \omega _{\frac{r}{N}}t\right] }\nonumber \\&\quad +\,\frac{1}{2N}\sum _{r=1}^{N-1} {\text {e}}^{-i\left[ \pi (2n-t)\frac{r}{N}+ \omega _{\frac{r}{N}}t\right] }\nonumber \\&\quad +\,\frac{1}{2N}\sum _{s=1}^{N-1} \frac{\cos \theta \cos \frac{\pi s}{N}}{\cos \omega _\frac{s}{N}} {\text {e}}^{i\left[ \pi (2n-t)\frac{s}{N}+ \omega _\frac{s}{N} t\right] }\nonumber \\&\quad +\,\frac{1}{2N}\sum _{s=1}^{N-1} {\text {e}}^{i\left[ \pi (2n-t)\frac{s}{N}+ \omega _\frac{s}{N} t\right] }, \end{aligned}$$

(82)

where the following identities

$$\begin{aligned} \omega _{\frac{r}{N}}&= \omega _{\frac{s}{N}},\\ \cos \frac{\pi r}{N}&= -\cos \frac{\pi s}{N}, \end{aligned}$$

have been taken into account. Finally, we find

$$\begin{aligned} \psi _{0}(n,t)&= \frac{1}{2N}\left[ \left( \cos \theta +1\right) - (-1)^t\left( \cos \theta -1\right) \right] \nonumber \\&\quad +\frac{1}{N}\sum _{r=1}^{N-1} \left[ 1+\frac{\cos \theta \cos \frac{\pi r}{N}}{\cos \omega _{\frac{r}{N}}}\right] \cos \bigg [\pi (2n-t)\frac{r}{N}+ \omega _{\frac{r}{N}}t\bigg ]. \end{aligned}$$

(83)

In the case of $\psi _{1}(n,t)$, we can proceed in a similar way. The result

$$\begin{aligned} \psi _{1}(n,t)&= \frac{{\text {e}}^{i \varphi }\sin \theta }{N}\Bigg \{\frac{1 - (-1)^t}{2}\nonumber \\&\quad +\sum _{r=1}^{N-1} \frac{1 }{\cos \omega _{\frac{r}{N}}} \cos \bigg [\pi (2n-t-1)\frac{r}{N}+ \omega _{\frac{r}{N}}t\bigg ]\Bigg \}, \end{aligned}$$

(84)

is almost immediate once one realizes that

$$\begin{aligned} \left( \lambda _{+}-\cos \theta \right) \left( \cos \theta - \lambda _{-}\right) {\text {e}}^{-i\pi r /N}=\sin ^2\theta \,{\text {e}}^{i\pi r /N}, \end{aligned}$$

an expression that combines Eqs. (78) and (80).

Analogously, when $\eta =\frac{\pi }{2}$ we obtain

$$\begin{aligned} \psi _{0}(n,t)&= \frac{{\text {e}}^{-i\varphi }\sin \theta }{N}\Bigg \{\frac{1 - (-1)^t}{2}\nonumber \\&\quad +\sum _{r=1}^{N-1} \frac{1}{\cos \omega _{\frac{r}{N}}} \cos \bigg [\pi (2n-t+1)\frac{r}{N}+ \omega _{\frac{r}{N}}t\bigg ]\Bigg \}, \end{aligned}$$

(85)

and

$$\begin{aligned} \psi _{1}(n,t)&= \frac{1}{2N}\left[ \left( 1-\cos \theta \right) + (-1)^t\left( 1+\cos \theta \right) \right] \nonumber \\&\quad +\frac{1}{N}\sum _{r=1}^{N-1} \left[ 1-\frac{\cos \theta \cos \frac{\pi r}{N}}{\cos \omega _{\frac{r}{N}}}\right] \cos \bigg [\pi (2n-t)\frac{r}{N}+ \omega _{\frac{r}{N}}t\bigg ], \end{aligned}$$

(86)

by simply applying the same ideas and intermediate formulas.

Finally, we can recover the general solution, Eqs. (20) and (21), through the superposition of these two cases.

