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

Answer: $|\psi {|}^2$.

• $\psi$ — negative ya complex ho sakta hai, aur probability kabhi negative nahi ho sakti. ✗
• ${\psi }^2$ — agar $\psi$ complex hai, toh ${\psi }^2$ bhi complex rahega (jaise $i^2=-1$). ✗
• $|\psi |$ — non-negative hai, toh lagta hai theek hai, lekin diffraction experiments (amplitudes add hote hain, phir square hota hai) $|\psi {|}^2$ se match karte hain, $|\psi |$ se nahi. ✗
• $|\psi {|}^2={\psi }^{\ast }\psi =a^2+b^2$ — real, $\ge 0$, aur experiment se match karta hai. ✓

$|\psi {|}^{2}dx$ ek pure probability hai (dimensionless). Kyunki $dx$ ke units metres (m) hain, $|\psi {|}^2$ ke units ${\text{m}}^{-1}$ hone chahiye (probability per metre). Kyunki magnitude mein $|\psi {|}^2=(\psi {)}^2$, $\psi$ ke units $({\text{m}}^{-1}{)}^{1/2}={\text{m}}^{-1/2}$ hain.

Kya: total probability $=1$ demand karo. Kyun: particle box mein kahin na kahin rehna chahiye. ${\int }_0^L|A{|}^{2}dx=|A{|}^{2}L=1\Rightarrow A=\frac{1}{\sqrt{L}}.$ Flat $|\psi {|}^2=1/L$ ka matlab hai box mein har jagah equally likely — uniform distribution.

$P={\int }_0^{L/4}\frac{1}{L}dx=\frac{1}{L}\cdot \frac{L}{4}=\frac{1}{4}.$ Box ka ek chauthai hissa probability ka ek chauthai rakhta hai — bilkul wahi jo "uniform" dena chahiye.

Kya: conjugate se multiply karo. Kyun: ${\psi }^{\ast }$ phase mein $i\to -i$ flip karta hai. $|\psi {|}^2={\psi }^{\ast }\psi =(A^{\ast }{e}^{-ikx}{e}^{-|x|/L}\left)\right(A{e}^{ikx}{e}^{-|x|/L}).$ Phases combine hote hain: $e^{-ikx}{e}^{ikx}=e^0=1$. Toh $|\psi {|}^2=|A{|}^2{e}^{-2|x|/L}.$ Oscillating factor $e^{ikx}$ (jo momentum carry karta hai, de Broglie Wavelength ke zariye) probability density ke liye invisible hai.

Normalize. Hume chahiye ${\int }_0^L{A}^2{\sin }^2\frac{\pi x}{L}dx=1$. ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$ use karo: ${\int }_0^L{\sin }^2\frac{\pi x}{L}dx=\frac{1}{2}{\int }_0^L(1-\cos \frac{2\pi x}{L})dx=\frac{1}{2}(L-0)=\frac{L}{2}.$ (Cosine ek full period pe zero integrate hota hai.) Toh $A^2\cdot \frac{L}{2}=1\Rightarrow A=\sqrt{2/L}$.

Forecast (symmetry). ${\sin }^2\frac{\pi x}{L}$ ek hump hai jo box centre $x=L/2$ ke baare mein symmetric hai. Toh left half aur right half dono equal probability carry karenge $\Rightarrow P=\frac{1}{2}$.

Integrate karke verify karo. $P=\frac{2}{L}{\int }_0^{L/2}{\sin }^2\frac{\pi x}{L}dx=\frac{2}{L}\cdot \frac{1}{2}[x-\frac{L}{2\pi }\sin \frac{2\pi x}{L}{]}_0^{L/2}=\frac{1}{L}\cdot \frac{L}{2}=\frac{1}{2}.✓$

$P=\frac{2}{L}{\int }_0^{L/4}{\sin }^2\frac{\pi x}{L}dx=\frac{2}{L}\cdot \frac{1}{2}[x-\frac{L}{2\pi }\sin \frac{2\pi x}{L}{]}_0^{L/4}.$ $x=L/4$ pe: bracket hai $\frac{L}{4}-\frac{L}{2\pi }\sin \frac{\pi }{2}=\frac{L}{4}-\frac{L}{2\pi }$. $P=\frac{1}{L}(\frac{L}{4}-\frac{L}{2\pi })=\frac{1}{4}-\frac{1}{2\pi }\approx \mathrm{0.0908.}$ Dhyan do yeh $\frac{1}{4}$ se kam hai — hump probability ko centre ki taraf push karta hai, edges ko starve karta hai.

