(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider a particle confined in a 1D potential well of lenght L. Derive the normalized wavefunctions and energies of the eigenstates of the system in terms of the quantum number n, working from the time independent Schördinger equation.

With the particle initially in the ground state (n=1) in the potential, the potential well instantaneously expands to twice it's original size.

Work out the probability, immediately after this change takes place, of measuring the system in (i) its new ground state and (ii) its new first excited state.

2. Relevant equations

The wavefunction can be written in terms of the eigenstates as follows:

[tex]\Psi = \sum_n a_n \phi_m[/tex]

If we change to a different se of basis states we can write:

[tex]\Psi = \sum_m b_m \chi_m[/tex]

where [tex]b_m = \sum_n a_n \int \chi_m^* \phi_n dx[/tex]

3. The attempt at a solution

[tex]\phi_1 = \sqrt{\frac{2}{L}} \sin{\left(\frac{\pi x}{L}\right)} [/tex]

[tex]\chi_m = \sqrt{\frac{1}{L}} \sin{\left(\frac{\pi x m}{2 L}\right)} [/tex]

I tried to express the probability as [tex]\left|a_m\right|^2[/tex] where [tex]a_m = \int \chi_m^* \phi_1 dx = \int_0^L \frac{\sqrt{2}}{L} \sin{\left(\frac{\pi x}{L}\right)} \sin{\left(\frac{\pi x m}{2L}\right)} dx = \frac{4\sqrt{2}\sin{\frac{m\pi}{2}}}{4\pi -m^2 \pi}[/tex]

Do you think this approach could be right? a_m^2 equals a nice expression - but unfortunately its infinite sum from m=1 to infinity does not converge ... (it should equal 1).

Thanks for the help!

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# Homework Help: Quantum mechanics: 1D potential well instananeously expanded to twice its size

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