<p>What if instead of a probabilistic TM, the verifier in the definition of SZK was a quantum TM?</p>
<p>How would this affect its relation to other classes? Would Statistical Difference still be a complete problem for this class? More generally, how does changing the model of computation of a class change its relationship to other classes?</p>

CC BY-SA 4.0
2023-05-10T19:20:08.773Z
This paper by John Watrous might contain some answers: https://ieeexplore.ieee.org/abstract/document/1181970
lamontap
CC BY-SA 4.0
2023-05-11T02:27:51.760Z
Thanks for the reference -- I have in fact already read Watrous' paper. His definition of QSZK has very nice properties and seems to be very analogous to classical SZK. I'm wondering if my version of QSZK above (i.e. the verifier being a quantum TM) is a special case of Watrous' QSZK?
Irna Mosa
<cc.complexity-theory><quantum-computing><zero-knowledge>
Is SZK dependent on the verifier’s model of computation?
Irna Mosa
# Is SZK dependent on the verifier’s model of computation?

(Asked by: username_0 on 2023-05-06)

What if instead of a probabilistic TM, the verifier in the definition of SZK was a quantum TM?

How would this affect its relation to other classes? Would Statistical Difference still be a complete problem for this class? More generally, how does changing the model of computation of a class change its relationship to other classes?

**Tags:** cc.complexity-theory, quantum-computing, zero-knowledge
> **Comments:**
> **username_1**: This paper by John Watrous might contain some answers: https://ieeexplore.ieee.org/abstract/document/1181970  
> **username_0**: Thanks for the reference -- I have in fact already read Watrous' paper. His definition of QSZK has very nice properties and seems to be very analogous to classical SZK. I'm wondering if my version of QSZK above (i.e. the verifier being a quantum TM) is a special case of Watrous' QSZK?  

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## 0 answers

