Generating a state t-design by diagonal quantum circuits

Applying a diagonal-unitary t-design on any pure state achieves a t-design of the corresponding phase-random states by definition. If we choose an appropriate initial state and a basis of the diagonal-unitary t-design, we can also achieve a good approximation of a t-design of random states, as stated in the following proposition:

Proposition 1. A t-design of phase-random states obtained by applying a diagonal-unitary t-design in the computational basis onto an initial state ${{\left| + \right\rangle }^{\otimes N}}$, where $\left| + \right\rangle =\frac{1}{\sqrt{2}}\left( \left| 0 \right\rangle +\left| 1 \right\rangle \right)$, is an $\eta \left( N,t \right)$-approximate state t-design, where $\eta \left( N,t \right)=\frac{t\left( t-1 \right)}{d}+O\left( \frac{1}{{{d}^{2}}} \right)$ and $d={{2}^{N}}$.

As we will see in the next subsection, a diagonal-unitary t-design for a small t can be achieved by using only diagonal gates acting on a small number of qubits, where the order of the applications of gates does not matter due to the commutativity of diagonal gates, i.e., there is no inherent temporal structure in the circuit. This is a big advantage in experimental implementations of the circuit. In particular, a diagonal-unitary t-design for $t\leqslant 3$ is implementable by using two- and one-qubit diagonal gates and the total number of gates is $O\left( {{N}^{2}} \right)$. This means that the protocol in proposition 1 generates an $\eta \left( N,t \right)$-approximate state t-design for $t\leqslant 3$ by a quantum circuit composed of $O\left( {{N}^{2}} \right)$ two- and one-qubit diagonal gates that has no temporal structure. This should be contrasted to a previously known protocol using a local random circuit [22], which is composed of two-qubit gates randomly chosen from $U\left( 4 \right)$ acting only on neighboring qubits. Although it achieves the same degree of approximation by using at most $O\left( {{t}^{5}}{\rm log} \left( t \right){{N}^{2}} \right)$ gates, the circuit is necessarily temporally structured. Thus, our protocol has a practical advantage for large N as long as the required degree of approximation is up to $\eta \left( N,t \right)$, particularly when $t\leqslant 3$.

To achieve a better degree of approximation, it may help to combine a diagonal-unitary t-design with other procedures. In [7], it was shown that an exact state 2-design is obtained if we combine a diagonal-unitary 2-design with a classical probabilistic procedure. However, we can show that adding a classical probabilistic procedure improves the degree of approximation by $O\left( {{d}^{1-t}} \right)$ in general, so that it is not effective for $t\geqslant 3$ (see appendix A). Here, we examine a method to apply a local random circuit following the application of a diagonal-unitary t-design on ${{\left| + \right\rangle }^{\otimes N}}$. Since an $\eta \left( N,t \right)$-approximate state t-design is already achieved by a diagonal-unitary t-design, it is natural to expect that this protocol generates an -approximate state t-design more efficiently than the one that uses only a local random circuit. We numerically check this. In contrast to the previous results about a local random circuit, where an input state is arbitrary, input states of the local random circuit in our protocol are determined by output states obtained by applying a diagonal-unitary t-design on ${{\left| + \right\rangle }^{\otimes N}}$. Hence, the necessary length of the local random circuit in our protocol is not directly obtained from the previous results.

Let T be the length of a parallelized local random circuit after applying a diagonal-unitary t-design, where we mean by a parallelized local random circuit that unitary gates acting on different qubits are applied simultaneously. We denote by D(T) the trace distance between an expectation of $\left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}}$ over the resulting ensemble and that over a state t-design (see definition 4). We numerically check how D(T) scales with T for t = 2 and $N=3,4,5,6$. In the numerics, we randomly generate a unitary matrix representing a parallelized local random circuit, and apply it to the states obtained by applying a diagonal-unitary t-design on ${{\left| + \right\rangle }^{\otimes N}}$. By repeating this and averaging the resulting states, we evaluate the expectation of $\left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}}$ over the ensemble obtained by our protocol. The result is shown in figure 1. We observe that the distance exponentially decreases for each N. Although the exponent of the exponential decrement depends on N for up to N = 6, we expect from the right figure in figure 1 that it converges to some value $\alpha \left( t \right)$ that depends on t but not on N. Accordingly, we conjecture the following.

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Conjecture 1. Let D(T) be the trace distance as defined above. Then,

$$\begin{eqnarray*}&&D(T)\sim \eta (N,t){{2}^{-T/\alpha (t)}}\sim \frac{t(t-1)}{d}{{2}^{-T/\alpha (t)}}.\end{eqnarray*}$$

If this is the case, a parallelized random circuit of length $T\left( \epsilon \right)$ following a diagonal-unitary t-design achieves an -approximate state t-design, where

$$\begin{eqnarray*}T(\epsilon )=\left\{ \begin{array}{ccccccccccccccc} 0 & {\rm{for}}\;\epsilon \geqslant \eta (N,t), \\ \alpha (t)[{{{\rm log} }_{2}}1/\epsilon -N+{{{\rm log} }_{2}}t(t-1)] & {\rm{for}}\;\epsilon <\eta (N,t). \\ \end{array} \right.\end{eqnarray*}$$

We present our result that an r-qubit phase-random circuit achieves a diagonal-unitary t-design, where r is determined by t. An r-qubit phase-random circuit is an extension of a phase-random circuit [7, 26]. An r-qubit phase-random circuit for an N-qubit system is a quantum circuit consisting of r-qubit diagonal gates in the computational basis with random phases applied on all combinations of r qubits out of N qubits (see also figure 2 ). Note that each r-qubit gate cannot be decomposed into a sequence of s-qubit diagonal gates ($s<r$) since the phases of r-qubit diagonal gates should be chosen independently and randomly, which cannot be achieved by randomizing the phases of gates acting only on $s<r$ qubits.

