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

2.3.10 · D2 · Physics › Modern Physics › Particle in a box — solving TISE, energy levels, wavefunctio

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Step 1 — Trap banao

KYA HAI. Hum ek chhota sa particle (socho ek single electron, "quantum stuff" ka ek tukda) ek line par rakhte hain, aur do walls banate hain: ek $x=0$ position par, ek $x=L$ position par. Unke beech particle free hai; walls infinitely hard hain, isliye particle kabhi bhi bahar ya bilkul wall par nahi paaya ja sakta.

KYUN. Har quantum problem ki shuruaat hoti hai jahan particle hone ki permission hai — yahi potential energy $V(x)$ hai. Wall bas ek aisi jagah hai jahan $V$ bahut zyada hai: wahan rehna infinite energy maangta hai, isliye particle simply wahan ja hi nahi sakta.

PICTURE.

Ek physical rule jo hum is picture se lete hain: particle ki presence har wall par exactly zero tak fadni chahiye. Hum iska use do baar karenge, aur wahi saara kaam karta hai.

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Step 2 — Particle ko ek wave se describe karo, $\psi$

KYA HAI. Quantum mechanics mein particle ko ek dot se nahi balki ek phaili hui wave se describe kiya jaata hai, jise $\psi (x)$ likhte hain (Greek letter "psi", bolo "sigh"). Har position $x$ par $\psi$ ki ek height hoti hai — positive, negative, ya zero.

$\psi$ kyun, sirf "position" kyun nahi? Kyunki ek quantum particle ki koi ek definite position nahi hoti. Wave tumhe batati hai ki usse yahan ya wahan paane ka chance kya hai: probability density hoti hai $|\psi (x){|}^2$ (height ka square, hamesha positive). Dekho Wavefunction and Born Rule. Abhi sirf ek baat matter karti hai:

PICTURE.

Do laal dots dekho: wave wahan zero par naili hui hai. Aage jo bhi hoga woh yahi sawal hoga: "kaun si wave shapes yeh follow kar sakti hain?"

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Step 3 — Andar wave ko jo rule follow karna hai (TISE)

KYA HAI. Box ke andar wave time-independent Schrödinger equation (Schrödinger Equation (TISE)) se govern hoti hai. Andar $V=0$ hone par, yeh equation kuch aisi banti hai:

$\underset{\text{ek positive number}}{\underbrace{-\frac{{\hslash }^2}{2m}}}\underset{\text{ψ ki curvature}}{\underbrace{\frac{d^2\psi }{d{x}^2}}}=\underset{\text{energy}}{\underbrace{E}}\underset{\text{wave}}{\underbrace{\psi }}$

KYUN yeh equation, aur har symbol kya kar raha hai?

Equation words mein yeh kehti hai: "wave ki curvature wave aur uski energy ke (minus) product ke proportional hai." Jo wave zyada bend karti hai uski energy zyada hoti hai. Yahi poori physics hai.

KYUN rearrange karein. Curvature isolate karne ke liye divide karo:

$\frac{d^2\psi }{d{x}^2}=-\underset{\text{ek positive number, isko }{k}^2\text{ bolo}}{\underbrace{\frac{2mE}{{\hslash }^2}}}\psi =-k^2\psi ,\boxed{k^2\equiv \frac{2mE}{{\hslash }^2}}$

PICTURE.

Figure ${\psi }^{\prime \prime }=-k^2\psi$ ka matlab dikhata hai: jahan bhi wave axis ke upar ho, woh neeche curve karti hai (zero ki taraf wapas kheenchi jaati hai); jahan bhi neeche ho, upar curve karti hai. Ek shape jo hamesha axis ki taraf bend karti rahe woh ek wiggle hai — ek sine ya cosine. Yahi agle step mein hai.

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Step 4 — Shape guess karo: sine aur cosine

KYA HAI. Hume ek aisa function chahiye jiska second derivative khud wahi function ho, lekin flip aur $-k^2$ se scale kiya hua. Do functions yeh karte hain, aur unka combination sabse general answer hai:

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

Sine aur cosine kyun (aur $e^x$, parabola, etc. kyun nahi)?

PICTURE.

$A$ aur $B$ sirf heights hain — hum dono shapes ko kitna mix karte hain. Boundary conditions ab inhe fix karengi.

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Step 5 — Left wall cosine ko khatam karta hai

KYA HAI. $\psi (0)=0$ apply karo. $\psi (x)=A\sin (kx)+B\cos (kx)$ mein $x=0$ daalo:

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

Toh $B=0$, aur wave sirf yeh reh jaati hai:

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

KYUN. $\sin$ pehle se zero par start karta hai, isliye woh automatically left wall follow karta hai. $\cos$ height $1$ par start hota hai — woh $x=0$ par kabhi zero nahi ho sakta, isliye jo bhi cosine hum add karein woh left wall se bahar nikal jaayega. Left wall isliye cosine ko poori tarah forbid karta hai.

PICTURE.

