The trial wave function of a system is expanded as $\psi_{\mathrm{t}}=\mathrm{c}_{1} \phi_{1}+\mathrm{c}_{2} \phi_{2}$. The matrix elements of the Hamiltonian are $\phi_{1} \mathrm{H} \phi_{1}=0 ; \phi 1 \mathrm{H} \phi 2=20=\phi_{2} \mathrm{H} \phi_{1}$ and $\phi_{2} \mathrm{H} \phi_{2} \mathrm{H} \phi_{2}=3.0$. The approximate ground-state energy of the system from the linear variational principle is
(a)-1.0
(b)-2.0
(c)+4.0
(d)+5.0
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Secular equation (0 − E)(3 − E) − 2² = 0 ⇒ E² − 3E − 4 = 0 ⇒ E = −1 or 4; the ground state is −1.0.