Solving Particle Motion in a Rigid Box: t=0

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To solve the problem of a particle in a rigid box when one wall is destroyed at t=0, the initial wavefunction must be expanded in terms of free particle plane waves with positive momentum, ensuring a node at the remaining wall. The lack of translation symmetry means momentum is not a good quantum number, but momentum squared remains valid. By combining eigenfunctions of both positive and negative momentum, one can derive eigenfunctions that meet the necessary boundary conditions. The evolution of the wavefunction ψ(x, t) can then be determined using these combined eigenfunctions. This approach effectively addresses the changes in the system due to the destruction of the wall.
Svyatoslav
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I have question, how can I solve problem of particle in rigid box when one of the wall gets completely destroyed? At time t = 0 the right wall of box gets completely destroyed, left wall is still here( ψ(0) = 0 ), also at t = 0 we know that particle is in ground state.
How can I search for evolution ψ(x, t)?
 

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Expand the t=0 wavefunction (presumably the g.s. of a particle in a box) in free particle plane waves with positive momentum and having a node at the left wall.
 
Christopher Grayce said:
Expand the t=0 wavefunction (presumably the g.s. of a particle in a box) in free particle plane waves with positive momentum and having a node at the left wall.
Due to the wall, there is no translation symmetry and momentum is not a good quantum number. However, momentum squared is. Combining eigenfunctions of positive and negative momentum you can get eigenfunctions of the Hamiltonian that satisfy the appropriate boundary condition.
 
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