We know that the Pauli matrices are:

$$\sigma_x = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}, \sigma_y = \begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}, \sigma_z = \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$$

I was wondering what would these matrices would represent in the physical world. I know that they can measure quantum state and get certain results. Would I be wrong if I assume that $\sigma_x$ measures the Diagonal polarization of a photon or $\sigma_z$ represents the measurement of vertical/horizontal polarization of a photon?


2 Answers 2


Your understanding is correct.

In the theory of photon polarization, the parametrization of the Bloch sphere (or its surface) has traditionally another name. On the wikipedia page for the Jones calculus (the parametrization of the Bloch sphere surface), you'll find a table for the correspondence between kets and polarizations.

To summarize, eigenstates of the Pauli matrix $\sigma_Z$ can be seen as the horizontal and vertical polarizations (commonly written as $|H\rangle$ and $|V\rangle$), eigenstates of $\sigma_X$ correspond to diagonal and anti-diagonal polarizations, and eigenstates of $\sigma_Y$ correspond to the left and right-handed polarizations of light.

Thus a measurement corresponding to $\sigma_X$ can be seen as putting a beam splitter in front of the light path, that splits the light into the horizontal and vertical polarization components. (You can also put a polarization filter, but as such filter measurements are destructive, the post-measurement will be different. And because of that, destructive measurements technically don't quite correspond to the Pauli's. Do it with beam splitters instead of filters, and the analogy is fine again.)

Of course, you can always choose a different basis (by a unitary transformation). Just make sure that choice made gives the correct commutation relations between the Pauli matrices.

PS: A similar concept to the Jones vector, but which also covers the interior of the Bloch Ball, are the so-called Stokes parameters.


I see the heart of your question. I'd like to clarify a bit, before answering your question though. Matrices (aka operators) do not measure quantum states--they operate on them. Specifically, they project the state into the matrix's eigenvectors. We can then measure that projected state in a particular basis that may be the same or different than what the quantum state was projected into. But, those are technical details...

A photon does not quite capture the whole picture. Let's see why not.

If you were to use photon polarization to ''visualize'' the Pauli matrices, $\sigma_z$ would represent the up and down possibilities and basis while $\sigma_x$ would be right and left. It gets a little tricky with $\sigma_y$ because we need a third dimension for polarization. As gIS pointed out, circular polarization can be paired with $\sigma_y$. Specifically, the right- and left-handed circular polarizations would be paired with $\sigma_y$.

A very standard way to ''visualize'' the Pauli matrices is with the Bloch sphere. It's a unit sphere with a vector pointing from the center to a point on the sphere. The z-axis of the sphere is paired with $\sigma_z$; the x-axis is paired with $\sigma_x$; and, the y-axis is paired with $\sigma_y$. This video from IBM can explain it better it a bit further and gives some nice images of the Bloch sphere.

  • 3
    $\begingroup$ $\sigma_y$ corresponds to circular polarization of the underlying EM field, that is, the situation in which the polarization rotates in the $xz$ plane $\endgroup$
    – glS
    Oct 10, 2019 at 10:20
  • $\begingroup$ Oh yes, thank you! I forgot about circular polarization. I will edit my response. $\endgroup$ Oct 10, 2019 at 13:26

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