Quantum version of MMSE-based massive MIMO uplink detection

Abstract

Regarding the massive multiple-input multiple-output (MIMO) uplink, the quantum version detection based on the minimum mean square error criterion is developed for complexity consideration. Specifically, a comprehensive quantum algorithm for encoding the estimated signal into a quantum state is presented, synthesizing the subroutines of state preparation, Hamiltonian simulation, phase estimation and so on. Indispensable assumptions along with the quantum algorithm are summarized and assumption testing in a MIMO scenario is conducted. Both theoretical analyses from the mathematical point of view and simulated realizations in massive MIMO systems basically confirm the applicability of the quantum algorithm. With desired precision, the quantum algorithm achieves almost quadratic speedup over all classical counterparts. Whereas, with weaker precision accompanied by undetermined performance loss, the quantum algorithm eventually breaks through polynomial-time detection complexity and achieves exponential speedup. Future work will be directed toward the error dependence improvement, classical form outputs and experimental applications in quantum communication systems.

Appendices

Appendix A

We start with the following lemma.

Lemma 1

Given a complex linear system of equations ${\mathbf {A}}{\mathbf {x}}={\mathbf {b}}$ where ${\mathbf {A}}$ is $M\times M$, write ${\mathbf {b}}$ as ${\mathbf {b}}=\sum _{j=1}^M\beta _j\varvec{\eta }_j$ where $\varvec{\eta }_j$ is the eigenvector of ${\mathbf {A}}$ corresponding to the eigenvalue $\lambda _j$ and $\beta _j=\varvec{\eta }_j^H\cdot {\mathbf {b}}$. Then ${\mathbf {x}}$ can be rephrased as ${\mathbf {x}}=\sum _{j=1}^M\beta _j\lambda _j^{-1}\varvec{\eta }_j$.

Proof

Since $\lambda _j$ and $\varvec{\eta }_j$ satisfy ${\mathbf {A}}\varvec{\eta }_j=\lambda _j\varvec{\eta }_j$, multiply ${\mathbf {A}}^{-1}$ at both left sides of this equation and obtain $\lambda _j^{-1}\varvec{\eta }_j={\mathbf {A}}^{-1}\varvec{\eta }_j$. Analogously, multiply ${\mathbf {A}}^{-1}$ at both left sides of ${\mathbf {A}}{\mathbf {x}}=\sum _{j=1}^M\beta _j\varvec{\eta }_j$ and obtain ${\mathbf {x}}=\sum _{j=1}^M\beta _j\lambda _j^{-1}\varvec{\eta }_j$. Hereto, Lemma 1 has been proved. $\square$

Defining ${\mathbf {W}}={\mathbf {H}}^H{\mathbf {H}}+\alpha {\mathbf {I}}_M$ and applying Eq. (4), we can obtain

$$\begin{aligned} \begin{aligned} {\mathbf {W}}&=\left( \sum _{j=1}^r\lambda _j{\mathbf {v}}_j{{\mathbf {u}}_j}^H\right) \cdot \left( \sum _{j=1}^r\lambda _j{\mathbf {u}}_j{{\mathbf {v}}_j}^H\right) +\alpha {\mathbf {I}}_M\\&=\sum _{j=1}^r\lambda _j^2{\mathbf {v}}_j{\mathbf {v}}_j^H+\alpha \sum _{j=1}^r{\mathbf {v}}_j{\mathbf {v}}_j^H =\sum _{j=1}^r(\lambda _j^2+\alpha ){\mathbf {v}}_j{\mathbf {v}}_j^H. \end{aligned} \end{aligned}$$

(38)

Defining ${\mathbf {y}}^{\text {MF}}={\mathbf {H}}^H{\mathbf {y}}$ and applying Eq. (5), we can obtain

$$\begin{aligned} \begin{aligned} {\mathbf {y}}^{\text {MF}}&=\left( \sum _{j=1}^r\lambda _j{\mathbf {v}}_j{{\mathbf {u}}_j}^H\right) \cdot \left( \Vert {\mathbf {y}}\Vert \sum _{j=1}^N\beta _j{\mathbf {u}}_j\right) =\Vert {\mathbf {y}}\Vert \sum _{j=1}^r\beta _j\lambda _j{\mathbf {v}}_j. \end{aligned} \end{aligned}$$

