Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ - 2: Worked examples

2.3.7 · D3 · Physics › Modern Physics › Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ

Yeh note Heisenberg parent note ka "sleeves roll up karo" wala child hai. Parent ne bataya tha kyun. Yahan hum har tarah ke problem ko hit karte hain jo ye do relations

$Δ x Δ p \geq \frac{ℏ}{2}, Δ E Δ t \geq \frac{ℏ}{2}$

tumhare samne phenk sakti hain. Koi bhi number daalne se pehle, aao har symbol ka matlab pin down kar lete hain taaki kuch bhi unexplained na rahe.

Scenario matrix

Har uncertainty problem inhi cells mein se ek hai. Neeche ke examples mein cell label laga hai jise woh cover karta hai, toh ant tak har row bhar jaati hai.

Cell	Kya distinct banata hai	Example
A. $Δ p$ solve karo	$Δ x$ diya, momentum spread nikalao	Ex 1
B. $Δ x$ solve karo	$Δ p$ (ya $Δ v$ ) diya, position spread nikalao	Ex 2
C. Macroscopic limit	badi mass ⇒ effect gayab ho jaata hai	Ex 3
D. Degenerate input $Δ x \to 0$	perfect position ⇒ $Δ p$ ka kya hoga?	Ex 4
E. Energy–time: lifetime ⇒ linewidth	$τ$ diya, $Δ E$ nikalao	Ex 5
F. Energy–time: linewidth ⇒ lifetime	ulta chalao	Ex 6
G. Zero-point energy (real-world word problem)	confinement ⇒ minimum energy $\neq = 0$	Ex 7
H. Unit / exam-twist trap	galat constant, eV↔J, "disturbance" red herring	Ex 8

Yahan koi "quadrants" ya signs nahi hain — charon quantities spreads hain, isliye woh construction se $\geq 0$ hain. Interesting extremes balki limits hain (mass $\to$ large, $Δ x \to 0$ , $τ \to 0$ ), jo cells C, D, aur F explicitly cover karte hain.

Example 1 — Cell A: momentum spread nikalao

Worked example Electron atom mein trapped hai,

$Δ x = 1 \overset{˚}{A} = 1.0 \times 1 0^{- 10} m$ . Sabse chhota possible $Δ p$ ? Forecast: Compute karne se pehle andaaza lagao — kya ek electron ke liye $Δ p$ tiny hoga ya large? Koi number joto.

1. Relation ko uske floor par use karo (equality), kyunki "sabse chhota possible" matlab hum wall par baithe hain: $Δ x Δ p = \frac{ℏ}{2} \Rightarrow Δ p = \frac{ℏ}{2 Δ x} .$ Yeh step kyun? "Minimum" word $\geq$ ko $=$ bana deta hai; koi bhi bada $Δ p$ allowed hai lekin forced nahi.

2. Values daalo: $Δ p = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 10} )} \approx 5.3 \times 1 0^{- 25} kg\cdotm/s .$ Yeh step kyun? Sirf arithmetic hai — $ℏ$ ko SI mein rakho taaki answer SI momentum units mein aaye.

3. $Δ v = Δ p / m_{e}$ use karke speed spread mein convert karo, $m_{e} = 9.11 \times 1 0^{- 31} kg$ : $Δ v \approx \frac{5.3 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}} \approx 5.8 \times 1 0^{5} m/s .$ Yeh step kyun? Speed momentum se zyada intuitive hai — yeh dikhata hai kyun electrons kabhi "still nahi baithe".

Verify: Units: $\frac{J\cdots}{m} = \frac{kg\cdotm ^{2} / s ^{2} \cdot s}{m} = kg\cdotm/s$ ✓. Aur $5.8 \times 1 0^{5} m/s$ light speed ka ek fraction hai — bada hai lekin sub-relativistic, physically theek hai.

Example 2 — Cell B: doosri taraf jao, position spread nikalao

Worked example Ek proton ke baare mein pata hai ki woh velocity spread

$Δ v = 1.0 \times 1 0^{3} m/s$ se chal raha hai. Sabse chhota region $Δ x$ kya hai jisme use confined kiya ja sakta hai? ( $m_{p} = 1.67 \times 1 0^{- 27} kg$ ) Forecast: Kya yeh ek atom (~ $1 0^{- 10}$ m) se bada ya chhota hoga?

