Digital representation of continuous observables in quantum mechanics

Abstract

To simulate quantum systems on classical or quantum computers, the continuous observables (e.g., coordinate and momentum or energy and time) must be reduced to discrete ones. In this paper, we consider the continuous observables represented in the positional systems as power series in the radix multiplied over the summands (“digits”), which turn out to be Hermitian operators with discrete spectrum. We investigate the obtained quantum mechanical operators of digits, the commutation relations between them, and the effects of the choice of a numeral system on lattices and representations. Renormalizations of diverging sums naturally occur in constructing the digital representation.

Appendix A

Rectangular function on a lattice

We consider a periodic rectangular function $c(\alpha)$, $\alpha\in\mathbb{Z}(T)$ such that

$$ c(\alpha) = \begin{cases} \mathscr{A}, & L\leqslant \alpha \leqslant R, \\ 0 & \text{in other cases}, \end{cases} \qquad\alpha\in \mathbb{Z}(T).$$

(A.1)

We use the orthonormal basis $E_k(\alpha)=e^{-2\pi i\alpha k/T}/\sqrt T$, $k\in\mathbb{Z}(T)$:

$$\begin{aligned} \, \tilde c(0)&=\langle E_0|c\rangle=\mathscr{A}\frac{R-L+1}{\sqrt T}, \end{aligned}$$

(A.2)

$$\begin{aligned} \, \tilde{c}(k)&=\langle E_k|c\rangle = \frac{1}{\sqrt T} \sum_{\alpha = L}^{R} \mathscr{A} e^{2 \pi i k \alpha/T} = {} \nonumber \\ &=\frac{\mathscr{A}}{\sqrt T} e^{2 \pi i k L/T}\biggl(\frac{1 - e^{2 \pi i k (R - L + 1)/T}} {1 - e^{2 \pi i k L/T}} \biggr), \quad k\not=0, \end{aligned}$$

(A.3)

$$\begin{aligned} \, c(\alpha) &=\sum_{k \in \mathbb{Z}_{T}}\tilde c(k)\cdot E_k(\alpha) ={} \nonumber \\ &= \mathscr{A}\frac{R-L+1}{T}+\sum_{k=1}^{T-1} \frac{\mathscr{A}}{T} e^{2 \pi i k L/T}\biggl(\frac{1 - e^{2 \pi i k (R - L + 1)/T}}{1 - e^{2 \pi i k L/T}}\biggr) e^{- 2 \pi i\alpha k/T}. \end{aligned}$$

(A.4)

$q$-nary system on the lattice

We now consider one ($r$th) digit of a $q$-nary system with digits $\{0,1, \dots, q\}$. It can be represented as a sequence of $q$ rectangular functions $c_{j}$, for which we can write

$$L_j = q^{n_{+}+r} \cdot (j - 1), \quad R_j = (q^{n_{+}+r} \cdot j) - 1, \quad \mathscr{A}_j = j - 1, \quad T_j=q^{n_{+}+r+1}, \quad j \in \{1, \dots, q\}.$$

Hence, we obtain

$$ p_r=\mathbf{d}(r,p)= \sum_{j = 1}^{q}c_{j}\biggl(\frac{p}{\Delta p}\biggr).$$

(A.5)

Using formula (A.4), we obtain the following expression for the momentum digit decomposition:

$$ p_r = \frac{q - 1}{2} - \Delta p q^{-r} \sum_{D \in \mathbb{Z}_{q^r/\Delta p}} \sum_{\sigma = 1}^{q - 1}\frac{e^{-2\pi i Ap}}{1 - e^{2\pi i \Delta p A}}, \qquad A = q^{-r}\biggl(D + \frac{\sigma}{q}\biggr).$$

(A.6)

Similarly, for the coordinate digit, we have

$$ x_s = \frac{q - 1}{2} - \Delta x q^{-s} \sum_{D \in \mathbb{Z}_{q^s/\Delta x}} \sum_{\sigma = 1}^{q - 1}\frac{e^{2\pi i Bx}}{1 - e^{2\pi i \Delta x B}}, \qquad B = q^{-s}\biggl(D + \frac{\sigma}{q}\biggr).$$

(A.7)

Appendix B

We consider the matrices of momentum digits and operators in the cases where they are compact enough to be put on paper.

Everywhere in this section, $\Delta x = 1$, $x \in \mathbb{Z}_{N} \in \{0,1, \dots,N - 1 \}$, the coordinates and momenta are labeled by ternary numbers, which are marked with the subscript 3, and $\Delta p = 3^{-n} = 1/N$.

