Answer on Question #51224, Physics, Other
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 = \langle x^2\rangle -\langle x\rangle^2 ( Δ x ) 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 ( Δ p ) 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 (\Delta p)^2 = \langle p^2\rangle -\langle p\rangle^2 ( Δ p ) 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2
Solution:
The wave function of a one-dimensional potential well is given by Eq.(1)
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 L L L
The momentum operator
P ^ X = − i ℏ ∂ ∂ x \hat{P}_X = -i \hbar \frac{\partial}{\partial x} P ^ X = − i ℏ ∂ x ∂
Find the average value of P X P_X P X
⟨ P X ⟩ = ∫ 0 L ψ n ∗ ( x ) P ^ X ψ n ( x ) d x = ∫ 0 L 2 L sin ( π n x L ) ( − i ℏ ∂ ∂ x ) 2 L sin ( π n x L ) d x = − 2 i ℏ L ∫ 0 L sin ( π n x L ) ( ∂ ∂ x ) sin ( π n x L ) d x = − 2 i ℏ L π n L ∫ 0 L sin ( π n x L ) cos ( π n 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 = − i ℏ 2 L ( cos ( 2 π n L L ) − cos 0 ) = = − i ℏ 2 L ( 1 − 1 ) = 0 \begin{aligned}
\langle P_X \rangle &= \int_0^L \psi_n^*(x) \hat{P}_X \psi_n(x) \, dx = \int_0^L \sqrt{\frac{2}{L}} \sin \left(\frac{\pi n x}{L}\right) \left(-i \hbar \frac{\partial}{\partial x}\right) \sqrt{\frac{2}{L}} \sin \left(\frac{\pi n x}{L}\right) dx = \\
&- \frac{2i\hbar}{L} \int_0^L \sin \left(\frac{\pi n x}{L}\right) \left(\frac{\partial}{\partial x}\right) \sin \left(\frac{\pi n x}{L}\right) dx = - \frac{2i\hbar}{L} \frac{\pi n}{L} \int_0^L \sin \left(\frac{\pi n x}{L}\right) \cos \left(\frac{\pi n 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) \Bigg|_0^L = - \frac{i\hbar}{2L} \left(\cos \left(\frac{2\pi n L}{L}\right) - \cos 0\right) = \\
&= - \frac{i\hbar}{2L} (1 - 1) = 0
\end{aligned} ⟨ P X ⟩ = ∫ 0 L ψ n ∗ ( x ) P ^ X ψ n ( x ) d x = ∫ 0 L L 2 sin ( L πn x ) ( − i ℏ ∂ x ∂ ) L 2 sin ( L πn x ) d x = − L 2 i ℏ ∫ 0 L sin ( L πn x ) ( ∂ x ∂ ) sin ( L πn x ) d x = − L 2 i ℏ L πn ∫ 0 L sin ( L πn x ) cos ( L πn 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 = − 2 L i ℏ ( cos ( L 2 πn L ) − cos 0 ) = = − 2 L i ℏ ( 1 − 1 ) = 0
Find the average value of P X 2 P_X^2 P X 2
⟨ P X 2 ⟩ = ∫ 0 L ψ n ∗ ( x ) P ^ X 2 ψ n ( x ) d x = ∫ 0 L 2 L sin ( π n x L ) ( − i ℏ ∂ ∂ x ) 2 2 L sin ( π n x L ) d x = − 2 ℏ 2 L ∫ 0 L sin ( π n x L ) ( ∂ 2 ∂ x 2 ) sin ( π n x L ) d x = 2 ℏ 2 L ( π n L ) 2 ∫ 0 L sin ( π n x L ) sin ( π n x L ) d x = = ℏ 2 n 2 π 2 2 L 2 ( 2 − sin [ 2 n π ] n π ) = ℏ 2 n 2 π 2 L 2 \begin{aligned}
