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The stationary state wavefunctions for the infinite square-well of width $a$ are given by $$\psi_n(x)=\sqrt{\frac{2}{a}}\sin{(\frac{n\pi x}{a})}.$$ These correspond to energies, $$E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}.$$ Suppose we are to modify the width of the well, such that the new width is given by $2a$. Then, the new stationary state wavefunctions become, $$\psi_n(x)=\sqrt{\frac{2}{a}}\sin{(\frac{n\pi x}{2a})},$$ corresponding to energies, $$E_n=\frac{n^2\pi^2\hbar^2}{2m(2a)^2}.$$ Evidently, the energies decrease as the width of the well increases.

Classically, this is perfectly logical to me. If we confine a discrete particle to a small region in one-dimensional space, it makes sense that it would bounce around (against the walls of its confines), much more than if we were to confine it to a larger region in one-dimensional space.

However, in quantum mechanics, we can't think of the particle as a point in space, but must rather think of it as a wave of sorts. If such is the case, then what could be a possible physical explanation for the decrease in energy as we increase the width of the well?

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    $\begingroup$ Actually, the classical analogy in your penultimate paragraph is incorrect. The frequency with which a classical point particle bounces with the walls does not affect its energy since kinetic energy is conserved in elastic collisions. $\endgroup$
    – lcortesh
    Oct 26, 2017 at 2:43

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In the infinite well, the kinetic energy $p^2/2m$ is the only quantity that matters because $V=0$ inside the well. Since $$ \frac{p^2}{2m}\to -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\, , $$ the kinetic enery is proportional to the curvature of the wave function.

If the well is narrow, the wavefunctions will have higher curvature since they must accommodate an integer number of sinusoïdal half-cycles inside the narrow well. Contrariwyse if the well is broad, the same solutions will have comparatively much less curvature and will thus have lower energy.

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    $\begingroup$ could you please explain why are you relating curvature with energy,physically? $\endgroup$
    – user157588
    Oct 26, 2017 at 11:47
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    $\begingroup$ The only argument is as above, i.e. $H=p^2/(2m)+V$, with $V=0$ inside so that $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$. Thus, $\psi''(x)=\frac{-2mE}{\hbar^2}\psi$ shows that the curvature of $\psi$, contained in $\psi''(x)$, is explicitly dependent on $E$. $\endgroup$ Oct 26, 2017 at 12:32
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The standard heuristic story is based on the Heisenberg uncertainty principle $\Delta x \cdot \Delta p \geq \hbar / 2$. The lower the uncertainty in a particle's position, the greater the uncertainty in its momentum (and vice versa). If a particle is confined to a box, then the uncertainty in its position certainly can't be greater than the size $L$ of the box, so $\Delta p \geq \hbar / (2L)$. Increasing the size of the box allows it to spread out more and increases the uncertainty in its position, thereby decreasing the spread of its momenta about the average value $p = 0$, so the particle "slows down" and its energy decreases.

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You showed that $E_n \propto \frac{1}{a^2}$. That is a physical explanation to why the energy decreases as $a$ increases. In QM it is usually not helpfull to look for analogies on how particles 'are moving' as in classical mechanics.

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Actually this must be so, since in the $a \rightarrow \infty$ limit, you better recover the quantum mechanics of a free particle moving on the real line. If you define $k_n = n\pi/a$, one can see that for $a$ very large, the gap between successive $k_n$'s becomes really small and so in the $a\rightarrow \infty$ limit it basically becomes a continuous variable. So one gets the energy spectrum $E_k = \hbar^2 k^2/2m$ with wave functions $\psi(x) \propto sin(kx)$, the usual free particle wave.

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