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One dimensional infinite potential well problem

hi,
I am not getting idea to solve below problem
A particle of mass m is in a one-dimensional ,rectangular potential well for which V(x)=0 for 0<x< L and V(x)=infinite elsewhere. The particle is intially prepared in the ground state ψ1 with eigen energy E1. Then , at time t=0, the potential is very rapidly changed so that the original wave function remains the same but V(x)=0 for 0<x<2L and V(x)=infinite elsewhere.Find the probability that the particle is in the first,second,third and fourth excited state of the system when t ≥ 0.
could you help me please.
 

Answers and Replies

Dick
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When the potential is changed suddenly the original wavefunction stays the same. To compute the amplitudes of being in any other state then just compute the overlap integral <psi1|phi> where phi is the wavefunction of the excited state. To get the probability find the modulus squared of the amplitude.
 
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When the potential is changed suddenly the original wavefunction stays the same. To compute the amplitudes of being in any other state then just compute the overlap integral <psi1|phi> where phi is the wavefunction of the excited state. To get the probability find the modulus squared of the amplitude.

the |phi> is the excited states in the new potential, right??
 
Dick
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compute the overlap integral <psi1|phi> where phi is the wavefunction of the excited state.
Two quick questions:

1. Is this 'overlap integral' the convolution of the wavefunctions in each potential?

2. Is taking this 'overlap integral' in such a situation generally the way to tackle problems such as this?
 
Dick
Science Advisor
Homework Helper
26,258
618
Two quick questions:

1. Is this 'overlap integral' the convolution of the wavefunctions in each potential?

2. Is taking this 'overlap integral' in such a situation generally the way to tackle problems such as this?
It's not a 'convolution'. That's something else. It's just the integral conjugate(psi1(x))*psi2(x) over the domain of the wavefunctions. And yes, if everything is properly normalized that's all you have to do.
 

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