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Our first result on an implementation of diagonal-unitary t-designs is given in the following theorem.

Theorem 1. An r-qubit phase-random circuit is a diagonal-unitary t-design if and only if $r\geqslant {{{\rm log} }_{2}}t+1$ for $t\leqslant {{2}^{N}}-1$, and r = N for $t\geqslant {{2}^{N}}$.

This result implies that we cannot achieve a diagonal-unitary t-design for $t\geqslant 4$ if we use only two- and one-qubit diagonal gates.

Our second result is that each r-qubit gate for implementing a diagonal-unitary t-design can be replaced by a sequence of multi-qubit controlled-phase-type gates that act on s qubits ($s\leqslant r$), where a multi-qubit controlled-phase-type gate is a unitary operation represented by ${\rm{diag}}\left( 1,1,\cdots ,1,{{e}^{i\alpha }} \right)$ in the computational basis. We also prove that a phase of a multi-qubit controlled-phase gate acting on s qubits can be randomly selected from a $\left( {\left\lfloor {t/{2^{s - 1}}} \right\rfloor + 1} \right)$-valued discrete set of phases. Thus, the smaller number of discrete phases is required for the gates acting on the larger number of qubits. This result also shows that a phase-random circuit achieves a diagonal-unitary t-design with a finite number of elements.

These results enable us to analyze implementations of diagonal-unitary t-designs by using two- and one-qubit non-diagonal gates. An explicit construction of a multi-qubit controlled-phase-type gate acting on r qubits is known and it requires $O\left( {{r}^{2}} \right)$ two-qubit gates [29], although it is unlikely to be optimal. By decomposing the multi-qubit controlled-phase gates in an r-qubit phase-random circuit, we can show that a diagonal-unitary t-design for an N-qubit system is obtained after applying M two-qubit non-diagonal gates, where

$$\begin{eqnarray*}&&M=\left( \begin{array}{ccccccccccccccc} N \\ r \\ \end{array} \right)\mathop{\mathop{\sum }\nolimits }\limits_{r-1}^{s=1}O({{s}^{2}})\left( \begin{array}{ccccccccccccccc} r \\ s \\ \end{array} \right),\end{eqnarray*}$$

and r is given in theorem 1. For a constant t, $M\sim O\left( {{N}^{{{{\rm log} }_{2}}t}} \right)$ and the construction is efficient. For larger t such as $t={\rm{poly}}\left( N \right)$, this provides a sub-exponential implementation of a diagonal-unitary t-design.

We show proposition 1 given in section 3.

Proof 1. The expectation of $\left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}}$ over random states is given by

$$\begin{eqnarray*}&&{{\mathbb{E}}_{\left| \phi \right\rangle \in {{\Upsilon }_{{\rm{Haar}}}}}}\left[ \left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}} \right]=\frac{1}{\left( \begin{array}{ccccccccccccccc} t+d-1 \\ t \\ \end{array} \right)}\mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}}\prime \in \mathcal{D}/{{\mathcal{P}}_{t}}}\left| {{\pi }^{-1}}({\mathbf{n}}\prime ) \right\rangle \left\langle {{\pi }^{-1}}({\mathbf{n}}\prime ) \right|.\end{eqnarray*}$$

This is obtained simply by applying Schurʼs lemma [30]. For a t-design of phase-random states ${{\Upsilon }_{{\rm{phase}}}}$ obtained by applying a diagonal-unitary t-design on ${{\left| + \right\rangle }^{\otimes N}}$, the expectation is given by

$$\begin{eqnarray*}&&{{\mathbb{E}}_{\left| \phi \right\rangle \in {{\Upsilon }_{{\rm{phase}}}}}}\left[ \left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}} \right]=\frac{1}{{{d}^{t}}}\mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}}\prime \in \mathcal{D}/{{\mathcal{P}}_{t}}}\left| {{\pi }^{-1}}({\mathbf{n}}\prime ) \right|\left| {{\pi }^{-1}}({\mathbf{n}}\prime ) \right\rangle \left\langle {{\pi }^{-1}}({\mathbf{n}}\prime ) \right|.\end{eqnarray*}$$

