Visual walkthrough — Particle in a box — solving TISE, energy levels, wavefunctions

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Step 1 — Draw the trap

WHAT. We put one tiny particle (imagine a single electron, a speck of "quantum stuff") on a line, and we build two walls: one at position $x=0$, one at position $x=L$. Between them the particle is free; the walls are infinitely hard, so the particle can never be found outside or exactly on a wall.

WHY. Every quantum problem starts with where the particle is allowed to be — that is the potential energy $V(x)$. A wall is just a place where $V$ is enormous: it costs infinite energy to be there, so the particle simply cannot.

PICTURE.

The single physical rule we take from this picture: the particle's presence must fade to exactly zero at each wall. We will use that twice, and it does all the work.

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Step 2 — Describe the particle with a wave, $\psi$

WHAT. In quantum mechanics a particle is described not by a dot but by a spread-out wave, written $\psi (x)$ (Greek letter "psi", say "sigh"). At each position $x$, $\psi$ has a height — positive, negative, or zero.

WHY $\psi$ and not just "position"? Because a quantum particle does not have a single definite position. The wave tells you the chance of finding it here or there: the probability density is $|\psi (x){|}^2$ (height squared, always positive). See Wavefunction and Born Rule. For now, only one fact matters:

PICTURE.

Look at the two red dots: the wave is nailed down to zero there. Everything that follows is: "which wave shapes can obey that?"

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Step 3 — The rule the wave must obey inside (the TISE)

WHAT. Inside the box the wave is governed by the time-independent Schrödinger equation (Schrödinger Equation (TISE)). With $V=0$ inside, it reads:

$\underset{\text{a positive number}}{\underbrace{-\frac{{\hslash }^2}{2m}}}\underset{\text{curvature of }\psi }{\underbrace{\frac{d^2\psi }{d{x}^2}}}=\underset{\text{energy}}{\underbrace{E}}\underset{\text{the wave}}{\underbrace{\psi }}$

WHY this equation, and what is each symbol doing?

The equation says, in words: "the curvature of the wave is proportional to (minus) the wave times its energy." A wave that bends more has more energy. That is the whole physics.

WHY rearrange. Divide through to isolate the curvature:

$\frac{d^2\psi }{d{x}^2}=-\underset{\text{a positive number, call it }{k}^2}{\underbrace{\frac{2mE}{{\hslash }^2}}}\psi =-k^2\psi ,\boxed{k^2\equiv \frac{2mE}{{\hslash }^2}}$

PICTURE.

The figure shows the meaning of ${\psi }^{\prime \prime }=-k^2\psi$: wherever the wave is above the axis, it curves downward (pulled back toward zero); wherever it is below, it curves upward. A shape that always bends back toward the axis is a wiggle — a sine or cosine. That is the next step.

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Step 4 — Guess the shape: sine and cosine

WHAT. We need a function whose second derivative is itself, flipped and scaled by $-k^2$. Two functions do this, and their combination is the most general answer:

$\psi (x)=\underset{\text{sine’s height}}{\underbrace{A}}\sin (kx)+\underset{\text{cosine’s height}}{\underbrace{B}}\cos (kx)$

WHY sine and cosine (and why not $e^x$, a parabola, etc.)?

PICTURE.

$A$ and $B$ are just heights — how much of each shape we mix in. The boundary conditions will now fix them.

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Step 5 — Left wall kills the cosine

WHAT. Apply $\psi (0)=0$. Plug $x=0$ into $\psi (x)=A\sin (kx)+B\cos (kx)$:

$\psi (0)=A\underset{=0}{\underbrace{\sin (0)}}+B\underset{=1}{\underbrace{\cos (0)}}=B\stackrel{!}{=}0.$

So $B=0$, and the wave collapses to just:

$\boxed{\psi (x)=A\sin (kx)}.$

WHY. $\sin$ already starts at zero, so it automatically obeys the left wall. $\cos$ starts at height $1$ — it can never be zero at $x=0$, so any cosine we add would poke through the left wall. The left wall therefore forbids the cosine entirely.

PICTURE.

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Step 6 — Right wall quantises $k$

WHAT. Now apply the second boundary condition, $\psi (L)=0$, to the surviving wave $\psi (x)=A\sin (kx)$:

$\psi (L)=A\sin (kL)=0.$

We refuse $A=0$ (that would mean no wave, no particle). So the sine itself must be zero:

$\sin (kL)=0.$

WHY this is the magic step. $\sin (\text{angle})=0$ only at the special angles $0,\pi ,2\pi ,3\pi ,\dots$ — every multiple of $\pi$ (half a full circle-turn). So $kL$ must be one of those:

$kL=n\pi ,n=1,2,3,\dots ⟹\boxed{k_n=\frac{n\pi }{L}}$

Only discrete values of $k$ survive — the right wall rejects every other wave. This is quantisation, appearing for the first time.

