# User:Guy vandegrift/Bell correlation attempt 3

ejp correlation.......ejp sandbox

## The basic wave in quantum mechanics

Consider a wave travelling in the ±z direction. In quantum mechanics travelling waves take the form,

${\displaystyle \psi _{\pm }(x)=e^{i(\pm kz-\omega t)},}$

where ω is positive if the energy is positive, and f± each represent a complex waveform. We also need notation for objects that rotate in the complex plane. Traditionally such rotations are associated with "left/right" or "plus/minus", but we need to reserve these terms for other attributes. Hence we use "clockwise" and "counter clockwise":

### Single particle states

First we consider single particles

${\displaystyle \phi _{\text{counter}}=\left({\hat {x}}-i{\hat {y}}\right)e^{-i\omega t}}$ and ${\displaystyle \phi _{\text{clock}}=\left({\hat {x}}+i{\hat {y}}\right)e^{-i\omega t}}$
(1)

For example the real part of ${\displaystyle \phi _{\text{counter}}}$ is ${\displaystyle \cos(\omega t){\hat {x}}+\sin(\omega t){\hat {y}}}$ which rotates in the counter-clockwise direction, as does the imaginary part of ${\displaystyle \phi _{\text{counter}}}$. Both real and imaginary part of ${\displaystyle \phi _{\text{clock}}}$ rotate clockwise. If the rotation is counter clockwise in the xy plane as viewed from +z, the motion is right-handed if the object is moving in the +z direction, but left-handed if moving in the −z direction:

${\displaystyle \psi _{+}^{R}=\left({\hat {x}}-i{\hat {y}}\right)e^{ikz-i\omega t}}$ is a Right-handed photon travelling in the +z direction.
${\displaystyle \psi _{-}^{R}=\left({\hat {x}}+i{\hat {y}}\right)e^{-ikz-i\omega t}}$ is a Right-handed photon travelling in the −z direction.
${\displaystyle \psi _{+}^{L}=\left({\hat {x}}+i{\hat {y}}\right)e^{ikz-i\omega t}}$ is a Left-handed photon travelling in the +z direction.
${\displaystyle \psi _{-}^{L}=\left({\hat {x}}-i{\hat {y}}\right)e^{-ikz-i\omega t}}$ is a Left-handed photon travelling in the −z direction.
(3)

It the parent atom does not change angular momentum during the two-photon decay, either both photons are left-hand polarized or both are right-hand polarized. For complicated reasons[1] we need a mixture of both states to achive photon states of even or odd parity:

${\displaystyle \psi _{\text{even}}=\psi _{+}^{R}\otimes \psi _{-}^{R}+\psi _{+}^{L}\otimes \psi _{-}^{L}}$
${\displaystyle \psi _{\text{even}}=\psi _{+}^{R}\otimes \psi _{-}^{R}-\psi _{+}^{L}\otimes \psi _{-}^{L}}$

stop

{\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[2] 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>.