Intuition The big picture (WHY this matters)Trap a quantum particle between two infinitely high walls. Like a guitar string clamped at both ends, only certain standing waves fit perfectly. Each allowed wave = one allowed energy. So confinement → quantisation . This is the simplest model that shows why energy levels are discrete in atoms, quantum dots, and nuclei.
Definition The infinite square wellA particle of mass m m m 0 < x < L 0 < x < L 0 < x < L V ( x ) = { 0 0 < x < L ∞ otherwise V(x) = \begin{cases} 0 & 0 < x < L \\ \infty & \text{otherwise} \end{cases} V ( x ) = { 0 ∞ 0 < x < L otherwise V = ∞ V=\infty V = ∞ ψ ( x ) = 0 \psi(x)=0 ψ ( x ) = 0
WHAT we want: the allowed wavefunctions ψ ( x ) \psi(x) ψ ( x ) E E E HOW : solve the time-independent Schrödinger equation (TISE) inside the box with boundary conditions.
Rearrange:
d 2 ψ d x 2 = − 2 m E ℏ 2 ψ = − k 2 ψ , k 2 ≡ 2 m E ℏ 2 \frac{d^2\psi}{dx^2} = -\frac{2mE}{\hbar^2}\,\psi = -k^2\psi, \qquad k^2 \equiv \frac{2mE}{\hbar^2} d x 2 d 2 ψ = − ℏ 2 2 m E ψ = − k 2 ψ , k 2 ≡ ℏ 2 2 m E
Why define k k k Because E > 0 E>0 E > 0 k k k k k k
So the wave number is quantised :
k n = n π L \boxed{k_n = \frac{n\pi}{L}} k n = L nπ
Intuition Why quantisation appearsThe wall at x = L x=L x = L half-wavelengths fit a whole number of times survive: L = n ⋅ λ n 2 L = n\cdot\frac{\lambda_n}{2} L = n ⋅ 2 λ n k k k
Intuition Zero-point energyThe lowest energy is E 1 = h 2 8 m L 2 ≠ 0 E_1 = \dfrac{h^2}{8mL^2} \ne 0 E 1 = 8 m L 2 h 2 = 0 can never be perfectly at rest . WHY: ψ = \psi= ψ = Δ x \Delta x Δ x Δ p \Delta p Δ p
ψ n \psi_n ψ n n − 1 n-1 n − 1 interior nodes (zeros) and n n n n n n ψ ′ ′ \psi'' ψ ′′
Worked example Example 1 — Electron in a 0.10 nm box (atomic scale)Find E 1 E_1 E 1 L = 1.0 × 10 − 10 L=1.0\times10^{-10} L = 1.0 × 1 0 − 10 E 1 = ( 6.63 × 10 − 34 ) 2 8 ( 9.11 × 10 − 31 ) ( 1.0 × 10 − 10 ) 2 E_1=\frac{(6.63\times10^{-34})^2}{8(9.11\times10^{-31})(1.0\times10^{-10})^2} E 1 = 8 ( 9.11 × 1 0 − 31 ) ( 1.0 × 1 0 − 10 ) 2 ( 6.63 × 1 0 − 34 ) 2
Why these numbers? m = m e m=m_e m = m e n = 1 n=1 n = 1 = 4.40 × 10 − 67 =4.40\times10^{-67} = 4.40 × 1 0 − 67 = 8 × 9.11 × 10 − 31 × 1.0 × 10 − 20 = 7.29 × 10 − 50 =8\times9.11\times10^{-31}\times1.0\times10^{-20}=7.29\times10^{-50} = 8 × 9.11 × 1 0 − 31 × 1.0 × 1 0 − 20 = 7.29 × 1 0 − 50 E 1 = 6.03 × 10 − 18 J ≈ 37.6 eV E_1 = 6.03\times10^{-18}\text{ J} \approx 37.6\text{ eV} E 1 = 6.03 × 1 0 − 18 J ≈ 37.6 eV Why eV? Divide by 1.6 × 10 − 19 1.6\times10^{-19} 1.6 × 1 0 − 19
Worked example Example 2 — Energy gap1 → 2 1\to2 1 → 2 E 2 − E 1 = ( 4 − 1 ) E 1 = 3 E 1 = 3 ( 37.6 ) = 113 eV E_2-E_1 = (4-1)E_1 = 3E_1 = 3(37.6) = 113\text{ eV} E 2 − E 1 = ( 4 − 1 ) E 1 = 3 E 1 = 3 ( 37.6 ) = 113 eV Why this step? E n ∝ n 2 E_n\propto n^2 E n ∝ n 2 2 2 − 1 2 = 3 2^2-1^2=3 2 2 − 1 2 = 3 λ = h c / Δ E ≈ 11 \lambda = hc/\Delta E \approx 11 λ = h c /Δ E ≈ 11
