Wave function ψ — probability density - ψ - ²: Question bank

2.3.8 · D5 · Physics › Modern Physics › Wave function ψ — probability density - ψ - ²

Shuru karne se pehle, teen words yaad rakhne laayak hain:

Amplitude — kisi point par $ψ$ ki raw complex value (negative ya imaginary ho sakti hai).
Density — probability per unit length, yaani $∣ ψ ∣^{2}$ ; ek actual probability paane ke liye tumhe length se multiply karna hoga (integrate karna hoga).
Global phase — poore $ψ$ ko ek fixed factor $e^{i θ}$ se multiply karna (har jagah same $θ$ ).

True or false — justify

True or false:

∣ ψ (x) ∣^{2}

ek single point

x

par, particle ko exactly

x

par paane ki probability hai.

False.

∣ ψ ∣^{2}

ek density hai; ek exact point par probability

∣ ψ ∣^{2} d x \to 0

hoti hai. Real probability paane ke liye tumhe ek interval par integrate karna hoga.

True or false: Agar

\int_{- \infty}^{\infty} ∣ ψ ∣^{2} d x = 4

hai, to

ψ

4

se divide karke normalize kiya jaata hai.

False. Normalization

∣ ψ ∣^{2}

par act karta hai, jo

ψ

mein quadratic hai.

ψ

4 = 2

se divide karo taaki

∣ ψ ∣^{2}

4

se divide kiya jaa sake.

True or false: Do wave functions

ψ

aur

e^{i θ} ψ

(same fixed

θ

) identical measurement outcomes predict karte hain.

True.

∣ e^{i θ} ψ ∣^{2} = e^{- i θ} e^{i θ} ∣ ψ ∣^{2} = ∣ ψ ∣^{2}

, isliye density se banaye gaye har prediction mein koi farq nahi aata. Ek global phase physically invisible hota hai.

True or false: Ek position-dependent phase

e^{ik x}

bhi physically invisible hota hai.

False. Yeh

∣ ψ ∣^{2}

mein ek single state ke liye cancel ho jaata hai, lekin yeh momentum information carry karta hai aur momentum aur interference mein matter karta hai. Sirf ek constant (global) phase invisible hota hai.

True or false:

ψ

woh physical wave hai jo tum principle mein directly measure kar sakte ho, jaise paani ki ripple.

False.

ψ

ek complex probability amplitude hai, koi physical displacement nahi. Sirf

∣ ψ ∣^{2}

aur

ψ

se bane expectation values measurable hote hain.

True or false: Agar

ψ

har jagah real hai, to

∣ ψ ∣^{2} = ψ^{2}

hoga.

True. Real

ψ

ke liye,

ψ^{*} = ψ

, isliye

ψ^{*} ψ = ψ^{2}

. Lekin complex

ψ

ke liye,

ψ^{2} \neq = ∣ ψ ∣^{2}

(e.g.

ψ = i

deta hai

ψ^{2} = - 1

lekin

∣ ψ ∣^{2} = 1

True or false: Ek function jo

x \to + \infty

par bound ke bina badhti jaati hai, woh phir bhi ek valid physical wave function ho sakta hai.

False. Yeh square-integrability fail kar deta hai:

\int ∣ ψ ∣^{2} d x

diverge ho jaata, isliye isse total probability 1 par normalize nahi kiya ja sakta.

True or false: Poore infinite space par ek constant function

ψ (x) = c \neq = 0

ek legal normalized wave function hai.

False.

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

, isliye isse normalize nahi kiya ja sakta. (Yeh plane wave ke liye ek useful idealization hai, lekin genuine physical state nahi hai.)

True or false:

ψ

ko real constant

2

se multiply karna physics ko unchanged chhod deta hai, jaise ek phase karta hai.

False. Real factor

2

∣ ψ ∣^{2}

4

se multiply ho jaata hai, total probability 1 se 4 ho jaati hai — yeh normalization tod deta hai. Sirf unit-modulus factors (

∣ e^{i θ} ∣ = 1

) physics ko intact chhod te hain.

Spot the error

Galti dhundo: "Kyunki

ψ

ek probability hai, isko

0 \leq ψ \leq 1

satisfy karna chahiye."