Appendix 2: Asymptotic expressions

In this appendix, we obtain asymptotic expressions for $\psi _{0,1}(n,t)$ and $\rho (n,t)$, formulas with a restricted validity but which in turn are more compact and readable than the exact ones.

We begin with a close analysis of the inner structure of the different pieces that compose Eqs. (20) and (21). The conclusion is that, in essence, we must find a way to approximate functions like $h(n,t)$,

$$\begin{aligned} h(n,t)\equiv \frac{\Xi (t)}{N} +\frac{1}{N}\sum _{r=1}^{N-1} g(r/N) \cos \left[ \varPhi (n,r,t;T)+\upsilon \pi r/N\right] , \end{aligned}$$

(87)

where

$$\begin{aligned} \varPhi (n,r,t;T)\equiv \pi (2n-t)r/N+ \omega _{\frac{r}{N}}t, \end{aligned}$$

(88)

and $\upsilon \in \{-1,0,1\}$. In every case, $g(\cdot )$ is a smooth function, and therefore, the behavior of the cosine terms does determine the overall result of the sum. Due to the presence of $\omega _{\frac{r}{N}}$ within $\varPhi (n,r,t;T)$, the argument of these cosine functions does not change linearly with $r$ but exhibits a maximum, and then, the use an adapted version of the method of the stationary phase is the one most indicated in this case [30, 32]: Only those terms for which $\varPhi (n,r,t;T)$ attains its maximum are relevant, whereas the rest of them are negligible.

To this end, let us first define $u\equiv r/N$, and $\nu \equiv n/t$, in terms of which we can rewrite $\varPhi (n,r,t;T)$,

$$\begin{aligned} \varPhi (\nu t,u (T-1) ,t;T)=\phi (\nu , u)t, \end{aligned}$$

(89)

with

$$\begin{aligned} \phi (\nu ,u)\equiv \pi (2\nu -1) u+ \omega _u. \end{aligned}$$

(90)

Our next step is to consider the function $h(n,t)$ in the continuum limit, $N\rightarrow \infty$,

$$\begin{aligned} h(n,t)&\sim \int _0^{1}g(u) \cos \left[ \phi (\nu ,u) t+\upsilon \pi u\right] {\text {d}}u\nonumber \\&\sim \mathrm{Re}\left\{ \int _0^{1}g(u){\text {e}}^{i \upsilon \pi u} {\text {e}}^{i\phi (\nu ,u) t} {\text {d}}u\right\} , \end{aligned}$$

(91)

and to expand $\phi (\nu ,u)$ in the vicinity of $u_0$,

$$\begin{aligned} \phi (\nu ,u)&\sim \phi (\nu ,u_0)+\frac{1}{2}\frac{\partial ^2\phi (\nu ,u_0)}{\partial u^2}(u-u_0)^2\\&= \phi _0(\nu )+\frac{1}{2}\phi ''_0(\nu )(u-u_0)^2, \end{aligned}$$

where $u_0$ is the point for which, given $\nu$, $\phi (\nu ,u)$ has its maximum:

$$\begin{aligned} \frac{\partial \phi (\nu ,u_0)}{\partial u}= \pi (2\nu -1) +\frac{\pi \cos \theta \cos \pi u_0}{\sqrt{1-\cos ^2\theta \sin ^2 \pi u_0}}=0. \end{aligned}$$

(92)

From Eq. (92), we have

$$\begin{aligned} \cos \pi u_0 =\frac{1-2\nu }{2\sqrt{\nu (1-\nu )}}\tan \theta , \end{aligned}$$

(93)

and

$$\begin{aligned} \sin \pi u_0 =\frac{1}{2\cos \theta }\sqrt{\frac{\cos ^2\theta -(2\nu -1)^2}{\nu (1-\nu )}}. \end{aligned}$$

(94)

Equation (94) shows us that the validity of the present approximation is restricted to values of $\nu$ for which one has $\cos ^2\theta -(2\nu -1)^2\ge 0$; that is,

$$\begin{aligned} \frac{1}{2}\left( 1-\cos \theta \right) \le \nu \le \frac{1}{2} \left( 1+\cos \theta \right) , \end{aligned}$$

since, by construction, $\cos \theta \ge 0$. Also from Eqs. (22) and (94), we get

$$\begin{aligned} \sin \omega _0&\equiv \sin \omega _{u_0}=\frac{1}{2}\sqrt{\frac{\cos ^2\theta -(2\nu -1)^2}{\nu (1-\nu )}}, \end{aligned}$$

(95)

$$\begin{aligned} \cos \omega _0&\equiv \cos \omega _{u_0}=\frac{\sin \theta }{2\sqrt{\nu (1-\nu )}}, \end{aligned}$$

(96)

expressions that will be helpful in forthcoming derivations.

