Question 705050

SCQMEDIUM

Using a normalized trial wavefunction ψ(x)=ae−a∣x∣\psi(x) = \sqrt{a}e^{-a|x|} (where aa is a positive real constant) for a particle of mass mm in the potential V(x)=−λδ(x)V(x) = -\lambda\delta(x), (λ>0\lambda > 0), the estimated ground state energy is

(A)

mλ2h2\frac{m\lambda^2}{h^2}

(B)

mλ2h2\frac{m\lambda^2}{h^2}

(C)

mλ22h2\frac{m\lambda^2}{2h^2}

(D)

−mλ22h2-\frac{m\lambda^2}{2h^2}

Detailed Solution

The ground state energy of a delta potential V(x)=−λδ(x)V(x) = -\lambda\delta(x) is given by E=−mλ22ℏ2E = -\frac{m\lambda^2}{2\hbar^2}. Given the trial wavefunction matches the exact form for this potential, the expectation value of the energy will result in the exact eigenvalue, which is −mλ22ℏ2-\frac{m\lambda^2}{2\hbar^2}.

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