\langle P_X^2 \rangle &= \int_0^L \psi_n^*(x) \hat{P}_X^2 \psi_n(x) \, dx = \int_0^L \sqrt{\frac{2}{L}} \sin \left(\frac{\pi n x}{L}\right) \left(-i \hbar \frac{\partial}{\partial x}\right)^2 \sqrt{\frac{2}{L}} \sin \left(\frac{\pi n x}{L}\right) dx = \\
&- \frac{2\hbar^2}{L} \int_0^L \sin \left(\frac{\pi n x}{L}\right) \left(\frac{\partial^2}{\partial x^2}\right) \sin \left(\frac{\pi n x}{L}\right) dx = \frac{2\hbar^2}{L} \left(\frac{\pi n}{L}\right)^2 \int_0^L \sin \left(\frac{\pi n x}{L}\right) \sin \left(\frac{\pi n x}{L}\right) dx = \\
&= \frac{\hbar^2 n^2 \pi^2}{2L^2} \left(2 - \frac{\sin [2n\pi]}{n\pi}\right) = \frac{\hbar^2 n^2 \pi^2}{L^2}
\end{aligned} ⟨ P X 2 ⟩ = ∫ 0 L ψ n ∗ ( x ) P ^ X 2 ψ n ( x ) d x = ∫ 0 L L 2 sin ( L πn x ) ( − i ℏ ∂ x ∂ ) 2 L 2 sin ( L πn x ) d x = − L 2 ℏ 2 ∫ 0 L sin ( L πn x ) ( ∂ x 2 ∂ 2 ) sin ( L πn x ) d x = L 2 ℏ 2 ( L πn ) 2 ∫ 0 L sin ( L πn x ) sin ( L πn x ) d x = = 2 L 2 ℏ 2 n 2 π 2 ( 2 − nπ sin [ 2 nπ ] ) = L 2 ℏ 2 n 2 π 2
The coordinate operator
x ^ = x \hat{x} = x x ^ = x
Find the average value of x ^ \hat{x} x ^
⟨ x ⟩ = ∫ 0 L ψ n ∗ ( x ) x ^ ψ n ( x ) d x = ∫ 0 L 2 L sin ( π x L ) x 2 L sin ( π x L ) d x = 2 L ∫ 0 L sin ( π x L ) x sin ( π x L ) d x = L / 2 \begin{array}{l}
\left\langle x \right\rangle = \int_{0}^{L} \psi_{n}^{*}(x) \hat{x} \psi_{n}(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 = \\
\frac{2}{L} \int_{0}^{L} \sin \left(\frac{\pi x}{L}\right) x \sin \left(\frac{\pi x}{L}\right) dx = L / 2
\end{array} ⟨ x ⟩ = ∫ 0 L ψ n ∗ ( x ) x ^ ψ n ( x ) d x = ∫ 0 L L 2 sin ( L π x ) x L 2 sin ( L π x ) d x = L 2 ∫ 0 L sin ( L π x ) x sin ( L π x ) d x = L /2
Find the average value of x 2 x^{2} x 2 n = 1 n = 1 n = 1
⟨ x 2 ⟩ = ∫ 0 L ψ n ∗ ( x ) x 2 ψ n ( x ) d x = ∫ 0 L 2 L sin ( π x L ) x 2 2 L sin ( π x L ) d x = 2 L ∫ 0 L sin ( π x L ) x 2 sin ( π x L ) d x = L 2 6 ( 2 − 3 / π 2 ) \begin{array}{l}
\left\langle x^{2} \right\rangle = \int_{0}^{L} \psi_{n}^{*}(x) x^{2} \psi_{n}(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{2}{L} \int_{0}^{L} \sin \left(\frac{\pi x}{L}\right) x^{2} \sin \left(\frac{\pi x}{L}\right) dx = \frac{L^{2}}{6} \left(2 - 3 / \pi^{2}\right)
\end{array} ⟨ x 2 ⟩ = ∫ 0 L ψ n ∗ ( x ) x 2 ψ n ( x ) d x = ∫ 0 L L 2 sin ( L π x ) x 2 L 2 sin ( L π x ) d x = L 2 ∫ 0 L sin ( L π x ) x 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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#339914 on Dec 2023