Answer on Question #54035, Physics Quantum Mechanics
Consider a quantum particle confined in a well of width a a a D x D p DxDp D x D p ( ) 222 D x = x − x () 222Dx = x - x ( ) 222 D x = x − x ( ) () ( )
222 D p = p − p 222Dp = p - p 222 D p = p − p Solution:
The average value of P X P_{X} P X
⟨ P X ⟩ = ∫ 0 a ψ 1 ∗ ( x ) P ^ X ψ 1 ( x ) d x = ∫ 0 a 2 a sin ( π x a ) ( − i ℏ ∂ ∂ x ) 2 a sin ( π x a ) d x = − 2 i ℏ a π n a ∫ 0 a sin ( π x a ) cos ( π x a ) d x = − i ℏ π a 2 ∫ 0 a sin ( 2 π x a ) d x = 0 \begin{aligned}
\langle P_{X} \rangle &= \int_{0}^{a} \psi_{1}^{*}(x) \hat{P}_{X} \psi_{1}(x) dx = \int_{0}^{a} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \left(-i\hbar \frac{\partial}{\partial x}\right) \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) dx = -\frac{2i\hbar}{a} \frac{\pi n}{a} \int_{0}^{a} \sin\left(\frac{\pi x}{a}\right) \cos\left(\frac{\pi x}{a}\right) dx = \\
&- \frac{i\hbar \pi}{a^{2}} \int_{0}^{a} \sin\left(\frac{2\pi x}{a}\right) dx = 0
\end{aligned} ⟨ P X ⟩ = ∫ 0 a ψ 1 ∗ ( x ) P ^ X ψ 1 ( x ) d x = ∫ 0 a a 2 sin ( a π x ) ( − i ℏ ∂ x ∂ ) a 2 sin ( a π x ) d x = − a 2 i ℏ a πn ∫ 0 a sin ( a π x ) cos ( a π x ) d x = − a 2 i ℏ π ∫ 0 a sin ( a 2 π x ) d x = 0
where a a a ψ 1 = 2 a sin ( π x a ) \psi_{1} = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) ψ 1 = a 2 sin ( a π x )
The average value of P X 2 P_{X}^{2} P X 2
⟨ P X 2 ⟩ = ∫ 0 a ψ 1 ∗ ( x ) P ^ X 2 ψ 1 ( x ) d x = ∫ 0 a 2 a sin ( π x a ) ( − i ℏ ∂ ∂ x ) 2 2 a sin ( π x a ) d x = 2 ℏ 2 a ( π n a ) 2 ∫ 0 a sin ( π x a ) sin ( π x a ) d x = ℏ 2 π 2 a 2 \begin{aligned}
\langle P_{X}^{2} \rangle &= \int_{0}^{a} \psi_{1}^{*}(x) \hat{P}_{X}^{2} \psi_{1}(x) dx = \int_{0}^{a} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \left(-i\hbar \frac{\partial}{\partial x}\right)^{2} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) dx = \\
&\frac{2\hbar^{2}}{a} \left(\frac{\pi n}{a}\right)^{2} \int_{0}^{a} \sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi x}{a}\right) dx = \frac{\hbar^{2} \pi^{2}}{a^{2}}
\end{aligned} ⟨ P X 2 ⟩ = ∫ 0 a ψ 1 ∗ ( x ) P ^ X 2 ψ 1 ( x ) d x = ∫ 0 a a 2 sin ( a π x ) ( − i ℏ ∂ x ∂ ) 2 a 2 sin ( a π x ) d x = a 2 ℏ 2 ( a πn ) 2 ∫ 0 a sin ( a π x ) sin ( a π x ) d x = a 2 ℏ 2 π 2
The average value of x ^ \hat{x} x ^
⟨ x ⟩ = ∫ 0 a ψ 1 ∗ ( x ) x ^ ψ 1 ( x ) d x = ∫ 0 a 2 a sin ( π x L ) x 2 L sin ( π x L ) d x = 2 a ∫ 0 a sin ( π x a ) x sin ( π x a ) d x = a / 2 \langle x \rangle = \int_{0}^{a} \psi_{1}^{*}(x) \hat{x} \psi_{1}(x) dx = \int_{0}^{a} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{L}\right) x \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) dx = \frac{2}{a} \int_{0}^{a} \sin\left(\frac{\pi x}{a}\right) x \sin\left(\frac{\pi x}{a}\right) dx = a / 2 ⟨ x ⟩ = ∫ 0 a ψ 1 ∗ ( x ) x ^ ψ 1 ( x ) d x = ∫ 0 a a 2 sin ( L π x ) x L 2 sin ( L π x ) d x = a 2 ∫ 0 a sin ( a π x ) x sin ( a π x ) d x = a /2
The average value of x 2 x^{2} x 2
⟨ x 2 ⟩ = ∫ 0 a ψ 1 ∗ ( x ) x 2 ψ 1 ( x ) d x = ∫ 0 a 2 a sin ( π x a ) x 2 2 a sin ( π x a ) d x = 2 a ∫ 0 a sin 2 ( π x a ) x 2 d x = a 2 6 ( 2 − 3 / π 2 ) \left\langle x ^ {2} \right\rangle = \int_ {0} ^ {a} \psi_ {1} ^ {*} (x) x ^ {2} \psi_ {1} (x) d x = \int_ {0} ^ {a} \sqrt {\frac {2}{a}} \sin \left(\frac {\pi x}{a}\right) x ^ {2} \sqrt {\frac {2}{a}} \sin \left(\frac {\pi x}{a}\right) d x = \frac {2}{a} \int_ {0} ^ {a} \sin^ {2} \left(\frac {\pi x}{a}\right) x ^ {2} d x = \frac {a ^ {2}}{6} \left(2 - 3 / \pi^ {2}\right) ⟨ x 2 ⟩ = ∫ 0 a ψ 1 ∗ ( x ) x 2 ψ 1 ( x ) d x = ∫ 0 a a 2 sin ( a π x ) x 2 a 2 sin ( a π x ) d x = a 2 ∫ 0 a sin 2 ( a π x ) x 2 d x = 6 a 2 ( 2 − 3/ π 2 )
Then
Δ x Δ p = ( ℏ 2 π 2 a 2 − 0 ) ( a 2 6 ( 2 − 3 / π 2 ) − a 2 4 ) = ℏ 2 3 π 2 − 6 \Delta x \Delta p = \sqrt {\left(\frac {\hbar^ {2} \pi^ {2}}{a ^ {2}} - 0\right) \left(\frac {a ^ {2}}{6} \left(2 - 3 / \pi^ {2}\right) - \frac {a ^ {2}}{4}\right)} = \frac {\hbar}{2 \sqrt {3}} \sqrt {\pi^ {2} - 6} Δ x Δ p = ( a 2 ℏ 2 π 2 − 0 ) ( 6 a 2 ( 2 − 3/ π 2 ) − 4 a 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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