- #1
boyu
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I am trying to prove the variational principle on 1st excited state, but have some questions here.
The theory states like this: If [tex]<\psi|\psi_{gs}>=0[/tex], then [tex]<H>\geq E_{fe}[/tex], 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., [tex]|\psi_{1}>=|\psi_{gs}>[/tex], [tex]|\psi_{2}>=|\psi_{fe}>[/tex]
[tex]<\psi|\psi_{1}>=0[/tex] ---> [tex]|\psi>=\sum^{\infty}_{n=2}c_{n}\psi_{n}[/tex]
[tex]<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}[/tex]
Since [tex]n\geq 2[/tex], [tex]E_{n}\geq E_{2}=E_{fe}[/tex]
so [tex]<H>\geq \sum^{\infty}_{n=2}E_{fe}|c_{n}|^{2}=E_{fe}\sum^{\infty}_{n=2}|c_{n}|^{2}[/tex]
Then how to get the conclusion of [tex]<H>\geq E_{fe}[/tex], since [tex]\sum^{\infty}_{n=2}|c_{n}|^{2}=1-|c_{1}|^{2}\leq 1[/tex]?
Edit: I've got the answer. [tex]c_{1}=0[/tex]. Thx.
The theory states like this: If [tex]<\psi|\psi_{gs}>=0[/tex], then [tex]<H>\geq E_{fe}[/tex], 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., [tex]|\psi_{1}>=|\psi_{gs}>[/tex], [tex]|\psi_{2}>=|\psi_{fe}>[/tex]
[tex]<\psi|\psi_{1}>=0[/tex] ---> [tex]|\psi>=\sum^{\infty}_{n=2}c_{n}\psi_{n}[/tex]
[tex]<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}[/tex]
Since [tex]n\geq 2[/tex], [tex]E_{n}\geq E_{2}=E_{fe}[/tex]
so [tex]<H>\geq \sum^{\infty}_{n=2}E_{fe}|c_{n}|^{2}=E_{fe}\sum^{\infty}_{n=2}|c_{n}|^{2}[/tex]
Then how to get the conclusion of [tex]<H>\geq E_{fe}[/tex], since [tex]\sum^{\infty}_{n=2}|c_{n}|^{2}=1-|c_{1}|^{2}\leq 1[/tex]?
Edit: I've got the answer. [tex]c_{1}=0[/tex]. Thx.
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