The case $n = 1$ and $N = 3^1 = 3$. Symmetric system

In this case, we have

$$\begin{aligned} \, &\hat{x} = \hat{x}_0 = \begin{pmatrix} +1&0&0 \\ 0&0&0 \\ 0&0&-1 \end{pmatrix} = \hat{s}_z, \\ &\hat p_{-1} =\frac{1}{\sqrt{3}} \begin{pmatrix} 0 & i & -i \\ -i & 0 & i \\ i & -i & 0 \end{pmatrix}= \frac{1}{\sqrt{3}} (\sqrt{2}\hat{s}_y + 2\hat{s}_y\hat{s}_x + i\hat{s}_z). \end{aligned}$$

The case $n = 1$ and $N = 3^1 = 3$. Nonsymmetric system

Here, we obtain results

$$\begin{aligned} \, &\hat{x} = \hat{x}_0 = \begin{pmatrix} 2&0&0 \\ 0&1&0 \\ 0&0&0 \end{pmatrix} = \hat{1} + \hat{s}_z, \\ &\hat p_{-1} =\frac{1}{6} \begin{pmatrix} 6& 1 - \sqrt{3}i& 1 + \sqrt{3}i \\ 1 + \sqrt{3}i& 6 & 1 - \sqrt{3}i \\ 1 - \sqrt{3}i& 1 + \sqrt{3}i& 6 \end{pmatrix}=\hat{1} + \frac{\sqrt{2}}{6}\hat{s}_{x} + \frac{1}{\sqrt{6}}\hat{s}_{y} - \frac{\sqrt{3}}{6}(2\hat{s}_{y}\hat{s}_{x} + i\hat{s}_{z}). \end{aligned}$$

The case $n = 2$ and $N = 3^2 = 9$. Nonsymmetric system

In this case, we have

$$\begin{aligned} \, &x = x_0 + 3\cdot x_1=\operatorname{diag}(8;7;6;5;4;3;2;1;0),\\ &x_0=\operatorname{diag}(2;1;0;2;1;0;2;1;0),\qquad x_1=\operatorname{diag}(2;2;2;1;1;1;0;0;0). \end{aligned}$$

We set

$$E_n = \frac{1}{e^{-2 \pi i n/9}- 1},$$

Then

$$\begin{aligned} \, &\hat{p}_{-1}=\frac{1}{3}\begin{pmatrix} 3&E_{8} &E_{7} & 0& E_{5}& E_{4}&0 &E_{2} & E_{1}\\ E_{1} & 3 &E_{8} &E_{7} & 0& E_{5}& E_{4}&0 &E_{2}\\ E_{2} &E_{1} & 3 &E_{8} &E_{7} & 0& E_{5}& E_{4}&0\\ 0&E_{2} &E_{1} & 3 &E_{8} &E_{7} & 0& E_{5}& E_{4}\\ E_{4} &0&E_{2} &E_{1} & 3 &E_{8} &E_{7} & 0& E_{5}\\ E_{5} &E_{4} &0&E_{2} &E_{1} & 3 &E_{8} &E_{7} & 0\\ 0 &E_{5} &E_{4} &0&E_{2} &E_{1} & 3 &E_{8} &E_{7} \\ E_{7} &0 &E_{5} &E_{4} &0&E_{2} &E_{1} & 3 &E_{8} \\ E_{8} & E_{7} &0 &E_{5} &E_{4} &0&E_{2} &E_{1} & 3 \end{pmatrix}, \\ &\hat{p}_{-2}=\begin{pmatrix} 1&0 &0 &3 E_{6} & 0&0 &3E_{3} &0 &0 \\ 0 & 1&0 &0 &3 E_{6} & 0&0 &3E_{3} &0 \\ 0 & 0& 1&0 &0 &3 E_{6} & 0&0 &3E_{3} \\ 3E_{3} &0 & 0& 1&0 &0 &3 E_{6} & 0&0 \\ 0& 3E_{3} &0 & 0& 1&0 &0 &3 E_{6} &0 \\ 0 & 0& 3E_{3} &0 & 0& 1&0 &0 &3 E_{6}\\ 3 E_{6} &0 & 0& 3E_{3} &0 & 0& 1&0 &0\\ 0 & 3 E_{6} &0 & 0& 3E_{3} &0 & 0& 1&0\\ 0 &0 & 3 E_{6} &0 & 0& 3E_{3} &0 & 0& 1\\ \end{pmatrix}, \end{aligned}$$

