Exercises — Schrödinger equation — time-dependent, time-independent

(a) mein ${\partial }_t$ aur $\Psi (x,t)$ hai → yeh TDSE hai. Bracket $\hat{H}$ hai, yaani Hamiltonian. (b) mein ordinary derivative $d^2/d{x}^2$ hai, sirf $\psi (x)$ hai, aur right side par constant $E$ hai → yeh TISE hai, eigenvalue equation $\hat{H}\psi =E\psi$. Kaise pehchana: time derivative hai ⇒ TDSE; ek bare number $E$ jo $\psi$ ko multiply kare ⇒ TISE.

Sirf $|\Psi {|}^2$ observable hai ($\Psi$ aur $\psi$ complex hain, directly measurable nahi). $|\Psi {|}^2={\Psi }^{\ast }\Psi =\left({\psi }^{\ast }{e}^{+iEt/\hslash }\right)\left(\psi e^{-iEt/\hslash }\right)=|\psi {|}^2{e}^0=|\psi {|}^2.$ Dono phase factors complex conjugates hain; unka product hai $e^{iEt/\hslash -iEt/\hslash }=e^0=1$. Toh $|\Psi {|}^2=|\psi (x){|}^2$ mein koi $t$ nahi — isliye ise stationary state kaha jaata hai.

$n=1$ ke saath numbers plug karo: $E_1=\frac{{\pi }^2(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})(0.50\times 10^{-9}{)}^2}.$ Numerator: ${\pi }^2(1.113\times 10^{-68})=1.099\times 10^{-67}$. Denominator: $2(9.11\times 10^{-31})(2.5\times 10^{-19})=4.555\times 10^{-49}$. $E_1=2.41\times 10^{-19}\text{J}=\frac{2.41\times 10^{-19}}{1.602\times 10^{-19}}=1.51\text{eV}.$ Yeh kyun important hai: ek atom-sized box mein trapped room-temperature electron ki energies order of 1 eV hoti hain — real atomic/optical scales se match karti hain.

Kyunki $E_n\propto n^2$ hai: $E_3/E_1=3^2/1^2=9$. $E_3=9E_1=9(1.51)=13.6\text{eV}$, toh $E_3-E_1=13.6-1.51=12.1\text{eV}$. Kyun: sirf $n^2$ factor badalta hai; baaki sab ratio mein cancel ho jaata hai.

Normalisation demand karti hai ki total probability $1$ ho: ${\int }_0^a|C{|}^{2}dx=|C{|}^{2}a=1\Rightarrow C=\frac{1}{\sqrt{a}}.$ Pehle units check (zaroori hai): $|\psi {|}^{2}dx$ ek pure probability hai (dimensionless), aur $dx$ ki units metres hain, toh $|\psi {|}^2$ ko ${\text{m}}^{-1}$ carry karni chahiye aur isliye $\psi$ ko ${\text{m}}^{-1/2}$ carry karni chahiye. Exactly yahi $1/\sqrt{a}$ deta hai. Numerically: $C=\frac{1}{\sqrt{2.0\times 10^{-9}\text{m}}}=2.24\times 10^4{\text{m}}^{-1/2}.$ Kyun: $|\psi {|}^2$ height $C^2$ (units ${\text{m}}^{-1}$) ka ek flat block hai, width $a$ (units m) par; uska area $C^{2}a$ dimensionless hai aur $1$ hona forced hai. $a$ ki units drop karne se $C$ ke liye ek meaningless bare number milega.

(a) ${\psi }^{\prime \prime }=-k^2\psi$ ⇒ $-\frac{{\hslash }^2}{2m}(-k^2)\psi =\frac{{\hslash }^2{k}^2}{2m}\psi$. Haan, $E={\hslash }^2{k}^2/2m=p^2/2m>0$ jahan $p=\hslash k$. ✓ (b) ${\psi }^{\prime \prime }=-k^2\psi$ bhi ⇒ same $E>0$. Haan ✓ ($e^{\pm ikx}$ ka ek standing combination, yaani equal aur opposite momenta $\pm \hslash k$). (c) ${\psi }^{\prime \prime }=+k^2\psi$ ⇒ $E=-\frac{{\hslash }^2{k}^2}{2m}<0$. Yeh ek valid eigenfunction hai lekin negative $E$ ke saath aur yeh $x\to +\infty$ par blow up karta hai, toh free particle ke liye normalise nahi ho sakta. Physically valid nahi yahan. Yeh kyun important hai: ${\psi }^{\prime \prime }/\psi$ ka sign sab kuch decide karta hai. Negative (${\psi }^{\prime \prime }=-k^2\psi$) ⇒ oscillation ⇒ real wavenumber $k$ ⇒ real momentum $p=\hslash k$ ⇒ ek genuine travelling/standing wave. Positive (${\psi }^{\prime \prime }=+k^2\psi$) ⇒ exponential growth/decay ⇒ imaginary $k$ ⇒ koi real momentum nahi ⇒ yeh sirf classically forbidden barrier ke andar survive karta hai. Neeche ki figure dono draw karti hai, aur dikhati hai ki momentum arrow $p=\hslash k$ sirf oscillating case se kaise attach hota hai.