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Step 6 — Right wall $k$ ko quantise karta hai

KYA HAI. Ab doosra boundary condition apply karo, $\psi (L)=0$, surviving wave $\psi (x)=A\sin (kx)$ par:

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

Hum $A=0$ refuse karte hain (iska matlab hoga koi wave nahi, koi particle nahi). Toh sine khud zero hona chahiye:

$\sin (kL)=0.$

KYUN yeh magic step hai. $\sin (\text{angle})=0$ sirf khaas angles $0,\pi ,2\pi ,3\pi ,\dots$ par hota hai — $\pi$ ka har multiple (circle ke adhe chakkar). Isliye $kL$ unhi mein se ek hona chahiye:

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

Sirf discrete values of $k$ bachti hain — right wall baaki sab waves ko reject karta hai. Yahi quantisation hai, pehli baar dikha.

PICTURE.

Blue wave ($n=2$, do arches) right wall par perfectly land karti hai — allowed. Dashed grey wave beech mein wall par land karti hai — forbidden, $\psi (L)=0$ se delete.

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Step 7 — Allowed $k$ ko allowed energy $E$ mein badlo

KYA HAI. Step 3 se yaad karo ki $k^2=\frac{2mE}{{\hslash }^2}$. Isse $E$ ke liye solve karo:

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

Ab sirf allowed $k$ values daalo, $k_n=\frac{n\pi }{L}$:

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

$\hslash =h/2\pi$ swap karo (toh ${\hslash }^2{\pi }^2=h^2/4$) aur textbook form milega:

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

KYUN har term kya contribute karta hai:

PICTURE.

Rungs evenly spaced nahi hain — $n$ se $n+1$ ka gap $(2n+1)E_1$ hai, isliye gaps upar jaate hue badhte hain. (Quantum Harmonic Oscillator ke evenly spaced ladder se compare karo.)

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Step 8 — Normalise karke height $A$ fix karo

KYA HAI. Particle poori certainty $1$ ke saath box ke andar kahin zaroor milna chahiye. Kyunki $|\psi {|}^2$ probability density hai, box par uska total area $1$ hai:

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

KYUN. Ab tak $A$ ek unknown height thi. Probability ka total $1$ hona zaroori hai — yahi ek demand $A$ ko pin down karti hai.

Iss fact ka use karte hue ki ${\sin }^2$ poore half-waves pe average $\frac{1}{2}$ hota hai, ${\int }_0^L{\sin }^2\left(\frac{n\pi x}{L}\right)dx=\frac{L}{2},$ toh $A^2\cdot \frac{L}{2}=1$, se milta hai $A=\sqrt{2/L}$ aur final wavefunction:

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

PICTURE.

$|{\psi }_1{|}^2$ ke neeche shaded area exactly $1$ ke barabar hai — yahi $\sqrt{2/L}$ guarantee karta hai.

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Step 9 — Edge cases (koi gap mat chhhodo)

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Ek-picture summary

Sab kuch ek frame mein: do walls (Step 1) ek wave (Step 2) ko pin karti hain jo curvature $\propto -\psi$ (Step 3) follow karti hai, sine/cosine force hoti hai (Step 4); left wall cosine delete karta hai (Step 5); right wall sirf $k_n=n\pi /L$ rakhti hai (Step 6); woh $k$'s $n^2$ energy ladder ban jaate hain (Step 7); normalisation height set karti hai (Step 8).

Recall Feynman retelling — poori walk simple words mein

Ek skipping rope socho jo do walls se bandhi ho. Dono ends par bilkul still rehna zaroori hai — koi choice nahi, ends naile hain. Toh jab tum use hilate ho, sirf saaf patterns dikhai dete hain: ek bada arch, phir do, phir teen, kabhi koi ulajha hua half-arch nahi. Har saaf pattern ke liye ek khaas shaking speed chahiye, aur tezi se hilana = zyada energy. Ek box mein quantum particle bilkul wahi rope hai. Schrödinger equation sirf yeh rule hai: "wave jitna zyada bend karti hai, utni zyada energy carry karti hai." Sine akeli smooth shape hai jo sahi tarah bend karti hai aur zero par start hoti hai, isliye left wall baki sab kuch throw away kar deta hai. Phir right wall kehta hai "tum yahan bhi zero hone chahiye," jo sirf kuch waves kar sakti hain — aur unhi kuch waves ki kuch energies hain, $1,4,9,16$ par spaced. Sabse calm allowed pattern (ek arch) mein abhi bhi energy hai, kyunki dono ends par pinned rope kabhi bilkul flat aur still nahi leti — aur isliye ek trapped particle kabhi poori tarah ruk nahi sakta.

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Recall checkpoint

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Connections

• Parent: Particle in a box — woh results jo yeh page visually derive karta hai.
• Schrödinger Equation (TISE) — Step 3 mein use hone wala master rule.
• Wavefunction and Born Rule — kyun $|\psi {|}^2$ probability hai (Steps 2 & 8).
• Heisenberg Uncertainty Principle — kyun zero-point energy unavoidable hai (Step 9).
• Standing Waves on a String — clamped-rope analogy jo Steps 2 & 6 drive karti hai.
• Quantum Harmonic Oscillator — soft walls wala box: evenly spaced levels.
• Quantum Tunnelling and Finite Well — jab walls finite hon toh kya badlta hai (Step 9).
• Quantum Dots — $1/L^2$ law tunable colour ke roop mein visible (Step 7).