Exercises — Particle in a box — solving TISE, energy levels, wavefunctions

WHAT an interior node is: a point where ${\psi }_n=0$ but that is not a wall. ${\psi }_n=\sqrt{2/L}\sin (n\pi x/L)$ is zero when $n\pi x/L=j\pi$, i.e. $x=jL/n$ for integer $j$. WHY we exclude the walls: $j=0$ gives $x=0$ and $j=n$ gives $x=L$ — those are the clamped ends, not interior. The interior zeros are $j=1,2,\dots ,n-1$, so there are $n-1$ of them. For $n=4$: $==3==$ interior nodes (at $x=L/4,L/2,3L/4$) and $4$ antinodes. Look at the figure below — the $n=4$ curve crosses the axis three times between the walls.

WHY just a ratio: $E_n=\frac{n^2{h}^2}{8m{L}^2}$, and the whole clumsy factor $\frac{h^2}{8m{L}^2}$ is the same for every level. Dividing kills it: $\frac{E_3}{E_1}=\frac{3^2}{1^2}=\frac{9}{1}.$ So $E_3:E_1===9:1==$. WHAT it means physically: the third level sits nine times higher than the ground level — energy grows like $n^2$, not like $n$.

WHAT we plug in: $n=1$, $m=m_e=9.11\times 10^{-31}$ kg, $L=2.0\times 10^{-10}$ m. $E_1=\frac{(6.63\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(2.0\times 10^{-10}{)}^2}.$ Numerator: $(6.63\times 10^{-34}{)}^2=4.396\times 10^{-67}$. Denominator: $8\times 9.11\times 10^{-31}\times (4.0\times 10^{-20})=2.915\times 10^{-49}$. $E_1=\frac{4.396\times 10^{-67}}{2.915\times 10^{-49}}=1.508\times 10^{-18}\text{J}.$ Convert: divide by $1.6\times 10^{-19}$: $E_1=\frac{1.508\times 10^{-18}}{1.6\times 10^{-19}}\approx ==9.4\text{eV}==.$ Sanity check: doubling $L$ from the parent's $0.10$ nm to $0.20$ nm should divide energy by $4$ (since $E\propto 1/L^2$): $37.6/4=9.4$ eV. ✓

WHY we compare to $E_1$: since $E_n=n^2{E}_1$, the ratio $E_n/E_1$ is $n^2$. $n^2=25\Rightarrow n=\sqrt{25}===5==.$ WHAT to reject: $n=-5$ also solves $n^2=25$, but negative $n$ just flips the sign of $\sin (n\pi x/L)$ — same physical standing wave — and $n$ must be a positive integer. So $n=5$.

WHAT a transition releases: the energy difference becomes a photon. $\Delta E=E_3-E_2=(3^2-2^2)E_1=(9-4)E_1=5E_1.$ $\Delta E=5\times 37.6=188\text{eV}.$ WHY use $hc=1240$ eV·nm: the photon relation $\Delta E=hc/\lambda$ rearranges to $\lambda =hc/\Delta E$, and in these mixed units the arithmetic is clean: $\lambda =\frac{1240\text{eV⋅nm}}{188\text{eV}}\approx ==6.6\text{nm}==.$ This is extreme ultraviolet — far more energetic than visible light (which needs $\lambda \sim 400$–700 nm). Tiny boxes ⟹ huge gaps ⟹ short wavelengths.

WHAT we expand: $E_{n+1}-E_n=[(n+1{)}^2-n^2]E_1.$ WHY the algebra simplifies: $(n+1{)}^2=n^2+2n+1$, so the $n^2$ cancels: $(n+1{)}^2-n^2=2n+1.$ $\boxed{E_{n+1}-E_n=(2n+1)E_1.}$ The gaps go $3E_1,5E_1,7E_1,\dots$ — odd multiples, growing with $n$. For $n=4$: $E_5-E_4=(2\cdot 4+1)E_1===9E_1==.$ WHAT it looks like: on an energy ladder the rungs spread apart as you climb — the opposite of the evenly spaced Quantum Harmonic Oscillator.

WHAT we integrate: probability is the area under $|{\psi }_1{|}^2$. $P={\int }_{L/3}^{2L/3}\frac{2}{L}{\sin }^2\frac{\pi x}{L}dx.$ WHY substitute $u=\pi x/L$: it turns the messy argument into a clean ${\sin }^{2}u$. Then $du=(\pi /L)dx$, so $dx=(L/\pi )du$, and the limits become $u=\pi /3$ to $u=2\pi /3$: $P=\frac{2}{L}\cdot \frac{L}{\pi }{\int }_{\pi /3}^{2\pi /3}{\sin }^{2}udu=\frac{2}{\pi }{\left[\frac{u}{2}-\frac{\sin 2u}{4}\right]}_{\pi /3}^{2\pi /3}.$ Evaluate: at $u=2\pi /3$, $\sin (4\pi /3)=-\frac{\sqrt{3}}{2}$; at $u=\pi /3$, $\sin (2\pi /3)=+\frac{\sqrt{3}}{2}$. $\left[\frac{u}{2}-\frac{\sin 2u}{4}\right]=\left(\frac{\pi }{3}+\frac{\sqrt{3}}{8}\right)-\left(\frac{\pi }{6}-\frac{\sqrt{3}}{8}\right)=\frac{\pi }{6}+\frac{\sqrt{3}}{4}.$ $P=\frac{2}{\pi }\left(\frac{\pi }{6}+\frac{\sqrt{3}}{4}\right)=\frac{1}{3}+\frac{\sqrt{3}}{2\pi }\approx ==0.609==.$ WHY bigger than 1/3: the density $|{\psi }_1{|}^2$ humps in the middle (see figure), so the central third holds far more than the flat-distribution guess of $0.333$.

