Problem Statement
Uncertainty Relation in a 1D Infinite Square Well Potential Problem Statement
Consider a one-dimensional infinite square well potential of width L L L V ( x ) V(x) V ( x ) V ( x ) = 0 V(x) = 0 V ( x ) = 0 0 ≤ x ≤ L 0 \le x \le L 0 ≤ x ≤ L V ( x ) = ∞ V(x) = \infty V ( x ) = ∞
The stationary states of a particle of mass m m m
In the ground state of this particle, the square of the product of the position uncertainty Δ x \Delta x Δ x Δ p \Delta p Δ p
( Δ x Δ p ) 2 = ( A π 2 + B C ) ℏ 2 (\Delta x \Delta p)^2 = \left( \frac{A\pi^2 + B}{C} \right) \hbar^2 ( Δ x Δ p ) 2 = ( C A π 2 + B ) ℏ 2 Here, Δ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 \Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} Δ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 Δ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 \Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} Δ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 ⟨ ⋅ ⟩ \langle \cdot \rangle ⟨ ⋅ ⟩ ℏ \hbar ℏ
Given that A A A C C C B B B A × ∣ B ∣ × C A \times |B| \times C A × ∣ B ∣ × C
Constraints
A A A C C C B B B
Input Format
Give the result of A × ∣ B ∣ × C A \times |B| \times C A × ∣ B ∣ × C
This problem requires computing quantum mechanical expectation values and completing advanced calculus.
The normalized wave function of the ground state of the 1D infinite square well is:
ψ ( x ) = 2 L sin ( π x L ) ( 0 ≤ x ≤ L ) \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) \quad (0 \le x \le L) ψ ( x ) = L 2 sin ( L π x ) ( 0 ≤ x ≤ L ) 1. Position Expectation Values ⟨ x ⟩ \langle x \rangle ⟨ x ⟩ ⟨ x 2 ⟩ \langle x^2 \rangle ⟨ x 2 ⟩
By symmetry, ⟨ x ⟩ = L 2 \langle x \rangle = \frac{L}{2} ⟨ x ⟩ = 2 L ⟨ x 2 ⟩ \langle x^2 \rangle ⟨ x 2 ⟩
⟨ x 2 ⟩ = ∫ 0 L x 2 ∣ ψ ( x ) ∣ 2 d x = 2 L ∫ 0 L x 2 sin 2 ( π x L ) d x \langle x^2 \rangle = \int_0^L x^2 |\psi(x)|^2 dx = \frac{2}{L} \int_0^L x^2 \sin^2\left(\frac{\pi x}{L}\right) dx ⟨ x 2 ⟩ = ∫ 0 L x 2 ∣ ψ ( x ) ∣ 2 d x = L 2 ∫ 0 L x 2 sin 2 ( L π x ) d x Using the half-angle formula sin 2 θ = 1 − cos 2 θ 2 \sin^2\theta = \frac{1 - \cos 2\theta}{2} sin 2 θ = 2 1 − c o s 2 θ
⟨ x 2 ⟩ = 1 L ∫ 0 L x 2 ( 1 − cos ( 2 π x L ) ) d x = L 2 3 − 1 L ∫ 0 L x 2 cos ( 2 π x L ) d x \langle x^2 \rangle = \frac{1}{L} \int_0^L x^2 \left(1 - \cos\left(\frac{2\pi x}{L}\right)\right) dx = \frac{L^2}{3} - \frac{1}{L} \int_0^L x^2 \cos\left(\frac{2\pi x}{L}\right) dx ⟨ x 2 ⟩ = L 1 ∫ 0 L x 2 ( 1 − cos ( L 2 π x ) ) d x = 3 L 2 − L 1 ∫ 0 L x 2 cos ( L 2 π x ) d x Applying integration by parts twice on the second term:
