Aharonov–Jones–Landau algorithm

The first idea behind the algorithm is to find a more tractable description for the operation of evaluating the Jones polynomial. This is done by means of the Markov trace.

The "Markov trace" is a trace operator defined on the Temperley–Lieb algebra ${\displaystyle TL_{n}(d)}$ as follows: given a ${\displaystyle T\in TL_{n}(d)}$ which is a single Kauffman diagram, let ${\displaystyle tr(T)=d^{a-n}}$ where ${\displaystyle a}$ is the number of loops attained by identifying each point in the bottom of ${\displaystyle T}$'s Kauffman diagram with the corresponding point on top. This extends linearly to all of ${\displaystyle TL_{n}(d)}$.

The Markov trace is a trace operator in the sense that ${\displaystyle \operatorname {tr} (1)=1}$ and ${\displaystyle \operatorname {tr} (XY)=\operatorname {tr} (YX)}$ for any ${\displaystyle X,Y\in TL_{n}(d)}$. It also has the additional property that if ${\displaystyle X}$ is a Kauffman diagram whose rightmost strand goes straight up then ${\displaystyle \operatorname {tr} (XE_{n-1})={\frac {1}{d}}\operatorname {tr} (X)}$.

A useful fact exploited by the AJL algorithm is that the Markov trace is the unique trace operator on ${\displaystyle TL_{n}(d)}$ with that property.[3]

For a complex number ${\displaystyle A}$ we define the map ${\displaystyle \rho _{A}:B_{n}\to TL_{n}(d)}$ via ${\displaystyle \sigma _{i}\mapsto AE_{i}+A^{-1}I}$. It follows by direct calculation that if ${\displaystyle A}$ satisfies that ${\displaystyle d=-A^{2}-A^{-2}}$ then ${\displaystyle \rho _{A}}$ is a representation.

Given a braind ${\displaystyle B\in B_{n}}$ let ${\displaystyle B^{tr}}$ be the link attained by identifying the bottom of the diagram with its top like in the definition of a Markov trace, and let ${\displaystyle V_{B^{tr}}}$ be the result link's Jones polynomial. The following relation holds:

${\displaystyle V_{B^{tr}}(A^{-4})=(-A)^{3w(B^{tr})}d^{n-1}\operatorname {tr} (\rho _{A}(B))}$

where ${\displaystyle w}$ is the writhe. As the writhe can be easily calculated classically, this reduces the problem of approximating the Jones polynomial to that of approximating the Markov trace.

We wish to construct a complex representation ${\displaystyle \tau }$ of ${\displaystyle TL_{n}(d)}$ such that the representation ${\displaystyle \tau \circ \rho _{A}}$ of ${\displaystyle B_{n}}$ will be unitary. We also wish that our representation will have a straightforward encoding into qubits.

Let

${\displaystyle Q_{n,k}=\left\{q\in \left\{1,\ldots ,k-1\right\}^{n+1}\mid q(1)=1,\left|q(i)-q(i+1)\right|=1\right\}}$

and let ${\displaystyle V_{n,k}=\mathbb {C} [Q_{n,k}]}$ be the vector space which has ${\displaystyle Q_{n,k}}$ as an orthonormal basis.

We choose define a linear map ${\displaystyle \tau :TL_{n}(d)\to V_{n,k}}$ by defining it on the base of generators ${\displaystyle \{1,E_{1},\ldots ,E_{n-1}\}}$. To do so we need to define the matrix element ${\displaystyle \tau (E_{i})_{q,q'}}$ for any ${\displaystyle q,q'\in Q_{n,k}}$.

We say that ${\displaystyle q}$ and ${\displaystyle q'}$ are 'compatible' if ${\displaystyle q(j)=q'(j)}$ for any ${\displaystyle j\neq i+1}$ and ${\displaystyle q(i)=q(i+2)}$. Geometrically this means that if we put ${\displaystyle q}$ and ${\displaystyle q'}$ below and above the Kauffman diagram in the gaps between the strands then no connectivity component will touch two gaps which are labeled by different numbers.

If ${\displaystyle q}$ and ${\displaystyle q'}$ are incompatible set ${\displaystyle \tau (E_{i})_{q,q'}=0}$. Else, let ${\displaystyle l}$ be either ${\displaystyle q(i)}$ or ${\displaystyle q(i+2)}$ (at least one of these number must be defined, and if both are defined they must be equal) and set

${\displaystyle \tau \left(E_{i}\right)_{q,q'}={\begin{cases}{\frac {\lambda _{l+1}}{\lambda _{l}}}&q\left(i+1\right)=q'(i+1)>l\\{\frac {\sqrt {\lambda _{l-1}\lambda _{l+1}}}{\lambda _{l}}}&q\left(i+1\right)\neq q'(i+1)\\{\frac {\lambda _{l-1}}{\lambda _{l}}}&q\left(i+1\right)=q'(i+1)<l\end{cases}}}$

where ${\displaystyle \lambda _{l}=\sin(\pi l/k)}$. Finally set ${\displaystyle d=2\cos(\pi /k)}$.

This representation, known as the path model representation, induces a unitary representation of the braid group.[4][5] Moreover, it holds that ${\displaystyle d=-A^{2}-A^{-2}}$ for ${\displaystyle A=ie^{-i\pi /2k}}$.

This implies that if we could approximate the Markov trace in this representation this will allow us to approximate the Jones polynomial in ${\displaystyle A^{-4}=e^{2\pi i/k}}$.

