Exercises — Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ - 2

KYA HOGA: Jaise jaise $\Delta x$ chhota hota hai, $\Delta p$ ka floor upar uthta hai. KYU: $\Delta x\Delta p\ge \hslash /2$ mein, right side $\hslash /2$ ek fixed positive number hai. Agar $\Delta x\to$ chhota ho, toh product ko $\ge \hslash /2$ rakhne ke liye $\Delta p$ ko kam se kam $1/\Delta x$ ki speed se badhna hoga. $\Delta p\ge \frac{\hslash }{2\Delta x}.$ Ek ko dabao, doosra uchhal jaata hai — balloon wali picture.

KYA: Energy–time relation $\Delta E\Delta t\ge \hslash /2$ use karo. KYU: Unknown ek energy spread hai; diya hua hai wo time jitni state exist karti hai. Lifetime $\tau$ hi $\Delta t$ hai — woh timescale jis par state appreciably change hoti hai (yahan, decay hokar). Toh $\Delta E\ge \hslash /(2\tau )$. (Number Q6 mein calculate karenge.)

Momentum: minimum matlab equality. $\Delta p=\frac{\hslash }{2\Delta x}=\frac{1.055\times 10^{-34}}{2(5.0\times 10^{-11})}=1.055\times 10^{-24}\text{kg⋅m/s}.$ Speed: electron mass se divide karo. $\Delta v=\frac{\Delta p}{m_e}=\frac{1.055\times 10^{-24}}{9.11\times 10^{-31}}\approx 1.16\times 10^6\text{m/s}.$ Ek million metres per second ka irreducible velocity spread — isliye electrons atoms ke andar still nahi baith sakte.

$\Delta v=\frac{\hslash }{2m\Delta x}=\frac{1.055\times 10^{-34}}{2(0.16)(1.0\times 10^{-6})}\approx 3.30\times 10^{-28}\text{m/s}.$ Comment: $\hslash$ bahut chhota hai, $m$ bahut bada — floor $10^{-28}\text{m/s}$ hai, bilkul measure nahi hoga. Quantum fuzziness hai, bas balls ke liye cosmically irrelevant hai.

$\Delta p$: $\Delta p_{\min }=\hslash /(2\Delta x)$ sirf $\Delta x$ par depend karta hai, mass par nahi. Toh dono ka $\Delta p$ equal hai. $\Delta v$: $\Delta v=\Delta p/m$. Same $\Delta p$, zyada mass ⇒ chhota $\Delta v$. Electron ka speed spread zyada hai, factor $1836$ se. Lesson: confinement momentum spread decide karta hai; mass phir decide karta hai ki woh kitna velocity banta hai.

$\Delta E=\frac{\hslash }{2\tau }=\frac{1.055\times 10^{-34}}{2(2.0\times 10^{-9})}\approx 2.64\times 10^{-26}\text{J}.$ Convert: $\frac{2.64\times 10^{-26}}{1.602\times 10^{-19}}\approx 1.65\times 10^{-7}\text{eV}.$ Yeh chhoti si spread state ki natural linewidth hai — dekho Natural linewidth and spectral broadening.

Setup (KYA & KYU): Confinement ka matlab hai particle box ke andar kahi hai, toh uska position spread box size ke barabar hai: $\Delta x\sim a$ (figure mein cyan bump ki width). Relation se, $\Delta p\sim \hslash /(2a)$. Kyunki box symmetric hai, average momentum zero hai: $⟨p⟩=0$. Phir typical momentum-squared sirf spread squared hai: $⟨p^2⟩\sim (\Delta p{)}^2$. Energy: kinetic energy $p^2/(2m)$ hai, toh $E_{\min }\sim \frac{(\Delta p{)}^2}{2m}\sim \frac{{\hslash }^2}{8m{a}^2}.$ Kyun zero nahi ho sakta: $E=0$ ke liye $p=0$ exactly chahiye, matlab $\Delta p=0$. Par phir $\Delta x\ge \hslash /(2\Delta p)\to \infty$ — particle poore space mein spread ho jaayega, jo "box mein trapped hai" se contradict karta hai. Isliye ground-state energy zero se upar forced hai: yahi zero-point energy hai (compare Particle in a box). Number (electron, $a=10^{-10}$ m): $E_{\min }\sim \frac{(1.055\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(10^{-10}{)}^2}\approx 1.53\times 10^{-19}\text{J}\approx 0.95\text{eV}.$ Atomic energies ke liye order-of-magnitude sahi hai — achha.

