5
$\begingroup$

In this paper Quantum Circuit Learning they say that the ability of a quantum circuit to approximate a function can be enhanced by terms like $x\sqrt{1-x^2}$ ($x\in[-1,1])$. Given inputs $\{x,f(x)\}$, it aims to approximate an analytical function by a polynomial with higher terms up to the $N$th order. the steps are similar to the following:

  1. Encoding $x$ by constructing a state $\frac{1}{2^N}\bigotimes_{i=1}^N \left[I+xX_i+\sqrt{1-x^2}Z_i\right]$

  2. Apply a parameterized unitary transformation $U(\theta)$.

  3. Minimize the cost function by tuning the parameters $\theta$ iteratively.

I am a little confused about how can terms like $x\sqrt{1-x^2}$ in the polynomial represented by the quantum state can enhance its ability to approximate the function. Maybe it's implemented by introducing nonlinear terms, but I can't find the exact mathematical representation.

Thanks for any help in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

The authors might have meant that cross-terms like $x \sqrt{1-x^2}$ can improve the ability of the quantum circuit to approximate $f$ beyond what could be done with just an $N$-th order polynomial.

Using the authors' notation we can write the input state as $\rho(x) = \sum_k a_k(x) P_k$ with $P_k \in \{I, X, Y, Z\}^N$. Specifically, $a_I(x) = 1$, $a_X(x) = x$, and $a_Z(x) = \sqrt{1-x^2}$, and we define $a_{PP'}(x) = a_P(x) a_{P'}(x)$. After applying a parameterized unitary $U(\theta)$ and measuring an arbitrary observable we'll have an output of the form $$ \hat{f}(x) = \sum_{k } b_{k}(\theta) a_k(x) \tag{1} $$

where $b_k(\theta)$ absorbs the components of the observable as well as the action of $U(\theta)$. For simplicity lets assume the coefficients $b_k$ can be arbitrary. An $N$-th order polynomial has the form

$$ g_N(x) = c_0 + c_1 x + c_2 x^2 + \cdots + c_N x^N \tag{2} $$

If we take the case where $N=2$, $g_2(x) = c_0 + c_1 x + c_2 x^2$ but Eq. (1) contains terms like $a_{XZ}(x) = x\sqrt{1-x^2}$ and $a_{IZ} = \sqrt{1-x^2}$. Writing $b_k := b_k(\theta)$ and rewriting subscripts $I$, $X$, $XZ$, as integers for brevity, the circuit therefore produces functions of the form

\begin{align} \hat{f}(x) &= b_0 + b_1 x + b_2 x^2 + b_3 \sqrt{1-x} + b_4 x\sqrt{1-x^2}\tag{3} \\&= b_0 + b_1 x + b_2x^2 + b_3 \left(1 - \frac{1}{2} x^2 - \frac{1}{8} x^4 +\cdots \right) + b_4 \left(x - \frac{1}{2} x^3 - \frac{1}{8} x^5 +\cdots \right) \tag{4} \\&= (b_0 + b_3) + (b_1 + b_4)x + \left(b_2 - \frac{b_3}{2}\right) x^2 -\frac{b_4}{2}x^3 - \frac{b_3}{8} x^4 + \cdots \tag{5} \end{align}

where I've used the Maclaurin series expansion for $\sqrt{1-x^2}$. So this function is strictly more expressive than Eq. (2) as we can recover an arbitrary second order polynomial $g_2(x)$ by choosing $\theta$ such that $b_3=b_4=0$ and $b_k = c_k$ otherwise. Conversely, keeping these additional terms might greatly improve the convergence rate (with respect to $N$) of $\hat{f}$ for approximating certain functions compared to $g_N$ (e.g. when $f(x)$ contains terms like $\sqrt{1-x^2}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.