Bell's theorem/Quantum correlation

Here it is shown that for an entangled pair of photons, the linear and circular polarization obey:

|LL> + |RR> = |xx> + |yy> ... if Ψ(−z)=Ψ(+z)

|RR> − |LL> = |yx> − |xy> ... if Ψ(−z)=−Ψ(+z)

The two photons, Alice and Bob in a Bell's theorem experiment, have the same linear polarization if entangled state has even parity, and perpendicular polarization if the parity is odd.

Circular motion of a phasor in the x-y plane

We begin by defining positive and negative helicity with respect to the +z direction

$\phi _{+}={\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}}e^{-i\omega t}$

rotates with a sign that is positive with respect to the z axis (counterclockwise if the z-axis is facing towards you). Both the real and imaginary parts rotate in this direction; for example the real part is $\Re \phi _{+}=\cos(\omega t){\hat {x}}+\sin(\omega t){\hat {y}}$. A signal with the opposite helicity,

$\phi _{-}={\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}e^{-i\omega t}$

rotates with a sign that is negative with respect to the z axis. It is convenient to define the two complex unit vectors:

${\hat {u}}_{+}={\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}}$ and ${\hat {u}}_{-}={\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}$

In order to ascertain whether whether the unit vectors ${\hat {u}}_{\pm }$ are associated with right or left-handed helicity, it is necessary to know which direction (along the z axis) the signal is propagating. Since there are two potons (called Bob and Alice), we need two different spatial variables (z_A,z_B along the z-axis. We shall temporarily assume that Alice is travelling in the −z direction, and Bob is travelling in the +z direction (and later find that we must entangle the two photon propagation directions due to parity considerations).

$\psi _{L}^{A}={\hat {u}}^{+}e^{i(-kz_{A}-\omega t)}$ represents a left-handed photon travelling in the −z direction

$\psi _{R}^{A}={\hat {u}}^{-}e^{i(-kz_{A}-\omega t)}$ is a right-handed photon travelling in the −z direction.

$\psi _{R}^{B}={\hat {u}}^{+}e^{i(kz_{B}-\omega t)}$ is a right-handed photon travelling in the +z direction.

$\psi _{L}^{B}={\hat {u}}^{-}e^{i(kz_{B}-\omega t)}$ is a left-handed photon travelling in the +z direction.

The Alice-first convention

When two quantum particles are considered simultaneously, it is necessary to create a composite, or tensor product of the two states. Taking a hint from Dirac's convenient Bra–ket notation, we simplify as much as possible by describing states with minimum notational description. For example, we may avoid labeling the unit vectors with A and B superscripts, provided we adopt a convention in which the polarization of the A (Alice) state is placed in front of the polarization of the B (Bob) state. It is also convenient to omit the time dependence e^−iωt:

$\psi _{RR}={\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}\cdot {\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}}e^{ik(z_{B}-z_{A})}$

$\psi _{LL}={\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}}\cdot {\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}e^{ik(z_{B}-z_{A})}$

For example, our "Alice first" convention allows us to interpret ${\hat {x}}{\hat {y}}$ as follows:

${\hat {x}}{\hat {y}}e^{ik(z_{B}-z_{A})}\equiv \left({\hat {x}}_{A}e^{-ikz_{A}}\right)\otimes \left({\hat {y}}_{B}e^{ikz_{B}}\right)$

Here, the location of ${\hat {x}}$ as the first unit vector implies that Alice is polarized in the x direction, while the location of ${\hat {y}}$ in the second position implies that Bob is polarized in the y direction. Alternatives to the composite notation $(\otimes )$ include the creation of a two-dimensional vector with components $[\phi _{A},\phi _{B}]$, or using the "ket", $|RL\rangle$, to represent $\psi _{RL}$.

Simplifying assumptions that reduce unnecessary complications

Even though $\psi _{RL}{\text{ and }}\psi _{LR}$ are allowed photon states, we are deliberately excluding them by restricting ourselves to transitions in which that parent atom's angular momentum does not change. We also exclude, for simplicity, the possibility that Alice and Bob have different energy. If the energies are different, expressions such as, $e^{ik(z_{B}-z_{A})}$, are replaced by, $e^{i(k_{B}z_{B}-k_{A}z_{A})}$, where k_A and k_B are the wavenumbers of Alice and Bob, respectively.

Expressing the circular polarized states in terms of linear polarization unit vectors

$\psi _{RR}=\left[{\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{2}}+i{\frac {{\hat {y}}{\hat {x}}-{\hat {x}}{\hat {y}}}{2}}\right]e^{ik(z_{B}-z_{A})}$

$\psi _{LL}=\left[{\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{2}}+i{\frac {{\hat {x}}{\hat {y}}-{\hat {y}}{\hat {x}}}{2}}\right]e^{ik(z_{B}-z_{A})}$

By restricting our discussion to a situation where the angular momentum of the parent atom does not change, we my ignore the two corresponding entangled photon states ($\psi _{RL}$ and $\psi _{LR}$) because they have net angular momentum. We now seek states of known parity, first by adding and subtracting:

${\frac {\psi _{RR}+\psi _{LL}}{\sqrt {2}}}={\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{\sqrt {2}}}e^{ik(z_{B}-z_{A})}\equiv \psi _{+}$

${\frac {\psi _{RR}-\psi _{LL}}{\sqrt {2}}}=i{\frac {{\hat {y}}{\hat {x}}-{\hat {x}}{\hat {y}}}{\sqrt {2}}}e^{ik(z_{B}-z_{A})}\equiv \psi _{-}$.

The states $\psi _{+}{\text{ and }}\psi _{-}$ are now almost states of even and odd parity, respectively. To finish the task, we take the complex conjugate of both wavefunctions and combine:

Entangled states of even and odd parity

The entangled photon states of parity, P=+1 and P=−1, are, respectively:

$\Psi _{+}={\frac {\psi _{+}+{\overline {\psi }}_{-}}{2}}={\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{\sqrt {2}}}\cos k(z_{B}-z_{A})$

$\Psi _{-}={\frac {\psi _{+}-{\overline {\psi }}_{-}}{2i}}={\frac {{\hat {y}}{\hat {x}}-{\hat {x}}{\hat {y}}}{\sqrt {2}}}\sin k(z_{B}-z_{A})$

The reader may wonder how the travelling photons Alice and Bob became standing waves. A standing wave is a superposition of travelling waves in both directions, and in order to create entangled photon states that were eigenstates of the parity operator, it was necessary to entangle the identity of two photons. If these photons have different energy, then each photon's energy takes an indeterminate value that can be established only after a measurement on one of the photons is made. The reader may verify the parity by applying the parity operator, P, to each state:

$P\Psi _{+}(z_{A},z_{B})=\Psi _{+}(-z_{A},-z_{B})=(+1)\Psi _{+}(z_{A},z_{B})$

$P\Psi _{-}(z_{A},z_{B})=\Psi _{-}(-z_{A},-z_{B})=(-1)\Psi _{-}(z_{A},z_{B})$

Footnotes and references

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