Worked example Example 3 — Probability in left thirdProbability the n = 1 n=1 n = 1 0 < x < L / 3 0<x<L/3 0 < x < L /3 P = ∫ 0 L / 3 2 L sin 2 π x L d x = 1 3 − 3 4 π ≈ 0.196 P=\int_0^{L/3}\frac{2}{L}\sin^2\!\frac{\pi x}{L}\,dx = \frac{1}{3}-\frac{\sqrt3}{4\pi}\approx 0.196 P = ∫ 0 L /3 L 2 sin 2 L π x d x = 3 1 − 4 π 3 ≈ 0.196
Why not exactly 1/3? The probability density ∣ ψ 1 ∣ 2 |\psi_1|^2 ∣ ψ 1 ∣ 2
Common mistake "Energy levels are equally spaced."Why it feels right: evenly spaced ladders (like SHM) are common. Fix: here E n ∝ n 2 E_n\propto n^2 E n ∝ n 2 grow : E 2 − E 1 = 3 E 1 E_2-E_1=3E_1 E 2 − E 1 = 3 E 1 E 3 − E 2 = 5 E 1 E_3-E_2=5E_1 E 3 − E 2 = 5 E 1 n n n n + 1 n+1 n + 1 ( 2 n + 1 ) E 1 (2n+1)E_1 ( 2 n + 1 ) E 1
Common mistake "Keep the cosine term too."Why it feels right: the general solution has both sin and cos. Fix: the boundary ψ ( 0 ) = 0 \psi(0)=0 ψ ( 0 ) = 0 kills the cosine (B = 0 B=0 B = 0
ψ \psi ψ Fix: ψ \psi ψ probability density is ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2 . You must integrate ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2
Common mistake "Ground state hasn = 0 n=0 n = 0 Why it feels right: counting usually starts at 0. Fix: n = 0 n=0 n = 0 ψ ≡ 0 \psi\equiv0 ψ ≡ 0 n = 1 n=1 n = 1
Recall Feynman: explain to a 12-year-oldImagine a jump rope tied to two walls. When you shake it, only certain neat patterns appear: one big arch, then two arches, then three — never a "one-and-a-half" arch, because the ends must stay still. Each neat pattern needs a certain shaking speed = a certain energy. A tiny particle in a box behaves exactly like that rope: only special wave patterns fit, so it can only have special energies. And it can never stop wiggling completely — the calmest pattern (one arch) still has some energy.
Mnemonic Remember the energy formula"n 2 n^2 n 2 8 m L 2 8mL^2 8 m L 2 h 2 h^2 h 2 Mnemonic: "En square = (nh)² / 8mL²" — because E n = ( n h ) 2 / ( 8 m L 2 ) E_n = (nh)^2/(8mL^2) E n = ( nh ) 2 / ( 8 m L 2 ) L ↑ L\uparrow L ↑
Why does B = 0 B=0 B = 0 ψ = A sin k x + B cos k x \psi=A\sin kx + B\cos kx ψ = A sin k x + B cos k x
How many nodes inside the box for n = 3 n=3 n = 3
What is E 3 / E 1 E_3/E_1 E 3 / E 1
What potential defines the infinite square well? V = 0 V=0 V = 0 for
0 < x < L 0<x<L 0 < x < L and
V = ∞ V=\infty V = ∞ outside (walls trap the particle).
What boundary conditions does ψ \psi ψ ψ ( 0 ) = ψ ( L ) = 0 \psi(0)=\psi(L)=0 ψ ( 0 ) = ψ ( L ) = 0 (continuity; particle can't be where
V = ∞ V=\infty V = ∞ ).
Why does the cosine term vanish? ψ ( 0 ) = 0 \psi(0)=0 ψ ( 0 ) = 0 forces
B = 0 B=0 B = 0 , leaving only
A sin ( k x ) A\sin(kx) A sin ( k x ) .
