Variation Method: Proving \int \phi^{*} \hat{H} \phi d\tau>E_1

[pedroobv]

Homework Statement

This is the problem 8.10 from Levine's Quantum Chemistry 5th edition:
Prove that, for a system with nondegenerate ground state, [tex]\int \phi^{*} \hat{H} \phi d\tau>E_{1}[/tex], if [tex]\phi[/tex] is any normalized, well-behaved function that is not equal to the true ground-state wave function. Hint: Let [tex]b[/tex] be a positive constant such that [tex]E_{1}+b<E_{2}[/tex]. Turn (8.4) into an inequality by replacing all [tex]E_{k}[/tex]'s except [tex]E_{1}[/tex] with [tex]E_{1}+b[/tex].

Homework Equations

Equation (8.4):

[tex]\int \phi^{*} \hat{H} \phi d\tau=\sum_{k}a^{*}_{k}a_{k}E_{k}=\sum_{k}|a_{k}|^{2}E_{k}[/tex]​

Other relevant equations:

[tex]\phi=\sum_{k}a_{k}\psi_{k}[/tex]​

where

[tex]\hat{H}\psi_{k}=E_{k}\psi_{k}[/tex]​

[tex]1=\sum_{k}|a_{k}|^{2}[/tex]


[tex]E_{1}<E_{2}<E_{3}...[/tex]​

The Attempt at a Solution

[tex]\int \phi^{*} \hat{H} \phi d\tau=|a_{1}|^{2}E_{1}+\sum^{\infty}_{k=2}|a_{k}|^{2}E_{k}>|a_{1}|^{2}E_{1}+\sum^{\infty}_{k=2}|a_{k}|^{2}\left(E_{1}+b\right)=|a_{1}|^{2}E_{1}+E_{1}\sum^{\infty}_{k=2}|a_{k}|^{2}+b\sum^{\infty}_{k=2}|a_{k}|^{2}=E_{1}\sum_{k}|a_{k}|^{2}+b\sum^{\infty}_{k=2}|a_{k}|^{2}[/tex]
[tex]\int \phi^{*} \hat{H} \phi d\tau>E_{1}+b\sum^{\infty}_{k=2}|a_{k}|^{2}[/tex]

I don't know how to apply the condition that [tex]\phi\neq \psi_{1}[/tex] to complete the proof, also I'm not sure if this is the right way to start but that's how I understand the hint given. If you need more information or something is not clear, please tell me so I can do the proper correction.

 

[Avodyne]

You just need to show that your final sum is not zero. If it was zero, what would that tell us about each [itex]a_k[/itex], [itex]k\ge 2[/itex]? And what would that tell us about [itex]\phi[/itex]?

Minor point: your [itex]>[/itex] sign should really be [itex]\ge[/itex] to account for this case.

 

[pedroobv]

But if the last sum is not zero that mean that there is a mistake somewhere since the purpose is to obtain [tex]\int \phi^{*} \hat{H} \phi d\tau>E_{1}[/tex] right?

 

[pedroobv]

I think that it is easy to show that the last sum is not zero because if it was zero that would mean that [tex]\phi = \psi_{1}[/tex] according to the equations

[tex]1=\sum_{k}|a_{k}|^{2}[/tex]

[tex]\phi = \sum_{k}a_{k}\psi_{k}[/tex]​

But as the problem statement says, [tex]\phi\neq \psi_{1}[/tex], so the sum can't be zero. So far, I have not been able to find the mistake (since as I said before the second term must be eliminated to complete the proof). Any help would be appreciated.