Normalize. $|\psi {|}^2=|A{|}^2{e}^{-2|x|/L}$. Integral fold karne ke liye even symmetry use karo: ${\int }_{-\infty }^{\infty }|A{|}^2{e}^{-2|x|/L}dx=2|A{|}^2{\int }_0^{\infty }{e}^{-2x/L}dx=2|A{|}^2\cdot \frac{L}{2}=|A{|}^{2}L.$ $=1$ set karo $\Rightarrow A=1/\sqrt{L}$.

Ek decay length ke andar probability. $P(|x|\le L)=\frac{2}{L}{\int }_0^L{e}^{-2x/L}dx=\frac{2}{L}\cdot \frac{L}{2}(1-e^{-2})=1-e^{-2}\approx \mathrm{0.8647.}$ Toh lagbhag $86\%$ probability $\pm L$ ke andar rehti hai — tails patli hain, jaisa ek square-integrable $\psi$ demand karta hai.

${\psi }_2$ ki density compute karo: $|{\psi }_2{|}^2=(e^{i\pi /3}A\sin \frac{\pi x}{L}{)}^{\ast }\left(e^{i\pi /3}A\sin \frac{\pi x}{L}\right)=e^{-i\pi /3}{e}^{i\pi /3}|A{|}^2{\sin }^2\frac{\pi x}{L}=|A{|}^2{\sin }^2\frac{\pi x}{L}=|{\psi }_1{|}^2.$ Kyunki kisi bhi real $\theta$ ke liye $|e^{i\theta }|=1$ hota hai, ek global phase $e^{i\theta }$ se $|\psi {|}^2$ $1$ se multiply hota hai. Saari probabilities — aur saare Expectation Values $⟨x⟩=\int x|\psi {|}^{2}dx$ — unchanged rehte hain. Principle: ek global phase physically invisible hai.

(a) Normalize. Density piecewise constant hai, toh integral bas height² × width hai: ${\int }_0^d{\alpha }^{2}dx+{\int }_d^{2d}{\beta }^{2}dx={\alpha }^{2}d+{\beta }^{2}d=1\Rightarrow d({\alpha }^2+{\beta }^2)=1.$

(b) Equally likely regions. Left region ki probability ${\alpha }^{2}d$ hai, right ki ${\beta }^{2}d$. Equal $\Rightarrow {\alpha }^{2}d={\beta }^{2}d=\frac{1}{2}$, toh $\alpha =\beta =\frac{1}{\sqrt{2d}}.$ (Check: $d({\alpha }^2+{\beta }^2)=d(\frac{1}{2d}+\frac{1}{2d})=1.✓$)

(c) Expectation value. $|\psi {|}^2=\frac{1}{2d}$ ke saath $[0,2d]$ pe (poore $2d$ pe uniform): $⟨x⟩={\int }_0^{2d}x\frac{1}{2d}dx=\frac{1}{2d}\cdot \frac{(2d{)}^2}{2}=\frac{1}{2d}\cdot 2d^2=d.$ Mean position midpoint $x=d$ hai — exactly symmetry centre. Sensible hai.

Do spreads ko multiply karo: $\Delta x\Delta p=\Delta x\cdot \frac{\hslash }{2\Delta x}=\frac{\hslash }{2}.$ Product exactly $\hslash /2$ ke barabar hai, toh Gaussian $\Delta x\Delta p\ge \hslash /2$ ki equality edge pe baitha hai — yeh minimum-uncertainty state hai. Koi bhi aur shape $|\psi {|}^2$ aur uske Fourier partner ko wider spread karta hai, product ko $\hslash /2$ se upar le jaata hai.

L1 ::: Recognize karo ki sirf $|\psi {|}^2$ ek valid density hai, aur uske units jaano. L2 ::: Normalization ($÷\sqrt{N}$) apply karo aur global phase strip karo. L3 ::: ${\sin }^2$ integrate karo aur symmetry use karo; equal lengths = equal probability nahi hota jab tak flat na ho. L4 ::: Decaying tails handle karo aur global-phase invariance prove karo. L5 ::: Superpositions build karo, $⟨x⟩$ compute karo, aur uncertainty bound se connect karo.