Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ - 2: Question bank

2.3.7 · D5 · Physics › Modern Physics › Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ

Is page mein parent note wali picture assume ki gayi hai: ek particle ek wave packet hai, aur bound $Δ x Δ p \geq ℏ/2$ Fourier ke is fact $Δ x Δ k \geq \frac{1}{2}$ se aata hai, plus de Broglie ka bridge $p = ℏ k$ . Har jawab ke liye in dono facts ko haath mein pakde rakho.

True or false — justify karo

True/false: Ek zyada precise ruler eventually

ℏ/2

ke floor ko beat kar lega.

False —

Δ x

ek hi tarah se prepared particles ka intrinsic statistical spread hai, jo state ki khud ki property hai; ek perfect instrument bhi usi spread ko measure karega.

True/false: Uncertainty product

Δ x Δ p

exactly

ℏ/2

ke barabar ho sakta hai.

True, lekin sirf ek Gaussian wave packet ke liye; har doosri shape strictly worse karti hai, isliye

ℏ/2

ek minimum hai jo bahut kam milta hai.

True/false: Ek plane wave

e^{ik x}

ka position uncertainty zero hota hai.

False — iska

Δ x

infinite hota hai; iska

∣ ψ ∣^{2}

har jagah flat hai, isliye particle kahin bhi ho sakta hai jabki uski momentum bilkul sharp hai.

True/false:

Δ E Δ t \geq ℏ/2

system ko thodi der ke liye energy conservation violate karne deta hai.

False — energy exactly conserved rehti hai;

Δ E

measured energy ka spread hai, jo state kitni der tak jiti hai us par depend karta hai, koi energy loan nahi hai.

True/false: Agar

Δ p = 0

set kar sako, toh particle ek exact point par baithega.

False —

Δ p = 0

Δ x \to \infty

force hota hai (ek single-wavelength wave poori space fill karti hai), toh particle maximally unlocalized ho jaayega.

True/false: Uncertainty principle tab hi matter karta hai jab particle ko measure karo.

False — yeh bina kisi measurement ke bhi hold karta hai, kyunki yeh is baat ka statement hai ki wave packet kya hai; measurement disturbance ek consequence hai, cause nahi.

True/false: Kyunki

ℏ

bahut chota hai, cricket ball ke liye uncertainty simply exist nahi karti.

False — exist karti hai lekin measure nahi ki ja sakti; macroscopic mass ke liye

Δ v = ℏ/ (2 m Δ x)

floor ~

1 0^{- 28}

m/s hai, jo har real effect mein dub jaati hai.

True/false: Energy–time relation mein

Δ t

tumhari stopwatch ki error hai.

False —

Δ t

woh timescale hai jisme state appreciably change hoti hai (uski lifetime); ek perfectly accurate clock isse nahi ghata sakti.

True/false: Ek stationary energy eigenstate (

Δ E = 0

) hamesha ke liye jeena chahiye.

True — sharp energy ka matlab

Δ t \to \infty

hai; sirf truly stable states ki energy exactly defined hoti hai, isliye finite lifetime wale excited states ki energies blurred hoti hain.

Spot the error

Flaw dhundho: "Hum

Δ x Δ p \geq ℏ

likhte hain kyunki yahi theorem hai."

\frac{1}{2}

ka factor drop ho gaya — RMS deviations ke liye exact theorem

\geq ℏ/2

hai; akela

ℏ

sirf order-of-magnitude shorthand hai aur "minimum value" ke sawaalon mein galat answers deta hai.

Flaw dhundho: "Uncertainty isliye aati hai kyunki jab hum dekhte hain toh photon electron ko kick karta hai."

Yeh ek real disturbance describe karta hai, lekin principle gehri hai: yeh Fourier width relation se follow karti hai aur zero interaction par bhi hold karti hai; disturbance wave nature ke baad ki cheez hai.

Flaw dhundho: "Ek short laser pulse ko perfectly brief aur perfectly single-colour dono banaya ja sakta hai."

Impossible —

Δ t

duration ki pulse mein frequency spread

Δ ω \geq 1/ (2Δ t)

hona hi chahiye; time mein brief hona color mein broad hone ko force karta hai, same Fourier trade jaisa

Δ x Δ k

Flaw dhundho: "Ek electron ko confine karo aur box ko sahi banake uski ground-state energy zero ki ja sakti hai."

Nahi — confinement deta hai

Δ x \sim a

, isliye

Δ p \sim ℏ/ (2 a) \neq = 0

, aur

E \sim (Δ p)^{2} /2 m > 0

; hamesha ek nonzero zero-point energy hoti hai.

