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I have read that circularly polarized light forms from the superposition of two linearly polarised light. Then is it true that polarisation can't happen with single Photon because it always need at least two photons to interfere with each other and superimpose the electric vectors? If polarisation can happen with single-photon then how does it happen?

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    $\begingroup$ This was already mentioned in an answer, but you can also view linearly polarized light as the superposition of two circularly polarized light waves. $\endgroup$ – Aaron Stevens Dec 6 '18 at 12:28
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Single photons are polarized.

The state of a circularly polarized photon can be described as a quantum mechanical superposition of two linearly polarized states. Note that this varies from your description in two ways. The photon does not form from two other photons, it's state does. And the combination is quantum mechanical superposition which is not the same as combining together two photons.

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Circular polarizations can be represented as superpositions of linear polarizations, and linear polarizations can be represented as superpositions of circular polarizations. This is because linear and circular polarizations describe the same collection of photon polarization states, and you can decompose that collection of states in terms of either linear or circular polarizations (formally, we say that linear and circular polarizations each form a basis of the vector space of polarization states). This is very similar to describing the coordinate plane in terms of two different sets of coordinates: for example, you can describe the full coordinate plane in terms of either ordinary Cartesian coordinates, or by Cartesian coordinates rotated by 45 degrees. Any point in the plane has well-defined coordinates in both of these systems, and which one is chosen is largely a matter of convenience.

For example, if you are intent on measuring the angular momentum of photons, then the circular-polarization basis would definitely be better to work in, as the left-handed and right-handed circular polarization states both have definite angular momenutum. If, on the other hand, you are intent on measuring the probability that a photon passes through a linear polarizer, then you would be better off using a linear polarization basis aligned with the polarizer, since the basis states have a 100 percent and 0 percent probability of passing through the polarizer, respectively.

Also, superposition does not require two photons. The classical definition of superposition (two waves being added together while occupying the same space) doesn't describe superposition as we mean it in quantum mechanics (where a single object can exist in a state that is not expressible in terms of a definite value of an observable).

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When I first learned quantum mechanics, I read this beautiful explanation in Dirac's book: see page 14 of this pdf, http://digbib.ubka.uni-karlsruhe.de/volltexte/wasbleibt/57355817/57355817.pdf or page four in the book.

In classical physics, polarisation defines the evolution of the electric field vector in a wave over time https://en.wikipedia.org/wiki/Polarization_(waves). In quantum mechanics, you can ascribe a polarisation to a single photons, but exactly what that "means" or "looks like" is tricky! I won't attempt to explain it better than Dirac does above (it's very lucid!) but perhaps a summary is that a photon is an object in quantum mechanics, which can be described by a quantum state. A quantum state can be thought of as a series of properties of the photon, along with a set of probabilities attached to physical measurements you might make of the photon. The polarisation of the photon is one such label associated to the state.

As was already mentioned by another answer, a single object can be in a superposition of two states so you don't need more than one photon to have a circular polarisation. Schrodinger's cat is a single object superimposed between being dead and alive - you don't need two cats to have the superposition.

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"I have read that circularly polarized light forms from the superposition of two linearly polarised light."

You can think of it that way. Similarly, when you have chosen a vector basis for your space, you can think of every vector as a sum of the basis vectors.

But your vector is not inherently the sum of your basis vectors. That's just a way to think of it, a way that might be useful to you after you choose your basis.

Similarly, linear polarization gives you vectors that are perpendicular to the direction of travel. Circular polarization is just a polarization vector that's inline with the direction of travel, and elliptical ones are everything else on the sphere.

When you multiply two linear polarizations you get a product that's intermediate between them, or something that's more circularly polarized, depending on the relative phases and amplitudes.

The polarization of each entity is just its polarization. How you choose to break it up into other polarizations is just something you do for your convenience.

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