Wave function ψ — probability density - ψ - ²: Worked examples

2.3.8 · D3 · Physics › Modern Physics › Wave function ψ — probability density - ψ - ²

Scenario matrix

Kuch bhi work karne se pehle, poora territory map karte hain. $ψ$ aur $∣ ψ ∣^{2}$ ke baare mein har problem in cells mein se kisi ek mein girti hai. Last column us worked example ka naam deta hai jo use cover karta hai.

#	Case class	Kya tricky banata hai ise	Covered by
A	Real, everywhere-decaying $ψ$ (Gaussian)	infinite domain, Gaussian integral	Example 1
B	Real $ψ$ on a finite box	domain limits, $sin^{2}$ average	Example 2
C	Complex $ψ$ with a phase $e^{ik x}$	phase must cancel in $ψ^{*} ψ$	Example 3
D	Piecewise $ψ$ (two regions glued)	integrate region by region, continuity	Example 4
E	Degenerate input: constant $ψ$ on all of $R$	area is infinite → NOT normalizable	Example 5
F	Sub-interval probability + a symmetry shortcut	compute $P (a \leq x \leq b)$	Example 6
G	Superposition of two states	cross terms, orthogonality	Example 7
H	Limiting behaviour: what happens as a width $\to 0$ or $\to \infty$	localisation vs spreading	Example 8
I	Word problem / real detector	translate physics to an integral	Example 9
J	Exam twist: normalize then a trap about "probability at a point"	density vs probability	Example 10

Ab hum har cell fill karte hain. Har example mein pehle tumhe Forecast karna hai — steps padhne se pehle answer guess karo.

Example 1 — Cell A: real decaying Gaussian

Worked example

$ψ (x) = A e^{- x^{2} / (2 σ^{2})}$ ko normalize karo $σ > 0$ ek fixed width hai. Positive constant $A$ nikalo.

Forecast: Kya $A$ badhega ya ghutega jab bell wider hoti hai (bada $σ$ )? Padhne se pehle guess karo.

Step 1. Density likho: kyunki $ψ$ real hai, $∣ ψ ∣^{2} = ψ^{*} ψ = ψ^{2} = A^{2} e^{- x^{2} / σ^{2}}$ . Yeh step kyun? Real function ke liye $ψ^{*} = ψ$ hota hai, isliye squaring yahan safe hai (dekho Complex Numbers).

Step 2. Total probability $1$ demand karo: $\int_{- \infty}^{\infty} A^{2} e^{- x^{2} / σ^{2}} d x = 1.$ Yeh step kyun? Particle kहीं na कहीं toh hoga — yahi normalization condition hai parent note se.

Step 3. Gaussian integral $\int_{- \infty}^{\infty} e^{- α x^{2}} d x = π / α$ use karo $α = 1/ σ^{2}$ ke saath: $\int_{- \infty}^{\infty} e^{- x^{2} / σ^{2}} d x = π σ^{2} = σ π .$ Yeh tool kyun? Humein bell curve ke neeche ka area chahiye; Gaussian integral woh ek standard result hai jo ise closed form mein deta hai — koi elementary antiderivative exist nahi karta, isliye yahi ek hi rasta hai.

Step 4. Toh $A^{2} σ π = 1 \Rightarrow A = (σ π)^{- 1/2} = \frac{1}{( σ π ) ^{1/2}}$ .

Verify: Jab $σ$ badhta hai, $A$ ghatta hai $σ^{- 1/2}$ ki tarah — ek wider bell ko area $= 1$ rakhne ke liye shorter hona padta hai. Units: $[σ] = m$ , toh $[A] = m^{- 1/2}$ ✓, bilkul 1D mein $ψ$ ki unit.

Figure — Wave function ψ — probability density - ψ - ²

Figure dekho: blue $ψ$ aur orange $∣ ψ ∣^{2}$ . Orange curve narrower hai (squaring tails ko zyada fast neeche kheenchta hai) aur uska total shaded area exactly $1$ hai.

Example 2 — Cell B: finite box mein real ψ

Worked example Box ke ground state ko normalize karo

$ψ (x) = A sin \frac{π x}{L}$ for $0 \leq x \leq L$ , aur bahar $ψ = 0$ . $A$ nikalo.

