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

WHAT interior node hota kya hai: woh point jahan ${\psi }_n=0$ ho lekin woh wall nahi hai. ${\psi }_n=\sqrt{2/L}\sin (n\pi x/L)$ zero hota hai jab $n\pi x/L=j\pi$, yaani $x=jL/n$ integer $j$ ke liye. WHY hum walls ko exclude karte hain: $j=0$ se $x=0$ milta hai aur $j=n$ se $x=L$ — ye clamped ends hain, interior nahi. Interior zeros $j=1,2,\dots ,n-1$ hain, toh inki sankhya $n-1$ hai. $n=4$ ke liye: $==3==$ interior nodes (at $x=L/4,L/2,3L/4$) aur $4$ antinodes. Neeche figure dekho — $n=4$ curve walls ke beech mein axis ko teen baar cross karta hai.

WHY sirf ratio: $E_n=\frac{n^2{h}^2}{8m{L}^2}$, aur poora clumsy factor $\frac{h^2}{8m{L}^2}$ har level ke liye same hota hai. Divide karne par woh cancel ho jaata hai: $\frac{E_3}{E_1}=\frac{3^2}{1^2}=\frac{9}{1}.$ Toh $E_3:E_1===9:1==$. WHAT ka physical matlab: teesra level ground level se naun guna upar hota hai — energy $n$ ki tarah nahi balki $n^2$ ki tarah badhti hai.

WHAT hum plug in karte hain: $n=1$, $m=m_e=9.11\times 10^{-31}$ 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: $1.6\times 10^{-19}$ se divide karo: $E_1=\frac{1.508\times 10^{-18}}{1.6\times 10^{-19}}\approx ==9.4\text{eV}==.$ Sanity check: parent ke $0.10$ nm se $L$ ko double karke $0.20$ nm karne par energy $4$ se divide ho jaani chahiye (kyunki $E\propto 1/L^2$): $37.6/4=9.4$ eV. ✓

WHY hum $E_1$ se compare karte hain: kyunki $E_n=n^2{E}_1$, ratio $E_n/E_1$ hi $n^2$ hai. $n^2=25\Rightarrow n=\sqrt{25}===5==.$ WHAT ko reject karein: $n=-5$ bhi $n^2=25$ solve karta hai, lekin negative $n$ sirf $\sin (n\pi x/L)$ ka sign flip karta hai — same physical standing wave — aur $n$ positive integer hona chahiye. Toh $n=5$.

WHAT transition release karta hai: energy difference ek photon ban jaati hai. $\Delta E=E_3-E_2=(3^2-2^2)E_1=(9-4)E_1=5E_1.$ $\Delta E=5\times 37.6=188\text{eV}.$ WHY $hc=1240$ eV·nm use karein: photon relation $\Delta E=hc/\lambda$ rearrange hoke $\lambda =hc/\Delta E$ deta hai, aur in mixed units mein arithmetic clean hai: $\lambda =\frac{1240\text{eV⋅nm}}{188\text{eV}}\approx ==6.6\text{nm}==.$ Ye extreme ultraviolet hai — visible light ($\lambda \sim 400$–700 nm) se kaafi zyada energetic. Chhote boxes ⟹ bade gaps ⟹ chhoti wavelengths.

WHAT hum expand karte hain: $E_{n+1}-E_n=[(n+1{)}^2-n^2]E_1.$ WHY algebra simplify hoti hai: $(n+1{)}^2=n^2+2n+1$, toh $n^2$ cancel ho jaata hai: $(n+1{)}^2-n^2=2n+1.$ $\boxed{E_{n+1}-E_n=(2n+1)E_1.}$ Gaps $3E_1,5E_1,7E_1,\dots$ jaati hain — odd multiples, $n$ ke saath badhte hain. $n=4$ ke liye: $E_5-E_4=(2\cdot 4+1)E_1===9E_1==.$ WHAT ye dikhta hai: energy ladder par rungs upar jaane ke saath zyada door hote jaate hain — evenly spaced Quantum Harmonic Oscillator ke bilkul ulta.