WHY this connects to Heisenberg Uncertainty Principle: confining the particle to a region of size $\Delta x\approx L$ forces a momentum spread $\Delta p\gtrsim h/L$. It literally cannot sit still. WHAT kinetic energy that implies: $E\sim \frac{(\Delta p{)}^2}{2m}\approx \frac{(h/L{)}^2}{2m}=\frac{h^2}{2m{L}^2}.$ Compare to the exact ground state $E_1=\frac{h^2}{8m{L}^2}$: $\frac{E_{\text{estimate}}}{E_1}=\frac{h^2/(2m{L}^2)}{h^2/(8m{L}^2)}=\frac{8}{2}===4==.$ Same $h^2/(m{L}^2)$ dependence, same order of magnitude (off by a factor of $4$ because $\Delta p\approx h/L$ is a rough bound). Conclusion: the nonzero $E_1$ is uncertainty made concrete — squeezing position inflates momentum, and moving costs kinetic energy.

WHAT orthogonal means here: the overlap integral of two different states is zero — they are "independent directions" in the space of wavefunctions. $I={\int }_0^L\frac{2}{L}\sin \frac{\pi x}{L}\sin \frac{2\pi x}{L}dx.$ WHY the product-to-sum identity: $\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$ turns a product (hard to integrate) into a difference of cosines (easy): $\sin \frac{\pi x}{L}\sin \frac{2\pi x}{L}=\frac{1}{2}\left[\cos \frac{\pi x}{L}-\cos \frac{3\pi x}{L}\right].$ Integrate each cosine over $0$ to $L$. Since ${\int }_0^L\cos (m\pi x/L)dx=\frac{L}{m\pi }\sin (m\pi )=0$ for any nonzero integer $m$ (because $\sin (m\pi )=0$): $I=\frac{2}{L}\cdot \frac{1}{2}\left[0-0\right]===0==.$ WHY it must be zero (deeper): the states are eigenfunctions of the same Hamiltonian (Schrödinger Equation (TISE)) with different energies, and such eigenfunctions are always orthogonal. Physically: measuring energy $E_2$ has zero chance of "leaking" into the $E_1$ pattern.

WHAT the symmetry is: $|{\psi }_1(x){|}^2=\frac{2}{L}{\sin }^2(\pi x/L)$ is a mirror image about the centre $x=L/2$. Check: replacing $x\to L-x$, $\sin \frac{\pi (L-x)}{L}=\sin \left(\pi -\frac{\pi x}{L}\right)=\sin \frac{\pi x}{L},$ so $|{\psi }_1(L-x){|}^2=|{\psi }_1(x){|}^2$. WHY that settles it: a distribution symmetric about $L/2$ puts equal area on each side. Since the total is $1$, each half is $P_{\text{left}}=P_{\text{right}}===\frac{1}{2}==.$ WHAT it looks like: fold the hump in the figure at the dashed centre line — the two halves land exactly on top of each other.

WHAT $k_n$ measures: it is the wave number from the toolbox, $k_n=n\pi /L$ — how fast the sine oscillates in space. (a) Plug $n=2$, $L=1.0\times 10^{-9}$ m: $k_2=\frac{2\pi }{1.0\times 10^{-9}}===6.283\times 10^9{\text{m}}^{-1}==.$ (b) WHY $p=\hslash k$: a sine wave $\sin (kx)$ carries momentum magnitude $\hslash k$ (this is the de Broglie relation dressed in wave-number form). So $p=\hslash k_2=(1.055\times 10^{-34})(6.283\times 10^9)===6.629\times 10^{-25}\text{kg⋅m/s}==.$ Confirm the energy: kinetic energy is $p^2/2m$: $\frac{p^2}{2m}=\frac{(6.629\times 10^{-25}{)}^2}{2(1.675\times 10^{-27})}=1.312\times 10^{-22}\text{J}.$ Compare to the formula $E_2=\frac{2^2{h}^2}{8m{L}^2}$ with $h=2\pi \hslash =6.629\times 10^{-34}$ J·s: it also gives $1.31\times 10^{-22}$ J. ✓ WHAT it shows: the toolbox symbol $k_n$ is not decoration — it directly encodes the particle's momentum, and squaring it recovers the energy ladder.

WHAT we track: only the $L$ dependence, since $E_n=\frac{n^2{h}^2}{8m{L}^2}\propto \frac{1}{L^2}$ affects every level identically. WHY we can ignore $n$, $m$, $h$: they are unchanged when we shrink the box, so they cancel in a ratio: $\frac{E_n(L/2)}{E_n(L)}=\frac{1/(L/2{)}^2}{1/L^2}=\frac{L^2}{L^2/4}===4==.$ Every level (and every gap, since gaps are also $\propto 1/L^2$) is multiplied by $4$. Larger gaps ⟹ higher-energy photons ⟹ shorter wavelength = blue shift. WHAT it means in the lab: this is exactly why smaller quantum dots glow bluer and larger ones glow redder — physical size is a dial for colour. Halving the dot pushes its light four times higher in photon energy, well toward the blue/UV end.

L1–L2 correct ::: You own the formulas $E_n=n^2{h}^2/8m{L}^2$ and ${\psi }_n=\sqrt{2/L}\sin (n\pi x/L)$. L3 correct ::: You handle transitions and growing gaps — square first, then subtract. L4 correct ::: You can integrate $|\psi {|}^2$ and link energy to uncertainty. L5 correct ::: You've reached orthogonality, symmetry, wave number/momentum and scaling — mastery level.