1. Di gayi velocity spread ko momentum spread mein convert karo: $Δ p = m_{p} Δ v = (1.67 \times 1 0^{- 27}) (1.0 \times 1 0^{3}) = 1.67 \times 1 0^{- 24} kg\cdotm/s .$ Yeh step kyun? Relation $Δ p$ mein bolta hai, $Δ v$ mein nahi — pehle hamesha convert karo.

2. $Δ x$ ke liye floor relation solve karo: $Δ x = \frac{ℏ}{2 Δ p} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.67 \times 1 0 ^{- 24} )} \approx 3.2 \times 1 0^{- 11} m .$ Yeh step kyun? Wohi wall equation rearrange ki; ab $Δ x$ unknown hai.

Verify: $3.2 \times 1 0^{- 11} m \approx 0.32 \overset{˚}{A}$ — sub-atomic, ek achhe se pinned proton ke liye sensible. Cross-check: $Δ x Δ p = (3.2 \times 1 0^{- 11}) (1.67 \times 1 0^{- 24}) \approx 5.3 \times 1 0^{- 35} \approx ℏ/2$ ✓.

Example 3 — Cell C: macroscopic limit (kyun tumhe kabhi pata nahi chalta)

Worked example Ek cricket ball,

$m = 0.16 kg$ , ki position $Δ x = 1.0 \times 1 0^{- 6} m$ tak jaani hai. Quantum theory iske liye kaunsa velocity spread force karti hai? Forecast: Kya $Δ v$ measurable hoga?

1. Floor relation, $Δ v = Δ p / m$ ke liye solve karke: $Δ v = \frac{ℏ}{2 m Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 0.16 ) ( 1.0 \times 1 0 ^{- 6} )} .$ Yeh step kyun? Wohi wall equation; bada $m$ denominator mein hai, isliye hum ek crushed-small answer expect karte hain.

2. Compute karo: $Δ v \approx 3.3 \times 1 0^{- 28} m/s .$ Yeh step kyun? Bade $m Δ x$ par tiny $ℏ$ ek aisa number deta hai jo Earth par koi bhi instrument detect nahi kar sakta.

Verify: $3.3 \times 1 0^{- 28} m/s$ par ball universe ki age se bhi kaafi zyada time mein ek atom ki width jitna drift karegi. Quantum fuzziness ball ke liye exist karti hai — yeh sirf fantastically irrelevant hai. Yeh classical limit hai: $ℏ \to 0$ effectively.

Example 4 — Cell D: degenerate input $Δ x \to 0$

Worked example Maan lo tum ek particle ko

exact point par pin kar sako, $Δ x = 0$ . Principle $Δ p$ ke baare mein kya kehta hai? Forecast: Finite? Zero? Infinite?

1. Rearrange karo: $Δ p \geq \frac{ℏ}{2 Δ x}$ . $Δ x \to 0$ hone do: $lim_{Δ x \to 0} \frac{ℏ}{2 Δ x} = + \infty.$ Yeh step kyun? Ek fixed positive number ko kisi aisi cheez se divide karo jo zero ki taraf shrink ho rahi hai, toh woh blow up ho jaata hai — yeh pura point ek limit ka hai, jab ek input extreme ki taraf jaata hai toh behaviour dekhna.

2. Interpret karo: ek perfectly located particle ka infinite momentum spread hoga — har momentum equally likely. Yeh step kyun? Yeh parent ke Step 1 se ulta match karta hai: position mein ek spike saari wavelengths ka superposition hai (Fourier), isliye saari momenta.

Verify: Reciprocally, ek plane wave (exact momentum, $Δ p = 0$ ) ko $Δ x = ℏ/ (2 \cdot 0) = \infty$ chahiye — saari space mein phela hua.

Neeche ki figure trade-off ko literal banati hai. Yeh do cases draw karti hai ek ke upar ek. Upar wale pair mein, magenta position bump narrow squeezed hai (chhota $Δ x$ ) aur uska orange momentum partner wide forced hai (bada $Δ p$ ). Neeche wale pair mein, violet position bump wide spread hone diya gaya hai (bada $Δ x$ ) aur uska navy momentum partner ek spike mein sharp ho jaata hai (chhota $Δ p$ ). Kisi bhi colour ko across trace karo: jab bhi ek curve narrow hoti hai, uska dashed partner flare karta hai — yeh $Δ x Δ p \geq ℏ/2$ ek picture ke roop mein drawn hai. Is example ka $Δ x \to 0$ limit sirf upar wala case hai apni extreme tak push kiya gaya, jahan orange curve infinitely wide ho jaati hai.