$n=0$ deta hai $\sin (0)=0$ har jagah ⇒ $\psi \equiv 0$ ⇒ $\int |\psi {|}^2=0\ne 1$: koi particle exist hi nahi karta, normalise nahi ho sakta. Negative $n$: $\sin (-n\pi x/L)=-\sin (n\pi x/L)$. Yeh sirf $+n$ state ka $-1$ times hai; wavefunction ko ek constant phase se multiply karna same physical state deta hai ($|\Psi {|}^2$ unchanged). Toh $n=1,2,3,\dots$ sab distinct states enumerate karte hain. Kyun: physical content $|\psi {|}^2$ mein rehta hai, jo overall sign ya phase se blind hai.

Time factor attach karo: ${\Psi }_1(x,t)=\sqrt{2/L}\sin (\pi x/L)e^{-i{E}_{1}t/\hslash }$. LHS (time derivative $-i{E}_1/\hslash$ neeche pull karta hai): $i\hslash {\partial }_t{\Psi }_1=i\hslash \left(-\frac{i{E}_1}{\hslash }\right){\Psi }_1=E_1{\Psi }_1.$ RHS (andar $V=0$ hai; ${\psi }_1^{\prime \prime }=-(\pi /L{)}^2{\psi }_1$ toh $\hat{H}{\psi }_1=\frac{{\hslash }^2}{2m}(\pi /L{)}^2{\psi }_1=E_1{\psi }_1$): $\hat{H}{\Psi }_1=E_1{\Psi }_1.$ LHS $=$ RHS. ✓ Yeh kyun important hai: yeh dikhata hai ki TISE solve karke $e^{-iEt/\hslash }$ se multiply karna automatically TDSE solve karta hai — yahi separation of variables ka poora point hai.

$\Psi =\frac{1}{\sqrt{2}}({\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash })$ likho jahan ${\psi }_1,{\psi }_2$ real hain. Tab $|\Psi {|}^2=\frac{1}{2}({\psi }_1^2+{\psi }_2^2+2{\psi }_1{\psi }_2\cos [(E_2-E_1)t/\hslash ]).$ Cross term mein $\cos ({\omega }_{21}t)$ rehta hai: probability cloud slosh karta hai aage peechhe. Ek single energy eigenstate stationary hai; do ka mix nahi hota. Yeh kyun important hai: saari real dynamics (ek electron oscillate karna, light emit karna) different energy ke states ko superpose karne se aati hai. Same-energy phases $|\Psi {|}^2$ mein cancel ho jaate hain; different-energy phases nahi hote.

(a) $E_2=4E_1=6.05\text{eV}$, toh $E_2-E_1=3E_1=4.54\text{eV}$. Convert: $4.54\times 1.602\times 10^{-19}=7.27\times 10^{-19}\text{J}$. (b) $\nu =\frac{7.27\times 10^{-19}}{6.626\times 10^{-34}}=1.097\times 10^{15}\text{Hz}$. (c) $\lambda =\frac{3.0\times 10^8}{1.097\times 10^{15}}=2.73\times 10^{-7}\text{m}=273\text{nm}$ (ultraviolet). Yeh kyun important hai: box model UV mein ek real spectral line predict karta hai — allowed levels ke beech quantum jumps woh discrete colours create karte hain jo atoms emit karte hain.

$\frac{E_{n+1}-E_n}{E_n}=\frac{(n+1{)}^2-n^2}{n^2}=\frac{2n+1}{n^2}.$ $n=1$ par: $\frac{3}{1}=3$ (300% jump — bahut quantum, levels bahut door). $n=100$ par: $\frac{201}{10000}=0.0201$ (lagbhag 2%). $n\to \infty$ par, $\frac{2n+1}{n^2}\to 0$: neighbouring levels near-continuum mein merge ho jaate hain. Yeh kyun important hai: yeh correspondence principle hai — quantum discreteness large quantum numbers par invisible ho jaati hai, classical physics recover hoti hai. Directly Energy Quantization se tied hai.

Box mein quantization boundary conditions $\psi (0)=\psi (L)=0$ se aayi, jo $kL=n\pi$ force karti thin — discrete $k$. Free particle mein koi confining walls nahi hain ⇒ koi boundary condition nahi jo $k$ select kare ⇒ koi bhi real $k$ (isliye koi bhi $E={\hslash }^2{k}^2/2m\ge 0$) allowed hai ⇒ ek continuous spectrum. Kyun: discreteness Schrödinger equation mein built-in nahi hai — yeh confinement se impose hoti hai. Box hatao, quantization hatao.