I Klein-Gordon Equation with boundary conditions

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The discussion focuses on finding solutions to the Klein-Gordon equation for a one-dimensional particle in an oscillating box with time-dependent boundary conditions. The proposed method involves using a homogeneous solution and Eigenfunction expansion, but progress has been difficult. An alternative approach suggests using an asymptotic expansion with a small parameter to simplify the boundary conditions. Additionally, applying a Laplace transform in time and changing variables may help streamline the calculations. It's noted that resonance could complicate the asymptotic assumptions if the oscillation frequency aligns with natural frequencies of the system.
dsaun777
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I am trying to find solutions for the Klien-Gordon equations in 1-d particle in a box. The difference here is the box itself oscillating and has boundary conditions that are time dependent, something like this L(t)=L0+ΔLsin(ωt). My initial approach is to use a homogeneous solution and use Eigenfunction expansion to get a solution. But I can't make progress. Are there other methods that are easier? maybe numerical methods...
 
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Do you mean that you have homogenous BCs at x = 0 and x = L(t)?

Set \epsilon = \Delta L/L_0 so that the boundary condition is applied at x = L_0(1 + \epsilon \sin \omega t). Then for small \epsilon, you can seek an asymptotic expansion \psi = \psi_0 + \epsilon \psi_1 + \dots where the boundary conditions on the \psi_n are shifted to x = L_0 by expanding \psi(L_0(1 + \epsilon \sin \omega t),t) in Taylor series in x about x = L_0 and comparing coefficients of powers of \epsilon. This gives you a series of problems which can be solved by eigenfunction expansion.

EDIT: It might be better to solve for each \psi_n using a Laplace transform in time. The change of variable \tilde x = x/L_0 may help to simplify the algebra. Also note that if \omega is a natural frequency of any of these problems then the resulting resonance will cause the asymptotic assumption to break down, since eventually the amplitude will exceed \epsilon^{-1}.
 
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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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