Schrödinger equation — time-dependent, time-independent: Question bank

2.3.9 · D5 · Physics › Modern Physics › Schrödinger equation — time-dependent, time-independent

True or false — justify

Schrödinger equation time mein second order hai, jaise light ke liye wave equation

False. Yeh time mein first order hai; ek akela

\partial_{t}

(saath mein

i

ke) exactly yehi banata hai ki

Ψ

ab poora future

Ψ

determine kar sake.

Wavefunction

Ψ

khud particle milne ki probability hai

False.

Ψ

complex hai; probability density

∣Ψ ∣^{2} = Ψ^{*} Ψ

hai, jo real aur non-negative hai —

Ψ

negative ya imaginary ho sakti hai.

Ek stationary state ka wavefunction time ke saath nahi badalta

False.

Ψ = ψ (x) e^{- i E t /ℏ}

phase mein ghoomta rehta hai; sirf observable

∣Ψ ∣^{2} = ∣ ψ ∣^{2}

time-independent hota hai.

Time-Independent Schrödinger Equation har quantum problem par apply hoti hai

False. TISE tabhi exist karti hai jab

V

time-independent ho, kyunki yeh variables separate karne se aati hai — ek time-varying

V (x, t)

ke liye full TDSE chahiye.

Box mein energies quantized hain kyunki humne assume kiya tha ki woh discrete hongi

False. Quantization boundary conditions

ψ (0) = ψ (L) = 0

se emerge hoti hai; humne

ψ

ki continuity assume ki thi,

E

ki discreteness nahi.

Separation of variables se aane wala constant

E

koi bhi complex number ho sakta hai

False.

E

real hona chahiye — yeh

\frac{1}{ψ} \hat{H} ψ

, yaani total energy, ke barabar hai, aur complex

E

∣Ψ ∣^{2}

grow ya decay karta, probability conservation tod kar.

Plane wave

A e^{i (k x - ω t)}

pure space mein normalizable hai

False.

∣Ψ ∣^{2} = ∣ A ∣^{2}

har jagah constant hai, isliye

\int_{- \infty}^{\infty} ∣ A ∣^{2} d x = \infty

— yeh ek idealized state hai, koi physical square-integrable state nahi (dekho Wavefunction and Born Interpretation).

Do alag quantum states ek hi energy share kar sakte hain

True. Alag eigenfunctions ek eigenvalue share kar sakti hain — isse degeneracy kehte hain aur yeh 2D/3D wells aur Quantum Harmonic Oscillator mein higher dimensions mein common hai.

Ek normalized

ψ

e^{i θ}

(constant phase) se multiply karne par physically alag state milti hai

False.

∣ e^{i θ} ψ ∣^{2} = ∣ ψ ∣^{2}

, isliye saari predictions identical hain — ek overall constant phase koi physical information carry nahi karta.

Spot the error

i ℏ \partial_{t}^{2} Ψ = \hat{H} Ψ

TDSE hai" — kya galat hai?

Time derivative first order hai, squared nahi:

i ℏ \partial_{t} Ψ = \hat{H} Ψ

. Ek second-order-in-time PDE ko future fix karne ke liye do initial conditions (

Ψ

aur

\partial_{t} Ψ

) chahiye, isliye ek akela diya hua

Ψ (t_{0})

evolution determine nahi kar paata — ek state se determinism khatam ho jaata.

"Infinite well ke liye main

cos

term rakhta hoon kyunki sine aur cosine dono

ψ^{''} = - k^{2} ψ

solve karte hain" — kya galat hai?

Boundary condition

ψ (0) = 0

cosine ko khatam kar deti hai, kyunki

cos 0 = 1 \neq = 0

; sirf

ψ = A sin (k x)

bachta hai (dekho Particle in a Box).

n = 0

box ka ek valid ground state hai, energy zero" — kya galat hai?

n = 0

ψ = A sin 0 = 0

har jagah milta hai, matlab koi particle hi nahi; sabse low allowed state

n = 1

hai (dekho Energy Quantization).

Ψ = ψ (x) + ϕ (t)

TDSE ko separate karta hai" — kya galat hai?

Separation of variables mein product

Ψ = ψ (x) ϕ (t)

use hota hai, sum nahi; sirf product se hi aap divide karke

t

-only aur

x

-only terms alag kar sakte ho.

"Kyunki

E = \frac{p ^{2}}{2 m}

hamesha hai, potential term zaroori nahi" — kya galat hai?

Yeh relation sirf free particle ke liye holds karta hai; ek real particle ke liye

E = \frac{p ^{2}}{2 m} + V (x)

hai, isliye

\hat{H}

mein

+ V

term hota hai.

"Hamiltonian operator sirf number

E

hai" — kya galat hai?

\hat{H} = - \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} + V

ek differential operator hai;

E

woh eigenvalue hai jo yeh produce karta hai jab yeh ek eigenfunction

ψ

par act karta hai (dekho Hamiltonian Operator).

"Probability

\int Ψ d x = 1

hai" — kya galat hai?

Normalization squared magnitude use karta hai:

\int ∣Ψ ∣^{2} d x = 1

Ψ

ko khud integrate karna ek complex field ke liye meaningless hai aur zaroorat nahi real bhi ho.