PICTURE.

The blue wave ($n=2$, two arches) lands perfectly on the right wall — allowed. The dashed grey wave lands partway up the wall — forbidden, deleted by $\psi (L)=0$.

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Step 7 — Turn allowed $k$ into allowed energy $E$

WHAT. Recall from Step 3 that $k^2=\frac{2mE}{{\hslash }^2}$. Solve it for $E$:

$E=\frac{{\hslash }^2{k}^2}{2m}.$

Now insert the only allowed $k$ values, $k_n=\frac{n\pi }{L}$:

$E_n=\frac{{\hslash }^2}{2m}{\left(\frac{n\pi }{L}\right)}^2=\frac{{\hslash }^2{n}^2{\pi }^2}{2m{L}^2}.$

Swap $\hslash =h/2\pi$ (so ${\hslash }^2{\pi }^2=h^2/4$) to get the textbook form:

$\boxed{E_n=\frac{n^2{h}^2}{8m{L}^2}},n=1,2,3,\dots$

WHY term by term:

PICTURE.

The rungs are not evenly spaced — the gap from $n$ to $n+1$ is $(2n+1)E_1$, so gaps grow going up. (Contrast the evenly spaced ladder of the Quantum Harmonic Oscillator.)

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Step 8 — Fix the height $A$ by normalising

WHAT. The particle must be found somewhere inside with total certainty $1$. Since $|\psi {|}^2$ is the probability density, its total area over the box is $1$:

${\int }_0^L|A\sin (k_{n}x){|}^{2}dx=1.$

WHY. Up to now $A$ was an unknown height. Probability must add to $1$ — that single demand pins $A$ down.

Using the fact that ${\sin }^2$ averages to $\frac{1}{2}$ over a whole number of half-waves, ${\int }_0^L{\sin }^2\left(\frac{n\pi x}{L}\right)dx=\frac{L}{2},$ so $A^2\cdot \frac{L}{2}=1$, giving $A=\sqrt{2/L}$ and the finished wavefunction:

$\boxed{{\psi }_n(x)=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)}.$

PICTURE.

The shaded area under $|{\psi }_1{|}^2$ equals exactly $1$ — that is what $\sqrt{2/L}$ guarantees.

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Step 9 — The edge cases (never leave a gap)

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The one-picture summary

Everything in one frame: two walls (Step 1) pin a wave (Step 2) that obeys curvature $\propto -\psi$ (Step 3), forcing sine/cosine (Step 4); the left wall deletes the cosine (Step 5); the right wall keeps only $k_n=n\pi /L$ (Step 6); those $k$'s become the $n^2$ energy ladder (Step 7); normalisation sets the height (Step 8).

Recall Feynman retelling — the whole walk in plain words

Picture a skipping rope tied to two walls. It has to be still at both ends — no choice, the ends are nailed. So when you shake it, only tidy patterns show up: one big arch, then two, then three, never a messy half-arch. Each tidy pattern needs a particular shaking speed, and faster shaking = more energy. A quantum particle in a box is exactly that rope. The Schrödinger equation is just the rule "the more the wave bends, the more energy it carries." Sine is the only smooth shape that bends the right way and starts at zero, so the left wall throws away everything else. Then the right wall says "you must also be zero here," which only certain waves manage — and those certain waves have certain energies, spaced $1,4,9,16$. The calmest allowed pattern (one arch) still has energy, because a rope pinned at both ends can never lie perfectly flat and still — and that is why a trapped particle can never fully stop.

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Recall checkpoint

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Connections

• Parent: Particle in a box — the results this page derives visually.
• Schrödinger Equation (TISE) — the master rule used in Step 3.
• Wavefunction and Born Rule — why $|\psi {|}^2$ is probability (Steps 2 & 8).
• Heisenberg Uncertainty Principle — why zero-point energy is unavoidable (Step 9).
• Standing Waves on a String — the clamped-rope analogy driving Steps 2 & 6.
• Quantum Harmonic Oscillator — a box with soft walls: evenly spaced levels.
• Quantum Tunnelling and Finite Well — what changes when walls are finite (Step 9).
• Quantum Dots — the $1/L^2$ law made visible as tunable colour (Step 7).