# User:Guy vandegrift/Bell correlation attempt 2

The following two complex signals rotate in opposite directions:

${\displaystyle \phi _{+}={\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}}e^{-i\omega t}}$ and ${\displaystyle \phi _{-}={\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}e^{-i\omega t}}$
(1)

For example the real part of ${\displaystyle \phi _{+}}$ is ${\displaystyle \cos(\omega t){\hat {x}}+\sin(\omega t){\hat {y}}}$ which rotates in the positive z direction (by the right hand rule), as does the imaginary part of ${\displaystyle \phi _{+}}$. Both real and imaginary part of ${\displaystyle \phi _{-}}$ rotate in the negative z direction. It is convenient to define the two complex unit vectors:

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

In order to ascertain whether whether the unit vectors ${\displaystyle {\hat {u}}_{\pm }}$ 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 (zA,zB) 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).

${\displaystyle \psi _{L}^{A}={\hat {u}}^{+}e^{i(-k_{A}z_{A}-\omega _{A}t)}}$ represents a left-handed photon travelling in the −z direction.
${\displaystyle \psi _{R}^{A}={\hat {u}}^{-}e^{i(-k_{A}z_{A}-\omega _{A}t)}}$ is a right-handed photon travelling in the −z direction.
${\displaystyle \psi _{R}^{B}={\hat {u}}^{+}e^{i(k_{B}z_{B}-\omega _{B}t)}}$ is a right-handed photon travelling in the +z direction.
${\displaystyle \psi _{L}^{B}={\hat {u}}^{-}e^{i(k_{B}z_{B}-\omega _{B}t)}}$ is a left-handed photon travelling in the +z direction.
(3)

We allow for the photons Alice and Bob to have different energy because this is the case for some Bell's theorem experiments, and also because it allows the reader to better follow the algebra. When two photons are considered, we need a convention for keeping track of their polarizations. A clumsy but transparent convention is to write them as composites of two wavefunctions. This notation is displayed in the next equation, but is replaced by a simpler notation where the order of the unit vectors keeps track of which unit vector corresponds to which photon:

{\displaystyle {\begin{aligned}\psi _{R}^{A}\otimes \psi _{R}^{B}&={\hat {u}}^{-}e^{i(-k_{A}z_{A}-\omega _{A}t)}\otimes {\hat {u}}^{+}e^{i(k_{B}z_{B}-\omega _{B}t)}\\&={\hat {u}}^{-}{\hat {u}}^{+}e^{i(k_{B}z_{B}-k_{A}z_{A})}e^{-\Sigma \omega t}=\psi _{RR}e^{-\Sigma \omega t}\end{aligned}}}

${\displaystyle \psi _{L}^{A}\otimes \psi _{L}^{B}={\hat {u}}^{+}{\hat {u}}^{-}e^{i(k_{B}z_{B}-k_{A}z_{A})}e^{-\Sigma \omega t}=\psi _{LL}e^{-\Sigma \omega t}}$

(4)

It is convenient to refer to the notation as the "Alice-first" convention, as we have allowed the spatial variaton to commute freely, but kept the order of the unit vectors intact so that the first refers to Alice and the second refers to Bob. Alternatives to the composite[1] notation ${\displaystyle (\otimes )}$ include the creation of a two-dimensional vector with components ${\displaystyle [\phi _{A},\phi _{B}]}$, or using the "ket", ${\displaystyle |RL\rangle }$, to represent ${\displaystyle \psi _{RL}}$. All of this complexity will seem strange to students that have only seen the Schrödinger equation for only a single scalar complex waveform, ψ. Here we need four components, not one: two describe the polarization in the x-y plane (perpendicular to motion), and two components to represent that we may wish to model them the entangled pair as separate particles.

As per the convention of the Schrödinger equation, the total energy is ${\displaystyle \hbar \Sigma \omega =\hbar (\omega _{A}+\omega _{B})}$. Equation (4) satisfies the condition that the photons have zero angular momentum. But it lacks even or odd parity with respect to the z direction. To remedy this, we first express our variables in terms of x and y unit vectors using (2):

${\displaystyle \psi _{RR}=\left[{\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{2}}+i{\frac {{\hat {y}}{\hat {x}}-{\hat {x}}{\hat {y}}}{2}}\right]e^{i(k_{B}z_{B}-k_{A}z_{A})}}$
${\displaystyle \psi _{LL}=\left[{\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{2}}+i{\frac {{\hat {x}}{\hat {y}}-{\hat {y}}{\hat {x}}}{2}}\right]e^{i(k_{B}z_{B}-k_{A}z_{A})}}$
(5)

Next we add and subtract the states:

${\displaystyle {\frac {\psi _{RR}+\psi _{LL}}{\sqrt {2}}}={\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{\sqrt {2}}}e^{i(k_{B}z_{B}-k_{A}z_{A})}\equiv \psi _{+}}$
${\displaystyle {\frac {\psi _{RR}-\psi _{LL}}{\sqrt {2}}}=i{\frac {{\hat {y}}{\hat {x}}-{\hat {x}}{\hat {y}}}{\sqrt {2}}}e^{i(k_{B}z_{B}-k_{A}z_{A})}\equiv \psi _{-}}$.
(6)

Now take the complex conjugate of both wavefunctions and combine to obtain states of even (P=+1) and odd (P=−1) parity:

${\displaystyle \Psi _{+}={\frac {\psi _{+}+{\overline {\psi }}_{+}}{2}}={\frac {{\hat {x}}{\hat {x}}+{\hat {y}}{\hat {y}}}{\sqrt {2}}}\cos(k_{B}z_{B}-k_{A}z_{A})}$
${\displaystyle \Psi _{-}={\frac {\psi _{-}-{\overline {\psi }}_{-}}{2}}={\frac {{\hat {x}}{\hat {y}}-{\hat {y}}{\hat {x}}}{\sqrt {2}}}\sin(k_{B}z_{B}-k_{A}z_{A})}$
(7)

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. Equations (6) and (7) are often stated in Dirac notation by expressing the even parity state as |LL>+|RR>=|xx>+|yy>, and the odd state as |RR>−|LL>=|xy>−|yx>.