Why questions

Schrödinger equation time mein first order kyun honi chahiye?

Taaki sirf

Ψ (x, t_{0})

jaanne se saara future

Ψ

determine ho sake — Newton's law ki tarah deterministic predictive power mile, aur randomness sirf

∣Ψ ∣^{2}

mein rahe.

TDSE mein imaginary

i

kyun aana chahiye?

Yeh

\hat{H}

ki action ko phase rotation

e^{- i E t /ℏ}

banata hai, growth/decay nahi, jisse

\int ∣Ψ ∣^{2} d x

time mein constant rehta hai — probability conserve hoti hai.

Time-independent

V

variables separate kyun karne deta hai?

Agar

V

t

par depend nahi karta, to spatial operator

\hat{H}

mein koi time nahi hai, isliye

t

-dependence cleanly

ϕ (t) = e^{- i E t /ℏ}

ki tarah factor out ho jaati hai.

Separation constant ko

E

(energy) kyun label karte hain?

Space side

\frac{1}{ψ} \hat{H} ψ

ke barabar hai; kyunki

\hat{H}

energy operator hai, yeh constant us stationary state ki total energy hai.

Stationary states ka probability cloud frozen kyun rehta hai jabki

Ψ

evolve karta rehta hai?

Unka time part ek pure phase

e^{- i E t /ℏ}

hai jiska magnitude 1 hai, isliye

∣Ψ ∣^{2} = ∣ ψ ∣^{2}

se saari time dependence chali jaati hai chahe

Ψ

ghoomta rahe.

Ek particle ko trap karna discrete energies kyun force karta hai?

Boundary conditions require karti hain ki wave exactly half-wavelengths ki ek integer number fit kare, aur sirf special

k

(isliye special

E

) unhe satisfy karte hain — jaise ek fixed string par standing waves (dekho de Broglie Hypothesis).

Derivation free-particle plane wave se kyun shuru karte hain?

Kyunki hum uska

E

aur

p

exactly jaante hain (

p = ℏ k

E = ℏ ω

), isliye hum

Ψ = A e^{i (k x - ω t)}

free TDSE mein substitute karke check kar sakte hain ki yeh kaam karta hai: left par

i ℏ \partial_{t} Ψ = ℏ ω Ψ

right par

- \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} Ψ = \frac{ℏ ^{2} k ^{2}}{2 m} Ψ

ke barabar hai, jo sahi dispersion

ℏ ω = \frac{ℏ ^{2} k ^{2}}{2 m}

, yaani

E = \frac{p ^{2}}{2 m}

, force karta hai.

Edge cases

Jab

L \to \infty

(walls infinitely door), well ki energies ka kya hoga?

E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} \to 0

aur levels paas aa jaate hain — spectrum effectively continuous ho jaata hai, free particle recover hota hai.

Infinite well ki ground-state energy kya hai, aur kya yeh zero ho sakti hai?

E_{1} = \frac{π ^{2} ℏ ^{2}}{2 m L ^{2}} > 0

; yeh zero nahi ho sakti kyunki ek confined particle ka nonzero momentum spread hona chahiye — ek zero-energy state Heisenberg Uncertainty Principle violate kar deta.

Agar potential har jagah constant

V = V_{0}

ho, to energies kaise badengi?

Constant sirf har eigenvalue shift kar deta hai:

E \to E + V_{0}

, jabki eigenfunctions

ψ (x)

waisi hi rehti hain — sirf energy differences physical hain.

Ek bound problem mein agar

E

kisi eigenvalue se match nahi karta to TISE kya deta hai?

Solution boundary conditions satisfy nahi kar pata (blow up kar jaata hai ya infinity par vanish nahi karta), isliye yeh normalizable nahi hai — aise

E

simply allowed nahi hain.

Jab

ℏ \to 0

(classical limit), box levels ki spacing ka kya hoga?

E_{n} \propto ℏ^{2}

isliye saare gaps zero ki taraf shrink karte hain, energies ek continuum mein blur ho jaati hain aur quantum discreteness gayab ho jaati hai — classical picture phir se aa jaata hai.

Ek normalized state ke liye, jab

x \to \pm \infty

tab

∣Ψ ∣^{2}

ko kya karna chahiye?

Yeh itni tezi se zero par girna chahiye ki

\int ∣Ψ ∣^{2} d x

finite (=1) rahe; jo probability density infinity par nonzero rehti hai use normalize nahi kiya ja sakta.

A visual map of the traps

Neeche diagram is page ke misconceptions ko us concept ke hisaab se group karta hai jis par woh attack karte hain — ise ek checklist ki tarah use karo "kahan slip ho sakta hai".

Figure — Schrödinger equation — time-dependent, time-independent

Connections

Wavefunction and Born Interpretation — $∣Ψ ∣^{2}$ ke traps.
Particle in a Box — boundary-condition aur $n = 0$ traps ka source.
Energy Quantization — discreteness kyun emerge hoti hai, assume nahi ki jaati.
Hamiltonian Operator — operator vs eigenvalue confusion.
Heisenberg Uncertainty Principle — zero ground-state energy kyun forbidden hai.
de Broglie Hypothesis — quantization ke peeche standing-wave picture.

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