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Suppose there's a particle in a box, initially in its ground state. Suppose that one chooses a system of coordinates such that the potential V(x) is 0 from 0 to L.

Suppose that one suddenly perturbate the system at a particular time so that V(x) becomes 0 from 0 to 2L.

I've calculated the probability for the particle to be in the ground state of the twice-bigger box as ##\frac{2^{3/2}}{3\pi}##.

Now suppose we solve a similar problem, where V(x) is initially worth 0 from -L/2 to L/2 and psi is in its ground state, and suddenly one expands the box so that V(x)=0 from -L to L.

Now the probability for the particle to be in the ground state of the 2L sized box is... ##\frac{2^3}{3\pi}##. Which is different from the 1st case. (I may not recall the exact values but I remember they differed).

However I'm having trouble understanding what's the difference physically in the 2 cases. To me it just looks like a 1 dimensional box of size L suddenly double its size, in both cases. So physically I don't understand why I get 2 different results as if there was a priviledged observer in the universe.

Mathematically I do understand that the overlap of the 2 wavefunctions (before and after the perturbation) are different in the 2 cases. It's just that physically I'm not grasping what's going on.

Could someone shed some light on this? Thanks.

P.S.: ##\psi _0(x)=\sqrt {\frac{2}{L}} \sin (\frac{\pi x}{L} )## for the ground state of the L sized box according to an observer whose coordinate system's origin is at the left wall.

If the origin of that coordinate system is set midway between the 2 walls, ##\psi _0(x)=\sqrt {\frac{2}{L}} \cos (\frac{\pi x}{L} )##.

##P=|\langle \psi _{\text {final}}| \psi _{\text {initial}} \rangle |^2##.

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# Sudden perturbation, particle in a box, ground state

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