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Consider a particle of mass m in the normal ground sate of an infinite square well potential of width a/2. Its normalized wave function at time t=0 is given by

[tex] \Psi(x,0) = \frac{2}{\sqrt{a}} \sin \frac{2 \pi x}{a} [/tex] for 0 <x <a/2

0 elsewhere

At this time the well suddenly changes to an infinite square well with width a without affectring the wave function.

By writing Psi(x,t) as a linear superposition of the energy eigenfunctions of the new potentaail find the probability taht a subsequency measurement of the enrgy will yield th result

[tex] E_{1} = \frac{\hbar^2 \pi^2}{2ma^2} [/tex]

{Hint: A linear superposition of square well eigenfunctions is a Fourier sine series and teh coefficients of teh series are given by simple integrals)

Now we know that

[tex] \Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) [/tex]

and the wave function didnt change

from teh hint

if i find the coefficients all i do is find an approximation to teh wave function. the new potentail would be

V(x) = 0 for 0<x<a

infinity elsewhere

and we know the wave function for that potential it is

[tex] \Psi_{2}(x,0) = \sqrt{\frac{2}{a}} \sin\frac{\pi x}{a} [/tex]

for the first exceited state

so am i to find Psi 1 in terms of Psi 2??

[tex] \Psi(x,0) = \frac{2}{\sqrt{a}} \sin \frac{2 \pi x}{a} [/tex] for 0 <x <a/2

0 elsewhere

At this time the well suddenly changes to an infinite square well with width a without affectring the wave function.

By writing Psi(x,t) as a linear superposition of the energy eigenfunctions of the new potentaail find the probability taht a subsequency measurement of the enrgy will yield th result

[tex] E_{1} = \frac{\hbar^2 \pi^2}{2ma^2} [/tex]

{Hint: A linear superposition of square well eigenfunctions is a Fourier sine series and teh coefficients of teh series are given by simple integrals)

Now we know that

[tex] \Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) [/tex]

and the wave function didnt change

from teh hint

if i find the coefficients all i do is find an approximation to teh wave function. the new potentail would be

V(x) = 0 for 0<x<a

infinity elsewhere

and we know the wave function for that potential it is

[tex] \Psi_{2}(x,0) = \sqrt{\frac{2}{a}} \sin\frac{\pi x}{a} [/tex]

for the first exceited state

so am i to find Psi 1 in terms of Psi 2??

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