# Why do the energies of the infinite square well decrease as width of the well increases?

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?

• 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. Oct 26, 2017 at 2:43

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.
• 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 $E$. Oct 26, 2017 at 12:32
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.
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.
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.