Now, we can fully evaluate Eq. (91) under the above premises:

$$\begin{aligned} h(n,t)&\sim \mathrm{Re}\left\{ \int _0^{1}g(u){\text {e}}^{i \upsilon \pi u} {\text {e}}^{i\phi (\nu ,u) t} {\text {d}}u\right\} \nonumber \\&\sim \mathrm{Re}\left\{ \int _0^{1}g(u_0){\text {e}}^{i \upsilon \pi u_0} {\text {e}}^{i t\left[ \phi _0(\nu )+\frac{1}{2}\phi ''_0(\nu )(u-u_0)^2\right] } {\text {d}}u\right\} \nonumber \\&\sim \mathrm{Re}\left\{ g(u_0){\text {e}}^{i \left[ \upsilon \pi u_0+\phi _0(\nu ) t\right] }\int _{-\infty }^{\infty } {\text {e}}^{\frac{i t}{2}\phi ''_0(\nu )(u-u_0)^2} {\text {d}}u\right\} \nonumber \\&= \sqrt{\frac{2 \pi }{t|\phi ''_0(\nu )|}} g(u_0) \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}+\upsilon \pi u_0\right] , \end{aligned}$$

(97)

with

$$\begin{aligned} \phi ''_0(\nu ) =-4 \pi ^2 \nu (1-\nu )\frac{\sqrt{\cos ^2\theta -(2\nu -1)^2}}{\sin \theta }. \end{aligned}$$

(98)

The approximate versions of Eqs. (20) and (21) are

$$\begin{aligned} \psi _{0}(n,t)&\sim \frac{\cos \eta }{\sqrt{t}} \sqrt{\frac{2(1-\nu )\sin \theta }{\pi \nu \sqrt{\cos ^2\theta -(2\nu -1)^2}}} \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] \nonumber \\&+\frac{{\text {e}}^{-i \varphi }\sin \eta }{\sqrt{t}} \sqrt{\frac{2\sin \theta }{\pi \sqrt{\cos ^2\theta -(2\nu -1)^2}}} \nonumber \\&\quad \times \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}+\pi u_0\right] , \end{aligned}$$

(99)

and

$$\begin{aligned} \psi _{1}(n,t)&\sim \frac{{\text {e}}^{i\varphi }\cos \eta }{\sqrt{t}} \sqrt{\frac{2\sin \theta }{\pi \sqrt{\cos ^2\theta -(2\nu -1)^2}}} \nonumber \\&\quad \times \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}-\pi u_0\right] \nonumber \\&+ \frac{\sin \eta }{\sqrt{t}} \sqrt{\frac{2\nu \sin \theta }{\pi (1-\nu )\sqrt{\cos ^2\theta -(2\nu -1)^2}}} \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] , \end{aligned}$$

(100)

and they follow from Eq. (97) once one realizes, cf. Eqs. (93) and (96), that if

$$\begin{aligned} g(u)= 1+\frac{\cos \theta \cos \pi u}{\cos \omega _u}, \end{aligned}$$

one gets

$$\begin{aligned} g(u_0)= 2(1-\nu ); \end{aligned}$$

if

$$\begin{aligned} g(u)= \frac{\sin \theta }{\cos \omega _u}, \end{aligned}$$

one has

$$\begin{aligned} g(u_0)= 2\sqrt{\nu (1-\nu )}; \end{aligned}$$

and finally, if

$$\begin{aligned} g(u)= 1-\frac{\cos \theta \cos \pi u}{\cos \omega _u}, \end{aligned}$$