The difference $\eta \left( N,t \right)$ between the expectations is given by

$$\begin{eqnarray*}\begin{array}{rcl} \eta (N,t) & = & \| {{\mathbb{E}}_{\left| \phi \right\rangle \in {{\Upsilon }_{{\rm{Haar}}}}}}[\left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}}]-{{\mathbb{E}}_{\left| \phi \right\rangle \in {{\Upsilon }_{{\rm{phase}}}}}}{{\left[ \left| \phi \right\rangle {{\left\langle \phi \right|}^{\otimes t}} \right]}\| _{1}} \\ {} & = & \mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}}\prime \in \mathcal{D}/{{\mathcal{P}}_{t}}}\left| \frac{1}{\left( \begin{array}{ccccccccccccccc} t+d-1 \\ t \\ \end{array} \right)}-\frac{1}{{{d}^{t}}} \right|{{\pi }^{-1}}({\mathbf{n}}\prime )\left| {} \right|. \\ \end{array}\end{eqnarray*}$$

We expand $\left( \begin{array}{ccccccccccccccc} t+d-1 \\ t \\ \end{array} \right)$ by $\frac{1}{t!}\mathop{\sum }\nolimits _{k=1}^{t}{{\alpha }_{k}}{{d}^{k}}$, where ${{\alpha }_{t}}=1$ and ${{\alpha }_{t-1}}=t\left( t-1 \right)/2$, and obtain

$$\begin{eqnarray}&&\eta \left( N,t \right)={{\left( \mathop{\mathop{\sum }\nolimits }\limits_{t}^{k=1}{{\alpha }_{k}}{{d}^{k-t}} \right)}^{-1}}\mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}}\prime \in \mathcal{D}/{{\mathcal{P}}_{t}}}\left| t!\frac{1}{{{d}^{t}}}-\left| {{\pi }^{-1}}\left( {\mathbf{n}}\prime \right) \right|\mathop{\mathop{\sum }\nolimits }\limits_{t}^{k=1}{{\alpha }_{k}}\frac{1}{{{d}^{2t-k}}} \right|.\end{eqnarray} \tag{ 1 }$$

We explicitly calculate this up to the order $1/d$. For $d\gg 1$ and a constant t, ${{\left( \mathop{\sum }\nolimits _{k=1}^{t}{{\alpha }_{k}}{{d}^{k-t}} \right)}^{-1}}$ is given by

$$\begin{eqnarray}&&{{\left( \mathop{\mathop{\sum }\nolimits }\limits_{t}^{k=1}{{\alpha }_{k}}{{d}^{k-t}} \right)}^{-1}}=1-\frac{t\left( t-1 \right)}{2}\frac{1}{d}+O\left( \frac{1}{{{d}^{2}}} \right).\end{eqnarray} \tag{ 2 }$$

The other term in equation (1), $|{{\pi }^{-1}}\left( {\mathbf{n}}\prime \right)|$, depends only on the number of the same N-bit sequences in ${\mathbf{n}}\prime$. For ${\mathbf{n}}\prime$ such that ${{\vec{n}}^{\left( i \right)}}\ne {{\vec{n}}^{\left( j \right)}}$ for $i\ne j$, $|{{\pi }^{-1}}\left( {\mathbf{n}}\prime \right)|=t!$ and the number of such ${\mathbf{n}}\prime$ is $\left( \begin{array}{ccccccccccccccc} d \\ t \\ \end{array} \right)$. For ${\mathbf{n}}\prime$ such that there exists only one pair (i, j), where $i\ne j$, satisfying ${{\vec{n}}^{\left( i \right)}}={{\vec{n}}^{\left( j \right)}}$, $|{{\pi }^{-1}}\left( {\mathbf{n}}\prime \right)|=t!/2$ and the number of such ${\mathbf{n}}\prime$ is $\left( t-1 \right)\left( \begin{array}{ccccccccccccccc} d \\ t-1 \\ \end{array} \right)$. In other cases, the number of each type of ${\mathbf{n}}\prime$ is at most ${{d}^{t-2}}$. Since the inside of the summation ${{\mathop{\sum }\nolimits }_{{\mathbf{n}}\prime \in \mathcal{D}/{{\mathcal{P}}_{t}}}}$ in equation (1) is at most $O\left( 1/{{d}^{t}} \right)$, we obtain that

$$\begin{eqnarray}&&\begin{array}{l} \mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}}\prime \in \mathcal{D}/{{\mathcal{P}}_{t}}}\left| t!\frac{1}{{{d}^{t}}}-|{{\pi }^{-1}}({\mathbf{n}}\prime )|\mathop{\mathop{\sum }\nolimits }\limits_{t}^{k=1}{{\alpha }_{k}}\frac{1}{{{d}^{2t-k}}} \right| \\ =\left( \begin{array}{ccccccccccccccc} d \\ t \\ \end{array} \right)\left( t!\frac{t(t-1)}{2}\frac{1}{{{d}^{t+1}}}+O(\frac{1}{{{d}^{t+2}}}) \right)+(t-1)\left( \begin{array}{ccccccccccccccc} d \\ t-1 \\ \end{array} \right)\left( \frac{t!}{2}\frac{1}{{{d}^{t}}}+O(\frac{1}{{{d}^{t+1}}}) \right)+O(\frac{1}{{{d}^{2}}}) \\ =\frac{t(t-1)}{d}+O(\frac{1}{{{d}^{2}}}), \\ \end{array}\end{eqnarray} \tag{ 3 }$$

where we have used relations such as $\left( \begin{array}{ccccccccccccccc} d \\ t \\ \end{array} \right)=\left( {{d}^{t}}+O\left( {{d}^{t-1}} \right) \right)/t!$ for a constant t. From equations (2) and (3), we obtain

$$\begin{eqnarray*}&&\eta (N,t)=\frac{t(t-1)}{d}+O(\frac{1}{{{d}^{2}}}).\end{eqnarray*}$$

□

We prove theorem 1 by restating it in a different way. Since the statement is obvious for r = N, we consider only $r\leqslant N-1$ in the following.

