Interference of polarized light

Question

Does polarized light interfere?

Answer 1

Let's do some math in order not to be unsubstantiated.

1. Perpendicular polarizations.

First wave $E_{1x} = E_0\,\cos\omega t$, second wave $E_{2y} = E_0\,\cos(\omega t+\Delta)$. Here $\Delta$ is a phase difference between waves.

Total field: $$\vec{E} = E_0\left(\vec{i}\cos\omega t+\vec{j}\cos(\omega t+\Delta)\right).$$

Intensity: $$I_\perp\sim \langle|\vec{E}|^2\rangle = E_0^2\langle\cos^2\omega t+\cos^2(\omega t+\Delta)\rangle,$$
where the average is defined as $\langle f(t)\rangle = \frac{1}{T}\int_0^{T}\,dt\,f(t)$ (so that $\langle \cos^2(\omega t+\Delta)\rangle = \frac{1}{2}$ for any $\Delta$).

Finally we've got $I_\perp\sim E_0^2$, which is independent on the phase difference between the waves.

2. Parallel polarizations.

First wave $E_{1x} = E_0\,\cos\omega t$,second wave $E_{2x} = E_0\,\cos(\omega t+\Delta)$.

Total field: $$\vec{E} = \vec{i}E_0\left(\cos\omega t+\cos(\omega t+\Delta)\right).$$

Intensity: $$I_\parallel\sim E_0^2\langle\cos^2\omega t+2\cos\omega t\cos(\omega t+\Delta)+\cos^2(\omega t+\Delta)\rangle=E_0^2(1+\cos\Delta),$$ which nicely depends on the phase shift between the waves.

Answer 2

Yes. In fact, light will only interfere with light of the same polarization. If you take a Mach–Zehnder interferometer, for example, and put a polarization rotating optic (a waveplate) in one of the arms, the interference pattern will lose contrast. If the polarization is rotated 90 degrees, the pattern will vanish completely.

Answer 3

As others have noted, you will not get any intensity modulation from the interference of two linearly polarized light beams with orthogonal polarizations. It's worth noting, though, that this does not mean that beams with perpendicular polarizations don't affect each other. In fact, a counter-propagating pair of beams with orthogonal linear polarizations-- the so-called "lin-perp-lin" configuration-- is the best system for understanding the Sisyphus cooling effect, the explanation of which was a big part of the 1997 Nobel Prize in Physics.

The superposition of two counter-propagating linearly polarized beams with orthogonal polarizations doesn't give you any modulation of intensity, but does create a polarization gradient. For the lin-perp-lin configuration, you get alternating regions of left- and right-circular polarization, and combined with optical pumping this lets you set up a scenario where you can cool atomic vapors to extremely low temepratures. This makes laser cooling vastly more useful than it would be otherwise, and allows all sorts of cool technologies like atomic fountain clocks.

It's not interference in the sense that is usually meant, but it is a cool phenomenon that results from overlapping beams with different polarizations. So you shouldn't think that just because it doesn't produce a pattern of bright and dark spots it's not interesting.

Answer 4

Your question is quite vague, but in short, the answer is: yes, look it up on wikipedia. But let's be more precise:

As long as the light intensity is low enough to not obtain nonlinear effects, the superposition principle of linear optics is valid. That means, the amplitudes of two electromagnetic (EM) fields sum up, thus yielding interference.

However, since the amplitudes are vectors (while the intensity, being related to the absolute squares of the amplitudes, is a scalar), the interference depends on the relative polarization, the total intensity for two linearly polarized EM waves is

$I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos(\Delta\varphi)$

where $\Delta\varphi$ denotes the angle between the two polarizations. You see that for perpendicular polarization the cosine term vanishes, the intensities just add up and you obtain no interference, while for antiparallel polarization ($\Delta\varphi = 180°$) you obtain destructive interference since the cosine becomes -1. In case you wonder about energy conservation (being proportional to the intensity) keep in mind that only the global energy is conserved while local fluctations are ok.

One final note: The whole thing only works for a well-defined phase relation between two EM waves. That is, only spectral components of the same wavelength can interfere, and the coherence length and time need to be large enough - you won't obtain perfect interference if your light source flickers due to heat for example.

Answer 5

There are plenty good answers already but I want to point out something that most other answers have missed and something Vladimir Kalitvianski was alluding to in their answer. If you want to skip to the example I give, head to the last section.

A comment on the semantics

Light interference has historically been assigned to the phenomena of waves adding up. Now, technically any linear differential equation allows you to generate new solutions by adding solutions so you can add any number and any kind (frequency, polarisation etc) of waves to get a new wave. This wave is technically born from the interference of waves.

That being said, semantically one almost always refers to interference as a visible evidence (fringe pattern for the double slit) of such an addition of waves especially when the properties of said waves can be deciphered from the visual evidence. Properties such as the propagation direction (in case of double slit, the slit width), the wavelength etc.

So when one asks "Do these waves interfere?" what they usually mean is, can I see a visible evidence of two or more waves overlapping/mixing/adding and that is what I mean from here on. And to answer that question we look to quantum mechanics (QM).

Necessary condition for interference

It is said that coherence is a necessary condition for interference. Coherence is just a convenient shorthand for definite phase relation between two waves. (A more rigorous definition for coherence came from Glauber in his seminal work in 1963 but we need not go there.)

You know what ensures definite phase relation? Make a wave interfere with itself! And if you "divide" the wave into parts that evolve differently over time (like from two different slits) and behold, an interference pattern! Bring in wave-particle duality to now conclude (as Dirac did in 1927) that each particle only interferes with itself.

Does that mean two different particles never interfere? No! Two different particles can still have a fixed phase relation. It's usually unlikely that two independent entities have a fixed phase relation but once again QM comes to the rescue. QM says that if two objects are identical then there is no way one can distinguish between them. Thus any state that represents the two identical objects (if they have overlapped) must necessarily be a joint state one that only picks up a phase under permutation. $\big(|\psi_{12}\rangle = \exp{i\theta}|\psi_{21}\rangle\big)$

Orthogonal yet Indistinguishable!

In the classic Hong-Ou-Mandel setup, two indistinguishable photons are fed into two different input ports of a beamsplitter and what is observed is that the two photons always exit out of either of the two but the same exit port. This happens due to a two-photon interference as the paths leading to the photons going to two different output ports can happen in two ways and they destructively interfere.

What happens if I send in two orthogonally polarised light, say $H$ from $a$ and $V$ from $b$? Since the two photons are distinguishable, the destructive interference no longer occurs and we see that the photons can come out from the same port or different ports.

Is it possible to send in orthogonally polarised light yet for them to be indistinguishable? Sounds counterintuitive right? But that's what entanglement lets us do! Say we send in one photon from each input port but we don't know which polarisation it is. All we know is that if the $a$ photon is $H$ then the $b$ one is $V$ and if the $a$ photon is $V$ then the $b$ one is $H$. One such state would be: $$|\psi_+\rangle = \frac{|HV\rangle_{ab}+|VH\rangle_{ab}}{\sqrt{2}}$$

Since the photon going in are now orthogonal yet indistinguishable, we see both photons going out of the same exit port!

Another such state would be: $$|\psi_-\rangle = \frac{|HV\rangle_{ab}-|VH\rangle_{ab}}{\sqrt{2}}$$

For this case, both photons go out of different exit ports and never the same! (This has to do with the fermionic nature of the state).

Answer 6

Yes, it does and this property is independent from a particular polarization. So non polarized light gives the same interference pattern.