Schrödinger equation — time-dependent, time-independent: Worked examples

2.3.9 · D3 · Physics › Modern Physics › Schrödinger equation — time-dependent, time-independent

Yeh page ek workout hai. Parent note ne do equations build ki thi; yahan hum common exam situations unpe apply karte hain taaki kum cases tumhe surprise karein.

Kuch bhi solve karne se pehle, teen vocabulary pieces hain jinpe poori page tikti hai. Main inhe abhi define karta hoon taaki koi symbol kabhi bina earn kiye use na ho.

Ab, ek "case" kya count hota hai?

The scenario matrix

Neeche har problem in cells mein se ek hai. Yeh matrix exhaustive nahi hai — teen bade topics (finite well & tunnelling, barrier scattering, harmonic oscillator) ke apne pages hain aur yahan sirf signpost kiye gaye hain — lekin yeh woh cases cover karta hai jo zyaadatar exams ek Schrödinger question mein pack karte hain.

Cell	Ise alag kya banata hai	Example
A. Diya hua $Ψ$ verify karo	TDSE mein plug karo, check karo dono sides match karein	Ex 1
B. $V = 0$ free particle	plane wave, continuous energy	Ex 2
C. Bound state, hard walls	$ψ = 0$ at edges se quantized $E_{n}$	Ex 3
D. Normalization	constant $A$ fix karo taaki total prob $= 1$	Ex 4
E. Excited / higher $n$	zyada wiggles, energy scales as $n^{2}$	Ex 5
F. Stationary vs moving fog	kyun $	\Psi
G. Degenerate / forbidden input	$n = 0$ , $E < V_{m i n}$ — kya toot ta hai	Ex 7
H. Real-world word problem	electron in a nanowire, numbers in eV	Ex 8
I. Exam twist	box shift karo, ya photon energy pucho	Ex 9

Hum teen constants poori jagah reuse karte hain. Inhe ek baar plain words mein define karte hain.

Cell A — Diya hua wavefunction verify karo

Ex 1 se pehle hume teen symbols earn karne honge jo woh use karta hai.

Worked example Ex 1 — Kya

$Ψ = A e^{i (k x - ω t)}$ free TDSE satisfy karta hai? (Cell A, B) Forecast: padhne se pehle guess karo — kya dono sides $Ψ$ ka same number denge, ya ek extra factor aayega? Apna bet likho.

Step 1. Left side $i ℏ \partial_{t} Ψ$ compute karo. Yeh step kyun? TDSE hi left $=$ right hai; humein har side alag banani hogi. $\partial_{t} Ψ = - iω Ψ \Rightarrow i ℏ \partial_{t} Ψ = i ℏ (- iω) Ψ = ℏ ω Ψ.$ $i \times (- i) = + 1$ , to $i$ 's cancel ho jaate hain aur hume ek real multiple $ℏ ω$ milta hai. Notice karo $A$ untouched chalta hai — isliye iska value yahan irrelevant hai.

Step 2. Right side $- \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} Ψ$ compute karo ( $V = 0$ ke saath, ek free particle). Yeh step kyun? Har space derivative ek factor $ik$ neeche kheenchti hai; kinetic energy ke liye humein dono chahiye. $\partial_{x}^{2} Ψ = (ik)^{2} Ψ = - k^{2} Ψ \Rightarrow - \frac{ℏ ^{2}}{2 m} (- k^{2}) Ψ = \frac{ℏ ^{2} k ^{2}}{2 m} Ψ.$

Step 3. Unhe equal set karo. Yeh step kyun? Plane wave ek solution hai sirf tab jab dono sides sab $x, t$ ke liye agree karein. $ℏ ω = \frac{ℏ ^{2} k ^{2}}{2 m} \Rightarrow E = \frac{p ^{2}}{2 m} (using E = ℏ ω, p = ℏ k) .$

Verify: Yeh exactly free-particle energy–momentum relation hai. Units: $ℏ ω$ hai (J·s)(1/s) = J; $\frac{ℏ ^{2} k ^{2}}{2 m}$ hai $\frac{( J\cdots ) ^{2} ( 1/ m ) ^{2}}{kg} = J$ . ✓ Dono sides energies hain. Plane wave kaam karta hai, aur yeh forces karta hai correct energy.

