When an infinite square well grows from $(0, a)$ to $(0, 2a)$, why do we integrate from $0$ to $2a$?

Question

Say we have a particle in an infinite square well where $V= infty$ if $x<0$ or $x>a$. Then, suppose the well expanded to $V= infty$ if $x<0$ or $x>2a$. While the wave function is momentarily undisturbed, to find the probability of new possible states of the particle, we compute

$$c_n = int_0^a sqrt{frac{2}{2a}}sinleft(frac{npi x}{2a}right)sqrt{frac{2}{a}}sinleft( frac{pi x}{a} right)dx.$$

I’m wondering why the bounds are from 0 to a and not 0 to 2a.

boundary conditions potential probability quantum mechanics

Chris · Accepted Answer

The key words are "the wavefunction is momentarily undisturbed." Before the perturbation, the wavefunction is zero outside the range $0$ to $a$. Immediately after the perturbation, the wavefunction is unchanged, so it's still zero in the range $a$ to $2a$. You can write this a little more explicitly as:
$$begin{align}
c_n &= int_0^{2a} sqrt{frac{2}{2a}}sinleft(frac{npi x}{2a}right)psi(x),dx &=int_0^a sqrt{frac{2}{2a}}sinleft(frac{npi x}{2a}right)sqrt{frac{2}{a}}sinleft( frac{pi x}{a} right)dx+int_a^{2a}sqrt{frac{2}{2a}}sinleft(frac{npi x}{2a}right) 0,dx
end{align}$$
where the second term clearly integrates to zero and thus can be left out entirely, leaving you with the expression in your question.

When an infinite square well grows from $(0, a)$ to $(0, 2a)$, why do we integrate from $0$ to $2a$?

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