∫ 0 L x 2 cos ( 2 π x L ) d x = L 3 2 π 2 \int_0^L x^2 \cos\left(\frac{2\pi x}{L}\right) dx = \frac{L^3}{2\pi^2} ∫ 0 L x 2 cos ( L 2 π x ) d x = 2 π 2 L 3 Therefore:
⟨ x 2 ⟩ = L 2 3 − L 2 2 π 2 = L 2 ( 1 3 − 1 2 π 2 ) \langle x^2 \rangle = \frac{L^2}{3} - \frac{L^2}{2\pi^2} = L^2 \left( \frac{1}{3} - \frac{1}{2\pi^2} \right) ⟨ x 2 ⟩ = 3 L 2 − 2 π 2 L 2 = L 2 ( 3 1 − 2 π 2 1 ) The position variance Δ x 2 \Delta x^2 Δ x 2
Δ x 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = L 2 ( 1 3 − 1 2 π 2 ) − L 2 4 = L 2 ( 1 12 − 1 2 π 2 ) \Delta x^2 = \langle x^2 \rangle - \langle x \rangle^2 = L^2 \left( \frac{1}{3} - \frac{1}{2\pi^2} \right) - \frac{L^2}{4} = L^2 \left( \frac{1}{12} - \frac{1}{2\pi^2} \right) Δ x 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = L 2 ( 3 1 − 2 π 2 1 ) − 4 L 2 = L 2 ( 12 1 − 2 π 2 1 ) 2. Momentum Expectation Values ⟨ p ⟩ \langle p \rangle ⟨ p ⟩ ⟨ p 2 ⟩ \langle p^2 \rangle ⟨ p 2 ⟩
Using the momentum operator p ^ = − i ℏ d d x \hat{p} = -i\hbar \frac{d}{dx} p ^ = − i ℏ d x d ψ ( x ) \psi(x) ψ ( x ) ⟨ p ⟩ = 0 \langle p \rangle = 0 ⟨ p ⟩ = 0
Computing ⟨ p 2 ⟩ \langle p^2 \rangle ⟨ p 2 ⟩
p ^ 2 = − ℏ 2 d 2 d x 2 \hat{p}^2 = -\hbar^2 \frac{d^2}{dx^2} p ^ 2 = − ℏ 2 d x 2 d 2 ⟨ p 2 ⟩ = π 2 ℏ 2 L 2 ∫ 0 L 2 L sin 2 ( π x L ) d x = π 2 ℏ 2 L 2 \langle p^2 \rangle = \frac{\pi^2 \hbar^2}{L^2} \int_0^L \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx = \frac{\pi^2 \hbar^2}{L^2} ⟨ p 2 ⟩ = L 2 π 2 ℏ 2 ∫ 0 L L 2 sin 2 ( L π x ) d x = L 2 π 2 ℏ 2 The momentum variance:
Δ p 2 = π 2 ℏ 2 L 2 \Delta p^2 = \frac{\pi^2 \hbar^2}{L^2} Δ p 2 = L 2 π 2 ℏ 2 3. Computing the Uncertainty Product
( Δ x Δ p ) 2 = Δ x 2 ⋅ Δ p 2 = L 2 ( 1 12 − 1 2 π 2 ) ⋅ π 2 ℏ 2 L 2 = ( π 2 12 − 1 2 ) ℏ 2 = ( π 2 − 6 12 ) ℏ 2 (\Delta x \Delta p)^2 = \Delta x^2 \cdot \Delta p^2 = L^2 \left( \frac{1}{12} - \frac{1}{2\pi^2} \right) \cdot \frac{\pi^2 \hbar^2}{L^2} = \left( \frac{\pi^2}{12} - \frac{1}{2} \right) \hbar^2 = \left( \frac{\pi^2 - 6}{12} \right) \hbar^2 ( Δ x Δ p ) 2 = Δ x 2 ⋅ Δ p 2 = L 2 ( 12 1 − 2 π 2 1 ) ⋅ L 2 π 2 ℏ 2 = ( 12 π 2 − 2 1 ) ℏ 2 = ( 12 π 2 − 6 ) ℏ 2 Comparing with the form ( A π 2 + B C ) ℏ 2 (\frac{A\pi^2 + B}{C}) \hbar^2 ( C A π 2 + B ) ℏ 2 A = 1 A = 1 A = 1 B = − 6 B = -6 B = − 6 C = 12 C = 12 C = 12 A = 1 A=1 A = 1 C = 12 C=12 C = 12
Therefore:
A × ∣ B ∣ × C = 1 × ∣ − 6 ∣ × 12 = 72 A \times |B| \times C = 1 \times |-6| \times 12 = 72 A × ∣ B ∣ × C = 1 × ∣ − 6∣ × 12 = 72 Answer : 72