Worked examples — Particle in a box — solving TISE, energy levels, wavefunctions

Everything here uses only the two boxed results above plus the meaning of $|\psi {|}^2$ from Wavefunction and Born Rule. No new physics — only complete coverage.

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The scenario matrix

Before solving, let us list what can vary. A "particle in a box" problem is fixed by three inputs — the mass $m$, the box length $L$, and the level $n$ — plus what is asked (energy, gap, wavefunction, probability, wavelength of emitted light). Each input can be "ordinary", "extreme", or "degenerate". The table below is the map; every cell gets covered by an example.

Cell	What is special about it	Covered by
A. Ground state energy	ordinary $m,L$, $n=1$	Example 1
B. Energy gap / photon	difference $E_{n^{\prime }}-E_n$, real emission	Example 2
C. Big quantum number	$n$ large → classical limit	Example 3
D. Probability in a region	integrate $|\psi {|}^2$, edge vs middle	Example 4
E. Probability at a node	degenerate zero-width / density-zero point	Example 5
F. Scaling limits	$L\to$ large (free particle), $L\to$ small (quantum dot)	Example 6
G. Heavy particle	$m$ large → levels vanish, classical	Example 7
H. Real-world word problem	quantum dot colour	Example 8
I. Exam twist	"find $n$ from a given energy", inverse problem	Example 9
J. Symmetry shortcut	probability by parity, no integral	Example 10

Two facts we will reuse constantly, so state them once with their units checked:

The figure above is the whole matrix in one picture: the energy ladder ($n^2$ spacing) on the left, the first three wave shapes on the right. Keep glancing back at it — every example below is a finger pointing at some part of it.

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Example 1 — Cell A: ground-state energy

Step 1 — Pick $n$. Lowest energy is $n=1$ (the smallest allowed level; $n=0$ is no particle). Why this step? "Lowest possible" always means the ground state, which is $n=1$ for the box.

Step 2 — Plug into $E_n$. $E_1=\frac{(6.626\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(2.0\times 10^{-10}{)}^2}$ Why this step? This is the only energy formula; we just substitute the three inputs.

Step 3 — Arithmetic. Numerator $=4.390\times 10^{-67}$. Denominator $=8\times 9.11\times 10^{-31}\times 4.0\times 10^{-20}=2.915\times 10^{-49}$. $E_1=1.506\times 10^{-18}\text{J}$

Step 4 — Convert to eV. Divide by $1.602\times 10^{-19}$: $E_1=9.40\text{eV}$ Why this step? Atomic energies are naturally quoted in eV; it makes the number human-sized.

Verify: doubling $L$ should divide $E$ by $2^2=4$. Parent's $37.6\text{eV}/4=9.40\text{eV}$. ✓ Matches the forecast exactly.

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Example 2 — Cell B: energy gap and emitted photon

Step 1 — Use the $n^2$ scaling for the gap. $\Delta E=E_3-E_1=(3^2-1^2)E_1=8E_1$ Why this step? Every level is $E_1$ times $n^2$, so differences are just $(n^{\prime 2}-n^2)$ times $E_1$ — no need to recompute the whole formula.

Step 2 — Numbers. $\Delta E=8\times 9.40=75.2\text{eV}$

Step 3 — Photon wavelength. A photon carries $E_{\gamma }=hc/\lambda$, so $\lambda =hc/\Delta E$. Why this step? Energy conservation: the lost quantum energy becomes one photon. Using $hc=1240\text{eV⋅nm}$ keeps units clean. $\lambda =\frac{1240\text{eV⋅nm}}{75.2\text{eV}}=16.5\text{nm}$

Verify: $16.5\text{nm}$ sits in the extreme-ultraviolet — matching the "large gap → short wavelength" forecast. Cross-check the gap: $E_3=9\times 9.40=84.6$, $E_1=9.40$, difference $75.2$. ✓

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Example 3 — Cell C: the classical limit (large $n$)

Step 1 — Write the gap. $E_{n+1}-E_n=[(n+1{)}^2-n^2]E_1=(2n+1)E_1$. Why this step? Expanding $(n+1{)}^2-n^2=2n+1$ is the algebra behind "spacings grow".

Step 2 — Form the fractional gap. $\frac{E_{n+1}-E_n}{E_n}=\frac{(2n+1)E_1}{n^2{E}_1}=\frac{2n+1}{n^2}$ Why this step? "How lumpy does the ladder feel?" is a relative question, so divide by the level itself.

Step 3 — Evaluate at $n=1$ and $n=100$. $n=1:\frac{3}{1}=3.00n=100:\frac{201}{10000}=0.0201$

Verify: the fractional gap falls from $300\%$ to about $2\%$. As $n\to \infty$, $\frac{2n+1}{n^2}\to 0$, so the discrete ladder blurs into a continuum — this is the correspondence principle. ✓

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Example 4 — Cell D: probability in a region

Step 1 — Write the probability integral. $P={\int }_{L/3}^{2L/3}\frac{2}{L}{\sin }^2\left(\frac{\pi x}{L}\right)dx$ Why this step? Probability = area under the density $|\psi {|}^2$, not under $\psi$ — the shaded red hump in the figure.