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

Example 5 — Cell E: lifetime ⇒ linewidth (energy–time forward)

Worked example Ek excited atomic state

$τ = 1.0 \times 1 0^{- 8} s$ tak jeeti hai. Uska minimum energy spread $Δ E$ nikalao, joules aur eV mein. Forecast: Kya yeh energy blur ek photon energy (~eV) se badi ya chhoti hogi?

1. $Δ t = τ$ set karo (lifetime wohi timescale hai jis par state change hoti hai): $Δ E = \frac{ℏ}{2 τ} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 8} )} \approx 5.3 \times 1 0^{- 27} J .$ Yeh step kyun? Energy–time relation mein $Δ t$ stopwatch error nahi hai — yeh exactly state ki lifetime hai. Woh identification physics hai.

2. $1 eV = 1.602 \times 1 0^{- 19} J$ use karke eV mein convert karo: $Δ E = \frac{5.3 \times 1 0 ^{- 27}}{1.602 \times 1 0 ^{- 19}} \approx 3.3 \times 1 0^{- 8} eV .$ Yeh step kyun? Spectroscopists line widths eV mein measure karte hain; conversion answer ko real data se comparable banata hai.

Verify: Sirf $3 \times 1 0^{- 8} eV$ se blurred ek $\sim$ eV transition ek extremely sharp line hai — allowed atomic transitions ki Natural linewidth and spectral broadening se consistent. Units: $J\cdots / s = J$ ✓.

Example 6 — Cell F: linewidth ⇒ lifetime (ulta chalao)

Worked example Ek spectral line measure ki gayi hai jiska natural energy width

$Δ E = 4.0 \times 1 0^{- 6} eV$ hai. Emitting state ki lifetime estimate karo. Forecast: Example 5 ke $1 0^{- 8}$ s se chhota ya lamba? (Wider line ka matlab kya hai?)

1. $Δ E$ ko joules mein convert karo: $Δ E = (4.0 \times 1 0^{- 6}) (1.602 \times 1 0^{- 19}) = 6.41 \times 1 0^{- 25} J .$ Yeh step kyun? Relation SI mein hai; divide karne se pehle convert karo.

2. $Δ t = τ$ ke liye floor relation solve karo: $τ = \frac{ℏ}{2 Δ E} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 6.41 \times 1 0 ^{- 25} )} \approx 8.2 \times 1 0^{- 11} s .$ Yeh step kyun? Wohi equation, unknown swap hua — yeh "blur measure karo, deduce karo kitna fast jeeta hai" direction hai jo short-lived resonances ke liye use hoti hai.

Verify: Line Example 5 ki se $\sim$ 100× wider hai, aur waakai $τ \approx 8 \times 1 0^{- 11}$ s $1 0^{- 8}$ s se $\sim$ 100× chhota hai. Wider line ⇒ shorter life — mnemonic "live fast, die fuzzy" se match karta hai ✓.

Example 7 — Cell G: real-world word problem, zero-point energy

Worked example Ek electron 1D "box" ke andar confined hai jis ki length

$a = 1.0 \times 1 0^{- 10} m$ hai (roughly ek atom). Uski minimum kinetic energy — uski zero-point energy — estimate karo, aur argue karo ki yeh zero nahi ho sakti. Forecast: Kya ek truly trapped electron kabhi perfectly at rest ( $E = 0$ ) ho sakta hai?

1. Confinement position spread set karta hai: $Δ x \approx a$ . Isliye $Δ p \approx \frac{ℏ}{2 a} .$ Yeh step kyun? Walls kahan electron ho sakta hai yeh limit karti hain, isliye $Δ x$ box size se zyada nahi ho sakta — "confined" ka yahi matlab hai.

2. Average momentum zero hai (dono taraf bounce karta hai), yaani $⟨ p ⟩ = 0$ , isliye typical momentum magnitude spread hi hai: $p_{m i n} \approx Δ p$ . Kinetic energy $E = p^{2} /2 m$ : $E_{m i n} \approx \frac{( Δ p ) ^{2}}{2 m _{e}} = \frac{ℏ ^{2}}{8 m _{e} a ^{2}} .$ Yeh step kyun? Kyunki $⟨ p ⟩ = 0$ (upar define kiya), saari "motion" spread mein rehti hai; use square karne se trapped hone ke consistent smallest energy milti hai.

