Two Player's Quantum Covert Lottery
日本語版: https://zenn.dev/yyu/articles/79c6c48226166aa0e875
Introduction
The word “covert lottery” is a security protocol introduced by CardBased Covert Lottery^{[1]}. In this protocol, there are two players and they input each secret request then:
 If these two requests are not the same, the protocol approve their requests
 Otherwise the requests are the same, the protocol decides either one player's request at random
For example, in Chess game we can use covert lottery to determine if the players take the first move or the second move. These players have their requests but since one's request would be known by the other player, the strategy maybe leak. Covert lottery protocol outputs the actual their requests as long as the requests are not the same, and otherwise it outputs a random result. Both players cannot know if the output reflects their requests or random so there is no information leak.
In the original paper, the protocol is implemented using cardbased cryptography^{[2]}.I read it and I think it also can be done using quantum technics. So I'll show that the quantum version of covert lottery protocol in this article.
Discussions and questions are very welcome! If you have something you want to ask, please let me know about it.
Two Players Covert Lottery Function
At first I explain about covert lottery in the original paper. There are two players: whose name are Alice and Bob. Alice input is
Note that NOT
operation for
Considering
In case of
Quantum Covert Lottery based on BB84
I show that the naive protocol using quantum computer at first.
Protocol

Alice generates 1 bit random
and chose her requestx \xleftarrow{\text{rnd}} \{0, 1\} a \in \{0, 1\} 
Alice decides 1 qubit
of the following four qubits by\ket{\psi_{a,x}} a, x \def\bra#1{\mathinner{\left\langle{#1}\right}} \def\ket#1{\mathinner{\left{#1}\right\rangle}} \left\{ \begin{array}{lcl} \ket{\psi_{0,0}} & \equiv & \ket{0} \\ \ket{\psi_{0,1}} & \equiv & \ket{1} \\ \ket{\psi_{1,0}} & \equiv & \ket{+} \\ \ket{\psi_{1,1}} & \equiv & \ket{} \end{array} \right.  where
\ket{\pm} \equiv \frac{1}{\sqrt{2}}\left(\ket{0} \pm \ket{1}\right)
 where

Alice sends
to Bob\ket{\psi_{a,x}} 
Bob receives
and chose his request\ket{\psi_{a,x}} b \in \{0, 1\} 
Bob select the ground state
, mesures\mathcal{B}_{\overline{b}} by it. He measures the qubit and the result store into\ket{\psi_{a,x}} y \mathcal{B}_{\overline{b}} \equiv \left\{\ket{\psi_{\overline{b},0}}, \ket{\psi_{\overline{b},1}}\right\}  Note that it requires
rather than\overline{b} b
 Note that it requires

Bob sends
to Alicey 
Alice receives
and she sendsy ^{[3]} as final result to Boba \oplus x \oplus y
Description
The following table shows that























The result
 Assuming that
, in (5) the masurementa = \overline{b}\,(\Rightarrow a \ne b) is the same as Alice's randomy x  On the other hand,
is a random value wherey soa \ne \overline{b} is a random value tooa \oplus x \oplus y
 On the other hand,
 If
thena = \overline{b} , thereforex = y ofx is canceled bya \oplus x \oplus y and the final result equals toy a
Security Assumption
The final result of this protocol is calculated by Alice. If Alice is not honest she can change the result into what she wants. So the protocol works well under assuming that at least Alice is semihonest.
Bob's request leakage
This naive protocol has a vulnerability that Alice can know the Bob's request in some cases. For example:
 Alice select
so she sendsa = 0, x = 0 to Bob\ket{0}  Bob choose
\overline{b} = 1\,(\Rightarrow b = 0)  In the case, he measures using
を\ket{0} then the output\left\{\ket{+}, \ket{}\right\} isy or0 under the same probability1  If
then there is no information leakagey = 0
 If
 But if
then Alice knows thaty = 1 because the output is\overline{b} = 1 in 100% when measuring usingy = 0 \left\{\ket{0}, \ket{1}\right\}
The protocol result is valid as
CZ Gate and 1 Qubit MeasurementBased Quantum Computation
In this section, I introduce
CZ Gate
 If the first qubit is
\ket{0} gate do nothingCZ  Otherwise
gate applyCZ gate to the second qubitZ
For example,
Note that: the
1 Qubit MeasurementBased Quantum Computation
Considering the measurement of
And
Finally using Hadamard gate
If the result of the measurement is
Definitions of Hadamard gate
Two Player's “Blind” Quantum Covert Lottery
This section is more complicated.
Protocol

