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I have the following question presented to me:

A particle of mass M is moving in a one-dimensional infinite potential well of length $a$, with walls at $x=0$ and $x=a$, described by the potential:

$ V(x)= \begin{cases} 0, &0\leq x\leq a\\ \infty, &\text{otherwise}\\ \end{cases} $

With this, I had to prove the energy levels, and find the corresponding wavefunctions. This didn't cause me too much grief; I got:

$$E_n = \frac{\hbar^2 \pi^2}{2Ma^2}n^2$$ $$\psi_{n}(x) = \sqrt(\frac{2}{a})\sin(\frac{n\pi x}{a})$$ so long as it is inside the well (0 outside)

The follow up to this is as follows:

The particle is initially in the ground state ($n=1$) of the well of length a. If the right wall of the well is suddenly moved from $x=a$ to $x=4a$, calculate the probability of finding the particle in

i) the ground state of the new well

ii) the first excited state of the new well.

So I began by rewriting the wavefunction in terms of this new length in the ground state (for part i)

which got me

$$\psi^{new}_{n}(x) = \sqrt(\frac{1}{2a})\sin(\frac{n\pi x}{4a})$$

I then proceeded to calculate the following: $$⟨\psi^{old}_{1}|\psi^{new}_1⟩$$

Which I think is (It's the bounds that I'm unsure of?):

$$\int_{0}^{4a} (\psi^{old}_{1}(x))^*\psi^{new}_{1}(x)dx$$

EDIT: Bound should be to just $a$ because the original wavefunction $\psi^{old}_{n}(x)$ does not exist beyond $a$. Thanks noah!

The thing is, I get $0$ when I calculate this, and I'm not too sure if I'm just wrong (conceptually or methodically), or if this is actually right?

The follow up to this, for part ii) of the above question; is it just $$⟨\psi^{old}_{1}|\psi^{new}_2⟩$$ in similar fashion?

Cheers!

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You are conceptually right. Technically, you have forgotten the complex conjugate if the first term in the integral, but everything is real here. But from your post it seems you might have forgotten to respect that $\psi_1^\text{old}$ is 0 outside the original interval. This effectively narrows the integration interval to $0$ to $a$, since the integrand is 0 everywhere else.

The way you have suggested to answer the second part is indeed correct.

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  • $\begingroup$ Perfect! I think that was the one bit of concept I was missing- Thank you so much!! $\endgroup$ – Frankie S. Palmer Apr 29 at 13:39

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