Forecast: Kya $A$ depend karega $L$ par? Agar box double width ka ho jaye, toh kya $A$ badhega ya ghutega?

Step 1. Sirf region $[0, L]$ contribute karta hai ( $ψ = 0$ elsewhere): $\int_{0}^{L} A^{2} sin^{2} \frac{π x}{L} d x = 1.$ Yeh step kyun? Box ke bahar $∣ ψ ∣^{2} = 0$ hai, toh woh parts kuch add nahi karte — integral collapse ho jaata hai $[0, L]$ par.

Step 2. $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ use karo: $\int_{0}^{L} sin^{2} \frac{π x}{L} d x = \int_{0}^{L} \frac{1}{2} d x - \frac{1}{2} \int_{0}^{L} cos \frac{2 π x}{L} d x = \frac{L}{2} - 0 = \frac{L}{2} .$ Yeh tool kyun? $sin^{2}$ ka apna koi simple antiderivative nahi hota; double-angle identity ise ek constant aur ek cosine mein convert kar deti hai jo ek full period mein zero integrate ho jaata hai. Cosine term vanish hota hai kyunki $0$ se $L$ tak exactly ek cycle complete hoti hai.

Step 3. Toh $A^{2} \cdot \frac{L}{2} = 1 \Rightarrow A = \frac{2}{L}$ .

Verify: Bada box $\Rightarrow$ chota $A$ ( $A \propto L^{- 1/2}$ ): same total probability zyada room mein pheli hui hai, isliye peak height girti hai. Yeh Particle in a Box se match karta hai. Units: $[A] = m^{- 1/2}$ ✓.

Example 3 — Cell C: complex ψ, phase cancel ho jaata hai

Worked example Ek travelling, decaying wave ki density

$ψ (x) = A e^{ik x} e^{- ∣ x ∣/ L}$ jahan $A$ real $> 0$ hai. $∣ ψ ∣^{2}$ nikalo, phir normalize karo.

Forecast: Kya momentum-carrying wiggle $e^{ik x}$ probability density mein dikhega?

Step 1. Conjugate: $ψ^{*} = A e^{- ik x} e^{- ∣ x ∣/ L}$ . Yeh step kyun? Conjugation $i \to - i$ flip karta hai, matlab $e^{ik x} \to e^{- ik x}$ (dekho Complex Numbers).

Step 2. Multiply karo: $∣ ψ ∣^{2} = ψ^{*} ψ = A^{2} e^{- ik x} e^{ik x} e^{- 2∣ x ∣/ L} = A^{2} e^{- 2∣ x ∣/ L} .$ Yeh step kyun? $e^{- ik x} e^{ik x} = e^{0} = 1$ : phase factor divide out ho jaata hai. Woh particle ka momentum carry kar raha tha, lekin $∣ ψ ∣^{2}$ ke liye invisible hai.

Step 3. Normalize karo. $e^{- 2∣ x ∣/ L}$ ki even symmetry se: $\int_{- \infty}^{\infty} A^{2} e^{- 2∣ x ∣/ L} d x = 2 A^{2} \int_{0}^{\infty} e^{- 2 x / L} d x = 2 A^{2} \cdot \frac{L}{2} = A^{2} L = 1.$ Factor 2 kyun? Integrand $x = 0$ ke baare mein symmetric hai, isliye puri line right half ke double ke equal hai — ek shortcut jo kaam aadha kar deta hai.

Step 4. $A = \frac{1}{L}$ .

Verify: $A^{2} L = (1/ L)^{2} \cdot L = 1$ ✓. Result $k$ se independent hai — do waves with different momenta lekin same envelope same $∣ ψ ∣^{2}$ share karte hain, exactly wahi "global-phase-invisible" lesson.

Example 4 — Cell D: piecewise ψ ek seam par glued

Worked example Two-region wave function

$ψ (x) = ⎩ ⎨ ⎧ A x, A (2 - x), 0, 0 \leq x \leq 1 1 \leq x \leq 2 otherwise$ (lengths metres mein). $A$ nikalo.

Forecast: Yeh ek triangular "tent" hai. Peak $x = 1$ par hai. Kya $A$ bada hoga ya chota $1$ se?