Let S be ${{\mathbb{E}}_{U\in {{\mathcal{U}}_{{\rm{diag}}}}}}\left[ {{U}^{\otimes t}}\otimes {{\left( {{U}^{\dagger }} \right)}^{\otimes t}} \right]$ and introduce expansion coefficients ${{S}_{{\mathbf{nm}}}}$ in the computational basis defined by

$$\begin{eqnarray*}&&S=\mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}},{\mathbf{m}}\in \mathcal{D}}{{S}_{{\mathbf{nm}}}}\left| {\mathbf{n}} \right\rangle \left\langle {\mathbf{n}} \right|\otimes \left| {\mathbf{m}} \right\rangle \left\langle {\mathbf{m}} \right|.\end{eqnarray*}$$

Note that S is diagonal in the computational basis since S is an expectation of ${{U}^{\otimes t}}\otimes {{\left( {{U}^{\dagger }} \right)}^{\otimes t}}$ for $U\in {{\mathcal{U}}_{{\rm{diag}}}}$, and $U\in {{\mathcal{U}}_{{\rm{diag}}}}$ is diagonal in the computational basis. A simple calculation leads to

$$\begin{eqnarray}{{S}_{{\mathbf{nm}}}}=\left\{ \begin{array}{ccccccccccccccc} 1 & {\rm{when}}\;\pi \left( {\mathbf{n}} \right)=\pi \left( {\mathbf{m}} \right), \\ 0 & {\rm{when}}\;\pi \left( {\mathbf{n}} \right)\ne \pi \left( {\mathbf{m}} \right). \\ \end{array} \right.\end{eqnarray} \tag{ 4 }$$

Our goal is to derive the value of r for which the r-qubit phase-random circuit achieves the coefficients ${{S}_{{\mathbf{nm}}}}$.

We denote by ${{I}_{s}}$ a subset of $\left\{ 1,\cdots ,N \right\}$ with s elements. For a given ${{I}_{s}}=\left\{ {{i}_{1}},\cdots ,{{i}_{s}} \right\}$, we denote s-bit subsequences in ${{\vec{n}}^{\left( k \right)}}$ and ${{\vec{m}}^{\left( k \right)}}$ at ${{I}_{s}}$ by $\vec{n}_{{{I}_{s}}}^{\left( k \right)}:=n_{{{i}_{1}}}^{\left( k \right)}\cdots n_{{{i}_{s}}}^{\left( k \right)}$ and $\vec{m}_{{{I}_{s}}}^{\left( k \right)}:=m_{{{i}_{1}}}^{\left( k \right)}\cdots m_{{{i}_{s}}}^{\left( k \right)}$, respectively, and $\left( \vec{n}_{{{I}_{s}}}^{\left( 1 \right)},\cdots ,\vec{n}_{{{I}_{s}}}^{\left( t \right)} \right)$ and $\left( \vec{m}_{{{I}_{s}}}^{\left( 1 \right)},\cdots ,\vec{m}_{{{I}_{s}}}^{\left( t \right)} \right)$ by ${{{\mathbf{n}}}_{{{I}_{s}}}}$ and ${{{\mathbf{m}}}_{{{I}_{s}}}}$, respectively. We generalize a canonical map π to the one mapping t s-bit sequences ${{\mathcal{D}}_{s}}$ to a quotient set ${{\mathcal{D}}_{s}}/{{\mathcal{P}}_{t}}$ by the permutation group ${{\mathcal{P}}_{t}}$. In this notation, ${{\mathcal{D}}_{N}}=\mathcal{D}$. We call the number of 1 in a bit sequence the weight of the sequence. By using these expressions, we first prove the following lemma.

Lemma 1. Let ${\mathbf{n}},{\mathbf{m}}\in {{\mathcal{D}}_{s}}$ be such that $\pi \left( {\mathbf{n}} \right)\ne \pi \left( {\mathbf{m}} \right)$ and $\pi \left( {{{\mathbf{n}}}_{{{I}_{s-1}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{s-1}}}} \right)$ for any ${{I}_{s-1}}\subset \left\{ 1,\cdots ,s \right\}$. Denote by ${{G}_{{\mathbf{n}}}}\left( {\vec{n}} \right)$ the number of $\vec{n}$ in ${\mathbf{n}}$. Then, for any s-bit sequence $\vec{n}$,

$$\begin{eqnarray}&&|{{G}_{{\mathbf{n}}}}\left( {\vec{n}} \right)-{{G}_{{\mathbf{m}}}}\left( {\vec{n}} \right)|=g,\end{eqnarray} \tag{ 5 }$$

where g is a constant positive integer. Moreover, $t\geqslant {{2}^{s-1}}g$.

