Can a two-levels photon pair be created either entangled or not entangled? [closed]

Question

I am learning about experiments on Quantum Optics and Quantum Tomography in order to understand how to measure two qubits with an arbitrary quantum state of their polarization degrees of freedom. Since I need to work with two-particle system, it is normally said that I have to work with a "black box" which provides me photon pairs. Something like this

As it seems like a SPDC device, I suspect is a sort of entangled photons device (also because of the Coincidence Detector). According to what I understand by entangled photons, are states expressed like this

$$ \frac{1}{\sqrt 2}(|0\rangle_A \otimes |1\rangle_B + \mathrm{e}^{i\phi}|1\rangle_A \otimes |0\rangle_B) $$

But, as I precise to work with a tensor product state (two systems), I believe that I must use states like this

$$ a|0\rangle_A \otimes |0\rangle_B + b|0\rangle_A \otimes |1\rangle_B + c|1\rangle_A \otimes |0\rangle_B + d|1\rangle_A \otimes |1\rangle_B$$

Can that "black box" produce both types? if not, Can I use entangled photons as a two-particle system with an arbitrary quantum state?

Answer 1

Spontaneous Parametric Down Conversion is a process that produces pairs of photons. The process goes like this: a strong beam of ultra-violet (UV) photons, is sent upon a down-conversion crystal (placed inside your black-box). Inside the crystal, the UV photon is SPLIT into two photons, named (I don't know why) SIGNAL and IDLER. In general the two photons have different frequencies, but sometimes they are of the same frequency. There is a wide literature about down-conversion. You can see also a description in Wikipedia

http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion

You'll see there that that each photon is emitted like a cone of light. If at some distance from the crystal we place a screen, we get on the screen two circles that intersect at two points. Let's label these points A and B. Now, if the two photons are of the same wavelength, and if we make holes in the screen exactly at A and B, through each hole exits one photon. But for the photon exiting A, we will not know if it is the signal, or it is the idler. What yes we know is that if the photon from A is the signal, than the photon from B is the idler, and vice-versa.

So we get the state

(1/sqrt(2)) (|$1_s$>|$1_i$> + |$1_i$>|$1_s$>),

where the subscript s means signal, and the subscript I means idler.

You have also to know that these photons are usually polarized, i.e. one of them is circularly right polarized ( |R> ), and the other circularly left polarized ( |L> ), in which case the state is

(1/sqrt(2)) (|L>|R> + |R>|L>).

About an arrangement to get a state as your second equality describes, the parametric down-conversion produces two photons. But, try to use beam-splitters on the ways of the two photons and see if you can get the desired combinations. It is known that if you bring on the each side of a beam splitter (BS) a photon, and the two photons are identical, they will both leave the BS on the same side.

Good Luck !

Answer 2

Here is a suggestion of how to produce the state in your second formula. Choose to make holes in a screen, NOT at the two points A and B of which I spoke in my first answer, but at two other points, e.g. on top of the upper cone, and on the bottom of the lower cone. Then, pass the upper beam through a beam splitter $BS_1$. Block the transmitted wave, let only the reflected one |$r_{upper}$>. The same with the lower beam, pass it through a beam-splitter, $BS_2$, and block the transmitted beam, leave only the reflected, |$r_{lower}$>. You get your desired second state, but I replace A and B in your formula by U (upper reflected beam) and L (lower reflected beam). Choosing beam-splitters with convenient absorption and transmission coefficients, you can control the parameters a, b, c, d.