Answer on Question #51444, Physics Solid State Physics
Consider a quantum particle confined in a well of width a a a Δ x Δ p \Delta x\Delta p Δ x Δ p ( Δ x ) 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 (\Delta x)^2 = \left\langle x^2\right\rangle -\left\langle x\right\rangle^2 ( Δ x ) 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 ( Δ p ) 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 (\Delta p)^2 = \left\langle p^2\right\rangle -\left\langle p\right\rangle^2 ( Δ p ) 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2
Solution:
Fig.1
The wave function of a one-dimensional potential well
ψ n = 2 L sin ( π n x L ) \psi_ {n} = \sqrt {\frac {2}{L}} \sin \left(\frac {\pi n x}{L}\right) ψ n = L 2 sin ( L πn x )
where x x x L L L n n n
The momentum operator
P ^ x = − i ℏ ∂ ∂ x \hat {P} _ {x} = - i \hbar \frac {\partial}{\partial x} P ^ x = − i ℏ ∂ x ∂
The average value of P x P_{x} P x
⟨ P X ⟩ = ∫ 0 L ψ 1 ∗ ( x ) P ^ X ψ 1 ( x ) d x = ∫ 0 L 2 L sin ( π x L ) ( − i ℏ ∂ ∂ x ) 2 L sin ( π x L ) d x = − 2 i ℏ L ∫ 0 L sin ( π x L ) ( ∂ ∂ x ) sin ( π x L ) d x = − 2 i ℏ L π L ∫ 0 L sin ( π x L ) cos ( π x L ) d x = − i ℏ π n L 2 ∫ 0 L sin ( 2 π n x L ) d x = − i ℏ π n L 2 ⋅ L 2 π n cos ( 2 π n x L ) 0 L = 0 \begin{array}{l}
\left\langle P_{X} \right\rangle = \int_{0}^{L} \psi_{1}^{*}(x) \hat{P}_{X} \psi_{1}(x) dx = \int_{0}^{L} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) \left(-i \hbar \frac{\partial}{\partial x}\right) \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) dx = - \frac{2i \hbar}{L} \int_{0}^{L} \sin \left(\frac{\pi x}{L}\right) \left(\frac{\partial}{\partial x}\right) \sin \left(\frac{\pi x}{L}\right) dx = \\
- \frac{2i \hbar}{L} \frac{\pi}{L} \int_{0}^{L} \sin \left(\frac{\pi x}{L}\right) \cos \left(\frac{\pi x}{L}\right) dx = - \frac{i \hbar \pi n}{L^{2}} \int_{0}^{L} \sin \left(\frac{2\pi n x}{L}\right) dx = - \frac{i \hbar \pi n}{L^{2}} \cdot \frac{L}{2\pi n} \cos \left(\frac{2\pi n x}{L}\right)_{0}^{L} = 0
\end{array} ⟨ P X ⟩ = ∫ 0 L ψ 1 ∗ ( x ) P ^ X ψ 1 ( x ) d x = ∫ 0 L L 2 sin ( L π x ) ( − i ℏ ∂ x ∂ ) L 2 sin ( L π x ) d x = − L 2 i ℏ ∫ 0 L sin ( L π x ) ( ∂ x ∂ ) sin ( L π x ) d x = − L 2 i ℏ L π ∫ 0 L sin ( L π x ) cos ( L π x ) d x = − L 2 i ℏ πn ∫ 0 L sin ( L 2 πn x ) d x = − L 2 i ℏ πn ⋅ 2 πn L cos ( L 2 πn x ) 0 L = 0
The average value of P X 2 P_{X}^{2} P X 2
⟨ P X 2 ⟩ = ∫ 0 L ψ 1 ∗ ( x ) P ^ X 2 ψ 1 ( x ) d x = ∫ 0 L 2 L sin ( π x L ) ( − i ℏ ∂ ∂ x ) 2 2 L sin ( π x L ) d x = − 2 ℏ 2 L ∫ 0 L sin ( π x L ) ( ∂ 2 ∂ x 2 ) sin ( π x L ) d x = 2 ℏ 2 L ( π L ) 2 ∫ 0 L sin ( π x L ) sin ( π x L ) d x = ℏ 2 π 2 L 2 \begin{array}{l}
