What is the Fidelity of quantum states?

Learn what State fidelity is and how to calculate it.

Source

In the last post, I shared with you 5 questions that will help you become great at linear algebra and ace Quantum computing.

Let’s dive in and solve the third one step by step.

Question 3

Given two quantum states as follows:

1. Compute their inner product

2. Compute their Fidelity

Solution 3

3.1: Computing the inner product

The inner product of two qubits |ψ> and |ϕ> is represented as <ϕ|ψ>.

This means that we first need to write |ψ> in its bra form as <ψ|.

\(\langle \psi| = \frac{1}{\sqrt{2}}(\langle 0|+\langle 1|) \)

Next:

\(\langle \psi|\phi\rangle = \frac{1}{\sqrt{2}}(\langle 0|+\langle 1|) \left(\frac{\sqrt{3}}{2}|0\rangle+\frac{1}{2}|1\rangle\right)\)

\(\langle \psi|\phi\rangle = \frac{1}{\sqrt{2}} \left[ \frac{\sqrt{3}}{2}\langle 0|0\rangle + \frac{1}{2}\langle 0|1\rangle + \frac{\sqrt{3}}{2}\langle 1|0\rangle + \frac{1}{2}\langle 1|1\rangle \right]\)

From previous lessons, we know that:

\(\begin{aligned} \langle 0|0\rangle &= 1 \\[6pt] \langle 1|1\rangle &= 1 \\[6pt] \langle 0|1\rangle &= 0 \\[6pt] \langle 1|0\rangle &= 0 \end{aligned}\)

This simplifies the above equation to:

\(\langle \psi|\phi\rangle = \frac{1}{\sqrt{2}} \left[ \frac{\sqrt{3}}{2}(1) + \frac{1}{2}(0) + \frac{\sqrt{3}}{2}(0) + \frac{1}{2}(1) \right]\)

\(= \frac{1}{\sqrt{2}} \left[ \frac{\sqrt{3}}{2} + \frac{1}{2} \right]\)

\(\langle \psi|\phi\rangle = \frac{\sqrt{3}+1}{2\sqrt{2}}\)

This is the inner product of these two qubits.

---

3.2: Computing the Fidelity

Fidelity is a measure of how similar two quantum states/ qubits are.

Fidelity tells how closely a practical, noisy quantum state/ operation matches its ideal theoretical counterpart. This helps to benchmark quantum computers and evaluate quantum state preparation.

The value of Fidelity lies between 0 and 1, where:

• 1 means that the two qubits are the same or their state is identical

• 0 means that the states are completely different or orthogonal

For two qubits |ψ> and |ϕ>, the Fidelity is calculated as:

\(F = |\langle \psi|\phi\rangle|^2\)

This means that we:

1. First, compute the inner product of the two qubits

2. Then, square the absolute value of this result

We already calculated the inner product of the two qubits as follows.

\(\langle \psi|\phi\rangle = \frac{\sqrt{3}+1}{2\sqrt{2}}\)

The Fidelity is calculated by taking the square of the absolute value of this result.

\(F = |\langle \psi|\phi\rangle|^2\)

\(F = \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)^2\)

\(F = \frac{(\sqrt{3}+1)^2}{(2\sqrt{2})^2}\)

\(F = \frac{3+2\sqrt{3}+1}{8} = \frac{4+2\sqrt{3}}{8} = \frac{2+\sqrt{3}}{4}\)

This is approximately equal to 0.933.

This tells us that these states are quite similar to one another (close to 1).

Thanks for being a curious reader of ‘Into Quantum’, a publication that teaches essential concepts in Quantum Computing from the ground up.

To get even more value from this publication, consider becoming a paid subscriber.