Variational Principle on First Excited State

[boyu]

I am trying to prove the variational principle on 1st excited state, but have some questions here.

The theory states like this: If <\psi|\psi_{gs}>=0, then <H>\geq E_{fe}, where 'gs' stands for 'grand state' and 'fe' for 'first excited state'.

Proof: Let ground state denoted by 1, and first excited state denoted by 2, i.e., |\psi_{1}>=|\psi_{gs}>, |\psi_{2}>=|\psi_{fe}>


<\psi|\psi_{1}>=0 ---> |\psi>=\sum^{\infty}_{n=2}c_{n}\psi_{n}

<H>=<\psi|H\psi>=<\sum^{\infty}_{m=2}c_{m}\psi_{m}|E_{n}\sum^{\infty}_{n=2}c_{n}\psi_{n}>=\sum^{\infty}_{m=2}\sum^{\infty}_{n=2}E_{n}c^{*}_{m}c_{n}<\psi_{m}|\psi_{n}>=\sum^{\infty}_{n=2}E_{n}|c_{n}|^{2}

Since n\geq 2, E_{n}\geq E_{2}=E_{fe}


so <H>\geq \sum^{\infty}_{n=2}E_{fe}|c_{n}|^{2}=E_{fe}\sum^{\infty}_{n=2}|c_{n}|^{2}


Then how to get the conclusion of <H>\geq E_{fe}, since \sum^{\infty}_{n=2}|c_{n}|^{2}=1-|c_{1}|^{2}\leq 1?

Edit: I've got the answer. c_{1}=0. Thx.

 

[DrClaude]

Correct. One has
$$
\sum_{n=2}^{\infty} |c_n|^2 = 1
$$