ψ

probability nahi hai; yeh ek amplitude hai jo negative ya complex ho sakta hai.

[0, 1]

mein constrained object probability hoti hai, jo sirf

\int ∣ ψ ∣^{2} d x

ke baad milti hai.

Galti dhundo: "

P (a \leq x \leq b) = ∣ ψ (b) ∣^{2} - ∣ ψ (a) ∣^{2}

, jaise ek antiderivative evaluate karna."

Tumhe density integrate karni hogi:

P = \int_{a}^{b} ∣ ψ ∣^{2} d x

∣ ψ ∣^{2}

integrand hai, antiderivative nahi, isliye endpoint values ko subtract karna meaningless hai.

Galti dhundo: "

ψ = A e^{- x^{2} / (2 σ^{2})}

ko normalize karne ke liye maine peak par

A e^{0} = 1

set kiya, jisse

A = 1

mila."

Normalization total area

\int ∣ ψ ∣^{2} d x = 1

par condition hai, peak value par nahi. Peak height wahi hogi jo honi chahiye taaki area 1 ho jaaye.

Galti dhundo: "

ψ = A e^{ik x}

ke liye, main

∣ ψ ∣^{2} = A^{2} e^{2 ik x}

compute karta hoon."

Tumne

ψ^{*} ψ

lene ki jagah

ψ

ko square kar diya. Sahi tarika:

∣ ψ ∣^{2} = A^{*} e^{- ik x} \cdot A e^{ik x} = ∣ A ∣^{2}

, ek constant — phase cancel ho jaata hai.

Galti dhundo: "1D mein

ψ

ke units probability ke same hain, dimensionless."

∣ ψ ∣^{2}

ke units

m^{- 1}

hote hain (probability per length), isliye

ψ

ke units

m^{- 1/2}

hote hain. Yeh dimensionless nahi hai.

Galti dhundo: "

∣ ψ ∣^{2} = ψ \cdot ψ^{*}

kabhi kabhi negative ho sakta hai agar

ψ

bahut negative ho."

Kabhi nahi.

ψ = a + ib

likhne par,

ψ^{*} ψ = a^{2} + b^{2} \geq 0

hamesha. Modulus ko square karne ka poora point hi non-negativity guarantee karna hai.

Galti dhundo: "Particle in a Box mein

ψ = A sin (π x / L)

ke saath, main

- \infty

\infty

tak integrate karke normalize karta hoon."

ψ

[0, L]

ke bahar zero hai, isliye integral sirf

[0, L]

par run karta hai:

\int_{0}^{L} A^{2} sin^{2} (π x / L) d x = A^{2} L /2 = 1

. Hamesha-zero tails ko integrate karne se kuch add nahi hota lekin domain ki galat samajh zaahir hoti hai.

Why questions

Hum

∣ ψ ∣^{2}

kyun use karte hain na ki

∣ ψ ∣

, jo bhi non-negative hai?

Interference experiments (e.g. electron diffraction) dikhate hain ki amplitudes pehle add hote hain, phir square hote hain. Predictions

∣ ψ ∣^{2}

se match karti hain,

∣ ψ ∣

se nahi; nature squared modulus select karta hai.

Ek bound particle ke liye

ψ \to 0

x \to \pm \infty

kyun hona chahiye?

Warna

\int ∣ ψ ∣^{2} d x

diverge ho jaata aur 1 ke barabar nahi ho sakta (square-integrability fail ho jaati). Ek localized particle ki probability infinitely door vanishing honi chahiye.

ψ

N

se divide karna (na ki

N

se) normalization kyun fix karta hai jab

\int ∣ ψ ∣^{2} d x = N

ho?

Kyunki

∣ ψ ∣^{2}

ψ

mein quadratic hai:

ψ \to ψ / N

scale karne se

∣ ψ ∣^{2} \to ∣ ψ ∣^{2} / N

scale hota hai, isliye total area

N / N = 1

ban jaata hai.

Phase factor

e^{ik x}

probability density se kyun vanish ho jaata hai lekin physically phir bhi matter karta hai?