Flaw dhundho: "Bound

Δ x Δ k \geq \frac{1}{2}

physics ka ek law hai."

Yeh kisi bhi wave/Fourier pair ka pure mathematics hai — sound, paani, light sab isko maante hain; physics tab aata hai jab de Broglie

Δ k

Δ p

mein convert karta hai.

Flaw dhundho: "Gaussian packet worst-case hai kyunki yeh spread out hai."

Ulta hai — Gaussian best case hai; yeh bound ko exactly

ℏ/2

par saturate karta hai, minimum possible joint uncertainty; koi bhi distortion extra spread add karta hai.

Why questions

Position bump ko narrow karne se

Δ p

spread kyun force ho jaata hai?

Ek narrow bump ko bahut saari wavelengths ki superposition chahiye taaki bahar cancel ho sake (Fourier), aur har wavelength

p = ℏ k

ke through ek momentum hai, isliye zyada wavelengths matlab zyada

Δ p

Sabhi shapes mein Gaussian wave packet kyun special hai?

Yeh unique shape hai jiska Fourier transform bhi ek Gaussian hai, width product ko exactly

ℏ/2

tak minimize karta hai; koi bhi distortion extra spread add karta hai.

Short-lived particles ki broad natural linewidth kyun hoti hai?

Short lifetime

τ

matlab chota

Δ t

, aur

Δ E \geq ℏ/ (2 τ)

, isliye unki energy — aur emitted photon frequency — intrinsically fuzzy hoti hai (dekho Natural linewidth and spectral broadening).

Electrons atom ke bottom par "still" kyun nahi baith sakte?

Still baithne ka matlab

Δ p \approx 0

hai, jiske liye

Δ x \to \infty

chahiye; lekin atom

Δ x \sim 1

Å confine karta hai, jo ek large momentum spread force karta hai aur isliye perpetual motion hoti hai.

Tunnelling energy conservation ko contradict kyun nahi karta, jabki particle ek forbidden barrier cross karta hai?

Energy poori tarah conserved rehti hai; wave amplitude simply barrier se leak hoti hai, aur energy–time relation sirf spread describe karta hai, koi temporary energy loan nahi (dekho Quantum tunnelling).

de Broglie wavelength crucial ingredient kyun hai, sirf Fourier relation kyun nahi?

Fourier sirf

Δ x Δ k \geq \frac{1}{2}

deta hai, jo waves ke baare mein ek statement hai;

p = ℏ k

woh physics hai jo abstract

k

ko measurable momentum mein badalta hai.

Edge cases

Jab

Δ x \to 0

(perfectly localized) ho toh

Δ p

ka kya hoga?

Δ p \to \infty

— ek point-like particle ke liye infinitely many wavelengths chahiye, isliye momentum completely undefined ho jaata hai; yeh plane wave ka extreme opposite hai.

Ek perfectly stable ground state ki energy uncertainty kya hai?

Exactly zero — yeh kabhi change nahi hoti, isliye

Δ t \to \infty

aur

Δ E \to 0

; sirf decaying states ko energy width milti hai.

Kya principle ek free particle par bhi apply hoti hai jiske liye koi walls ya forces nahi hain?

Haan — koi bhi wave packet, confined ho ya nahi,

Δ x Δ p \geq ℏ/2

follow karta hai; ek free Gaussian packet time ke saath spread bhi karta hai kyunki uske components alag speeds se travel karte hain.

Kya do alag quantities ko hamesha saath mein precisely measure kiya ja sakta hai?

Nahi, sirf complementary pairs ke liye (jaise

x

aur

p_{x}

, ya

E

aur

t

) jo Fourier conjugates hain; unrelated quantities simultaneously sharp ho sakti hain.

Size

a

ke box mein particle ki minimum possible energy kya hai?

Lagbhag

E_{m i n} \sim ℏ^{2} / (8 m a^{2}) > 0

— kabhi zero nahi, kyunki zero energy ka matlab

Δ p = 0

aur unbounded

Δ x

hoga, jo confinement ko contradict karta hai (dekho Particle in a box).

ℏ \to 0

ki limit mein (imaginary classical world), bound ka kya hoga?

Floor

ℏ/2 \to 0

ho jaata hai, isliye position aur momentum dono sharp ho sakte hain — exactly yahi wajah hai ki classical mechanics kaam karta hai aur quantum fuzziness khatam ho jaati hai.

Recall Ek-line self-test

Agar koi kahe "uncertainty sirf measurement error hai," toh kaunsa ek word unhe sahi kar dega? ::: "Intrinsic" — yeh state ki wave nature ki property hai, ek perfect instrument ke saath bhi present hai.