$\hat{p} = \hat{p}_{-1}/3 + \hat{p}_{-2}/9$,

$$\hat{p} = \frac{1}{9}\begin{pmatrix} 4&E_{8} &E_{7} & 3 E_{6}& E_{5}& E_{4}&3 E_{3} &E_{2} & E_{1}\\ E_{1} & 4 &E_{8} &E_{7} & 3 E_{6}& E_{5}& E_{4}&3 E_{3} &E_{2}\\ E_{2} &E_{1} & 4 &E_{8} &E_{7} & 3 E_{6}& E_{5}& E_{4}&3 E_{3}\\ 3 E_{3}&E_{2} &E_{1} & 4 &E_{8} &E_{7} & 3 E_{6}& E_{5}& E_{4}\\ E_{4} &3 E_{3}&E_{2} &E_{1} & 4 &E_{8} &E_{7} & 3 E_{6}& E_{5}\\ E_{5} &E_{4} &3 E_{3}&E_{2} &E_{1} & 4 &E_{8} &E_{7} & 3 E_{6}\\ 3 E_{6} &E_{5} &E_{4} &3 E_{3}&E_{2} &E_{1} & 4 &E_{8} &E_{7} \\ E_{7} &3 E_{6} &E_{5} &E_{4} &3 E_{3}&E_{2} &E_{1} & 4 &E_{8} \\ E_{8} & E_{7} &3 E_{6} &E_{5} &E_{4} &3 E_{3}&E_{2} &E_{1} & 4 \end{pmatrix}.$$

The case $n = 2$ and $N = 3^2 = 9$. Symmetric system

In this case, we have

$$\begin{aligned} \, x=x_0 + 3 x_1 = \operatorname{diag}(4;3;2;1;0;-1;-2;-3;-4),\\ x_0=\operatorname{diag}(1;0;-1;1;0;-1;1;0;-1),\qquad x_1=\operatorname{diag}(1;1;1;0;0;0;-1;-1;-1). \end{aligned}$$

We set

$$G_n = \frac{(-1)^n}{2\sqrt{3}\sin[(9-n)/9]}.$$

Then

$$\begin{aligned} \, &\hat{p}_{-1}=\frac{1}{\sqrt{3}i}\begin{pmatrix} 0&G_{8} &G_{7} & 0& G_{5}& G_{4}&0 &G_{2} & G_{1}\\ G_{1} & 0 &G_{8} &G_{7} & 0& G_{5}& G_{4}&0 &G_{2}\\ G_{2} &G_{1} & 0 &G_{8} &G_{7} & 0& G_{5}& G_{4}&0\\ 0&G_{2} &G_{1} & 0 &G_{8} &G_{7} & 0& G_{5}& G_{4}\\ G_{4} &0&G_{2} &G_{1} & 0 &G_{8} &G_{7} & 0& G_{5}\\ G_{5} &G_{4} &0&G_{2} &G_{1} & 0 &G_{8} &G_{7} & 0\\ 0 &G_{5} &G_{4} &0&G_{2} &G_{1} & 0 &G_{8} &G_{7} \\ G_{7} &0 &G_{5} &G_{4} &0&G_{2} &G_{1} & 0 &G_{8} \\ G_{8} & G_{7} &0 &G_{5} &G_{4} &0&G_{2} &G_{1} & 0 \end{pmatrix}, \\ &\hat{p}_{-2}=\frac{1}{\sqrt{3}i} \begin{pmatrix} 0&0 &0 &1 & 0&0 &-1 &0 &0 \\ 0 & 0&0 &0 &1 & 0&0 &-1 &0 \\ 0 & 0& 0&0 &0 &1 & 0&0 &-1 \\ -1 &0 & 0& 0&0 &0 &1 & 0&0 \\ 0& -1 &0 & 0& 0&0 &0 &1 &0 \\ 0 & 0& -1 &0 & 0& 0&0 &0 &1\\ 1 &0 & 0& -1 &0 & 0& 0&0 &0\\ 0 & 1 &0 & 0& -1 &0 & 0& 0&0\\ 0 &0 & 1 &0 & 0& -1 &0 & 0& 0\\ \end{pmatrix}, \end{aligned}$$

$\hat{p} = \hat{p}_{-1}/3 + \hat{p}_{-2}/9$,

$$\hat{p} = \frac{1}{9\sqrt{3}i}\begin{pmatrix} 0&3G_{8} &3G_{7} & 1& 3G_{5}& 3G_{4}&-1&3G_{2} & 3G_{1}\\ 3G_{1} & 0 &3G_{8} &3G_{7} & 1& 3G_{5}& 3G_{4}&-1 &3G_{2}\\ 3G_{2} &3G_{1} & 0 &3G_{8} &3G_{7} & 1& 3G_{5}& 3G_{4} & -1\\ -1&3G_{2} &3G_{1} & 0 &3G_{8} &3G_{7} & 1& 3G_{5}& 3G_{4}\\ 3G_{4} &-1&3G_{2} &3G_{1} & 0 &3G_{8} &3G_{7} & 1& 3G_{5}\\ 3G_{5} &3G_{4} &-1&3G_{2} &3G_{1} & 0 &3G_{8} &3G_{7} & 1\\ 1 &3G_{5} &3G_{4} &-1&3G_{2} &3G_{1} & 0 &3G_{8} &3G_{7} \\ 3G_{7} &1 &3G_{5} &3G_{4} &-1&3G_{2} &3G_{1} & 0 &3G_{8} \\ 3G_{8} & 3G_{7} &1 &3G_{5} &3G_{4} &-1&3G_{2} &3G_{1} & 0 \end{pmatrix}.$$