Variational Principle on First Excited State

  • Thread starter boyu
  • Start date
20
0
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.
 
Last edited:

DrClaude

Mentor
6,969
3,140
Correct. One has
$$
\sum_{n=2}^{\infty} |c_n|^2 = 1
$$
 

Want to reply to this thread?

"Variational Principle on First Excited State" You must log in or register to reply here.

Related Threads for: Variational Principle on First Excited State

Replies
8
Views
2K
  • Posted
Replies
1
Views
1K
  • Posted
Replies
6
Views
2K
  • Posted
Replies
2
Views
1K
  • Posted
Replies
3
Views
1K
  • Posted
Replies
1
Views
442
  • Posted
Replies
4
Views
3K
  • Posted
Replies
4
Views
12K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top