Step 1. Integral ko seam $x = 1$ par split karo: $\int_{0}^{2} ∣ ψ ∣^{2} d x = A^{2} \int_{0}^{1} x^{2} d x + A^{2} \int_{1}^{2} (2 - x)^{2} d x .$ Yeh step kyun? $ψ$ ka har side par alag formula hai, isliye tum har region ko alag integrate karo, phir add karo.

Step 2. Left piece: $\int_{0}^{1} x^{2} d x = \frac{1}{3}$ . Right piece: $u = 2 - x$ , $d u = - d x$ substitute karo; jab $x : 1 \to 2$ , $u : 1 \to 0$ , deta hai $\int_{0}^{1} u^{2} d u = \frac{1}{3}$ . Yeh step kyun? Tent symmetric hai, isliye dono halves same $\frac{1}{3}$ contribute karte hain.

Step 3. Total: $A^{2} (\frac{1}{3} + \frac{1}{3}) = \frac{2}{3} A^{2} = 1 \Rightarrow A = \frac{3}{2} = 1.5 \approx 1.2247$ .

Verify: Continuity check — $x = 1$ par left deta hai $A \cdot 1 = A$ aur right deta hai $A (2 - 1) = A$ : woh match karte hain, toh $ψ$ continuous hai, physical conditions mein se ek. $A > 1$ as forecast (ek narrow tent tall honi chahiye). Units: area metres mein tha, $[A] = m^{- 1/2}$ ✓.

Example 5 — Cell E: degenerate, NON-normalizable case

Worked example Har jagah constant — ek trap

Kya $ψ (x) = A$ (ek nonzero constant, sab $x$ ke liye) kabhi normalize ho sakta hai?

Forecast: Computing se pehle haan ya nahi bolo.

Step 1. Condition try karo: $\int_{- \infty}^{\infty} ∣ A ∣^{2} d x = ∣ A ∣^{2} \cdot (\infty) = \infty$ . Yeh step kyun? Ek infinite line par constant density infinite area enclose karta hai — koi finite $A$ $\infty$ ko rescale karke $1$ nahi bana sakta.

Step 2. Conclusion: not square-integrable ⇒ valid physical wave function nahi hai. Sirf woh "constant" normalize hoti hai jo $A = 0$ hai, jo koi particle nahi describe karta.

Verify: Yahi reason hai ki parent ki condition " $ψ \to 0$ as $x \to \pm \infty$ " exist karti hai. Plane wave $e^{ik x}$ (constant magnitude) same story hai — ek useful idealisation, lekin strictly normalize hone ke liye ek box ya wave packet chahiye. Cross-check karo de Broglie Wavelength se: ek infinitely long pure wave ka momentum perfectly known hota hai lekin position completely unknown — dekho Heisenberg Uncertainty Principle.

Example 6 — Cell F: sub-interval probability with a symmetry shortcut

Worked example Box ke left half mein probability

Normalized ground state $ψ = 2/ L sin \frac{π x}{L}$ (Example 2 se) use karke, $P (0 \leq x \leq L /2)$ nikalo.

Forecast: Density $sin^{2}$ box centre $x = L /2$ ke baare mein symmetric hai. Integrate karne se pehle $P$ guess karo.

Step 1. Symmetry argument: $sin^{2} \frac{π x}{L}$ mein $x = L /2$ ke baare mein mirror symmetry hai, toh left half ka area right half ke area ke barabar hai. Yeh step kyun? Agar do halves equal hain aur $1$ mein add hote hain, toh har ek exactly $\frac{1}{2}$ hai.

Step 2. Integration se confirm karo: $P = \int_{0}^{L /2} \frac{2}{L} sin^{2} \frac{π x}{L} d x = \frac{2}{L} [\frac{x}{2} - \frac{L}{4 π} sin \frac{2 π x}{L}]_{0}^{L /2} .$ Yeh step kyun? Example 2 jaisi hi double-angle trick; bracket $sin^{2}$ ka antiderivative hai.

Step 3. $x = L /2$ par: $\frac{L /2}{2} - \frac{L}{4 π} sin π = \frac{L}{4} - 0 = \frac{L}{4}$ . $x = 0$ par: $0$ . Toh $P = \frac{2}{L} \cdot \frac{L}{4} = \frac{1}{2}$ .

Verify: $P = \frac{1}{2}$ symmetry forecast se match karta hai ✓. Ek Forecast-then-Verify win: picture ne algebra se pehle hi answer bata diya.