Cell B — Free particle, real numbers

Worked example Ex 2 — Ek de Broglie electron ki momentum aur energy

(Cell B) Ek electron ki wavelength $λ = 1.0 \times 1 0^{- 10} m$ hai (1 Ångström, atomic size). $k$ , $p$ , aur $E$ dhundho. Forecast: guess karo — kya uski energy kuch eV hogi, ya hazaaron? (Atomic-scale wavelengths typically tens of eV deti hain.)

Step 1. Wavenumber $k = 2 π / λ$ nikalo. Yeh step kyun? $k$ (upar define kiya) radians of wave per metre count karta hai; yeh wavelength se momentum ka bridge hai $p = ℏ k$ ke zariye. Dekho de Broglie Hypothesis. $k = \frac{2 π}{1.0 \times 1 0 ^{- 10}} = 6.283 \times 1 0^{10} m^{- 1} .$

Step 2. Momentum $p = ℏ k$ . $p = (1.055 \times 1 0^{- 34}) (6.283 \times 1 0^{10}) = 6.63 \times 1 0^{- 24} kg\cdotm/s .$

Step 3. Energy $E = \frac{p ^{2}}{2 m}$ . Yeh step kyun? Free particle mein sirf kinetic energy hoti hai; TDSE Ex 1 ne prove kiya ki yahi sahi formula hai. $E = \frac{( 6.63 \times 1 0 ^{- 24} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} = 2.41 \times 1 0^{- 17} J = 150 eV .$

Verify: Convert back: $2.41 \times 1 0^{- 17} /1.602 \times 1 0^{- 19} = 150.6 eV$ . Yeh classic "electron microscope" energy hai — realistic. ✓

Cell C — Bound state, hard walls (quantization ka janam)

Worked example Ex 3 — 1 nm box mein electron ki ground-state energy

(Cell C) Box width $L = 1.0 \times 1 0^{- 9} m$ , infinite walls (to $V = 0$ andar aur $V = \infty$ bahar — vertical cliffs wali ek valley). $E_{1}$ dhundho. Forecast: 1 Å par free electron 150 eV tha. 1 nm mein trapped (das guna wide), guess karo kya $E_{1}$ 1 eV se upar ya neeche hai.

Step 1. Trapped energies yaad karo (parent note ne inhe derive kiya tha): $E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} .$ Yeh step kyun? Walls force karti hain $ψ (0) = ψ (L) = 0$ , jo sirf $k L = nπ$ allow karta hai — wahin se $n^{2}$ aata hai. Dekho Energy Quantization aur Particle in a Box.

Step 2. $n = 1$ set karo aur substitute karo. $E_{1} = \frac{π ^{2} ( 1.055 \times 1 0 ^{- 34} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} ) ( 1.0 \times 1 0 ^{- 9} ) ^{2}} .$

Step 3. Calculate karo. $E_{1} = 6.03 \times 1 0^{- 20} J = 0.376 eV .$

Verify: 1 eV se neeche, jaise forecast tha — wider box matlab longer allowed wavelength, chhoti momentum, chhoti energy. Units: $\frac{( J\cdots ) ^{2}}{kg\cdotm ^{2}} = J$ . ✓

Neeche ki figure yeh ground state draw karti hai. Cyan curve dekho: yeh ek single hump hai — exactly ek sine wave ka aadha — dono amber walls par zero par pinned. Woh "one half-wave fits the box" picture hi boundary condition $k L = π$ ko visible banata hai, aur geometric reason hai ki $n = 1$ lowest allowed state kyun hai.

