Let $H$ be a finite dimensional hilbert space. Let $L:H\otimes H\rightarrow H\otimes H$ be a unitary transformation. Then the equation $$(L\otimes I)(I\otimes L)(L\otimes I)=(I\otimes L)(L\otimes I)(I\otimes L)$$ where $I:H\rightarrow H$ is the identity mapping is known as the Yang-Baxter equation.

We shall call a linear transformation $L:H\otimes H\rightarrow H\otimes H$ a universal quantum gate if there is some $n$ such that the unitary transformations of the form $I^{\otimes i}\otimes L\otimes I^{\otimes (n-i-2)}$ along with the mappings $\lambda\cdot I^{\otimes n}$ where $|\lambda|=1$ generate a dense subgroup of the unitary group $U(2^{n})$.

Does there exist a universal quantum gate $L:H\otimes H\rightarrow H\otimes H$ which satisfies the Yang-Baxter equation? If so, then what is the least natural number $n$ so that there is a universal quantum gate $L:\mathbb{C}^{n}\otimes\mathbb{C}^{n}\rightarrow \mathbb{C}^{n}\otimes\mathbb{C}^{n}$ which satisfies the Yang-Baxter equation?


If $L$ is a quantum gate which satisfies the Yang-Baxter equation, then for all $n$, the function $\phi:B_{n}\rightarrow H^{\oplus n}$ defined by $\phi(\sigma_{i})=I^{\oplus(i-1)}\oplus L\oplus I^{\oplus(n-i-1)}$ is a group homomorphism where $B_{n}$ denotes the $n$-strand braid group.

If $L$ is also universal, then any quantum algorithm (called algorithm A) can be emulated by a unitary transformation $\phi(w)$ for some braid word $w$. However, if $\sigma_{i_{1}}^{e_{1}}...\sigma_{i_{n}}^{e_{n}}$ is one of the normal forms for the braid for $w$, then the sequence of quantum gates $(\phi(\sigma_{i_{1}})^{e_{1}},...,\phi(\sigma_{i_{n}})^{e_{n}})$ obfuscates the original quantum algorithm so that a person cannot figure out how algorithm A works simply by looking at the braid $\sigma_{i_{1}}^{e_{1}}...\sigma_{i_{n}}^{e_{n}}$.

The idea of using braid group representations to obfuscate quantum algorithms has been proposed in this recent paper where they have observed that the Fibonacci representation of braid groups could be used to obfuscate quantum algorithms. However, a universal unitary transformation that satisfies the Yang-Baxter transformation seems to be a more direct way of using braids to obfuscate quantum algorithms.

  • $\begingroup$ Carlo Beenakker's references show that there is most likely no universal Yang-Baxter unitary operator in the qubit case when $H$ is a 2-dimensional complex Hilbert space. It is unknown whether a universal Yang-Baxter unitary operator exists on a higher dimensional Hilbert space. $\endgroup$ Jun 30, 2016 at 8:01

1 Answer 1


Braiding Operators are Universal Quantum Gates, by Louis Kauffman and Samuel Lomonaco (2004)

In this paper, we prove that certain solutions to the Yang-Baxter equation together with local unitary two-dimensional operators form a universal set of quantum gates. In particular, we show that a single specific solution of the Yang-Baxter Equation, the Bell basis change matrix, is a universal gate.

If we wish to do away with the local unitaries, and only use unitary solutions of the Yang-Baxter equation, then universal quantum computation is not known to be possible, because a classical simulation is available, see Classical simulation of Yang-Baxter gates (2004).

  • $\begingroup$ In that paper, they show that the local unitary operators (all one qubit operators) along with the Yang-Baxter unitary operator give you universal quantum computation (the one qubit maps get in the way of obfuscating quantum algorithms). I am wondering if we can get universal quantum computation using the Yang-Baxter unitary operator alone. $\endgroup$ Jun 28, 2016 at 10:09
  • $\begingroup$ that is not known to be possible. $\endgroup$ Jun 28, 2016 at 11:03

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