Step 2 — Use ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$. $P=\frac{2}{L}{\int }_{L/3}^{2L/3}\frac{1}{2}\left(1-\cos \frac{2\pi x}{L}\right)dx$ Why this step? We cannot integrate ${\sin }^2$ directly; this identity turns it into a constant plus a cosine, both easy.

Step 3 — Integrate. $P=\frac{1}{L}{\left[x-\frac{L}{2\pi }\sin \frac{2\pi x}{L}\right]}_{L/3}^{2L/3}=\frac{1}{3}+\frac{\sqrt{3}}{2\pi }$ (The endpoints give $\sin \frac{4\pi }{3}-\sin \frac{2\pi }{3}=-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=-\sqrt{3}$, and the $-\frac{L}{2\pi }$ factor flips the sign.)

Step 4 — Number. $P=\frac{1}{3}+\frac{\sqrt{3}}{2\pi }\approx 0.333+0.276=0.609$

Verify: $0.609>0.333$, and it matches the "hump in the middle" forecast. Sanity: the middle third (0.609) plus the two edge thirds (from parent, $0.196$ each) gives $0.609+2(0.196)=1.001\approx 1$. ✓ The whole box holds probability 1.

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Example 5 — Cell E: probability at a node (degenerate zero-width)

Step 1 — Check the density at the point. ${\psi }_2\left(\frac{L}{2}\right)=\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi }{L}\cdot \frac{L}{2}\right)=\sqrt{\frac{2}{L}}\sin \pi =0$ Why this step? $x=L/2$ is where the second wave crosses zero — an interior node ($n-1=1$ node for $n=2$).

Step 2 — Recall probability needs an interval. $P(\{x=L/2\})={\int }_{L/2}^{L/2}|{\psi }_2{|}^{2}dx=0$ Why this step? Any integral over a zero-width interval is $0$, regardless of the wave. A point never has finite probability in a continuous distribution.

Verify: two independent reasons give $0$: (a) the density itself vanishes at the node, (b) a single point has measure zero. Both agree. This is the degenerate case the matrix demanded — "probability at a point" is always $0$; only regions carry probability. ✓

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Example 6 — Cell F: scaling limits (box grows / shrinks)

Step 1 — Isolate the $L$ dependence. $E_1=\frac{h^2}{8m}\cdot \frac{1}{L^2}$ Why this step? Everything except $L$ is a fixed constant, so all behaviour lives in $1/L^2$.

Step 2 — Large box. As $L\to \infty$, $\frac{1}{L^2}\to 0$, so $E_1\to 0$ and the gap $(2n+1)E_1\to 0$: the ladder becomes a continuum — an unconfined free particle can have any energy. This is why free electrons in large regions are not visibly quantised.

Step 3 — Small box. As $L\to 0$, $E_1\to \infty$: squeezing a particle costs escalating energy — the fingerprint of the Heisenberg Uncertainty Principle ($\Delta x$ small forces $\Delta p$, hence energy, large).

Verify: numerically, going from $L=1\text{nm}$ to $L=0.1\text{nm}$ (÷10) multiplies $E_1$ by $10^2=100$. $E_1(1\text{nm})$ for an electron is $0.376\text{eV}$; $\times 100=37.6\text{eV}$, exactly the parent's $0.1\text{nm}$ answer. ✓ The limit trend is real.

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Example 7 — Cell G: heavy particle (classical object)

Step 1 — Plug in. $E_1=\frac{(6.626\times 10^{-34}{)}^2}{8(1.0\times 10^{-9})(1.0\times 10^{-3}{)}^2}$ Why this step? Same single formula — the physics does not change for big objects, only the numbers.

Step 2 — Arithmetic. Numerator $=4.390\times 10^{-67}$. Denominator $=8\times 10^{-9}\times 10^{-6}=8.0\times 10^{-15}$. $E_1=5.49\times 10^{-53}\text{J}$

Step 3 — Interpret. In eV: $5.49\times 10^{-53}/1.602\times 10^{-19}\approx 3.4\times 10^{-34}\text{eV}$. Why this step? Comparing to any real energy (thermal $kT\sim 0.025\text{eV}$) shows the level gap is $\sim 10^{-32}$ times smaller — utterly undetectable.