What is the quantisation condition on k k k sin ( k L ) = 0 ⇒ k n = n π / L , n = 1 , 2 , 3 , … \sin(kL)=0\Rightarrow k_n=n\pi/L,\ n=1,2,3,\dots sin ( k L ) = 0 ⇒ k n = nπ / L , n = 1 , 2 , 3 , … Energy levels of a particle in a box? E n = n 2 h 2 8 m L 2 = ℏ 2 n 2 π 2 2 m L 2 E_n=\dfrac{n^2h^2}{8mL^2}=\dfrac{\hbar^2 n^2\pi^2}{2mL^2} E n = 8 m L 2 n 2 h 2 = 2 m L 2 ℏ 2 n 2 π 2 .
Normalised wavefunctions? ψ n ( x ) = 2 / L sin ( n π x / L ) \psi_n(x)=\sqrt{2/L}\,\sin(n\pi x/L) ψ n ( x ) = 2/ L sin ( nπ x / L ) .
Why is there a nonzero ground-state energy? Confinement + uncertainty: a flat/zero wave is forbidden, so
E 1 = h 2 / 8 m L 2 > 0 E_1=h^2/8mL^2>0 E 1 = h 2 /8 m L 2 > 0 (zero-point energy).
How do level spacings behave? They grow:
E n + 1 − E n = ( 2 n + 1 ) E 1 E_{n+1}-E_n=(2n+1)E_1 E n + 1 − E n = ( 2 n + 1 ) E 1 , not equal.
How many interior nodes does ψ n \psi_n ψ n n − 1 n-1 n − 1 nodes (and
n n n antinodes).
What is the probability density vs the wavefunction? Density is
∣ ψ n ∣ 2 |\psi_n|^2 ∣ ψ n ∣ 2 ; integrate it over a region for a probability.
Effect of larger box L L L E n ∝ 1 / L 2 E_n\propto 1/L^2 E n ∝ 1/ L 2 , so levels drop and crowd together.
Schrödinger Equation (TISE) — the master equation we solved.Wavefunction and Born Rule — meaning of ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2 Heisenberg Uncertainty Principle — explains zero-point energy.Quantum Harmonic Oscillator — another bound state, but equally spaced levels.Quantum Tunnelling and Finite Well — what changes when walls are finite.Standing Waves on a String — classical analogue of the quantisation.Quantum Dots — real-world particle-in-a-box (size tunes colour).
Infinite square well V=0 inside
Time-independent Schrodinger eqn
Define k squared = 2mE/hbar squared
General solution A sin kx + B cos kx
Boundary conditions psi=0 at walls
Energy levels E_n = n^2 h^2 / 8mL^2
Intuition Hinglish mein samjhoSocho ek particle ko do infinitely high deewaron ke beech trap kar diya — bilkul ek guitar string ki tarah jo dono taraf se bandhi hai. String par sirf kuch hi standing wave patterns bante hain, jahan ends hamesha still rehte hain. Isi tarah box mein particle ka wavefunction ψ ( x ) \psi(x) ψ ( x ) V = ∞ V=\infty V = ∞
TISE solve karte hain box ke andar jahan V = 0 V=0 V = 0 ψ ′ ′ = − k 2 ψ \psi'' = -k^2\psi ψ ′′ = − k 2 ψ ψ ( 0 ) = 0 \psi(0)=0 ψ ( 0 ) = 0 ψ ( L ) = 0 \psi(L)=0 ψ ( L ) = 0 sin ( k L ) = 0 \sin(kL)=0 sin ( k L ) = 0 k n = n π / L k_n=n\pi/L k n = nπ / L quantisation aata hai — sirf khaas values allowed hain. Energy E n = n 2 h 2 / 8 m L 2 E_n = n^2h^2/8mL^2 E n = n 2 h 2 /8 m L 2 1 , 4 , 9 , 16 1,4,9,16 1 , 4 , 9 , 16
Sabse important baat: ground state n = 1 n=1 n = 1 zero-point energy kehte hain. Particle kabhi bhi poori tarah rukti nahi, kyunki uncertainty principle aur boundary conditions dono allow nahi karte. Yeh model chhota lagta hai par yahi atoms, quantum dots aur nuclei mein discrete energy levels ka asli reason samjhata hai. Normalise karne par ψ n = 2 / L sin ( n π x / L ) \psi_n=\sqrt{2/L}\sin(n\pi x/L) ψ n = 2/ L sin ( nπ x / L ) ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2 ψ \psi ψ