In order to be able to act on elements of the path model representation by means of quantum circuits, we need to encode the elements of ${\displaystyle Q_{n,k}}$ into qubits in a way which allows us to easily describe the images of the generators ${\displaystyle E_{i}}$.

We represent each path as a sequence of moves, where ${\displaystyle 0}$ indicates a move to the right and ${\displaystyle 1}$ indicates a move to the left. For example, the path ${\displaystyle 1,2,3,2}$ will be represented by the state ${\displaystyle \left|001\right\rangle }$.

This encodes ${\displaystyle \mathbb {C} [Q_{n,k}]}$ as a subspace of the state space on ${\displaystyle k-1}$ qubits.

We define the operators ${\displaystyle \varphi _{i}}$ within this subspace we define

${\displaystyle \Phi _{i}\left|w\right\rangle ={\begin{cases}0&w_{i}=w_{i+1}\\{\frac {\lambda _{z_{i}-\left(-1\right)^{w_{i}}}}{\lambda _{z_{i}}}}\left|w\right\rangle +{\frac {\sqrt {\lambda _{z_{i}-1}\lambda _{z_{i}+1}}}{\lambda _{z_{i}}}}X_{i}X_{i+1}\left|w\right\rangle &w_{i}\neq w_{i+1}\end{cases}}}$

where ${\displaystyle X_{i}}$ is the Pauli matrix flipping the ${\displaystyle i}$th bit and ${\displaystyle z_{i}}$ is the position of the path represented by ${\displaystyle \left|w\right\rangle }$ after ${\displaystyle i}$ steps.

We arbitrarily extend ${\displaystyle \varphi _{i}}$ to be the identity on the rest of the space.

We note that mapping ${\displaystyle E_{i}\mapsto \varphi _{i}}$ retains all the properties of the path model representation. Specifically, it induces a unitary representation ${\displaystyle \varphi }$ of ${\displaystyle B_{n}}$. It is fairly straightforward to show that ${\displaystyle \varphi _{i}}$ can be implemented by ${\displaystyle {\text{poly}}\left(n,k\right)}$ gates, so it follows that ${\displaystyle \varphi (B)}$ can be implemented for any ${\displaystyle B\in B_{n}}$ using ${\displaystyle {\text{poly}}\left(m,n,k\right)}$ where ${\displaystyle m}$ is the number of crossings in ${\displaystyle B}$.

The benefit of this construction is that it gives us a way to represent the Markov trace in a way which can be easily approximated.

Let ${\displaystyle {\mathcal {H}}_{n,k}}$ be the subspace of paths we described in the previous clause, and let ${\displaystyle {\mathcal {H}}_{n,k,l}}$ be the subspace spanned by basis elements which represent walks which end on the ${\displaystyle l}$-th position.

Note that each of the operators ${\displaystyle \varphi _{i}}$ fix ${\displaystyle {\mathcal {H}}_{n,k,l}}$ setwise, and so this holds for any ${\displaystyle W\in {\text{Im}}\Phi \left(TL_{n}\left(d\right)\right)}$, hence the operator ${\displaystyle W|_{l}:=W|_{{\mathcal {H}}_{n,k,l}}}$ is well defined.

We define the following operator:

${\displaystyle \operatorname {Tr} _{n}\left(W\right)={\frac {1}{N}}\sum _{l=1}^{k-1}\lambda _{l}\operatorname {Tr} \left(W|_{l}\right)}$

where ${\displaystyle \operatorname {Tr} }$ is the usual matrix trace.

It turns out that this operator is a trace operator with the Markov property, so by the theorem stated above it has to be the Markov trace. This finishes the required reductions as it establishes that to approximate the Jones polynomial it suffices to approximate ${\displaystyle \operatorname {Tr} _{n}}$.

algorithm Approximate-Jones-Trace-Closure is
    input ${\displaystyle B\in B_{n}}$ with ${\displaystyle m}$ crossings
                An integer ${\displaystyle k}$
    output a number ${\displaystyle s}$ such that ${\displaystyle |s-V_{B^{tr}}(e^{2\pi /k})|<1/poly(n,k,m)}$ 
                 with all but exponentially small probability
    repeat for ${\displaystyle j=1}$ to ${\displaystyle poly(n,m,k)}$
        1. Pick a random ${\displaystyle l\in \{1,\ldots ,k-1\}}$ such that the probability
           to choose a particular ${\displaystyle l}$ is proportional to ${\displaystyle \lambda _{l}}$
        2. Randomly pick ${\displaystyle q\in Q_{n,k}}$ which ends in position ${\displaystyle l}$
        3. Using the Hadamard test create a random variable ${\displaystyle x_{j}}$ with
           ${\displaystyle \mathbb {E} \left[x_{j}\right]={\mathcal {Re}}\left\langle \alpha \mid Q\left(B\right)\mid \beta \right\rangle }$
    Do the same to create ${\displaystyle y_{j}}$ with ${\displaystyle \mathbb {E} \left[y_{j}\right]={\mathcal {Im}}\left\langle \alpha \mid Q\left(B\right)\mid \beta \right\rangle }$
    let ${\displaystyle r}$ be the average of ${\displaystyle x_{j}+iy_{j}}$
    return ${\displaystyle (-A)^{3w(B^{tr})}d^{n-1}r}$

Note that the parameter ${\displaystyle d}$ used in the algorithm depends on ${\displaystyle k}$.

The correctness of this algorithm is established by applying the Hoeffding bound to ${\displaystyle x_{j}}$ and ${\displaystyle y_{j}}$ separately.

1. D. Aharonov, V. Jones, Z. Landau - A Polynomial Quantum Algorithm for Approximating the Jones Polynomial