Quantum Secret Sharing using GHZ States: Difference between revisions

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Quantum secret sharing (QSS) is a quantum protocol that grants unconditional security for communication. The scheme allows a secret holder Alice can split her secret into <math>n</math> shares and send them to <math>n</math> others, when and only when <math>k (\frac{n}{2}< k \le n)</math> or more shares work together can recover the secret. In formal, we call the scheme ((k,n)) secret sharing.


== Quantum Secret Sharing using GHZ States ==
== Simplest Case ((2,2)) ==
The simplest case is ((2,2)). In this case, the secret holder Alice sends two receivers Bob and Charlie, to a qubit. When and only when Bob and Charlie work together, they can recover the secret message from Alice. Neither Bob nor Charlie can extract Alice’s secret on their own. The scheme has a  [https://github.com/Yichi-Lionel-Cheung/QuantumSecretSharingDEMO qiskit implementation].
 
'''Step1'''
 
Alice initiates the protocol by sharing with each of Bob and Charlie one particle from a GHZ triplet in the (standard) Z-basis: <math>|\mathrm{\Psi}\rangle_{\rm GHZ} = \frac{|000\rangle + |111\rangle}{\sqrt{2}}</math> through a quantum channel. Alice keeps the first qubit herself, Bob holds the second one and Charlie holds the third one. We denote the three qubits as <math>a,b,c</math>.
 
'''Step2'''
 
Alice, Bob, and Charlie measure their qubit on an X- or Y-basis chosen at random and share the basis via a public classical channel. Note they just share the information of the measurement basis, but don’t share the measurement results.
 
Revisiting the GHZ state can be written in the following 4 forms:
* <math>|\mathrm{\Psi}\rangle_{\rm GHZ} = \frac{1}{2\sqrt{2}}[({|+x\rangle_{a}|+x\rangle_{b} + |-x\rangle_{a}|-x\rangle_{b}})(|0\rangle_{c} + |1\rangle_{c})  +  ({|+x\rangle_{a}|-x\rangle_{b} + |-x\rangle_{a}|+x\rangle_{b}})(|0\rangle_{c} - |1\rangle_{c})]</math>
* <math>|\mathrm{\Psi}\rangle_{\rm GHZ} = \frac{1}{2\sqrt{2}}[({|+y\rangle_{a}|+x\rangle_{b} + |-y\rangle_{a}|-x\rangle_{b}})(|0\rangle_{c} - i|1\rangle_{c})  +  ({|+y\rangle_{a}|-x\rangle_{b} + |-y\rangle_{a}|+x\rangle_{b}})(|0\rangle_{c} + i|1\rangle_{c})]</math>
* <math>|\mathrm{\Psi}\rangle_{\rm GHZ} = \frac{1}{2\sqrt{2}}[({|+x\rangle_{a}|+y\rangle_{b} + |-x\rangle_{a}|-y\rangle_{b}})(|0\rangle_{c} - i|1\rangle_{c})  +  ({|+x\rangle_{a}|-y\rangle_{b} + |-x\rangle_{a}|+y\rangle_{b}})(|0\rangle_{c} + i|1\rangle_{c})]</math>
* <math>|\mathrm{\Psi}\rangle_{\rm GHZ} = \frac{1}{2\sqrt{2}}[({|+y\rangle_{a}|+y\rangle_{b} + |-y\rangle_{a}|-y\rangle_{b}})(|0\rangle_{c} - |1\rangle_{c})  +  ({|+y\rangle_{a}|-y\rangle_{b} + |-y\rangle_{a}|+y\rangle_{b}})(|0\rangle_{c} + |1\rangle_{c})]</math>
 
