Infinite Square Well — Eigenfunctions and Energy Levels: 17-Phys-A4 Quantum Mechanics, May 2018, Q1

Question 1: Infinite Square Well — Eigenfunctions and Energy Levels (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A particle of mass $m$ is confined to $0\le x\le L$ by an infinitely high potential step at each wall, $V(x)=0$ inside and $V(x)=+\infty$ outside.

Find. The normalized energy eigenfunctions $\psi_n(x)$ and the corresponding energy eigenvalues $E_n$.

Fig. 1 — infinite square well of width $L$; the first three eigenfunctions $\psi_n(x)$ are drawn about their own energy level $E_n=n^2E_1$, each vanishing at $x=0$ and $x=L$.

Approach. Solve the time-independent Schrödinger equation in the field-free interior, apply the two boundary conditions $\psi(0)=\psi(L)=0$ forced by the infinite walls, then normalize.

Write the TISE inside the well. For $0 < x < L$, $V=0$, so $$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \quad\Longrightarrow\quad \frac{d^2\psi}{dx^2} = -k^2\psi,\qquad k=\frac{\sqrt{2mE}}{\hbar}. $$ Outside the well $V\to+\infty$ forces $\psi=0$ there (a particle cannot penetrate an infinite barrier).
General solution and boundary conditions. The general solution is $\psi(x)=A\sin(kx)+B\cos(kx)$. Continuity at the walls (matching the $\psi=0$ exterior) requires $\psi(0)=0\Rightarrow B=0$, and $\psi(L)=0\Rightarrow A\sin(kL)=0$. Since $A\ne0$ (else no wavefunction), $\sin(kL)=0$, so $$ kL = n\pi,\qquad n=1,2,3,\dots $$ ($n=0$ is excluded — it gives $\psi\equiv0$, no particle).
Normalize. With $k_n=n\pi/L$, $\psi_n(x)=A\sin(n\pi x/L)$. Requiring $\int_0^L|\psi_n|^2dx=1$: $$ A^2\int_0^L\sin^2\!\left(\frac{n\pi x}{L}\right)dx = A^2\cdot\frac{L}{2}=1 \;\Longrightarrow\; A=\sqrt{\frac{2}{L}}. $$ $$ \boxed{\psi_n(x)=\sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right),\qquad n=1,2,3,\dots} $$
Quantized energies. Substituting $k_n=n\pi/L$ back into $E=\hbar^2k^2/2m$: $$ \boxed{E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}=\frac{n^2h^2}{8mL^2},\qquad n=1,2,3,\dots} $$ Energies scale as $n^2$ and are non-degenerate in 1-D — a direct consequence of confinement (zero-point energy $E_1=\pi^2\hbar^2/2mL^2\ne0$ is itself a signature of the uncertainty principle).

Final results
Quantity	Result
Eigenfunctions	$\psi_n(x)=\sqrt{2/L}\,\sin(n\pi x/L)$
Energy levels	$E_n=n^2\pi^2\hbar^2/(2mL^2)$, $n=1,2,3,\dots$

Topic: Bound states of the infinite square well.

Key relations: TISE $-\dfrac{\hbar^2}{2m}\psi''=E\psi$; quantization from the boundary condition $\sin(kL)=0$; $E_n\propto n^2$.

Why this works: An infinite potential step forces the wavefunction (and hence the probability of finding the particle) identically to zero outside the well, which is a much stronger constraint than a finite step; it is this hard boundary condition — not the Schrödinger equation itself — that quantizes the spectrum. The resulting standing-wave eigenfunctions are exactly the normal modes of a string fixed at both ends, with $k_n=n\pi/L$ playing the role of the mode wavenumber.

Common pitfall: Forgetting to discard $n=0$ (gives the null wavefunction, not a valid state) or keeping the cosine term (it cannot satisfy $\psi(0)=0$ unless its coefficient is zero).

Source: Griffiths-QM §2.2 (The Infinite Square Well).

Question 1 of 7: Infinite Square Well — Eigenfunctions and Energy Levels

Question 1: Infinite Square Well — Eigenfunctions and Energy Levels (20 marks)