Ab ek non-symmetric slice, yeh dikhane ke liye ki shortcut hamesha available nahi hota:

Example 7 — Cell G: superposition aur cross terms

Worked example Do box states ka equal mix

Maano $ϕ_{1} = 2/ L sin \frac{π x}{L}$ aur $ϕ_{2} = 2/ L sin \frac{2 π x}{L}$ do normalized box states hain. Consider karo $ψ = c (ϕ_{1} + ϕ_{2})$ . Constant $c$ nikalo jo $ψ$ ko normalize kare.

Forecast: Naively tum guess kar sakte ho ki norm $1 + 1 = 2$ hai toh $c = 1/ 2$ . Kya ek cross term yeh bigadega?

Step 1. $∣ ψ ∣^{2} = c^{2} (ϕ_{1} + ϕ_{2})^{2} = c^{2} (ϕ_{1}^{2} + 2 ϕ_{1} ϕ_{2} + ϕ_{2}^{2})$ expand karo. Yeh step kyun? Middle cross term $2 ϕ_{1} ϕ_{2}$ wahan hai jahan interference rehti hai — humein ise check karna hoga.

Step 2. Term by term integrate karo: $\int ϕ_{1}^{2} = 1, \int ϕ_{2}^{2} = 1, \int ϕ_{1} ϕ_{2} d x = \frac{2}{L} \int_{0}^{L} sin \frac{π x}{L} sin \frac{2 π x}{L} d x = 0.$ Cross term zero kyun hai? Do box states orthogonal hain — $sin \frac{π x}{L}$ aur $sin \frac{2 π x}{L}$ alag standing-wave modes hain; unka product box mein zero integrate hota hai. Yeh Particle in a Box se ek key fact hai.

Step 3. Toh $\int ∣ ψ ∣^{2} = c^{2} (1 + 0 + 1) = 2 c^{2} = 1 \Rightarrow c = \frac{1}{2}$ .

Verify: Naive guess sahi nikla — lekin sirf isliye kyunki states orthogonal hain. Agar nahi hote, toh cross term $c$ change kar deta. Hamesha check karo.

Example 8 — Cell H: spreading packet ka limiting behaviour

Worked example Gaussian width

$\to 0$ aur $\to \infty$ hone par kya hota hai? Example 1 ki normalized Gaussian density lo: $∣ ψ ∣^{2} = \frac{1}{σ π} e^{- x^{2} / σ^{2}}$ . Do limits study karo.

Forecast: Kis limit mein particle "sharply located" hai? Kis mein woh "everywhere aur nowhere" hai?

Step 1 — peak height at $x = 0$ : $∣ ψ (0) ∣^{2} = \frac{1}{σ π}$ . Yeh step kyun? Peak measure karta hai ki "fog" kitna concentrated hai.

Step 2 — limit $σ \to 0$ : peak $\to \infty$ , width $\to 0$ , lekin total area $1$ rehta hai. Yeh ek spike hai (Dirac-delta-like localisation): particle ki position almost certain ho jaati hai. Physically kyun? Perfect position knowledge $\Rightarrow$ (Heisenberg Uncertainty Principle se) totally unknown momentum.

Step 3 — limit $σ \to \infty$ : peak $\to 0$ , bell Example 5 ke constant ki taraf flatten ho jaata hai — non-normalizable case ki taraf ja raha hai: particle puri space par smeared hai, momentum sharply defined hai.

Verify: Kisi bhi $σ$ par area compute karo: $\int \frac{1}{σ π} e^{- x^{2} / σ^{2}} d x = \frac{1}{σ π} \cdot σ π = 1$ ✓ — hamesha exactly $1$ , chahe width kuch bhi ho. Trade-off height vs spread ka hai, kabhi total probability ka nahi.

Example 9 — Cell I: real-world detector word problem

Worked example Ek detector slit

Ek electron box ground state $ψ = 2/ L sin \frac{π x}{L}$ mein hai jahan $L = 1.0 nm$ hai. Ek detector central strip $x = 0.4 nm$ se $x = 0.6 nm$ tak cover karta hai. Electron ke pakde jaane ki probability kya hai?

Forecast: Central strip wahan hai jahan density peak karti hai. Kya tum $0.2/1.0 = 0.2$ "flat" guess se zyada expect karte ho ya kam?