(39)

Applying Lemma 1, we derive

$$\begin{aligned} {\hat{\mathbf {s}}}={\mathbf {W}}^{-1}{\mathbf {y}}^{\text {MF}} =\Vert {\mathbf {y}}\Vert \sum _{j=1}^r\frac{\lambda _j}{{\lambda _j}^2+\alpha }\beta _j{\mathbf {v}}_j. \end{aligned}$$

(40)

Hereto, Eq. (6) has been proved.

Appendix B

Proof of $Ch(\lambda _j,\alpha )={\varOmega }(1/\kappa )$ is adapted from [27] for the convenience of readers.

Based on $h(\lambda ,\alpha )=\frac{\lambda }{\lambda ^2+\alpha }$ with $\frac{\lambda _{\text {max}}}{\lambda _{\text {min}}}=\kappa$, it can be obtained that

$$\begin{aligned} \begin{aligned}&\frac{\mathop {\max }_{\lambda }h(\lambda ,\alpha )}{\mathop {\min }_{\lambda }h(\lambda ,\alpha )} =\mathop {\max }_{\lambda _1,\lambda _2}\frac{h(\lambda _1,\alpha )}{h(\lambda _2,\alpha )} =\frac{\lambda _1(\lambda _2^2+\alpha )}{\lambda _2(\lambda _1^2+\alpha )}\\&\quad \le \frac{\lambda _1}{\lambda _2}\left( \frac{\lambda _2^2}{\lambda _1^2}+1\right) =\frac{\lambda _2}{\lambda _1}+\frac{\lambda _1}{\lambda _2} \le \kappa +\frac{1}{\kappa }=O(\kappa ). \end{aligned} \end{aligned}$$

(41)

Hence, $C=O(\mathop {\max }_{\lambda _j}h(\lambda _j,\alpha ))^{-1}$ makes $Ch(\lambda _j,\alpha )={\varOmega }(1/\kappa )$.

Appendix C

Proof of $\epsilon _r=O(\kappa \epsilon )$ regardless of $\alpha$ is adapted from [27] for the convenience of readers.

The relative error of $h(\lambda ,\alpha )$ scales as $O(|g(\lambda )|\epsilon _{\lambda })$, where

$$\begin{aligned} g(\lambda )=\frac{h^{'}(\lambda ,\alpha )}{h(\lambda ,\alpha )}=\frac{\alpha -\lambda ^2}{\lambda (\lambda ^2+\alpha )} \end{aligned}$$

(42)

is defined and $\epsilon _{\lambda }=(N+M)\epsilon$ is the absolute error of $\lambda$. Since

$$\begin{aligned} g^2(\lambda )-\frac{1}{\lambda ^2}=\frac{-4\alpha }{(\lambda ^2+\alpha )^2}<0, \end{aligned}$$

(43)

$|g(\lambda )|<\frac{1}{\lambda }$ is promised. Thus, the relative error of $h(\lambda ,\alpha )$ roughly scales as $\epsilon _r=O(\frac{1}{\lambda }(N+M)\epsilon )=O(\frac{\epsilon }{\lambda /(N+M)})=O(\kappa \epsilon )$, where $O(\frac{1}{\lambda /(N+M)})=O(\kappa )$ is easily obtained according to Assumption 4.

