Potential well

Setting up the infinite well. A particle of mass $m$ moves freely inside $0 \le x \le L$ but is completely excluded from outside, where $V(x) = \infty$. Because the particle can never be found outside, the wavefunction must vanish at the edges:

where: $\phi(x)$ is the spatial wavefunction and $L$ is the well width. These are boundary conditions, not assumptions — they follow from $\phi = 0$ wherever $V = \infty$, plus continuity of $\phi$.

Solving inside the well. With $V = 0$ inside, the time-independent Schrödinger equation becomes $-\dfrac{\hbar^2}{2m}\dfrac{d^2\phi}{dx^2} = E\phi$, a simple harmonic-oscillator-type equation in space rather than time. Its general solution is $\phi(x) = A\sin(kx) + B\cos(kx)$ with $k = \sqrt{2mE}/\hbar$. The condition $\phi(0)=0$ kills the cosine term ($B=0$); the condition $\phi(L)=0$ then forces $\sin(kL)=0$, which only happens at $kL = n\pi$ for a positive integer $n$. Quantizing $k$ quantizes the energy through $E = \hbar^2k^2/2m$.

Why n cannot be zero. $n=0$ gives $\phi \equiv 0$ everywhere, meaning no particle at all, so it is excluded. The lowest physically allowed state is $n=1$, and it carries strictly positive energy — a zero-point energy forced by confinement plus the uncertainty principle: squeezing the particle to $\Delta x \sim L$ requires $\Delta p$, and hence kinetic energy, to be nonzero.

Finite well — a qualitative picture. Replace the infinite walls with walls of finite height $V_0$. Outside the well, where classically the particle has no business being ($E < V_0$), the wavefunction no longer cuts off sharply — it decays exponentially, the same mathematics as barrier tunneling. Two consequences follow directly: the well now supports only a finite number of bound states (past some $n$, the required energy would exceed $V_0$), and each allowed energy sits slightly below the corresponding infinite-well value, because the wavefunction effectively spreads over more than width $L$, lowering its curvature and kinetic energy.

An electron sits in an infinite square well of width $L = 0.20\ \text{nm}$, an atomic-scale trap. Find $E_1$ and $E_2$ in eV, the wavelength of a photon emitted in the $n=2 \to n=1$ transition, and the probability of finding the ground-state electron in the central third of the well.

Step 1 — ground-state energy. $E_1 = \dfrac{\pi^2\hbar^2}{2mL^2}$ with $\hbar = 1.055\times10^{-34}\ \text{J}\,\cdot\,\text{s}$, $m = 9.11\times10^{-31}\ \text{kg}$, $L = 0.20\times10^{-9}\ \text{m}$. The numerator $\pi^2\hbar^2 = 1.099\times10^{-67}$, the denominator $2mL^2 = 7.29\times10^{-50}$, so $E_1 = 1.51\times10^{-18}\ \text{J} = 9.4\ \text{eV}$.

Step 2 — first excited state and transition. Since $E_n \propto n^2$, $E_2 = 4E_1 = 37.6\ \text{eV}$, and the transition energy is $\Delta E = E_2 - E_1 = 3E_1 = 28.2\ \text{eV} = 4.51\times10^{-18}\ \text{J}$. Using $\Delta E = hc/\lambda$ with $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ and $c = 3\times10^8\ \text{m/s}$, $\lambda = hc/\Delta E = 4.41\times10^{-8}\ \text{m} = 44.1\ \text{nm}$, in the ultraviolet.

Step 3 — probability in the central third. $P = \dfrac{2}{L}\displaystyle\int_{L/3}^{2L/3}\sin^2\!\left(\frac{\pi x}{L}\right)dx$. Using $\sin^2\theta = \tfrac12(1-\cos2\theta)$ and evaluating the limits gives $P = \tfrac13 + \dfrac{\sqrt3}{2\pi} \approx 0.333 + 0.276 = 0.609$.

Final answers: $E_1 = 9.4\ \text{eV}$, $E_2 = 37.6\ \text{eV}$, emitted photon $\lambda = 44.1\ \text{nm}$ (UV), and about $61\%$ probability in the central third — above the classical $1/3$, since the $n=1$ state peaks at the well's centre.

Competitive stretch

If the same electron instead sat in a finite well of the same width $L=0.20\ \text{nm}$ and depth $V_0 = 50\ \text{eV}$, both $E_1$ and $E_2$ would come out somewhat below $9.4\ \text{eV}$ and $37.6\ \text{eV}$ respectively (wavefunction leakage lowers curvature and hence kinetic energy), and since $E_2 = 37.6\ \text{eV}$ from the infinite-well estimate is comfortably below $V_0 = 50\ \text{eV}$, both levels should still exist as bound states — but a third level near $4^2E_1 \approx 150\ \text{eV}$ would exceed $V_0$ and therefore not be bound at all, illustrating the finite well's limited state count without needing the transcendental matching equations.

• Including $n=0$ as an allowed state — it forces $\phi \equiv 0$ everywhere, so the true ground state is always $n=1$, never $n=0$, and $E_1 \ne 0$.
• Assuming a finite well has infinitely many bound states like the infinite well — a finite well of given depth and width supports only a limited number.
• Forgetting that outside a finite well $\phi(x)$ decays smoothly and exponentially rather than cutting off sharply — the walls are not perfectly reflecting.
• Misapplying $E_n \propto n^2$ to level spacing — the gap $E_{n+1}-E_n$ grows with $n$, it is not constant, unlike the evenly spaced harmonic oscillator.

Ask the tutor: Want to work through the normalization integral by hand, or check when a candidate energy level in a finite well would exceed $V_0$ and stop being bound — ask the AI tutor to walk it with you.