Proof 2. If $\exists q,q\prime \in \left\{ 1,\cdots ,t \right\}$ such that ${{\vec{n}}^{\left( q \right)}}={{\vec{m}}^{\left( {{q}^{\prime }} \right)}}$, we remove them from ${\mathbf{n}}$ and ${\mathbf{m}}$. As a result, we obtain $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$ and $\mathop{{\mathbf{m}}}\limits^{\tilde{}}$, which are composed of $t\prime$ s-bit sequences for $t\prime \leqslant t$. Note that $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$ and $\mathop{{\mathbf{m}}}\limits^{\tilde{}}$ still satisfy $\pi \left( {{\mathop{{\mathbf{n}}}\limits^{\tilde{}}}_{{{I}_{s-1}}}} \right)=\pi \left( {{\mathop{{\mathbf{m}}}\limits^{\tilde{}}}_{{{I}_{s-1}}}} \right)$ for any ${{I}_{s-1}}\subset \left\{ 1,\cdots ,s \right\}$.

Without loss of generality, we assume that the s-bit sequence with the most occurrences in $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$ is $00\cdots 0$, and let g be the number of occurrences. Since $\pi \left( {{\mathop{{\mathbf{n}}}\limits^{\tilde{}}}_{{{I}_{s-1}}}} \right)=\pi \left( {{\mathop{{\mathbf{m}}}\limits^{\tilde{}}}_{{{I}_{s-1}}}} \right)$ for any ${{I}_{s-1}}\subset \left\{ 1,\cdots ,s \right\}$ and all s-bit sequences in $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$ differ from those in $\mathop{{\mathbf{m}}}\limits^{\tilde{}}$, all sequences with weight one should be contained in $\mathop{{\mathbf{m}}}\limits^{\tilde{}}$. Moreover, the number of each sequence with weight one in $\mathop{{\mathbf{m}}}\limits^{\tilde{}}$ is g since the number of $00\cdots 0$ in $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$ is g. This in turn implies that all sequences with weight two should be contained in $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$. Similarly, the number of each of such sequences should be g. By repeating this, it follows that all sequences with zero or even weight are contained in $\mathop{{\mathbf{n}}}\limits^{\tilde{}}$, and those with odd weight are in $\mathop{{\mathbf{m}}}\limits^{\tilde{}}$. In addition, the number of each sequence is g. Thus, we obtain for any s-bit sequence $\vec{n}$,

$$\begin{eqnarray*}&&|{{G}_{\mathop{{\mathbf{n}}}\limits^{\tilde{}}}}(\vec{n})-{{G}_{\mathop{{\mathbf{m}}}\limits^{\tilde{}}}}(\vec{n})|=g,\end{eqnarray*}$$

and $t\prime ={{2}^{s-1}}g$. By construction, $|{{G}_{\mathop{{\mathbf{n}}}\limits^{\tilde{}}}}\left( {\vec{n}} \right)-{{G}_{\mathop{{\mathbf{m}}}\limits^{\tilde{}}}}\left( {\vec{n}} \right)|=|{{G}_{{\mathbf{n}}}}\left( {\vec{n}} \right)-{{G}_{{\mathbf{m}}}}\left( {\vec{n}} \right)|$, so that we obtain equation (5). Moreover, as $t\geqslant t\prime$, it follows that $t\geqslant {{2}^{s-1}}g$.□

By using lemma 1, we show the following proposition.

Proposition 2. For $r\leqslant N-1$, the following are equivalent;

• (A)  
  An r-qubit phase-random circuit achieves an exact diagonal-unitary t-design,
• (B)  
  For any ${\mathbf{n}}$, ${\mathbf{m}}\in \mathcal{D}$, if $\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right)$ for any ${{I}_{r}}\subset \left\{ 1,\cdots ,N \right\}$, then $\pi \left( {\mathbf{n}} \right)=\pi \left( {\mathbf{m}} \right)$,
• (C)  
  $r>{{{\rm log} }_{2}}t$.

From the equivalence of (A) and (C), we obtain theorem 1.

Proof 3. We first show the equivalence of (A) and (B), and then that of (B) and (C). The unitary matrix corresponding to an r-qubit phase-random circuit is given by

$$\begin{eqnarray*}&&{{W}_{\phi }}=\mathop{\mathop{\prod }\nolimits }\limits_{{{I}_{r}}\subset \{1,\cdots ,N\}}{{W}_{{{I}_{r}}}},\end{eqnarray*}$$