In the first expansion, if Charlie chooses the X-basis (with 50% probability) to perform the measurement, he will know whether Alice and Bob have correlated results. But Charlie does not know Alice’s actual result because he does not know Bob’s result. Also, Bob does not know Alice’s actual result because he does not know whether his result is correlated or anticorrelated to Alice’s result.
Also, if Charlie chooses Y-basis, he will get no information. Since Charlie has 50% probability to get $|+y\rangle$ and 50% probability to get $|-y\rangle$. Therefore they will cancel this turn and repeat.
The [https://arxiv.org/pdf/quant-ph/9806063.pdf below table] shows the relationship of Alice's and Bob's measurements on Charlie's state for the standard GHZ triplet:
{| class="wikitable"
|-
!
! colspan="5" | Alice
|-
|
|
| <math>+x</math>
| <math>-x</math>
| <math>+y</math>
| <math>-y</math>
|-
| rowspan="4" | Bob
| <math>+x</math>
| <math>|0\rangle + |1\rangle</math>
| <math>|0\rangle - |1\rangle</math>
| <math>|0\rangle - i |1\rangle</math>
| <math>|0\rangle + i |1\rangle</math>
|-
| <math>-x</math>
| <math>|0\rangle - |1\rangle</math>
| <math>|0\rangle + |1\rangle</math>
| <math>|0\rangle + i |1\rangle</math>
| <math>|0\rangle - i |1\rangle</math>
|-
| <math>+y</math>
| <math>|0\rangle - i |1\rangle</math>
| <math>|0\rangle + i |1\rangle</math>
| <math>|0\rangle - |1\rangle</math>
| <math>|0\rangle + |1\rangle</math>
|-
| <math>-y</math>
| <math>|0\rangle + i |1\rangle</math>
| <math>|0\rangle - i |1\rangle</math>
| <math>|0\rangle + |1\rangle</math>
| <math>|0\rangle - |1\rangle</math>
|}
 
'''Step3'''
 
Charlie works with Bob by telling Bob his measurement result if he chooses the correct basis.
 
* Charlie provides the information on whether Alice and Bob have correlated results
* Bob provides his measurement result
 
 
<div style='text-align: right;'>''*contributed by Yichi Lionel Zhang''</div>

Latest revision as of 12:39, 27 September 2024

Quantum secret sharing (QSS) is a quantum protocol that grants unconditional security for communication. The scheme allows a secret holder Alice can split her secret into shares and send them to others, when and only when or more shares work together can recover the secret. In formal, we call the scheme ((k,n)) secret sharing.

Simplest Case ((2,2))[edit]

The simplest case is ((2,2)). In this case, the secret holder Alice sends two receivers Bob and Charlie, to a qubit. When and only when Bob and Charlie work together, they can recover the secret message from Alice. Neither Bob nor Charlie can extract Alice’s secret on their own. The scheme has a qiskit implementation.

Step1

Alice initiates the protocol by sharing with each of Bob and Charlie one particle from a GHZ triplet in the (standard) Z-basis: through a quantum channel. Alice keeps the first qubit herself, Bob holds the second one and Charlie holds the third one. We denote the three qubits as .

Step2

Alice, Bob, and Charlie measure their qubit on an X- or Y-basis chosen at random and share the basis via a public classical channel. Note they just share the information of the measurement basis, but don’t share the measurement results.

Revisiting the GHZ state can be written in the following 4 forms:

In the first expansion, if Charlie chooses the X-basis (with 50% probability) to perform the measurement, he will know whether Alice and Bob have correlated results. But Charlie does not know Alice’s actual result because he does not know Bob’s result. Also, Bob does not know Alice’s actual result because he does not know whether his result is correlated or anticorrelated to Alice’s result. Also, if Charlie chooses Y-basis, he will get no information. Since Charlie has 50% probability to get $|+y\rangle$ and 50% probability to get $|-y\rangle$. Therefore they will cancel this turn and repeat. The below table shows the relationship of Alice's and Bob's measurements on Charlie's state for the standard GHZ triplet:

Alice
Bob

Step3

Charlie works with Bob by telling Bob his measurement result if he chooses the correct basis.

  • Charlie provides the information on whether Alice and Bob have correlated results
  • Bob provides his measurement result


*contributed by Yichi Lionel Zhang