Question

Q5. Consider a particle moving in one dimension whose Hamiltonian is \begin{equation} \hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x), \end{equation} with a potential \begin{equation} V(x) = \begin{cases} -\lambda x & x < 0 \\ \lambda x & x \ge 0 \end{cases} \end{equation} for a positive constant $\lambda$. Consider the wave function \begin{equation} \psi(x) = e^{-\alpha x^2/2} \end{equation} for a parameter $\alpha$. Normalize the state and then calculate the expectation value of the energy \begin{equation} E = \langle \psi | \hat{H} | \psi \rangle \end{equation} [Given: $\int_{-\infty}^{\infty} dx \, e^{-\alpha x^2} = (\pi/\alpha)^{1/2}$, $\int_{0}^{\infty} dx \, x^2 e^{-\alpha x^2} = \pi^{1/2}/(2\alpha^{3/2})$, $\int_{0}^{\infty} dx \, x e^{-\alpha x^2} = 1/(2\alpha)$]

          Q5. Consider a particle moving in one dimension whose Hamiltonian is
\begin{equation*}
\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x),
\end{equation*}
with a potential
\begin{equation*}
V(x) = \begin{cases}
-\lambda x & x < 0 \\
\lambda x & x \ge 0
\end{cases}
\end{equation*}
for a positive constant $\lambda$.
Consider the wave function
\begin{equation*}
\psi(x) = e^{-\alpha x^2/2}
\end{equation*}
for a parameter $\alpha$. Normalize the state and then calculate the expectation value of the
energy
\begin{equation*}
E = \langle \psi | \hat{H} | \psi \rangle
\end{equation*}
[Given: $\int_{-\infty}^{\infty} dx \, e^{-\alpha x^2} = (\pi/\alpha)^{1/2}$, $\int_{0}^{\infty} dx \, x^2 e^{-\alpha x^2} = \pi^{1/2}/(2\alpha^{3/2})$, $\int_{0}^{\infty} dx \, x e^{-\alpha x^2} = 1/(2\alpha)$]

Q5. Consider a particle moving in one dimension whose Hamiltonian is

Ĥ = -(ħ^2)/(2m)(d^2)/(dx^2) + V(x),

with a potential

V(x) =
-λ x x < 0
λ x x ≥ 0

for a positive constant λ.
Consider the wave function

ψ(x) = e^-α x^2/2

for a parameter α. Normalize the state and then calculate the expectation value of the
energy

E = ⟨ψ | Ĥ | ψ⟩

[Given: ∫-∞^∞ dx e^-α x^2 = (π/α)^1/2, ∫0^∞ dx x^2 e^-α x^2 = π^1/2/(2α^3/2), ∫0^∞ dx x e^-α x^2 = 1/(2α)]

Added by Frank T.

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