3. Numbers daalo: $E_{m i n} \approx \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 1.0 \times 1 0 ^{- 10} ) ^{2}} \approx 1.5 \times 1 0^{- 19} J \approx 0.95 eV .$ Yeh step kyun? eV-scale energy exactly atomic binding energies ka order hai — estimate physically believable hai. Particle in a box ke exact ground state se compare karo.

4. Zero kyun nahi? $E = 0$ ko $p = 0$ exactly chahiye, yaani $Δ p = 0$ , jo $Δ x = \infty$ force karta hai — electron box se escape kar jaata. Contradiction. Yeh step kyun? Yeh logical heart hai: confinement + uncertainty $\Rightarrow$ nonzero ground energy.

Verify: $E_{m i n} \approx 0.95 eV \neq = 0$ ✓. Exact particle-in-a-box ground state hai $E_{1} = π^{2} ℏ^{2} / (2 m_{e} a^{2}) \approx 37.6 eV$ — humara rough $Δ x \approx a$ estimate $\sim$ 40 factor low aata hai, yaani right order-of-magnitude ballpark aur correctly lower-side guess, jaisa ek crude uncertainty estimate hona chahiye.

Example 8 — Cell H: exam-twist / unit trap

Worked example Ek student likhta hai: "Electron jo

$Δ x = 2.0 \times 1 0^{- 10} m$ tak confined hai, uske liye minimum momentum spread $Δ p = ℏ/Δ x$ hai." Sahi $Δ p$ nikalao aur do traps ke naam batao. Forecast: Student ka answer kitne factor se galat hai?

1. Trap #1 pakdo: missing $\frac{1}{2}$ . Theorem hai $Δ x Δ p \geq ℏ/2$ , nahi $\geq ℏ$ . Toh: $Δ p = \frac{ℏ}{2 Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 2.0 \times 1 0 ^{- 10} )} \approx 2.6 \times 1 0^{- 25} kg\cdotm/s .$ Yeh step kyun? Student ka version exactly 2 factor se overshoot karta hai — ek classic order-of-magnitude slip jo parent ke "mistakes" mein point out ki gayi hai.

2. Trap #2 pakdo (red herring): agar question add karta hai "...kyunki measuring photon electron ko kick karta hai," toh woh reasoning galat mark karo. Bound intrinsic hai (Fourier/wave nature); disturbance sirf ek consequence hai. Yeh step kyun? Exams ko pasand hai sahi number ko galat cause ke saath pair karna. Relation ek perfect, non-disturbing measurement ke saath bhi hold karta hai, kyunki yeh particle ke hone se aata hai ek wave packet — Fourier math, jo tumhare dekhne se pehle hi present hai.

Verify: Student ki value $ℏ/Δ x = 5.3 \times 1 0^{- 25}$ ; sahi value $2.6 \times 1 0^{- 25}$ ; ratio $= 2$ ✓. Units $J\cdots/m = kg\cdotm/s$ ✓.

Active recall

Minimum problem mein "

\geq

" ko "

=

" mein kya convert karta hai?

Sabse chhoti possible value maangna — tum exactly floor

ℏ/2

par baithe ho.

Energy–time relation mein

Δ t

physically kya hai?

State ki lifetime — woh time jis mein woh appreciably change hoti hai, clock error nahi.

Δ x \to 0

hone par

Δ p

ka kya hota hai?

Yeh infinity tak diverge ho jaata hai (

Δ p \geq ℏ/2Δ x \to \infty

Confined particle ki ground energy nonzero kyun hoti hai?

E = 0

Δ p = 0

chahiye, jo

Δ x = \infty

force karta hai, confinement se contradict karta hai.

Wider spectral line lifetime ke baare mein kya imply karti hai?

Shorter lifetime, kyunki

τ = ℏ/ (2Δ E)

⟨ p ⟩

ka matlab kya hai, aur yeh

Δ p

se kaise alag hai?

⟨ p ⟩

average (mean) momentum hai;

Δ p

spread hai. Ek box particle ka

⟨ p ⟩ = 0

hota hai lekin

Δ p \neq = 0

Connections

Parent topic (Hinglish) — woh theory jise yeh examples exercise karte hain.
Particle in a box — Example 7 ka exact counterpart.
Natural linewidth and spectral broadening — Examples 5 aur 6 lab mein.
de Broglie wavelength — har momentum step ke peeche $p = ℏ k$ bridge.
Wave packets and Fourier analysis — Example 4 mein trade-off kyun exist karti hai.