where ${{W}_{{{I}_{r}}}}:={\rm{dia}}{{{\rm{g}}}_{{{I}_{r}}}}\left( {{e}^{i{{\phi }_{1}}}},\cdots ,{{e}^{i{{\phi }_{{{2}^{r}}}}}} \right)\otimes {{\mathcal{I}}_{\left\{ 1,\cdots ,N \right\}{{I}_{r}}}}$ is a diagonal unitary matrix with random phases $\left\{ {{\phi }_{1}},\cdots ,{{\phi }_{{{2}^{r}}}} \right\}$ acting non-trivially on the qubits at sites ${{I}_{r}}$. The matrix ${{\mathcal{I}}_{I}}$ represents the identity matrix acting on qubits at sites $I\subset \left\{ 1,\cdots ,N \right\}$. Since the random phases are independently chosen for each ${{W}_{{{I}_{r}}}}$, the expectation of $W_{\phi }^{\otimes t}\otimes {{\left( W_{\phi }^{\dagger } \right)}^{\otimes t}}$ over all random phases is given by

$$\begin{eqnarray*}\begin{array}{rcl} \mathbb{E}[W_{\phi }^{\otimes t}\otimes {{(W_{\phi }^{\dagger })}^{\otimes t}}] & = & \mathop{\mathop{\prod }\nolimits }\limits_{{{I}_{r}}\subset \{1,\cdots ,N\}}\mathbb{E}[W_{{{I}_{r}}}^{\otimes t}\otimes {{(W_{{{I}_{r}}}^{\dagger })}^{\otimes t}}] \\ {} & = & \mathop{\mathop{\prod }\nolimits }\limits_{{{I}_{r}}\subset \{1,\cdots ,N\}}\mathop{\mathop{\sum }\nolimits }\limits_{{\mathbf{n}},{\mathbf{m}}\in \mathcal{D}}W_{{{I}_{r}}}^{{\mathbf{nm}}}\left| {\mathbf{n}} \right\rangle \left\langle {\mathbf{n}} \right|\otimes \left| {\mathbf{m}} \right\rangle \left\langle {\mathbf{m}} \right|, \\ \end{array}\end{eqnarray*}$$

where

$$\begin{eqnarray}W_{{{I}_{r}}}^{{\mathbf{nm}}}=\left\{ \begin{array}{ccccccccccccccc} 1 & {\rm{when}}\;\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right), \\ 0 & {\rm{when}}\;\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)\ne \pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right). \\ \end{array} \right.\end{eqnarray} \tag{ 6 }$$

By comparing equation (6) with equation (4), we obtain the equivalence of (A) and (B).

Next, we show that (C) implies (B) by showing its contraposition, namely, if there exists ${\mathbf{n}}$, ${\mathbf{m}}\in \mathcal{D}$ that satisfies $\pi \left( {\mathbf{n}} \right)\ne \pi \left( {\mathbf{m}} \right)$ but $\forall {{I}_{r}}\subset \left\{ 1,\cdots ,N \right\}$, $\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right)$, then $t\geqslant {{2}^{r}}$. By assumption, there exists $r\prime \geqslant r$ and ${{I}_{r\prime +1}}$ such that $\pi \left( {{{\mathbf{n}}}_{{{I}_{r\prime +1}}}} \right)\ne \pi \left( {{{\mathbf{m}}}_{{{I}_{r\prime +1}}}} \right)$ and $\pi \left( {{{\mathbf{n}}}_{{{I}_{r\prime }}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{r\prime }}}} \right)$ for any ${{I}_{r\prime }}\subset {{I}_{r\prime +1}}$. It follows from lemma 1 that $t\geqslant {{2}^{r\prime }}g$, where $g\geqslant 1$. As $r\prime \geqslant r$, we obtain $t\geqslant {{2}^{r}}$.

Finally, we show that (B) implies (C). This is also obtained by showing its contraposition. Consider ${\mathbf{n}}$ and ${\mathbf{m}}\in \mathcal{D}$ such that, for a fixed ${{I}_{r+1}}$, ${{{\mathbf{n}}}_{{{I}_{r+1}}}}$ and ${{{\mathbf{m}}}_{{{I}_{r+1}}}}$ contain all $\left( r+1 \right)$-bit sequences with even or zero weight and those with odd weight, respectively, and ${{{\mathbf{n}}}_{\left\{ 1,\cdots ,N \right\}{{I}_{r+1}}}}={{{\mathbf{m}}}_{\left\{ 1,\cdots ,N \right\}{{I}_{r+1}}}}$. Such ${\mathbf{n}}$ and ${\mathbf{m}}$ exist if $t\geqslant {{2}^{r}}$. It is obvious that $\pi \left( {\mathbf{n}} \right)\ne \pi \left( {\mathbf{m}} \right)$. However, it is easy to see that $\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right)$ for any ${{I}_{r}}\subset \left\{ 1,\cdots ,N \right\}$. This shows the contraposition of the statement that (B) implies (C), and concludes the proof.□

We show how to decompose each r-qubit diagonal gate in an r-qubit phase-random circuit into a sequence of multi-qubit controlled-phase-type gates with discrete random phases. More precisely, we prove the following.

