# Polarization of light and quantum spin states

In Sakurai's Modern Quantum Mechanics Chapter 1, he makes an analogy between polarized light and the results of the Stern-Gerlach experiment.

He compares the existence of a $$S_{z}^{+}$$ and $$S_{z}^{-}$$ component in an $$S_{z}^{+}$$ beam passed through an apparatus that splits the beam as $$S_{x}^{+}$$ and $$S_{x}^{-}$$ to polarized light that passes first through an $$x$$ filter followed by a filter at $$45^{\circ}$$ and then through a $$y$$ filter. In both cases, the beam appears to 'forget' what its initial state was, after it passes through the intermediary.

Now, Sakurai extends this analogy to write $$S_{x}^{+}$$ and $$S_{x}^{-}$$ as a linear combination of $$S_{z}^{+}$$ and $$S_{z}^{-}$$, thus exhausting all possible linear combinations using $$S_{z}$$ as a basis. He now asks how $$S_{y}^{+}$$ and $$S_{y}^{-}$$ should be written, now that $$S_{z}$$ has been exhausted. For this, he alludes to how circularly polarized light and linearly polarized light at $$45^{\circ}$$ both have equal components along $$\hat{x}$$ and $$\hat{y}$$, but are both represented in terms of the same basis; the circularly polarized light having complex coefficients.

My question is as follows. How can we compare the $$y$$ component of the spin state to circularly polarized light? Aren't circularly and linearly polarized light different? As in, circularly polarized light gives equal components along any set of perpendicular axes, however, a $$45^{\circ}$$ linearly polarized light doesn't for any set of axes: it has to be at $$45^{\circ}$$ to the perpendicular axes. On that note, $$S_{x}$$ and $$S_{y}$$ are identical, as in, they do not have this kind of difference, right?

If we work in the circular basis for light: $$|+\rangle$$ and $$|-\rangle$$, then linear polarization is a linear combination:

$$|\leftrightarrow\rangle = \frac {-i} {\sqrt 2}[|+\rangle - |-\rangle]$$

$$|\updownarrow\rangle = \frac 1 {\sqrt 2}[|+\rangle + |-\rangle]$$

I used double arrows to indicate that it is a tensor polarization.

The orthogonal $$\pm 45^{\circ}$$ linear polarizations are then:

$$\frac 1 {\sqrt{2}}[|\updownarrow\rangle \pm |\leftrightarrow\rangle ]$$

and they are equal combinations of $$|+\rangle$$ and $$|-\rangle$$ with different phases.

This is very similar to the spin 1/2 case, in which:

$$|\uparrow\rangle_x = \frac 1 {\sqrt 2}[|\uparrow\rangle + |\downarrow\rangle]$$ $$|\downarrow\rangle_x = \frac 1 {\sqrt 2}[|\uparrow\rangle - |\downarrow\rangle]$$

are the $$x$$-basis and

$$|\uparrow\rangle_y = \frac 1 {\sqrt 2}[|\uparrow\rangle + i|\downarrow\rangle]$$ $$|\downarrow\rangle_y = \frac 1 {\sqrt 2}[|\uparrow\rangle - i|\downarrow\rangle]$$

are the $$y$$-basis.

Linear polarized light is a tensor polarization, so a 180 degree rotation changes the state by $$-1$$, leaving it unchanged in the lab. A 90 degree rotation just flips the definition of horizontal and vertical, and a 45 degree rotation is an orthogonal diagonal basis. This is all because the photon is spin-1.

Spin 1/2 behaves differently. A 360 degree rotation changes the states by a phase factor of -1, while a 180 degree rotation flips the definition of up and down. (Note: "minus up is not down). A 90 degree rotation changes between orthogonal basis in $$x$$ and $$y$$.

You can continue this pattern to spin-2 gravitational waves. One may stretch vertically and squish horizontally and then half a cycle latter squish vertically and stretch horizontally. Thus, a 90 degree rotation adds a phase factor of -1, and 45 degree rotation gives the state that completes the orthogonal basis for the 2 states of polarization (see figure).