Cell D — Normalization

Worked example Ex 4 — Ground state

$ψ_{1} = A sin (π x / L)$ ko normalize karo (Cell D) Ab hum finally fix karte hain amplitude $A$ jo humne Ex 1 mein free chhodh diya tha. Forecast: Particle box mein kahin to hai, to $\int_{0}^{L} ∣ ψ_{1} ∣^{2} d x = 1$ . Guess karo ki kya $A$ bada hoga ya chhota jab box $L$ wide ho jaata hai.

Step 1. Normalization demand likho. Yeh step kyun? $∣ ψ ∣^{2}$ ek probability density hai (Wavefunction and Born Interpretation); total probability 1 ke barabar honi chahiye — isliye $A$ pin hota hai. $\int_{0}^{L} A^{2} sin^{2} (\frac{π x}{L}) d x = 1.$

Step 2. $\int_{0}^{L} sin^{2} (π x / L) d x = L /2$ use karo. Yeh step kyun? $sin^{2}$ ek half-period pe $\frac{1}{2}$ average karta hai, aur box exactly ek half-period hold karta hai, to integral hai (average) $\times$ (width) $= \frac{1}{2} \cdot L$ . $A^{2} \cdot \frac{L}{2} = 1.$

Step 3. $A$ ke liye solve karo. $A = \frac{2}{L} .$

Verify: Jaise $L$ bada hota hai, $A$ chhota hota hai — wider box same total probability ko thinner spread karta hai, to peak height girta hai. Units: $1/ m$ , ek 1-D wavefunction ke liye correct hai taaki $∣ ψ ∣^{2} d x$ dimensionless ho. ✓ $L = 1 nm$ ke liye: $A = 2/1 0^{- 9} = 4.47 \times 1 0^{4} m^{- 1/2}$ .

Cell E — Excited state (higher $n$ )

Worked example Ex 5 — 1 nm box mein

$n = 1$ se $n = 2$ tak energy jump (Cell E) Forecast: kyunki $E_{n} \propto n^{2}$ , $1 \to 2$ jaane par $E$ char guna ho jaata hai. Gap $Δ E = E_{2} - E_{1}$ guess karo.

Step 1. Ratio $E_{n} = n^{2} E_{1}$ use karo. Yeh step kyun? $n^{2}$ ke alawa sab kuch levels ke beech identical hai, to hum constants dobara compute karne se bachte hain. $E_{2} = 4 E_{1} = 4 (0.376) = 1.505 eV .$

Step 2. Subtract karo. $Δ E = E_{2} - E_{1} = (4 - 1) E_{1} = 3 E_{1} = 1.129 eV .$

Verify: $Δ E$ ground energy ka teen guna hai, double nahi — kyunki energy $n^{2}$ ki tarah badhti hai, $n$ ki tarah nahi. ✓

Figure $n = 1$ (cyan) aur $n = 2$ (amber) ko overlay karta hai. White dot dekho centre mein: $n = 2$ curve wahaan se zero se guzarti hai — usne single-hump ground state ke comparison mein ek extra node gain kiya hai. Zyada nodes = tighter wiggles = bada $k$ = zyada energy; picture physical reason hai ki level spacing kyun badhta hai.

Cell F — Stationary vs moving fog

Worked example Ex 6 — Dikhao ki stationary state ke liye

$∣Ψ ∣^{2}$ time-independent hai (Cell F) Maano $Ψ (x, t) = ψ (x) e^{- i E t /ℏ}$ . Forecast: "stationary" — kya tumhe expect hai ki probability cloud drift karega, ya phase mein spin karte hue $Ψ$ ke saath still baithega?

