06:05
So some random thought of a quantum experiment:
Produce two beams of photons heading towards a screen connected to a computer
In the middle of the setup is some nonlinear crystal that function as a CNOT thus entangling the beams
and just before that, is some nonlinear crystal that has the effect of randomising whether beam 1, beam 2 or both can passes through classically
So you now basically have two product states, heading to the first crystal thus turning it into a mixed state of product states, and then whenever the outcome produced involve both beam 1 and 2, they are entangled before hitting the screen, otherwise, only beam 1 and beam 2 hits the screen
It's a simple setup suggested by last night dream, but I don't see how that will allow more complicated messages to be encoded
Let $\lvert 0\rangle, \lvert 1\rangle$ be beam 1 and beam 2 transmitted though the crystals respectively. Then the initial state is something like:
$\lvert 11\rangle = \begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$
After passing through the 1st crystal, it transforms into the mixed state:
(sorry, should be $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$)
$\frac{1}{3}\left( \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}\right)$
The initial state, where both beams are not blocked, is:
$\lvert 11\rangle = \begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}$
After passing through the first crystal, it becomes the mixed state:
$\frac{1}{3}\left( \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} + \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}\right)$
And then it passes through the CNOT to become
$\frac{1}{3}\left( \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix} + \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}\right)$
huh ok, so it actually does not do anything
ah... I need bell states for the CNOT to do something
So I might need to use other quantum gates instead
ok, I will try this gate instead:
$\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}\mapsto \frac{1}{\sqrt 2}\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$ and leave unchanged otherwise
Then we have the final state be:
$\frac{1}{3}\left( \frac{1}{\sqrt 2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}\right)$
so 1/3 of the time, beam 1 hits the screen while beam 2 is blocked, 1/3 of the time, beam 2 hits the screen while beam 1 is blocked
and then 1/9 of the time both beams are blocked and 1/9 of the time both beams passes through
sorry, 1/6 of the time both beams blocked and 1/6 both beams passes
A quick check with $\rho^2 \neq \rho$ showed it is a mixed state