one obtains

$$\begin{aligned} g(u_0)= 2\nu . \end{aligned}$$

To derive the asymptotic expression for $\rho (n,t)$, one has to calculate $\left| \psi _{0,1}(n,t)\right| ^2$ first:

$$\begin{aligned} \left| \psi _0(n,t)\right| ^2&\sim \frac{2}{\pi t} \sqrt{\frac{1-\nu }{\nu }}\frac{\sin \theta }{\sqrt{\cos ^2\theta -(2\nu -1)^2}} \nonumber \\&\quad \times \Bigg \{\sqrt{\frac{1-\nu }{\nu }}\cos ^2\eta \cos ^2\left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] \nonumber \\&+ \sqrt{\frac{\nu }{1-\nu }}\sin ^2\eta \cos ^2\left[ \phi _0(\nu ) t-\frac{\pi }{4}+ \pi u_0\right] \nonumber \\&+\sin 2\eta \cos \varphi \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}+ \pi u_0\right] \Bigg \},\qquad \quad \end{aligned}$$

(101)

and

$$\begin{aligned} \left| \psi _1(n,t)\right| ^2&\sim \frac{2}{\pi t} \sqrt{\frac{\nu }{1-\nu }}\frac{\sin \theta }{\sqrt{\cos ^2\theta -(2\nu -1)^2}} \nonumber \\&\quad \times \Bigg \{\sqrt{\frac{1-\nu }{\nu }} \cos ^2\eta \cos ^2\left[ \phi _0(\nu ) t-\frac{\pi }{4}-\pi u_0\right] \nonumber \\&+\sqrt{\frac{\nu }{1-\nu }}\sin ^2\eta \cos ^2\left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] \nonumber \\&+\sin 2\eta \cos \varphi \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}-\pi u_0\right] \Bigg \},\qquad \quad \end{aligned}$$

(102)

then use the following trigonometric identities,

$$\begin{aligned}&2\cos ^2\left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] = 1+ \sin \left[ 2\phi _0(\nu ) t\right] ,\\&2\cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}\right] \cos \left[ \phi _0(\nu ) t-\frac{\pi }{4}\pm \pi u_0\right] \\&=\cos \pi u_0 +\cos \pi u_0 \sin \left[ 2\phi _0(\nu ) t\right] \pm \sin \pi u_0 \cos \left[ 2\phi _0(\nu ) t\right] ,\\&2\cos ^2\left[ \phi _0(\nu ) t-\frac{\pi }{4}\pm \pi u_0\right] \\&=1+\cos 2\pi u_0 \sin \left[ 2\phi _0(\nu ) t\right] \pm \sin 2\pi u_0 \cos \left[ 2\phi _0(\nu ) t\right] , \end{aligned}$$

in coordination with Eqs. (93) to (96), to ultimately obtain

$$\begin{aligned} \rho (n,t)&\sim \frac{1}{2 \pi t} \frac{1}{\nu (1-\nu )}\frac{\sin \theta }{\sqrt{\cos ^2\theta -(2\nu -1)^2}} \nonumber \\&\quad \times \Bigg \{1-(2\nu -1) \bigg [\cos 2\eta +\sin 2\eta \tan \theta \cos \varphi \nonumber \\&+\left( \frac{2\nu -1}{\cos ^2\theta }-\cos 2\eta -\sin 2\eta \tan \theta \cos \varphi \right) \sin \left[ 2\phi _0(\nu ) t\right] \nonumber \\&+\sqrt{1-\frac{(2\nu -1)^2}{\cos ^2\theta }}\left( \cos 2\eta \tan \theta -\sin 2\eta \cos \varphi \right) \cos \left[ 2\phi _0(\nu ) t\right] \bigg ]\Bigg \}\nonumber \\&\equiv \bar{\rho }(n,t). \end{aligned}$$

(103)

This expression leads to Eq. (48) after a straightforward trigonometric transformation.

Note finally that, beyond being cornerstones in the calculation of $\bar{\rho }(n,t)$, Eqs. (99) and (100) are worth by themselves: There are problems that require precise information about the asymptotic behavior of each wave function, as in [33, 34].