Step 1. $∣Ψ ∣^{2} = Ψ^{*} Ψ$ likho. Yeh step kyun? Probability density modulus squared hai, aur modulus squared hamesha complex phases ko khatam karta hai. $Ψ^{*} = ψ^{*} (x) e^{+ i E t /ℏ} .$

Step 2. Multiply karo. $∣Ψ ∣^{2} = ψ^{*} ψ e^{+ i E t /ℏ} e^{- i E t /ℏ} = ∣ ψ (x) ∣^{2} \cdot e^{0} = ∣ ψ (x) ∣^{2} .$ Yeh step kyun? Do opposite phases cancel ho jaate hain: $e^{+ i θ} e^{- i θ} = 1$ . $t$ gayab ho jaata hai.

Verify: Koi $t$ nahi bachta — fog frozen hai chahe $Ψ$ rotate kare. Yahi "stationary state" ka exactly matlab hai. $E = 1.129 eV$ par phase $ω = E /ℏ = 1.71 \times 1 0^{15} rad/s$ ki speed se spin karta hai, phir bhi $∣Ψ ∣^{2}$ kabhi nahi hilta. ✓

Figure yeh directly dikhata hai: amber curve $∣Ψ ∣^{2}$ perfectly still baithti hai, jabki faded cyan curves teen instants par $Re (Ψ)$ hain — clearly upar aur neeche spin kar rahi hain. Observable (amber) frozen hai; phase (cyan) nahi.

Cell G — Degenerate / forbidden inputs

Worked example Ex 7 —

$n = 0$ aur $E < V_{m i n}$ par kya galat ho jaata hai? (Cell G — woh case jo log skip karte hain) Yahan $V_{m i n} = 0$ (box ka flat floor), to " $E < V_{m i n}$ " matlab " $E < 0$ " hai. Forecast: guess karo inme se kaun "allowed but boring" hai, aur kaun "physically impossible."

Case $n = 0$ . Yeh kyun check karein? Quantization $k L = nπ$ lagta hai $n = 0$ allow karta hai. Tab $k = 0$ , aur $ψ = A sin (0 \cdot x) = 0$ har jagah. Total probability $\int ∣ ψ ∣^{2} = 0 \neq = 1$ . Ek particle jo kahin nahi hai woh particle nahi hai. To $n = 0$ rejected hai — lowest real state $n = 1$ hai, aur uski already nonzero energy $E_{1} > 0$ hai.

Case negative $n$ . $sin (- θ) = - sin θ$ , to $ψ_{- n} = - ψ_{n}$ . Kyunki overall $- 1$ se $∣ ψ ∣^{2}$ nahi badlta, yeh same physical state hai. To $n = - 2$ kuch naya nahi deta; sirf $n = 1, 2, 3 \dots$ count karte hain.

Case $E < V_{m i n}$ (yahan $E < 0$ ). Yeh kyun check karein? Kya ek trapped particle ki energy uske potential landscape ke floor se neeche ho sakti hai? Tab $k = 2 m E /ℏ$ imaginary hai, to $ψ$ ek growing/decaying exponential ban jaata hai, sine nahi. Yeh dono walls par zero aur ek saath normalizable nahi ho sakta — koi valid state nahi. To bound energies hamesha $E > V_{m i n}$ hoti hain.

Verify: Forbidden inputs ( $n = 0$ , $E < V_{m i n}$ ) ya to zero function dete hain ya non-normalizable — mathematics khud unhe refuse karta hai. Ground energy $E_{1} > 0$ zero-point energy hai, Heisenberg Uncertainty Principle se forced: ek confined particle kabhi perfectly at rest nahi ho sakta. ✓

Cell H — Real-world word problem

Worked example Ex 8 — Nanowire electron

$n = 2 \to n = 1$ drop karne par photon emit karta hai (Cell H) Same 1 nm box. Electron excited state se ground state mein girta hai, ek photon emit karta hai. Photon wavelength dhundho. Forecast: visible light ~400–700 nm hai. Guess karo kya yeh photon visible hai, infrared hai, ya ultraviolet.