ψ^{*} ψ

mein yeh cancel ho jaata hai (

e^{- ik x} e^{ik x} = 1

), isliye position probabilities mein kabhi appear nahi karta. Lekin yeh momentum encode karta hai (

p = ℏ k

via de Broglie Wavelength), jo Expectation Values aur interference mein dikhta hai.

ψ

single-valued kyun hona chahiye?

Agar

ψ

ke ek point par do values hote, to

∣ ψ ∣^{2}

wahan do conflicting probability densities deta — physically nonsensical. Ek point, ek probability.

Hum

ψ

(aur aksar

ψ^{'}

) ko continuous kyun require karte hain?

ψ

mein jump

ψ^{'}

ko infinite bana deta, aur Schrödinger Equation

ψ^{''}

ko finite energy se link karta hai — isliye discontinuities generally infinite energy demand karti hain, jo unphysical hai (infinite potential walls par chhod kar).

Edge cases

Edge case: do boxes ka ek state,

ψ

nonzero

[0, L]

aur

[2 L, 3 L]

mein, beech mein zero. Kya yeh legal hai?

Haan, provided total

\int ∣ ψ ∣^{2} d x = 1

ho. Particle ki probability disconnected regions mein spread ho sakti hai; gaps of zero probability kuch nahi rok te.

Edge case:

ψ (x) = 0

kisi interior point

x_{0}

par (ek node). Kya particle us point ko "avoid" karta hai?

Probability density wahan zero hai, isliye

x_{0}

ke aas-paas ek infinitesimal window mein isko paane ki probability vanishing hoti hai. Yeh exactly wahi hota hai jo ek Particle in a Box excited state ke nodes par hota hai — ek real, observable feature.

Edge case: kya

ψ \equiv 0

har jagah ek valid wave function hai?

Nahi. Yeh

\int ∣ ψ ∣^{2} d x = 0 \neq = 1

deta hai aur isse rescale nahi kiya ja sakta (zero ko kisi se bhi divide karo, zero hi rahega). "Kahin koi particle nahi" ek normalizable state nahi hai.

Edge case: ek

ψ

jisme ek finite jump discontinuity ho lekin phir bhi square-integrable ho — kya yeh acceptable hai?

Integral

\int ∣ ψ ∣^{2} d x

exist kar sakta hai, lekin physical conditions continuity demand karti hain (taaki Schrödinger equation ke liye

ψ^{'}

finite rahe). Aisa

ψ

unphysical reject ho jaata hai square-integrable hone ke bawajood.

Edge case: kya

∣ ψ ∣^{2}

ka bahut tall, thin spike matlab hai ki particle "wahi hai" almost exactly?

Sirf high density ke sense mein. Actual probability spike ka area hai (

∣ ψ ∣^{2} d x

); ek tall spike ek tiny width par phir bhi kam probability carry kar sakta hai.

x

mein tightly localize karna momentum mein spread force karta hai (Heisenberg Uncertainty Principle).

Recall wrap-up

Recall Ek-line takeaways

Density vs probability ::: $∣ ψ ∣^{2}$ per-length hai; real probability ke liye $d x$ se multiply karo (integrate karo).
Normalize karo ::: $ψ$ ko $N$ se divide karke, kyunki $∣ ψ ∣^{2} \propto ψ^{2}$ .
Invisible cheez ::: sirf ek global phase $e^{i θ}$ ; position-dependent phase momentum carry karta hai.
Illegal states ::: $ψ \equiv 0$ , non-normalizable (constant / growing), multi-valued, discontinuous.

Connections

2.3.08 Wave function ψ — probability density - ψ - ² (Hinglish) — parent topic (Hinglish).
Schrödinger Equation — $ψ$ aur $ψ^{'}$ ki continuity kyun enforce hoti hai.
Heisenberg Uncertainty Principle — spike-in- $x$ ⇒ spread-in- $p$ edge case.
Particle in a Box — nodes, domain of integration, disconnected regions.
de Broglie Wavelength — $e^{ik x}$ mein woh $k$ jo phase mein chhupta hai.
Complex Numbers — kyun $ψ^{*} ψ = a^{2} + b^{2} \geq 0$ .
Expectation Values — jahan "invisible" phase dobara dikhta hai.