In linear variation method using two orthogonal basis functions, the two roots obtained are ϵ₀ and ϵ₁(ϵ₀ < ϵ₁). The correct relation of these with exact ground and first excited state energies, E₀ and E₁, respectively, is

Concept:-

It involves a variational method to measure energy levels of relatively difficult systems in quantum theory.

variational method explains pretty well that the calculated energy is higher than the actual energy if the trial wave function is taken into account.\(energy\: of\: trial\: wave\: function \geqslant exact\: energy\)

Trial wave function:

It's continuous and must possess the characteristics of a wave function like the shape and boundary conditions.

Furthermore, it's represented as linear combinations of single trial wave functions.

Explanation:-

Orthogonal basis functions are non-degenerate and their successive energy levels are also wave functions.

\(\epsilon _0 < \epsilon_1\) given, where \(\epsilon _0\) =lower energy function, \(\epsilon_1\)=higher energy wave function

So,\(\epsilon_0 \geqslant E_0\)

\(\epsilon_1 \geqslant E_1\)

Conclusion:-

In the linear variation method using two orthogonal basis functions, the two roots obtained are ϵ0 and ϵ1(ϵ0 < ϵ1). The correct relation of these with the exact ground and first excited state energies, E0 and E1, respectively, is \(\epsilon_0 \geqslant E_0\) , \(\epsilon_1 \geqslant E_1\).

Hence option 4 is correct.