= Solution
\b[Equation and interpretation.] The supplied equation is the <Time-independent Schrödinger equation> for a nonrelativistic particle with <mass> $m$ in a prescribed <potential energy> $V(x)$. Equivalently,
$$
\widehat H\psi=E\psi,\qquad
\widehat H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x).
$$
It is an <eigenvalue> problem for the <Hamiltonian operator>. A stationary energy state has full <wavefunction> $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$. By the <Born rule>, $|\psi(x)|^2dx$ is the probability of finding the particle in a small interval $dx$, with <wavefunction normalization> $\int_{-\infty}^{\infty}|\psi|^2dx=1$.
\b[Infinite barriers.] Infinite exterior <potential energy> excludes any finite-energy probability outside the interval. The <infinite square well> consequently has $\psi=0$ outside and the hard-wall boundary conditions
$$
\boxed{\psi(0)=\psi(a)=0.}
$$
The interior <wavefunction> must meet the zero exterior amplitude continuously. Its derivative need not be continuous across an infinite wall; imposing both zero value and zero derivative would eliminate every nonzero interior solution.
\b[Ground state.] Inside the well $V=0$, so $\psi''+k^2\psi=0$, with $k=\sqrt{2mE}/\hbar$. Its general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$, and the right wall gives $ka=n\pi$ for $n=1,2,\ldots$. Thus the discrete energy <eigenvalues> are $E_n=n^2\pi^2\hbar^2/(2ma^2)$. The smallest, $n=1$, is the <ground state>. Since
$$
\int_0^a\sin^2\frac{\pi x}{a}\,dx=\frac a2,
$$
its normalized <wavefunction>, choosing an irrelevant overall phase to make it positive inside, is
$$
\boxed{\psi_1(x)=\begin{cases}
\sqrt{2/a}\,\sin(\pi x/a),&0<x<a,\\
0,&x\le0\ \text{or}\ x\ge a,
\end{cases}\qquad E_1=\frac{\pi^2\hbar^2}{2ma^2}.}
$$
\b[Finite barriers.] For a finite step and constant particle <mass>, both $\psi$ and $\psi'$ are continuous at $x=0,a$. <Wavefunction matching at a finite potential step> follows by integrating the <Schrödinger equation> across a vanishingly small neighborhood of the step: a bounded potential cannot generate a finite jump in $\psi'$. The bound-state exterior solutions decay rather than vanish identically:
$$
\psi(x)\propto e^{\kappa x}\quad(x<0),\qquad
\psi(x)\propto e^{-\kappa(x-a)}\quad(x>a),\qquad
\kappa=\frac{\sqrt{2m(V_w-E)}}{\hbar}.
$$
Normalizability excludes the growing exterior exponentials. The equivalent interior boundary conditions are $\psi'(0)=\kappa\psi(0)$ and $\psi'(a)=-\kappa\psi(a)$; neither boundary value is zero for the nontrivial <ground state>.
The <ground state of a finite square well> remains symmetric about $a/2$ and has no nodes. It bends smoothly into exponential tails at both walls, extending over a larger region than the infinite-well state. Its maximum after normalization is lower for the deep well considered here. When $V_w\gg E$, the penetration length $1/\kappa$ is short, so the two shapes are close except near the walls and in the exterior tails.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018-ia-physics-q13-well-states.png]
{title=Normalized infinite- and finite-well ground states, with smooth exponentially decaying tails for the finite well}
The finite curve is an illustrative exact matched solution with $V_w=10E_{1,\infty}$; no numerical wall height is specified by the question.
\b[Energy comparison.] The finite barriers relax the confinement, allowing a less rapidly varying interior <wavefunction> and therefore smaller <kinetic energy>. More explicitly, the even interior solution is proportional to $\cos[k(x-a/2)]$, and matching to the tails gives $k\tan(ka/2)=\kappa$. The lowest root lies at $0<ka<\pi$, while the infinite well has $ka=\pi$. Hence
$$
\boxed{E_{1,\rm finite}<\frac{\pi^2\hbar^2}{2ma^2}=E_{1,\infty}.}
$$
As $V_w\to\infty$, the tails disappear and the finite-well energy approaches the infinite-well value from below. This comparison uses the same zero of <potential energy> inside both wells.
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