Step 1. Photon energy $=$ level gap. Yeh step kyun? Energy conservation: electron $Δ E$ lose karta hai, photon use carry karta hai. $E_{γ} = Δ E = 1.129 eV = 1.808 \times 1 0^{- 19} J .$

Step 2. Photon wavelength $E_{γ} = h c / λ$ se. Yeh formula kyun? Ek photon ki energy uski frequency se set hoti hai $E = h f = h c / λ$ ; hum $λ$ ke liye invert karte hain. $λ = \frac{h c}{E _{γ}} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{1.808 \times 1 0 ^{- 19}} .$

Step 3. Compute karo. $λ = 1.099 \times 1 0^{- 6} m \approx 1100 nm .$

Verify: ~1100 nm near-infrared hai — aankhon ke liye invisible, jaise forecast tha. Quick sanity shortcut: $λ (nm) \approx 1240/ E (eV) = 1240/1.129 = 1098 nm$ . ✓ Match karta hai.

Cell I — Exam twist

Worked example Ex 9 — Box shift karo: kya

$- \frac{L}{2} < x < \frac{L}{2}$ se energies badlti hain? (Cell I) Same infinite well, lekin origin par centred. Forecast: solve karne se pehle haan ya nahi guess karo — kya energies depend karti hain kahaan box rakha hai?

Step 1. Andar same TISE, naye boundaries $ψ (\pm L /2) = 0$ . Yeh step kyun? Physics unchanged hai; sirf coordinate labels move hue. Lekin $ψ$ ki form alag dikh sakti hai.

Step 2. Ab dono $sin$ aur $cos$ survive karte hain. Even states $ψ = B cos (k x)$ ko $cos (k L /2) = 0 \Rightarrow k L /2 = \frac{π}{2}, \frac{3 π}{2}, \dots$ chahiye; odd states $ψ = A sin (k x)$ ko $sin (k L /2) = 0 \Rightarrow k L /2 = π, 2 π, \dots$ chahiye. Yeh step kyun? Ek symmetric box symmetric (cos) aur antisymmetric (sin) solutions alag alag admit karta hai.

Step 3. Dono families combine karo: $k L /2 = nπ /2$ , yaani $k L = nπ$ , $n = 1, 2, 3 \dots$ — Ex 3 se identical. $E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} .$

Verify: Energies exactly same hain. Energy sirf box ki width $L$ par depend karti hai, kabhi uski position par nahi — origin shift karna sirf coordinates relabel karna hai, jo physics nahi badal sakta. Wavefunctions sirf all-sines se alternating cos/sin mein badlte hain. ✓

Active Recall

Recall Kaun sa cell toot ta hai, aur kyun?

$n = 0$ forbidden kyun hai? ::: yeh $ψ \equiv 0$ deta hai, ek particle jo kahin exist nahi karta — normalizable nahi.
Ek bound particle ki $E < V_{m i n}$ kyun nahi ho sakti? ::: $k$ imaginary ho jaata hai, $ψ$ blow up karta hai, dono walls aur normalization satisfy nahi kar sakta.
Kya box shift karne se $E_{n}$ badlta hai? ::: nahi — energy sirf width $L$ par depend karti hai.
Stationary state ke liye $∣Ψ ∣^{2}$ kyun nahi hilta? ::: $e^{\pm i E t /ℏ}$ phases $Ψ^{*} Ψ$ mein cancel ho jaate hain.
TISE ko kya chahiye jo TDSE ko nahi chahiye? ::: time-independent potential $V (x)$ .

Connections

Parent (Hinglish)
Particle in a Box — Cells C, E, G, I; aur finite wells & tunnelling.
Quantum Harmonic Oscillator — valley landscape $V = \frac{1}{2} m ω^{2} x^{2}$ .
Energy Quantization — $n^{2}$ ladder kyun appear hoti hai.
Wavefunction and Born Interpretation — Cells D, F.
de Broglie Hypothesis — Cell B, $p = ℏ k$ bridge.
Heisenberg Uncertainty Principle — $E_{1} > 0$ kyun (zero-point energy).
Hamiltonian Operator — woh $\hat{H}$ jo har case solve karta hai.

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