Answer on Question #54034, Physics Quantum Mechanics
Obtain the expectation value of the potential energy 221 V ( x ) = m w x 221\mathrm{V}(\mathrm{x}) = \mathrm{mw}\mathrm{x} 221 V ( x ) = mw x
Solution:
The expectation value of the potential energy
⟨ ψ 1 ∣ V ∣ ψ 1 ⟩ = ∫ − ∞ + ∞ ψ 1 ( x ) V ( x ) ψ 1 ( x ) d x = ∫ − ∞ + ∞ a 3 / 2 2 π 1 / 4 x exp [ − a 2 x 2 2 ] m ω 2 x 2 2 a 3 / 2 2 π 1 / 4 x exp [ − a 2 x 2 2 ] d x = 2 a 3 π ∫ − ∞ + ∞ m ω 2 x 4 2 exp [ − a 2 x 2 ] d x = m ω 2 a 2 π ∫ − ∞ + ∞ ( a x ) 4 exp [ − a 2 x 2 ] d ( a x ) = ∣ ξ = ( a x ) 2 d ξ = 2 a x d ( a x ) = 2 y ξ d ( a x ) ∣ = m ω 2 a 2 π ∫ − ∞ + ∞ ξ 2 exp [ − ξ ] d ξ 2 ξ = m ω 2 a 2 π ∫ − ∞ + ∞ ξ 2 exp [ − ξ ] d ξ 2 ξ = m ω 2 2 a 2 π ∫ − ∞ + ∞ ξ 3 / 2 exp [ − ξ ] d ξ = m ω 2 2 a 2 π Γ ( 5 / 2 ) = m ω 2 2 a 2 π ⋅ 3 2 ⋅ 1 2 ⋅ Γ ( 1 / 2 ) = 3 m ω 2 8 a 2 π ⋅ π = 3 m ω 2 8 a 2 \begin{array}{l}
\left\langle \psi_{1} \mid V \mid \psi_{1} \right\rangle = \int_{-\infty}^{+\infty} \psi_{1}(x) V(x) \psi_{1}(x) dx = \int_{-\infty}^{+\infty} \frac{a^{3/2} \sqrt{2}}{\pi^{1/4}} x \exp \left[ - \frac{a^{2} x^{2}}{2} \right] \frac{m \omega^{2} x^{2}}{2} \frac{a^{3/2} \sqrt{2}}{\pi^{1/4}} x \exp \left[ - \frac{a^{2} x^{2}}{2} \right] dx = \\
\frac{2a^{3}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \frac{m \omega^{2} x^{4}}{2} \exp \left[ - a^{2} x^{2} \right] dx = \frac{m \omega^{2}}{a^{2} \sqrt{\pi}} \int_{-\infty}^{+\infty} (a x)^{4} \exp \left[ - a^{2} x^{2} \right] d(a x) = \left| \begin{array}{l} \xi = (a x)^{2} \\ d \xi = 2 a x d(a x) = 2 \sqrt{y} \xi d(a x) \end{array} \right| \\
= \frac{m \omega^{2}}{a^{2} \sqrt{\pi}} \int_{-\infty}^{+\infty} \xi^{2} \exp [-\xi] \frac{d \xi}{2 \sqrt{\xi}} = \frac{m \omega^{2}}{a^{2} \sqrt{\pi}} \int_{-\infty}^{+\infty} \xi^{2} \exp [-\xi] \frac{d \xi}{2 \sqrt{\xi}} = \frac{m \omega^{2}}{2 a^{2} \sqrt{\pi}} \int_{-\infty}^{+\infty} \xi^{3/2} \exp [-\xi] d \xi = \\
\frac{m \omega^{2}}{2 a^{2} \sqrt{\pi}} \Gamma(5/2) = \frac{m \omega^{2}}{2 a^{2} \sqrt{\pi}} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \Gamma(1/2) = \frac{3 m \omega^{2}}{8 a^{2} \sqrt{\pi}} \cdot \sqrt{\pi} = \frac{3 m \omega^{2}}{8 a^{2}}
\end{array} ⟨ ψ 1 ∣ V ∣ ψ 1 ⟩ = ∫ − ∞ + ∞ ψ 1 ( x ) V ( x ) ψ 1 ( x ) d x = ∫ − ∞ + ∞ π 1/4 a 3/2 2 x exp [ − 2 a 2 x 2 ] 2 m ω 2 x 2 π 1/4 a 3/2 2 x exp [ − 2 a 2 x 2 ] d x = π 2 a 3 ∫ − ∞ + ∞ 2 m ω 2 x 4 exp [ − a 2 x 2 ] d x = a 2 π m ω 2 ∫ − ∞ + ∞ ( a x ) 4 exp [ − a 2 x 2 ] d ( a x ) = ∣ ∣ ξ = ( a x ) 2 d ξ = 2 a x d ( a x ) = 2 y ξ d ( a x ) ∣ ∣ = a 2 π m ω 2 ∫ − ∞ + ∞ ξ 2 exp [ − ξ ] 2 ξ d ξ = a 2 π m ω 2 ∫ − ∞ + ∞ ξ 2 exp [ − ξ ] 2 ξ d ξ = 2 a 2 π m ω 2 ∫ − ∞ + ∞ ξ 3/2 exp [ − ξ ] d ξ = 2 a 2 π m ω 2 Γ ( 5/2 ) = 2 a 2 π m ω 2 ⋅ 2 3 ⋅ 2 1 ⋅ Γ ( 1/2 ) = 8 a 2 π 3 m ω 2 ⋅ π = 8 a 2 3 m ω 2
where Γ ( ξ ) \Gamma(\xi) Γ ( ξ ) ψ 1 ( x ) = a 3 / 2 2 π 1 / 4 x exp [ − a 2 x 2 2 ] \psi_{1}(x) = \frac{a^{3/2} \sqrt{2}}{\pi^{1/4}} x \exp \left[ -\frac{a^{2} x^{2}}{2} \right] ψ 1 ( x ) = π 1/4 a 3/2 2 x exp [ − 2 a 2 x 2 ]
Answer: ⟨ ψ 1 ∣ V ∣ ψ 1 ⟩ = 3 m ω 2 8 a 2 \left\langle \psi_{1} \mid V \mid \psi_{1} \right\rangle = \frac{3 m \omega^{2}}{8 a^{2}} ⟨ ψ 1 ∣ V ∣ ψ 1 ⟩ = 8 a 2 3 m ω 2
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