# [Physics] Two questions about Variational Method of quantum mechanics

atomic-physicsground-statequantum mechanicsvariational-principle

I have two question about variational method of quantum mechanics.

1. Why we always find the ground state energy by this approach. Why not the other excited states?

2. When we find the ground state energy of helium by this method, at one stage we consider $Z$ (charge of nucleus) is a variable parameter in stead of using "2". But we have done it because our analytical value was not matching with experimental value. My question is, if experimental value always guide us to modify the Hamiltonian, I mean if we use experimental value of ground state energy of helium at first place, then what is the benefit of this method? it sounds quite silly to me. If anybody can not understand me then I will edit my question. No problem. But give some response at least.

While the excited states are usually not extrema of the variation functional, they are stationary points! That is, by varying the functional you get the excited states as solutions, by requiring the variation vanish. Consider the variation functional with a Lagrange multiplier to enforce the norm of the states: $$E\big[\left|\psi\right> , \left<\psi\right|\big] = \left< \psi \middle| H \middle| \psi \right> - \lambda \big( \left< \psi \middle| \psi \right> -1 \big).$$ Computing the variation one gets: $$\delta E = \left< \delta \psi \middle| H \middle| \psi \right> + \left< \psi \middle| H \middle| \delta \psi \right> - \lambda \left< \delta \psi \middle| \psi \right> - \lambda \left< \psi \middle| \delta \psi \right>$$ (Note, that we consider $\left| \psi \right>$ and $\left< \psi \right|$ to be independently varied.)
From this we can see, that the functional is stationary for: $$H \left| \psi \right> = \lambda \left| \psi \right>.$$ Which is the time independent Schrödinger equation, which gives all the stationary states. The Langrange multiplier turns out to be the eigenenergy $E$ of the state.