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?


2 Answers 2


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.

  • 1
    $\begingroup$ Forgot about the normalization part. Cheers! $\endgroup$
    – aikky
    Jan 11, 2021 at 6:23

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

  • $\begingroup$ Hi and welcome to Quantum Computing SE. What do you mean by eigenvalue1 and eigenvalue2. $\endgroup$ Jan 11, 2021 at 8:04

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