Proposition 3. To implement a diagonal-unitary t-design by an r-qubit phase-random circuit, every gate ${\rm{dia}}{{{\rm{g}}}_{{{I}_{r}}}}\left( 1,{{e}^{i{{\phi }_{1}}}},\cdots ,{{e}^{i{{\phi }_{{{2}^{r}}-1}}}} \right)\otimes {{\mathcal{I}}_{\left\{ 1,\cdots ,N \right\}{{I}_{r}}}}$ non-trivially acting on r qubits at ${{I}_{r}}$ with random phases ${{\phi }_{k}}\in \left[ 0,2\pi \right)$ for $k=1,\cdots ,{{2}^{r}}-1$ can be replaced by

$$\begin{eqnarray*}&&{{D}_{{{I}_{r}}}}=\mathop{\mathop{\prod }\nolimits }\limits_{{{I}_{s}}\subset {{I}_{r}},1\leqslant s\leqslant r}{\rm{dia}}{{{\rm{g}}}_{{{I}_{s}}}}(1,1,\cdots ,1,{{e}^{i{{\alpha }_{{{I}_{s}}}}}})\otimes {{\mathcal{I}}_{\{1,\cdots ,N\}{{I}_{s}}}},\end{eqnarray*}$$

where ${{\alpha }_{{{I}_{s}}}}$ are randomly and independently chosen from

$$\begin{eqnarray}&&{{\left\{ \frac{2\pi k}{t/{{2}^{s-1}}+1} \right\}}_{k=0,1,\cdots ,t/{{2}^{s-1}}}}.\end{eqnarray} \tag{ 7 }$$

Proof 4. We consider an r-qubit diagonal gate acting on qubits at ${{I}_{r}}$. Since it is sufficient to consider a nontrivial part of the matrix, we investigate a diagonal matrix given by

$$\begin{eqnarray*}&&{{W}_{{{I}_{r}}}}=\mathop{\mathop{\sum }\nolimits }\limits_{{{{\vec{n}}}_{{{I}_{r}}}}}{{e}^{i{{\phi }_{{{{\vec{n}}}_{{{I}_{r}}}}}}}}\left| {{{\vec{n}}}_{{{I}_{r}}}} \right\rangle \left\langle {{{\vec{n}}}_{{{I}_{r}}}} \right|.\end{eqnarray*}$$

Its tensor product $W_{{{I}_{r}}}^{\otimes t}\otimes {{\left( W_{{{I}_{r}}}^{\dagger } \right)}^{\otimes t}}$ is given by

$$\begin{eqnarray*}&&W_{{{I}_{r}}}^{\otimes t}\otimes {{(W_{{{I}_{r}}}^{\dagger })}^{\otimes t}}=\mathop{\mathop{\sum }\nolimits }\limits_{{{{\mathbf{n}}}_{{{I}_{r}}}},{{{\mathbf{m}}}_{{{I}_{r}}}}\in {{\mathcal{D}}_{r}}}{{e}^{i{{\phi }_{{{{\mathbf{n}}}_{{{I}_{r}}}}{{{\mathbf{m}}}_{{{I}_{r}}}}}}}}\left| {{{\mathbf{n}}}_{{{I}_{r}}}} \right\rangle \left\langle {{{\mathbf{n}}}_{{{I}_{r}}}} \right|\otimes \left| {{{\mathbf{m}}}_{{{I}_{r}}}} \right\rangle \left\langle {{{\mathbf{m}}}_{{{I}_{r}}}} \right|,\end{eqnarray*}$$

where ${{\phi }_{{{{\mathbf{n}}}_{{{I}_{r}}}}{{{\mathbf{m}}}_{{{I}_{r}}}}}}=\mathop{\sum }\nolimits _{k=1}^{t}\left( {{\phi }_{\vec{n}_{{{I}_{r}}}^{\left( k \right)}}}-{{\phi }_{\vec{m}_{{{I}_{r}}}^{\left( k \right)}}} \right)$. If every phase is randomly chosen from $\left[ 0,2\pi \right)$,

$$\begin{eqnarray}\mathbb{E}\left[ {{e}^{i{{\phi }_{{{{\mathbf{n}}}_{{{I}_{r}}}}{{{\mathbf{m}}}_{{{I}_{r}}}}}}}} \right]=\left\{ \begin{array}{ccccccccccccccc} 1 & {\rm{when}}\;\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right), \\ 0 & {\rm{when}}\;\pi \left( {{{\mathbf{n}}}_{{{I}_{r}}}} \right)\ne \pi \left( {{{\mathbf{m}}}_{{{I}_{r}}}} \right). \\ \end{array} \right.\end{eqnarray} \tag{ 8 }$$