\left\langle P_{X}^{2} \right\rangle = \int_{0}^{L} \psi_{1}^{*}(x) \hat{P}_{X}^{2} \psi_{1}(x) dx = \int_{0}^{L} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) \left(-i \hbar \frac{\partial}{\partial x}\right)^{2} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) dx = \\
- \frac{2 \hbar^{2}}{L} \int_{0}^{L} \sin \left(\frac{\pi x}{L}\right) \left(\frac{\partial^{2}}{\partial x^{2}}\right) \sin \left(\frac{\pi x}{L}\right) dx = \frac{2 \hbar^{2}}{L} \left(\frac{\pi}{L}\right)^{2} \int_{0}^{L} \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{\pi x}{L}\right) dx = \frac{\hbar^{2} \pi^{2}}{L^{2}}
\end{array} ⟨ P X 2 ⟩ = ∫ 0 L ψ 1 ∗ ( x ) P ^ X 2 ψ 1 ( x ) d x = ∫ 0 L L 2 sin ( L π x ) ( − i ℏ ∂ x ∂ ) 2 L 2 sin ( L π x ) d x = − L 2 ℏ 2 ∫ 0 L sin ( L π x ) ( ∂ x 2 ∂ 2 ) sin ( L π x ) d x = L 2 ℏ 2 ( L π ) 2 ∫ 0 L sin ( L π x ) sin ( L π x ) d x = L 2 ℏ 2 π 2
The coordinate operator
x ^ = x \hat{x} = x x ^ = x
The average value of x ^ \hat{x} x ^
⟨ x ⟩ = ∫ 0 L ψ 1 ∗ ( x ) x ^ ψ 1 ( x ) d x = ∫ 0 L 2 L sin ( π x L ) x 2 L sin ( π x L ) d x = L / 2 \langle x \rangle = \int_{0}^{L} \psi_{1}^{*}(x) \hat{x} \psi_{1}(x) dx = \int_{0}^{L} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) x \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) dx = L / 2 ⟨ x ⟩ = ∫ 0 L ψ 1 ∗ ( x ) x ^ ψ 1 ( x ) d x = ∫ 0 L L 2 sin ( L π x ) x L 2 sin ( L π x ) d x = L /2
The average value of x 2 x^{2} x 2
⟨ x 2 ⟩ = ∫ 0 L ψ 1 ∗ ( x ) x 2 ψ 1 ( x ) d x = ∫ 0 L 2 L sin ( π x L ) x 2 2 L sin ( π x L ) d x = L 2 6 ( 2 − 3 / π 2 ) \left\langle x^{2} \right\rangle = \int_{0}^{L} \psi_{1}^{*}(x) x^{2} \psi_{1}(x) dx = \int_{0}^{L} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) x^{2} \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L}\right) dx = \frac{L^{2}}{6} \left(2 - 3 / \pi^{2}\right) ⟨ x 2 ⟩ = ∫ 0 L ψ 1 ∗ ( x ) x 2 ψ 1 ( x ) d x = ∫ 0 L L 2 sin ( L π x ) x 2 L 2 sin ( L π x ) d x = 6 L 2 ( 2 − 3/ π 2 )
Then
Δ x Δ p = ( ℏ 2 π 2 L 2 − 0 ) ( L 2 6 ( 2 − 3 / π 2 ) − L 2 4 ) = ℏ 2 3 π 2 − 6 \Delta x \Delta p = \sqrt{\left(\frac{\hbar^{2} \pi^{2}}{L^{2}} - 0\right) \left(\frac{L^{2}}{6} \left(2 - 3 / \pi^{2}\right) - \frac{L^{2}}{4}\right)} = \frac{\hbar}{2\sqrt{3}} \sqrt{\pi^{2} - 6} Δ x Δ p = ( L 2 ℏ 2 π 2 − 0 ) ( 6 L 2 ( 2 − 3/ π 2 ) − 4 L 2 ) = 2 3 ℏ π 2 − 6
Answer: Δ x Δ p = ℏ 2 3 π 2 − 6 \Delta x\Delta p = \frac{\hbar}{2\sqrt{3}}\sqrt{\pi^2 - 6} Δ x Δ p = 2 3 ℏ π 2 − 6
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