Step 1. Ek integral mein translate karo (jahan $L = 1$ nm toh $\frac{π x}{L} = π x$ , $x$ nm mein): $P = \int_{0.4}^{0.6} 2 sin^{2} (π x) d x .$ Yeh step kyun? Ek physical "detection chance over a region" exactly $\int_{a}^{b} ∣ ψ ∣^{2} d x$ hai.

Step 2. Antiderivative: $2 \int sin^{2} (π x) d x = x - \frac{sin ( 2 π x )}{2 π}$ . Yeh tool kyun? Double-angle identity phir se — $sin^{2}$ ke liye workhorse.

Step 3. Evaluate karo: $P = [x - \frac{s i n 2 π x}{2 π}]_{0.4}^{0.6} = (0.6 - \frac{s i n 1.2 π}{2 π}) - (0.4 - \frac{s i n 0.8 π}{2 π}) .$ Numerically $sin 1.2 π \approx - 0.5878$ , $sin 0.8 π \approx 0.5878$ , toh $P \approx (0.6 + 0.09355) - (0.4 - 0.09355) = 0.69355 - 0.30645 \approx 0.3871.$

Verify: $P \approx 0.39$ , jo flat guess $0.20$ se zyada hai — sahi hai, kyunki density middle mein peak karti hai jahan detector rakha hai. Probability dimensionless hai (nm units $x$ mein aur $ψ^{2} \sim nm^{- 1}$ mein cancel ho jaate hain) ✓.

Example 10 — Cell J: exam twist (density vs point-probability)

Worked example Classic trap

Ek student $ψ = 2/ L sin \frac{π x}{L}$ normalize karta hai aur puchha jaata hai: "Particle ko exactly $x = L /2$ par paane ki probability kya hai?" Woh answer deta hai $∣ ψ (L /2) ∣^{2} = \frac{2}{L} sin^{2} \frac{π}{2} = \frac{2}{L}$ . Ise mark karo.

Forecast: Kya $\frac{2}{L}$ koi probability hai bhi?

Step 1. Units check: $\frac{2}{L}$ ke units $m^{- 1}$ hain. Probability dimensionless hoti hai. Toh $\frac{2}{L}$ probability ho hi nahi sakta — yeh ek density hai. Yeh step kyun? Units physics mein sabse fast lie-detector hain.

Step 2. Sahi answer: ek single exact point par probability hai $P ({L /2}) = \int_{L /2}^{L /2} ∣ ψ ∣^{2} d x = 0.$ Kyun? Zero width ka ek interval zero area enclose karta hai. Nonzero number sirf ek strip $[x_{0} - \frac{Δ}{2}, x_{0} + \frac{Δ}{2}]$ par milta hai, jahan small $Δ$ ke liye $P \approx ∣ ψ (x_{0}) ∣^{2} Δ = \frac{2}{L} Δ$ hota hai.

Verify: Box ke liye, agar centre ke around $Δ = 0.01 L$ ka strip ho, toh $P \approx \frac{2}{L} \cdot 0.01 L = 0.02$ — small lekin nonzero, aur dimensionless ✓. Student ne density ( $\frac{2}{L}$ ) ko probability (jisme width factor chahiye) ke saath confuse kiya.

Coverage check

Recall Kya humne har cell fill ki?

A Gaussian ::: Example 1 B finite box ::: Example 2 C complex phase ::: Example 3 D piecewise ::: Example 4 E degenerate/non-normalizable ::: Example 5 F sub-interval probability ::: Example 6 G superposition cross terms ::: Example 7 H limiting behaviour ::: Example 8 I word problem ::: Example 9 J exam density-vs-point trap ::: Example 10

Connections

Parent — Hinglish version
Particle in a Box — Examples 2, 6, 7, 9, 10 sab yahan rehte hain.
Complex Numbers — Example 3 mein conjugation aur modulus.
Heisenberg Uncertainty Principle — Example 8 ka width trade-off.
de Broglie Wavelength — Example 5 ke non-normalizable case ke peeche plane wave.
Schrödinger Equation — woh equation jise yeh $ψ$ follow karte hain.
Expectation Values — natural next step jab $∣ ψ ∣^{2}$ normalize ho jaaye.