We investigate the coefficient of $\left| {{{\mathbf{n}}}_{{{I}_{r}}}} \right\rangle \left\langle {{{\mathbf{n}}}_{{{I}_{r}}}} \right|\otimes \left| {{{\mathbf{m}}}_{{{I}_{r}}}} \right\rangle \left\langle {{{\mathbf{m}}}_{{{I}_{r}}}} \right|$ in $D_{{{I}_{r}}}^{\otimes t}\otimes {{\left( D_{{{I}_{r}}}^{\dagger } \right)}^{\otimes t}}$. Our goal is to show that the choice of the phases defined in equation (7) achieves the same average as equation (8). For this purpose, it is sufficient to prove the following two properties: (a) For any ${{I}_{1}}\subset {{I}_{r}}$, the average over ${{\alpha }_{{{I}_{1}}}}$ makes the coefficient of $\left| {{{\mathbf{n}}}_{{{I}_{1}}}} \right\rangle \left\langle {{{\mathbf{n}}}_{{{I}_{1}}}} \right|\otimes \left| {{{\mathbf{m}}}_{{{I}_{1}}}} \right\rangle \left\langle {{{\mathbf{m}}}_{{{I}_{1}}}} \right|$ vanish if $\pi \left( {{{\mathbf{n}}}_{{{I}_{1}}}} \right)\ne \pi \left( {{{\mathbf{m}}}_{{{I}_{1}}}} \right)$. (b) For any ${{I}_{s}}\subset {{I}_{r}}$ with any $1<s\leqslant r$, the average over ${{\alpha }_{{{I}_{s}}}}$ makes the coefficients of $\left| {{{\mathbf{n}}}_{{{I}_{s}}}} \right\rangle \left\langle {{{\mathbf{n}}}_{{{I}_{s}}}} \right|\otimes \left| {{{\mathbf{m}}}_{{{I}_{s}}}} \right\rangle \left\langle {{{\mathbf{m}}}_{{{I}_{s}}}} \right|$ vanish if $\pi \left( {{{\mathbf{n}}}_{{{I}_{s}}}} \right)\ne \pi \left( {{{\mathbf{m}}}_{{{I}_{s}}}} \right)$ and $\pi \left( {{{\mathbf{n}}}_{{{I}_{s-1}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{s-1}}}} \right)$ for all ${{I}_{s-1}}\subset {{I}_{s}}$.

The property (a) holds since the coefficients of $\left| {{{\mathbf{n}}}_{{{I}_{1}}}} \right\rangle \left\langle {{{\mathbf{n}}}_{{{I}_{1}}}} \right|\otimes \left| {{{\mathbf{m}}}_{{{I}_{1}}}} \right\rangle \left\langle {{{\mathbf{m}}}_{{{I}_{1}}}} \right|$ in $D_{{{I}_{r}}}^{\otimes t}\otimes {{\left( D_{{{I}_{r}}}^{\dagger } \right)}^{\otimes t}}$ includes the factor ${{e}^{it\prime {{\alpha }_{{{I}_{1}}}}}}$ with $1\leqslant t\prime \leqslant t$, which vanishes after taking the average of ${{\alpha }_{{{I}_{1}}}}$ over ${{\left\{ \frac{2\pi k}{t+1} \right\}}_{k=0,1,\cdots ,t}}$. The property (b) is obtained from lemma 1. When ${{{\mathbf{n}}}_{{{I}_{s}}}}$ and ${{{\mathbf{m}}}_{{{I}_{s}}}}$ satisfy $\pi \left( {{{\mathbf{n}}}_{{{I}_{s}}}} \right)\ne \pi \left( {{{\mathbf{m}}}_{{{I}_{s}}}} \right)$ and $\pi \left( {{{\mathbf{n}}}_{{{I}_{s-1}}}} \right)=\pi \left( {{{\mathbf{m}}}_{{{I}_{s-1}}}} \right)$ for all ${{I}_{s-1}}\subset {{I}_{s}}$, it follows from lemma 1 that, for an s-bit sequence ${{\vec{1}}_{s}}:=11\cdots 1$,

$$\begin{eqnarray*}&&|{{G}_{{{{\mathbf{n}}}_{{{I}_{s}}}}}}({{\vec{1}}_{s}})-{{G}_{{{{\mathbf{m}}}_{{{I}_{s}}}}}}({{\vec{1}}_{s}})|=g.\end{eqnarray*}$$

This implies that the coefficient of $\left| {{{\mathbf{n}}}_{{{I}_{s}}}} \right\rangle \left\langle {{{\mathbf{n}}}_{{{I}_{s}}}} \right|\otimes \left| {{{\mathbf{m}}}_{{{I}_{s}}}} \right\rangle \left\langle {{{\mathbf{m}}}_{{{I}_{s}}}} \right|$ for such ${{{\mathbf{n}}}_{{{I}_{s}}}}$ and ${{{\mathbf{m}}}_{{{I}_{s}}}}$ contains ${{e}^{ig{{\alpha }_{{{I}_{s}}}}}}$ or ${{e}^{-ig{{\alpha }_{{{I}_{s}}}}}}$. We also obtain from lemma 1 that $t\geqslant {{2}^{s-1}}g$, namely, $g\leqslant t/{{2}^{s-1}}$, where the equality holds for ${{{\mathbf{n}}}_{{{I}_{s}}}}$ and ${{{\mathbf{m}}}_{{{I}_{s}}}}$ that do not share the same s-bit sequences. Thus, for these terms to be zero by taking the average over ${{\alpha }_{{{I}_{s}}}}$, it is sufficient to take ${{\alpha }_{{{I}_{s}}}}$ randomly and independently from

$$\begin{eqnarray*}&&{{\left\{ \frac{2\pi k}{t/{{2}^{s-1}}+1} \right\}}_{k=0,\cdots ,t/{{2}^{s-1}}}}.\end{eqnarray*}$$

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