Alice choose a random angle
\theta \xleftarrow{\text{rnd}} \left\{0, \frac{1\pi}{4}, \dots, \frac{7\pi}{4}\right\} 
Alice sends the following 1 qubit
to Bob\ket{+_\theta} \ket{+_\theta} \equiv \frac{1}{\sqrt{2}}\left(\ket{0} + e^{i\theta}\ket{1}\right) 
Bob apply
gate toCZ received from Alice and\ket{+_\theta} \ket{+} CZ\ket{+_\theta}_1\ket{+}_2 
can be simplifiedCZ\ket{+_\theta}_1\ket{+}_2 CZ\ket{+_\theta}_1\ket{+}_2 = \frac{1}{\sqrt{2}}\ket{0}_1\ket{+}_2 + \frac{e^{i\theta}}{\sqrt{2}}\ket{1}_1\ket{}_2


Alice choose her request
and a random valuea \in {0, 1} . And she sends the following anglex \xleftarrow{\text{rnd}} \{0, 1\} to Bob\delta \delta := \left\{ \begin{array}{ll} \theta + x\pi& \text{if}\; a = 0 \\ \theta + \frac{\pi}{2} + x\pi & \text{if}\; a = 1 \end{array} \right. 
Bob prepares ground state
using\left\{\ket{0} \pm e^{i\delta}\ket{1}\right\} from Alice. And he measures the first qubit, and the result is\delta y 
Bob choose his request
and decides a ground stateb \in \{0, 1\} depending on his request\mathcal{B}_{\overline{b}} b \left\{ \begin{array}{lcl} \mathcal{B}_{0} & \equiv & \left\{\ket{\pm}\right\} \\ \mathcal{B}_{1} & \equiv & \left\{\frac{1}{\sqrt{2}}\left(\ket{0} \pm i\ket{1}\right)\right\} \end{array} \right. 
Bob measures the second qubit using
and the result is\mathcal{B}_{\overline{b}} . Then he sendsz to Alicez 
Alice outputs
as the final result to Boba \oplus x \oplus z
Protocol Behavior
It's difficult to understand the procedure (5). The second qubit is as follows after Bob measured the first qubit using

In case
y = 0 \begin{align*} HR_Z(\delta)\ket{+_\theta} &= H\left(\ket{0} + e^{i\theta}\cdot e^{i\delta}\ket{1}\right) \\ &= H\left(\ket{0} + e^{i(\delta  \theta)}\ket{1}\right) \end{align*} 
In case
y = 1 XHR_Z(\delta)\ket{+_\theta} = XH\left(\ket{0} + e^{i(\delta  \theta)}\ket{1}\right)
This means that what the second qubit would be depends on
Second qubit  

If

Bob's ground state is
\{\ket{\pm}\} \ket{\pm}\bra{\pm}\left(\frac{1}{\sqrt{2}}(i\ket{0} + \ket{1})\right) = \left(\frac{1}{2}i \pm \frac{1}{2}\right)\ket{\pm}  The probability that the result would be
is\ket{\pm} \left\frac{1}{2}i \pm \frac{1}{2}\right^2 = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\pm 1}{2}\right)^2}^2 = \frac{1}{2}
 The probability that the result would be

Bob's ground state is
\{\frac{1}{\sqrt{2}}\left(\ket{0} \pm i\ket{1}\right)\} \left(\frac{1}{\sqrt{2}}\left(\ket{0} \pm i\ket{1}\right)\right)\left(\frac{1}{\sqrt{2}}\left(\bra{0} \pm i\bra{1}\right)\right)\left(\frac{1}{\sqrt{2}}(i\ket{0} + \ket{1})\right)\\ \,\\ = \left\{ \begin{array}{ll} \left(\frac{1}{2}i + \frac{1}{2}i = i\right)\frac{1}{\sqrt{2}}\left(\ket{0} + i\ket{1}\right) & \text{if}\; \frac{1}{\sqrt{2}}\left(\ket{0} + i\ket{1}\right) \\ \\ \left(\frac{1}{2}i  \frac{1}{2}i = 0\right)\frac{1}{\sqrt{2}}\left(\ket{0}  i\ket{1}\right) & \text{if}\; \frac{1}{\sqrt{2}}\left(\ket{0}  i\ket{1}\right) \end{array} \right.  So the probability that the result would be
is\frac{1}{\sqrt{2}}\left(\ket{0} + i\ket{1}\right) , otherwisei^2 = 1 0^2 = 0
 So the probability that the result would be
In case Alice's request
In procedure (6) Bob choose
Finally the result of this protocol is
x
Random value In procedure (4), you maybe doubted whether a random value
Second qubit  

The second qubit is either one of the red parts of the following figure.
Similar to the naive protocol, if the result would be
Conclusion
I explained two quantum covert lottery protocols. In the original paper, the author extends twoparty protocol to manyparty. So I want to extend this quantum version protocol to manyparty.
References
 量子コンピュータで2人の“Covert”⁉️ガチャ (Japanese)
 CardBased Covert Lottery
 観測に基づく量子計算 (Japanese)
 量子計算理論 量子コンピュータの原理 (Japanese)
 量子コンピューティング: 基本アルゴリズムから量子機械学習まで (Japanese)
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