Verify: the classical limit is consistent — massive, large-scale objects have $E_1\approx 0$ and levels so dense they merge. Quantum effects hide only at the small-$m$, small-$L$ corner. ✓

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Example 8 — Cell H: real-world word problem (quantum dot)

Step 1 — Ground-state energy for $L=5.0\text{nm}$. $E_1=\frac{h^2}{8m_e{L}^2}=\frac{(6.626\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(5.0\times 10^{-9}{)}^2}$ Numerator $=4.390\times 10^{-67}$, denominator $=8\times 9.11\times 10^{-31}\times 2.5\times 10^{-17}=1.822\times 10^{-46}$. $E_1=2.409\times 10^{-21}\text{J}=0.01504\text{eV}$ Why this step? The colour is set by the gap, and the gap is a multiple of $E_1$, so compute $E_1$ first.

Step 2 — Gap $2\to 1$. $\Delta E=(2^2-1^2)E_1=3E_1=0.04513\text{eV}$. Why this step? Same $n^2$ trick as Example 2: $4-1=3$.

Step 3 — Wavelength. $\lambda =\frac{hc}{\Delta E}=\frac{1240\text{eV⋅nm}}{0.04513\text{eV}}=2.75\times 10^4\text{nm}$

Verify: $\approx 27,000\text{nm}$ is far infrared — not visible. This teaches a real lesson: the 1-D box is a toy; real quantum dots are 3-D with much tighter effective confinement, which is why they actually glow in the visible. Our number is self-consistent, and it honestly flags the model's limits. ✓ ($\Delta E$ check: $3\times 0.01504=0.04513\text{eV}$.)

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Example 9 — Cell I: exam twist (inverse problem)

Step 1 — Compute $E_1$ for this box. $E_1=\frac{h^2}{8m_e{L}^2}=0.376\text{eV}(\text{as in Example 6, }L=1\text{nm})$ Why this step? Every level is a known multiple of $E_1$, so we need $E_1$ to decode $n$.

Step 2 — Solve for $n^2$. $n^2=\frac{E}{E_1}=\frac{2.256}{0.376}=6.0$ Why this step? $E=n^2{E}_1\Rightarrow n^2=E/E_1$ — pure rearrangement.

Step 3 — Test if $n$ is a valid integer. $\sqrt{6.0}\approx 2.449$ — not an integer. Why this step? Only integer $n$ are allowed levels. A non-integer means the stated energy is not a genuine box level.

Verify: $n=2$ gives $4\times 0.376=1.504\text{eV}$; $n=3$ gives $9\times 0.376=3.384\text{eV}$. The measured $2.256\text{eV}$ lies between them, so no level matches — the exam trap is that a plausible-looking number can be forbidden. Lesson: always confirm $E/E_1$ is a perfect square. ✓

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Example 10 — Cell J: symmetry shortcut (no integral)

Step 1 — Establish the symmetry. Reflecting $x\to L-x$: $\sin \left(\frac{n\pi (L-x)}{L}\right)=\sin \left(n\pi -\frac{n\pi x}{L}\right)=(-1{)}^{n+1}\sin \left(\frac{n\pi x}{L}\right)$ Squaring kills the sign: $|{\psi }_n(L-x){|}^2=|{\psi }_n(x){|}^2$. Why this step? Symmetry of the density (not the wave) is what shares probability equally.

Step 2 — Conclude by symmetry. Equal density on mirrored halves + total probability $1$ ⟹ each half has $\frac{1}{2}$. Why this step? No integral needed: symmetry does the accounting for us.

Verify: for $n=1$, direct integration ${\int }_0^{L/2}\frac{2}{L}{\sin }^2\frac{\pi x}{L}dx=\frac{1}{2}$. For $n=2$, likewise $\frac{1}{2}$. Symmetry holds for every $n$. ✓

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Recall Scenario self-test

Which matrix cell is each question?

1. "Electron drops $4\to 2$, find the photon." ::: Cell B (energy gap / photon).
2. "As $L\to 0$ what happens to $E_1$?" ::: Cell F (scaling limits).
3. "Probability the $n=1$ particle sits exactly at $x=L/4$." ::: Cell E (point → probability 0).
4. "An electron has $E=1.5\text{eV}$ in a $1\text{nm}$ box; which $n$?" ::: Cell I (inverse; here $n=2$ exactly, since $1.504\approx 1.5$).
5. "Chance in the right half for $n=5$." ::: Cell J (symmetry → $\frac{1}{2}$).

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Connections

• Particle in a box — solving TISE, energy levels, wavefunctions — the parent that derives $E_n$ and ${\psi }_n$ used throughout.
• Wavefunction and Born Rule — why probability is $\int |\psi {|}^{2}dx$ (Examples 4, 5, 10).
• Heisenberg Uncertainty Principle — the "small box → huge energy" limit (Example 6).
• Quantum Dots — the real-world word problem (Example 8).
• Quantum Tunnelling and Finite Well — what changes when the walls are finite.
• Standing Waves on a String — the classical mirror-image of these standing wave patterns.
• Quantum Harmonic Oscillator — contrast: equally spaced levels, unlike the box's $n^2$ ladder.