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The stationary state solutions to the infinite square well potential are given by, $$\psi_n(x)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right).$$ The energies corresponding to these states are $$E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}.$$ In the limit that the width of the well tends to infinity ($a\to\infty$), we are essentially left with an unconfined free particle. Is it possible to somehow recover at least some bits and pieces of the free-particle solution from the solutions to the infinite square well in the limit $a\to\infty$?

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    $\begingroup$ There is (at least) one subtle issue I think. The solution given is for a well that extends from $x=0$ to $x=a$ so letting $a\rightarrow\infty$ does not remove the infinite potential for $x < 0$. This is easy to fix but there's another issue I'm thinking of related to the fact that the stationary states have zero expectation for momentum. $\endgroup$ – Alfred Centauri Oct 27 '17 at 1:42
  • $\begingroup$ Let's note that the eigenfunctions do become continuous in this limit, which is indeed an element of the known solution. $\endgroup$ – psitae Oct 27 '17 at 5:05
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This is complex, rather than answering immediately let me give a construction that I just came up with. Take a look at the energy spectrum. $E_n$ is discrete, however it can be arbitrarily large. This means that taking the limit $a\to\infty$ regarding $n$ constant is not justified. Sending both to infinity makes you lose information regarding the ground state, maybe even more. You can however, very conveniently rename what is involved in the following fashion: $$ E_n=\frac{1}{2m}\left(\frac{n\pi \hbar}{a}\right)^2=\frac{p_n^2}{2m} $$ where $p_n=n\pi\hbar/a$. Your wavefunction is then: $$ \psi_n(x)=\sqrt{\frac{2}{a}}\sin{\left(\frac{p_nx}{\hbar}\right)}=\frac{-i}{\sqrt{2a}}\left(e^{ip_nx/\hbar}-e^{-ip_nx/\hbar}\right) $$ If $a$ is finite, you can always find $n$ so that $n\pi\hbar>a$ (this is called the Archimedean property). We can state that in such situation the spectrum is not bounded from above (however it is clearly bounded from below).

On the other hand for every $p\in \mathbb{R}$ we can find an $n/a$ such that $p_n$ is arbitrarily close to it (can you see why?).

Almost as good as the last statement, assume that $a$ is very big but finite, then as $\|p_{n+1}-p_n\|\propto 1/a$. It is then easy to see that given $p$, one can find $p_n$ such that $\|p-p_n\|\propto 1/2a$.

The point here is that as $a$ gets larger, the energy spectrum remains unbounded above and you can guarantee that you can get really close to any value of $p$ you want. This heuristic argument is to justify the following prescription: $$ \psi_n(x)\to \psi(x,p)=\psi(x,p_{n(p)}) $$ where $$ \psi(x,p)=\frac{-i}{\sqrt{2a}}\left(e^{ip_{n(p)}x/\hbar}-e^{-ip_{n(p)}x/\hbar}\right) $$ and $n$ is the integer that guarantees that $p_n$ lies closer to $p$ than any other integer. We are forced to admit that this implies that there is degeneracy in the now continuum index $p$. However we can establish that the solution is a superposition of two free eigenstates of the momentum operator $i\hbar\partial_x$, with positive and negative eigenvalues, whose absolute value are equal and in general really close to $p$.

Obviously the wavefunction remains normalized, we have merely rewritten it in terms of $p$. This owes to the fact that we have retained the factor $1/\sqrt{a}$. This makes sense because $a$ remains finite.

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Sure. But you have to be more careful.

The right limit to use is $a \to \infty$ and $n \to \infty$ such that $\frac{n}{a}$ remains constant.

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  • $\begingroup$ Are we simply taking $n\to\infty$, because the index vanishes in the case of the free particle? $\endgroup$ – quanticbolt Oct 27 '17 at 1:41
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    $\begingroup$ Notice that you wrote that energy is proportional to $n^2/a^2$. in order to have free solutions we need to have a varying energies that that isn't going to happen is you allow $a$ to grow without bound while retaining a fixed $n$. By choosing the constant value of $n/a$ you select an energy and different values give you different energies. $\endgroup$ – dmckee Oct 27 '17 at 2:06

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