Appendix D

Denote X and Y as the length and the length squared of a complex Gaussian variable, respectively. Since the length of a complex Gaussian variable with zero mean and unit variance obeys Rayleigh distribution whose pdf is

$$\begin{aligned} f_{X}(x)= {\left\{ \begin{array}{ll} 2xe^{-x^2}&{}x\ge 0\\ 0&{} \text {otherwise} \end{array}\right. }, \end{aligned}$$

(44)

the pdf of the length squared can be straightly derived as

$$\begin{aligned} f_{Y}(y)= {\left\{ \begin{array}{ll} e^{-y}&{}y\ge 0\\ 0&{} \text {otherwise} \end{array}\right. }. \end{aligned}$$

(45)

Then $E\{Y\}$ can be easily obtained by

$$\begin{aligned} E\{Y\}=\int _{-\infty }^{+\infty }yf_{Y}(y)\mathrm{d}y=\int _{0}^{+\infty }ye^{-y}\mathrm{d}y=1. \end{aligned}$$

(46)

Denote $Y_m$ as the maximal length squared during an observation consisting of N i.i.d. complex Gaussian variables. The exact pdf of $Y_m$ is hard to determine, which means $E\{Y_m\}$ cannot be solved by $E\{Y_m\}=\int _{-\infty }^{+\infty }y_mf_{Y_m}(y_m)\mathrm{d}y_m$.

We can treat this problem from another perspective. Primarily, the CDF of Y can be easily derived as

$$\begin{aligned} F_{Y}(y)=\int _{-\infty }^{y}f_{Y}(y)\mathrm{d}y= {\left\{ \begin{array}{ll} 1-e^{-y}&{}y\ge 0\\ 0&{} \text {otherwise} \end{array}\right. }. \end{aligned}$$

(47)

Typically, for $y_0>0$, $p_0$ is defined as

$$\begin{aligned} p_0=P(Y\le y_0)=1-e^{-y_0}. \end{aligned}$$

(48)

During an observation consisting of N variables, the probability of the event that all of N variables have a length squared no more than $y_0$ is given by

$$\begin{aligned} p={p_0}^N=(1-e^{-y_0})^N. \end{aligned}$$

(49)

Naturally, p will decrease as $y_0$ becomes large. If $y_0$ is large enough to make the probability p reduce to 1/2, i.e.,

$$\begin{aligned} (1-e^{-y_0})^N=\frac{1}{2}, \end{aligned}$$

(50)

then $y_0$ can be roughly viewed as an approximation of $E\{Y_m\}$, which reveals the expression

$$\begin{aligned} E\{Y_m\}=-\ln {\left[ 1-\left( \frac{1}{2}\right) ^{\frac{1}{N} }\right] }. \end{aligned}$$

(51)

Combining Eqs. (46), (51) and (25), the empirical formula is yielded as

$$\begin{aligned} E\{P\}=-\frac{1}{\ln {\left[ 1-\left( \frac{1}{2}\right) ^{\frac{1}{N} }\right] }}. \end{aligned}$$

(52)

Hereto, Eq. (26) has been proved.

Appendix E

By reviewing the previous part of the paper, it can be observed that only the proof in “Appendix C” is based on Assumption 4, which is directly relevant to the total complexity. Thus, we need to modify the proof in “Appendix C” based on the realistic case $\frac{\lambda _j}{\sqrt{N+M}}\in [k_a/\kappa ,k_a]$.

Since $\epsilon _r=O(\frac{\epsilon }{\lambda /(N+M)})=O(\kappa \epsilon \sqrt{N+M})$ is updated, the absolute error of $h(\lambda ,\alpha )$, which satisfies $\epsilon _r=\epsilon _h/h(\lambda ,\alpha )$, scales as

$$\begin{aligned} \begin{aligned} \epsilon _h&=h(\lambda ,\alpha )\epsilon _r=O\left( \kappa \epsilon \frac{\sqrt{N+M}\lambda }{\lambda ^2+\alpha }\right) \\&= O\left( \frac{\kappa \epsilon }{\lambda /\sqrt{N+M}+\alpha /(\lambda \sqrt{N+M})}\right) =O(\kappa ^2\epsilon ). \end{aligned} \end{aligned}$$

(53)

As a consequence, the total complexity is modified as $O\left( \text {poly} \log {(N+M)}\kappa ^5/\epsilon ^3\right)$ where $\epsilon$ is the absolute error of $h(\lambda ,\alpha )$. Hereto, Theorem 3 has been proved.