Analysis: Relevant relation $\Delta E\Delta t\ge \hslash /2$ hai jisme $\Delta t=\tau$, lifetime hai. $0.99c$ ki velocity ek red herring hai — yeh batata hai muon kitna door jaata hai, uski energy blur nahi. $\Delta E=\frac{\hslash }{2\tau }=\frac{1.055\times 10^{-34}}{2(2.2\times 10^{-6})}\approx 2.40\times 10^{-29}\text{J}.$ eV mein: $\frac{2.40\times 10^{-29}}{1.602\times 10^{-19}}\approx 1.50\times 10^{-10}\text{eV}.$ Ek famously razor-sharp energy — kyunki muon comparatively "lambe" samay tak jeeta hai.

Exact value: $E_1=\frac{{\pi }^2(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})(10^{-10}{)}^2}\approx 6.03\times 10^{-18}\text{J}\approx 37.6\text{eV}.$ Ratio: $E_1/E_{\min }=\frac{{\pi }^2{\hslash }^2/(2m{a}^2)}{{\hslash }^2/(8m{a}^2)}=4{\pi }^2\approx \mathrm{39.5.}$ Estimate chhota kyun hai: Uncertainty argument ne loosest bound $\Delta x\sim a$ aur $\Delta x\Delta p=\hslash /2$ mein equality use ki. Real confined wavefunction ka effective $\Delta p$ thoda zyada hota hai (use dono walls par zero hona hota hai), isliye true energy crude floor se kaafi upar hoti hai. Uncertainty principle sahi predict karta hai "nonzero aur roughly atomic-scale," exact number nahi.

(i) $p=\frac{h}{\lambda }=\frac{2\pi \hslash }{\lambda }=\frac{2\pi (1.055\times 10^{-34})}{1.0\times 10^{-10}}\approx 6.63\times 10^{-24}\text{kg⋅m/s}.$ (ii) $\Delta p=p$ set karo aur $\Delta x\ge \hslash /(2\Delta p)$ use karo: $\Delta x=\frac{\hslash }{2p}=\frac{1.055\times 10^{-34}}{2(6.63\times 10^{-24})}\approx 7.96\times 10^{-12}\text{m}\approx 0.080\text{A˚}.$ Matlab: momentum spread ko ek poore de Broglie momentum jitna bada banane ke liye, electron ko ek ångström ke dasve hisse se bhi chhote space mein localize karna padega. Momentum spread aur de Broglie momentum ek hi wave nature ke do chehere hain (de Broglie wavelength).

(i) $\Delta E=\frac{\hslash }{2\tau }=\frac{1.055\times 10^{-34}}{2(1.6\times 10^{-8})}\approx 3.30\times 10^{-27}\text{J}.$ (ii) $|\Delta E|=\frac{hc}{{\lambda }^2}\Delta \lambda$ rearrange karo $\Rightarrow \Delta \lambda =\frac{{\lambda }^2\Delta E}{hc}$: $\Delta \lambda =\frac{(589\times 10^{-9}{)}^2(3.30\times 10^{-27})}{(6.626\times 10^{-34})(3.0\times 10^8)}\approx 5.76\times 10^{-15}\text{m}\approx 5.8\times 10^{-6}\text{nm}.$ Spectral line intrinsically lagbhag $6\times 10^{-6}$ nm wide hai — yahi natural width hai, kisi bhi Doppler ya collision broadening se pehle (Natural linewidth and spectral broadening).

Verdict: Galat. Quantum mechanics mein energy exactly conserved hoti hai. Symbols ka matlab: $\Delta E$ ek state ke identically prepared copies par measurements ki statistical spread hai; $\Delta t$ woh timescale hai jis par state appreciably change hoti hai (uski lifetime). Yeh dono koi "loan" nahi hain. Argument kahan toot ta hai: Jo state sirf $\Delta t$ tak rehti hai woh koi single sharp energy nahi hai — woh ek superposition hai jiske energies genuinely $\Delta E\ge \hslash /(2\Delta t)$ se spread hain. Measurement ek value deta hai us spread ke saath; average conserved rehta hai aur koi bhi single measurement kabhi "extra" energy appear-then-vanish nahi dikhata. "Borrowing" phrase virtual particles ke liye ek heuristic hai, real violation nahi.

Plug in: $E_{\min }\sim \frac{{\hslash }^2}{8m_p{a}^2}=\frac{(1.055\times 10^{-34}{)}^2}{8(1.67\times 10^{-27})(5.0\times 10^{-15}{)}^2}\approx 3.33\times 10^{-14}\text{J}.$ MeV mein convert: $E_{\min }\approx \frac{3.33\times 10^{-14}}{1.602\times 10^{-13}}\approx 0.21\text{MeV}.$ Comment: $E_{\min }\propto 1/a^2$. Nucleus atom se $\sim 10^5$ times chhota hai, isliye confinement energies $\sim 10^{10}$ times badi hain — eV ki jagah MeV. Isliye nuclear processes chemical ones se kahin zyada powerful hote hain: ek proton ko nucleus mein squeeze karne mein enormous zero-point energy lagti hai, jo measured nuclear binding energies ke MeV order ki hi hoti hai.