The Wave Function Squared of a Wave Equation with Phase Shifts


Posted by doug
summary: 
The product of one wave equation with another whose phase has been shifted results in an interference pattern.
description: 
The product of one wave function with another that has been shifted is animated. The order of this product is done both ways, resulting in the red and blue animations. In the 3 complex planes, we see 3 patterns. The circle in the t-y plane indicates that a wave equation with harmonic boundary conditions is on display. The straight line in the t-z plane shows nothing is happening except the march of time. Really the same happens in the t-x plane, but the change in phase has been included.
command: 
q_graph -out interference_ty_red_blue -dir int14 -box 1.6 -command '221 "t_function -t_function cos -x_function zero -y_function sin -z_function zero -n_step 1799 -pi 124 -n_t_cycles 1000 -n_t_step 0 1 0 1 0 | q_conj" "t_function -t_function cos -x_function zero -y_function sin -z_function zero -n_step 1799 -pi 124 -n_t_cycles 1000 -n_t_step 0.031 1 0 1 0" | q_x | q_add 0 -1.5 0 0 | q_add_n_m 0 0.003 0 0 1800 1000' -color red -command '221 "t_function -t_function cos -x_function zero -y_function sin -z_function zero -n_step 1799 -pi 124 -n_t_cycles 1000 -n_t_step 0.031 1 0 1 0 | q_conj" "t_function -t_function cos -x_function zero -y_function sin -z_function zero -n_step 1799 -pi 124 -n_t_cycles 1000 -n_t_step 0 1 0 1 0" | q_x | q_add 0 -1.5 0 0 | q_add_n_m 0 0.003 0 0 1800 1000' -color blue
equation: 

\phi'^* \phi = (cos(\omega t + \delta), 0, -sin(\omega t + \delta), 0)(cos(\omega t), 0, sin(\omega t), 0) + (0, k \delta, 0, 0) \, \textrm{with} \, \delta: 0 \to 10 \pi in red
\phi^* \phi'= (cos(\omega t), 0, -sin(\omega t), 0)(cos(\omega t + \delta), 0,  sin(\omega t + \delta), 0) + (0, k \delta, 0, 0) \, \textrm{with} \, \delta: 0 \to 10 \pi in blue



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