WHAT hum integrate karte hain: probability $|{\psi }_1{|}^2$ ke neeche ka area hai. $P={\int }_{L/3}^{2L/3}\frac{2}{L}{\sin }^2\frac{\pi x}{L}dx.$ WHY $u=\pi x/L$ substitute karein: iska messy argument ek clean ${\sin }^{2}u$ ban jaata hai. Phir $du=(\pi /L)dx$, toh $dx=(L/\pi )du$, aur limits ban jaati hain $u=\pi /3$ se $u=2\pi /3$: $P=\frac{2}{L}\cdot \frac{L}{\pi }{\int }_{\pi /3}^{2\pi /3}{\sin }^{2}udu=\frac{2}{\pi }{\left[\frac{u}{2}-\frac{\sin 2u}{4}\right]}_{\pi /3}^{2\pi /3}.$ Evaluate karo: $u=2\pi /3$ par, $\sin (4\pi /3)=-\frac{\sqrt{3}}{2}$; $u=\pi /3$ par, $\sin (2\pi /3)=+\frac{\sqrt{3}}{2}$. $\left[\frac{u}{2}-\frac{\sin 2u}{4}\right]=\left(\frac{\pi }{3}+\frac{\sqrt{3}}{8}\right)-\left(\frac{\pi }{6}-\frac{\sqrt{3}}{8}\right)=\frac{\pi }{6}+\frac{\sqrt{3}}{4}.$ $P=\frac{2}{\pi }\left(\frac{\pi }{6}+\frac{\sqrt{3}}{4}\right)=\frac{1}{3}+\frac{\sqrt{3}}{2\pi }\approx ==0.609==.$ WHY 1/3 se zyada: density $|{\psi }_1{|}^2$ beech mein hump karti hai (figure dekho), toh central third mein flat-distribution guess $0.333$ se kaafi zyada probability hoti hai.

WHY ye Heisenberg Uncertainty Principle se connect hota hai: particle ko $\Delta x\approx L$ size ke region mein confine karne par momentum spread forced hoti hai $\Delta p\gtrsim h/L$. Ye literally still nahi baith sakta. WHAT kinetic energy iska imply karta hai: $E\sim \frac{(\Delta p{)}^2}{2m}\approx \frac{(h/L{)}^2}{2m}=\frac{h^2}{2m{L}^2}.$ Exact ground state $E_1=\frac{h^2}{8m{L}^2}$ se compare karo: $\frac{E_{\text{estimate}}}{E_1}=\frac{h^2/(2m{L}^2)}{h^2/(8m{L}^2)}=\frac{8}{2}===4==.$ Same $h^2/(m{L}^2)$ dependence, same order of magnitude ($4$ ke factor se off kyunki $\Delta p\approx h/L$ ek rough bound hai). Conclusion: nonzero $E_1$ uncertainty ka concrete form hai — position squeeze karo toh momentum inflate hoga, aur move karne mein kinetic energy lagti hai.

WHAT orthogonal yahan matlab: do alag states ka overlap integral zero hai — ye wavefunctions ke space mein "independent directions" hain. $I={\int }_0^L\frac{2}{L}\sin \frac{\pi x}{L}\sin \frac{2\pi x}{L}dx.$ WHY product-to-sum identity: $\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$ ek product (integrate karna mushkil) ko cosines ke difference (aasaan) mein badal deta hai: $\sin \frac{\pi x}{L}\sin \frac{2\pi x}{L}=\frac{1}{2}\left[\cos \frac{\pi x}{L}-\cos \frac{3\pi x}{L}\right].$ Har cosine ko $0$ se $L$ tak integrate karo. Kyunki ${\int }_0^L\cos (m\pi x/L)dx=\frac{L}{m\pi }\sin (m\pi )=0$ kisi bhi nonzero integer $m$ ke liye (kyunki $\sin (m\pi )=0$): $I=\frac{2}{L}\cdot \frac{1}{2}\left[0-0\right]===0==.$ WHY ye zero hona chahiye (deeper): ye states usi Hamiltonian ke eigenfunctions hain (Schrödinger Equation (TISE)) alag-alag energies ke saath, aur aise eigenfunctions hamesha orthogonal hote hain. Physically: energy $E_2$ measure karne ka chance zero hai ki woh $E_1$ pattern mein "leak" ho.

