Which matrix gives the first-order shifts of an isolated degenerate energy?

• V restricted to the degenerate eigenspace
• H0 restricted to an unrelated non-degenerate eigenspace
• V with its off-diagonal entries discarded before calculation

Why can averaging V in arbitrary degenerate basis states miss the first-order energy shifts?

• Every first-order energy expectation divides by the gap between those basis states.
• A shared unperturbed energy forces every perturbation matrix element to vanish.
• Off-diagonal perturbation elements can mix those basis states.

For an admissible nonzero trial vector, the Rayleigh quotient gives a lower bound on the ground-state energy.

For H=H0+lambda V, E_d=8 eV and lambda=0.2. The restricted V eigenvalues are -3 eV and 1 eV. What are the first-order energies?

• 5.0 eV and 9.0 eV
• 7.8 eV and 8.2 eV
• 7.4 eV and 8.2 eV

The restricted perturbation matrix W has rows (1, 3) and (3, 1). What are its eigenvalues w?

• -3 and 3
• -2 and 4
• 1 and 3

Two admissible trial states for the same Hamiltonian give 4.8 eV and 5.2 eV. Which conclusion follows?

• E0<=4.8 eV is the tighter upper bound.
• E0>=5.2 eV is the tighter lower bound.
• E0=4.8 eV because minimizing two trials proves exactness.

For H=diag(4,9) eV and phi=(1/2,0), Mira reports <phi|H|phi>=1 eV as a trial energy below E0=4 eV. How should Mira correct the calculation?

• Multiply 1 eV by the norm squared 1/4 to obtain 0.25 eV.
• Treat the small coefficient as changing H's ground energy to 1 eV.
• Divide 1 eV by the norm squared 1/4 to obtain 4 eV.

In an orthonormal basis, H=diag(2,6) eV and phi=(1,1). What is the variational energy of this unnormalized vector?

• 2 eV
• 4 eV
• 8 eV