Variation Method

[pedroobv]

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


2. Relevant equations

Equation (8.4):
\int \phi^{*} \hat{H} \phi d\tau=\sum_{k}a^{*}_{k}a_{k}E_{k}=\sum_{k}|a_{k}|^{2}E_{k}

Other relevant equations:

\phi=\sum_{k}a_{k}\psi_{k}

where

\hat{H}\psi_{k}=E_{k}\psi_{k}


1=\sum_{k}|a_{k}|^{2}


E_{1}<E_{2}<E_{3}...

3. The attempt at a solution

\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}
\int \phi^{*} \hat{H} \phi d\tau>E_{1}+b\sum^{\infty}_{k=2}|a_{k}|^{2}

I don't know how to apply the condition that \phi\neq \psi_{1} 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 a_k, k\ge 2? And what would that tell us about \phi?

Minor point: your > sign should really be \ge 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 \int \phi^{*} \hat{H} \phi d\tau>E_{1} 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 \phi = \psi_{1} according to the equations
1=\sum_{k}|a_{k}|^{2}

\phi = \sum_{k}a_{k}\psi_{k}

But as the problem statement says, \phi\neq \psi_{1}, 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.