# Why is the matrix obtained from the coefficients of orthogonal states unitary?

I'm having troubles in understanding a statement in Box 2.7 at page 113 in the Nielsen & Chuang.

Firstly, it assumed to be working with a two-qubits quantum system in state $$|\psi\rangle = \frac{|01\rangle - |10\rangle}{\sqrt2}$$.

We perform a measurement of the observable $$\vec{v}\cdot \vec{\sigma}$$ on each qubit in the system. Assuming $$|a\rangle$$ and $$|b\rangle$$ being the eigenstates of $$\vec{v}\cdot \vec{\sigma}$$, we can write: $$|0\rangle = \alpha |a\rangle + \beta |b\rangle \\ |1\rangle=\gamma |a\rangle + \delta |b\rangle$$

Now, here comes my doubt. In the text, the authors says that the matrix: $$\begin{bmatrix}\alpha &\beta\\\gamma&\delta\end{bmatrix}$$ is unitary.

Why is this matrix unitary? I'm having troubles understanding the reason behind this argument.

• Because $\langle 0|0\rangle = 1$, $\langle 0|1\rangle = 0$ and so on Sep 4, 2023 at 8:20
• @EvgeniyZh thanks for your answer! I did the calculations and now the argument is clear :) Sep 4, 2023 at 8:30
• Unitarity is akin to orthogonality, in that it means that its column vectors are all orthonormal, and vice versa for row vectors. In the matrix you wrote down, the row vectors are the components of $|0\rangle$ and $|1\rangle$ in the $\{ |a\rangle, |b\rangle\}$ basis, so they must be orthonormal!
– 이희원
Sep 5, 2023 at 2:03
• generally speaking, a matrix is unitary iff its columns are orthonormal (equivalently, iff its row are orthonormal)
– glS
Feb 1 at 13:37

The key thing is that the two eigenvectors of $$\vec{v}\cdot\vec{\sigma}$$ are orthogonal. So if you create a matrix such as $$U=|a\rangle\langle 0|+|b\rangle\langle 1|,$$ (this is technically the complex conjugate of the matrix you wrote down) it satisfies $$U^\dagger U=|0\rangle\langle 0|+|1\rangle\langle 1|=I$$, proving that $$U$$ is unitary.