I Infinite Square Well with an Oscillating Wall (Klein-Gordon Equation)

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The discussion focuses on solving a relativistic infinite square well problem with an oscillating wall using the Klein-Gordon equation in Mathematica. The user transformed the spatial coordinate to make the wall appear static, but encountered issues with the wavefunction blowing up when the oscillation frequency was set to 1000. This blow-up was linked to the wall's velocity exceeding the speed of light in their unit system. The user identified the error and corrected it by adjusting the wall's function to ensure the velocity remained within acceptable limits. The resolution prevented the previously encountered issues with the wavefunction.
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Numerical solution of the wavefunction blows up when the frequency of oscillation of the wall is high. What does this mean?
I am trying to numerically solve (with Mathematica) a relativistic version of infinite square well with an oscillating wall using Klein-Gordon equation. Firstly, I transform my spatial coordinate ## x \to y = \frac{x}{L[t]} ## to make the wall look static (this transformation is used a lot in solving non-static boundary condition in the non-relativistic case), which brings Klein-Gordon equation to :
Input :
1644917229990.png

Output :
1644917258275.png

All constants have been set to 1
1644917266102.png

I tried to solve this system where ##L(t)=2+sin(1000 t)## using NDSolve :
1644917349008.png

Then I plot my result as a function of ##(y,t)## :
1644917373977.png

The wavefunction ##\psi## blows up. This doesn't happen when I tune the frequency down to 1, ##L(t)=2+sin(t)##
1644917399498.png

My question is why does ##\psi## blow up when the frequency of the oscillation is high and what does it mean? Does it have anything to do with particle production? Or did I just mess up my code?
 

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I have located my error. It was the fact that when I set ##\omega## to 1000, the velocity of the wall, ##\dot{L} = \omega cos(\omega t) \ge 1##, which is not allowed in the units that I am working with ##(c=1)##.
The solution to this is just to set ##L(t) = \frac{1}{\omega} (2+sin(\omega t))##, and this disastrous result doesn't happen anymore.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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