WHAT symmetry hai: $|{\psi }_1(x){|}^2=\frac{2}{L}{\sin }^2(\pi x/L)$ centre $x=L/2$ ke baare mein mirror image hai. Check karo: $x\to L-x$ replace karne par, $\sin \frac{\pi (L-x)}{L}=\sin \left(\pi -\frac{\pi x}{L}\right)=\sin \frac{\pi x}{L},$ toh $|{\psi }_1(L-x){|}^2=|{\psi }_1(x){|}^2$. WHY isse settle hota hai: $L/2$ ke baare mein symmetric distribution dono sides par equal area deta hai. Kyunki total $1$ hai, har half hai $P_{\text{left}}=P_{\text{right}}===\frac{1}{2}==.$ WHAT ye dikhta hai: figure mein hump ko dashed centre line par fold karo — dono halves exactly ek doosre ke upar aate hain.

WHAT $k_n$ measure karta hai: ye toolbox ka wave number hai, $k_n=n\pi /L$ — space mein sine kitni fast oscillate karti hai. (a) $n=2$, $L=1.0\times 10^{-9}$ m plug karo: $k_2=\frac{2\pi }{1.0\times 10^{-9}}===6.283\times 10^9{\text{m}}^{-1}==.$ (b) WHY $p=\hslash k$: ek sine wave $\sin (kx)$ momentum magnitude $\hslash k$ carry karta hai (ye de Broglie relation hai wave-number form mein). Toh $p=\hslash k_2=(1.055\times 10^{-34})(6.283\times 10^9)===6.629\times 10^{-25}\text{kg⋅m/s}==.$ Energy confirm karo: kinetic energy $p^2/2m$ hai: $\frac{p^2}{2m}=\frac{(6.629\times 10^{-25}{)}^2}{2(1.675\times 10^{-27})}=1.312\times 10^{-22}\text{J}.$ Formula $E_2=\frac{2^2{h}^2}{8m{L}^2}$ se compare karo jahan $h=2\pi \hslash =6.629\times 10^{-34}$ J·s: woh bhi $1.31\times 10^{-22}$ J deta hai. ✓ WHAT ye dikhata hai: toolbox symbol $k_n$ decoration nahi hai — ye directly particle ka momentum encode karta hai, aur ise square karne par energy ladder milti hai.

WHAT hum track karte hain: sirf $L$ dependence, kyunki $E_n=\frac{n^2{h}^2}{8m{L}^2}\propto \frac{1}{L^2}$ har level ko identically affect karta hai. WHY hum $n$, $m$, $h$ ignore kar sakte hain: jab hum box shrink karte hain toh ye unchanged hain, toh ratio mein cancel ho jaate hain: $\frac{E_n(L/2)}{E_n(L)}=\frac{1/(L/2{)}^2}{1/L^2}=\frac{L^2}{L^2/4}===4==.$ Har level (aur har gap, kyunki gaps bhi $\propto 1/L^2$ hain) $4$ se multiply ho jaata hai. Bade gaps ⟹ zyada energy wale photons ⟹ chhoti wavelength = blue shift. WHAT iska lab mein matlab: yahi reason hai ki chhote quantum dots bluer glow karte hain aur bade wale redder — physical size colour ka dial hai. Dot ko half karne par uski light photon energy mein chaar guna higher ho jaati hai, blue/UV end ki taraf.

L1–L2 correct ::: Tumhare paas formulas $E_n=n^2{h}^2/8m{L}^2$ aur ${\psi }_n=\sqrt{2/L}\sin (n\pi x/L)$ hain. L3 correct ::: Tum transitions aur growing gaps handle kar sakte ho — pehle square karo, phir subtract karo. L4 correct ::: Tum $|\psi {|}^2$ integrate kar sakte ho aur energy ko uncertainty se link kar sakte ho. L5 correct ::: Tum orthogonality, symmetry, wave number/momentum aur scaling tak pahunch gaye ho — mastery level.