Answer on Question #51032, Physics, Quantum Mechanic
Find the average value of P X P_{X} P X ⟨ P X ⟩ \langle P_{X} \rangle ⟨ P X ⟩ n = 1 n = 1 n = 1 L L L
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 ⟩ = ∫ 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{array}{l} \left\langle P _ {X} \right\rangle = \int_ {0} ^ {L} \psi_ {n} ^ {*} (x) \hat {P} _ {X} \psi_ {n} (x) d x = \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) d x = \\ - \frac {2 i \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) d x = - \frac {2 i \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) d x = \\ - \frac {i \hbar \pi n}{L ^ {2}} \int_ {0} ^ {L} \sin \left(\frac {2 \pi n x}{L}\right) d x = - \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}{2 L} \left(\cos \left(\frac {2 \pi n L}{L}\right) - \cos 0\right) = \\ = - \frac {i \hbar}{2 L} (1 - 1) = 0 \\ \end{array} ⟨ 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
The projection of the momentum is the value fluctuates near the equilibrium position, like a pendulum, and varying in magnitude and direction. Therefore, the average value of the projection of the momentum is zero.
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