## How to find the normalization factor of the eigenvectors of the $\sigma_x$ Pauli gate?

2

I'm trying to calcaute the eigenstates for the $$\sigma_x$$ gate, and I can follow the process up to finding eigenvalues $$\pm 1$$, but I don't understand where the $$\frac{1}{\sqrt{2}}$$ coefficient comes from for the answer:

$$\begin{bmatrix} -\lambda & 1\\ 1 & -\lambda \end{bmatrix}v = 0 \implies v = \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1 \end{bmatrix}$$

For the solution $$\lambda = 1$$, why does that $$\frac{1}{\sqrt{2}}$$ show up?

4

The $$\dfrac{1}{\sqrt{2}}$$ is the normalization constant to make sure the state/eigenvector is a unit vector.

Note that: if $$|\psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ then $$\bigg| \bigg| |\psi \rangle \bigg| \bigg| = |1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1$$.

The reason for this is because in quantum mechanics, states are always normalized. It is one of the postulates of quantum mechanics.

1Forgot about the normalization part. Cheers! – Markian Hromiak – 2021-01-11T06:23:35.690

0

It is normalized by dividing with the modulus or magnitude which is sqrt of (eigenvalue1 * 2 + eigenvalue2 * 2) = sqrt(2)

Hi and welcome to Quantum Computing SE. What do you mean by eigenvalue1 and eigenvalue2. – Martin Vesely – 2021-01-11T08:04:38.603