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

[boooster]

Homework Statement

Consider a particle confined in a 1D potential well of length 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.

Homework Equations

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

$$\Psi = \sum_n a_n \phi_m$$

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

$$\Psi = \sum_m b_m \chi_m$$

where $$b_m = \sum_n a_n \int \chi_m^* \phi_n dx$$

The Attempt at a Solution

$$\phi_1 = \sqrt{\frac{2}{L}} \sin{\left(\frac{\pi x}{L}\right)}$$
$$\chi_m = \sqrt{\frac{1}{L}} \sin{\left(\frac{\pi x m}{2 L}\right)}$$

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

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!

 

[nrqed]

Homework Statement

Consider a particle confined in a 1D potential well of length 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.

Homework Equations

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

$$\Psi = \sum_n a_n \phi_m$$

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

$$\Psi = \sum_m b_m \chi_m$$

where $$b_m = \sum_n a_n \int \chi_m^* \phi_n dx$$

The Attempt at a Solution

$$\phi_1 = \sqrt{\frac{2}{L}} \sin{\left(\frac{\pi x}{L}\right)}$$
$$\chi_m = \sqrt{\frac{1}{L}} \sin{\left(\frac{\pi x m}{2 L}\right)}$$

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

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!

I di dnot check your integral. But it's sure that the sum of a_m^2 you gave from 1 to infinity does converge. But again, I did not check the integration.

Note that you need only a_1 and a_2 for your assignmnent

 

[Avodyne]

First of all, the problem is ambiguous, because it does not tell you how the walls move. You have assumed that one wall stays put at $x=0$, and the other moves from $x=L$ to $x=2L$. If the walls instead moved from $x=\pm L/2$ to $x=\pm L$, the answer would be different.

Given your assumption (which is perfectly reasonable), your answer is correct. Note that $a_2$ has to be obtained by taking a limit, and that the result is $a_2=1/\sqrt{2}$. All other $a_m$'s with $m$ even are zero. Then, the sum over odd $m$ of $|a_m|^2$ is $1/2$, and adding $|a_2|^2$ yields at total of one.