# 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.

Last edited:
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#### DrClaude

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

"Variational Principle on First Excited State"

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