Constructing Quantum Circuit

Constructing Quantum Circuit in Amplitude Estimation

Why This Quantum Circuit? Let's Start Over

You may wonder, in the Paper [1]., why we are using such a specific structure for our quantum circuit. Let’s step back and unpack the logic from the ground up.

Measuring the Amplitude

In quantum mechanics, consider a single-qubit state of the form $$ |\psi\rangle = \alpha |0\rangle + \beta |1\rangle $$ where $\alpha, \beta \in \mathbb{C}$ and the normalization condition holds: $$ |\alpha|^2 + |\beta|^2 = 1. $$ According to the Born rule, the amplitude of the state $|1\rangle$ is $\beta$, but what we can actually measure is the probability: $$ P(|1\rangle) = |\beta|^2. $$ Probabilities must be real and non-negative. So we compute the square of the magnitude: $$ P = |\text{amplitude}|^2 = (\text{Re}[\beta])^2 + (\text{Im}[\beta])^2. $$

---

Encoding $f(x)$ into a Quantum State

Now here’s the key idea. We want to encode a classical function $f(x)$ into a quantum circuit in such a way that we can extract information about it via measurement. Specifically, we want the measurement probability of a certain basis state (say $|1\rangle$) to be equal to $f(x)$.

So we aim to construct a quantum state such that: $$ |\beta|^{2} = \sin^2(p(x)) = f(x) $$ In this setup, $p(x)$ is some function that encodes $f(x)$ via the sine-squared map. Inverting this, we get: $$ p(x) = \sin^{-1}\left(\sqrt{f(x)}\right) $$ This is what we mean when we say "encoding" $f(x)$ — we are constructing a rotation angle $p(x)$ such that the probability of measuring $|1\rangle$ becomes exactly $f(x)$.

---

But Why Polynomial Approximation?

The challenge is: arbitrary functions $p(x)$ are generally hard to implement in quantum circuits directly. Quantum gates are discrete, hardware has finite resolution, and arbitrary-angle rotations are not feasible.

So instead of directly implementing: $$ p(x) = \sin^{-1}(\sqrt{f(x)}) $$ we approximate it with a low-degree polynomial — something we can construct in a circuit using basic gates.

That’s the whole idea: translate classical functions into quantum-measurable quantities through amplitude encoding, using inverse trigonometric maps and approximations that play nicely with quantum hardware.

---

Ok Polynomial Approximation, but why it looks ugly?

You certainly will have this type of question, why its must be this form $c$

The form $$ c(f(x)-\frac{1}{2})+\frac{1}{2} $$ is choosen carefully. We want to measure using $\text{sin}^{2}$, so we manipulate the sine function to behave linearly. Let's start with the identity of $\text{sin}^{2}(\theta)$: $$ \text{sin}^{2}(\theta) = \frac{1-\text{cos}(2\theta)}{2} $$ therefore, $$ \text{sin}^{2}(y+\frac{\pi}{4}) = \frac{1-\text{cos}(2y+\frac{\pi}{2})}{2} $$ and you must know that $\text{cos}(2y+\frac{\pi}{2}) = -\text{sin}(2y)$, therefore $$ \text{sin}^{2}(\theta) = \frac{1-\text{cos}(2\theta)}{2} = \frac{1+\text{sin}(2y)}{2}. $$ Now we expand $\text{sin}(2y)$ at $y=0$ using Taylor series expansion: $$ \text{sin}(2y) = 2y - \frac{(2y)^{3}}{3!} + \frac{(2y)^{5}}{5!} - \cdots = 2y - \frac{4y^{3}}{3} + \frac{8y^{5}}{15} - \cdots $$ therefore $$ \frac{1+\text{sin}(2y)}{2} = \frac{1}{2} + y - \frac{2y^{3}}{3} + \frac{4y^{5}}{15}. $$ now we only consider the linear term and omit the high order term, we have $$ \frac{1+\text{sin}(2y)}{2} = \frac{1}{2} + y $$ By replacing $y$ into our $f$, we approximate $$ \frac{1+\text{sin}(2y)}{2} = \frac{1}{2} + y = f(x) + \frac{1}{2} $$

Ok but why do we shift function

You might wonder — doesn't shifting by $\frac{1}{2}$ push things into a nonlinear region?

Actually, we do it on purpose.

Taylor expansion is a local approximation — it's most accurate around the point it's expanded at.
Since we expand $\sin^2(y + \frac{\pi}{4})$ around $y = 0$, we need the input to stay close to 0.

But in our case, $f(x) \in [0, 1]$, so we center it by subtracting $\frac{1}{2}$: $$ f(x) - \frac{1}{2} \in [-0.5, 0.5] $$ Then, we scale with a small constant $c$ to shrink the range: $$ c(f(x) - \frac{1}{2}) \in [-0.5c, 0.5c] $$ This keeps the input near 0 and improves the accuracy of the Taylor expansion.

Finally, we add $\frac{1}{2}$ back to ensure the result is still within $[0,1]$.

Linearization of \(\sin^2(y + \pi/4)\) near \(y=0\)

---

Now you may understand the key poltnominal approximation trick in paper [1]..

Reference

1. Quantum Risk Analysis