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, $$\int \phi^{*} \hat{H} \phi d\tau>E_{1}$$, 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 $$E_{1}+b$$.

Homework 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

$$\hat{H}\psi_{k}=E_{k}\psi_{k}$$​

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


$$E_{1}<E_{2}<E_{3}...$$​

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.