# First-Order Perturbation of Energy Eigenfunction

I have a homework questions where I'm struggling to understand the methodology to use. We derive first the energy functional for the energy eigenfunction equation (this is fine, I used some vector identities):

$$E = \int d^3r \left[ \frac{\hbar}{2m} \nabla\Psi\cdot\nabla\Psi + V(\vec r)\Psi^2\right].$$ Then we are told to make the transformation

$$\Psi \to \Psi + \delta\Psi .$$

Noting the perturbation is conventionally normalised

$$\int d^3r (\Psi + \delta\Psi)^2 = 1.$$

The question asks to show the change in the energy functional vanishes to first order in $$\delta\Psi$$. I'm assuming that means show the $$\delta$$ terms of order 1 vanish but I have no idea how to go about that. If I straight up plug in the transformation into the energy functional I think we get the below expression on RHS:

$$\int d^3r \left[ \frac{\hbar}{2m} \nabla(\Psi+\delta\Psi)\cdot\nabla(\Psi+\delta\Psi) + V(\vec r)(\Psi+\delta\Psi)^2\right].$$

Any tips from here? We haven't covered perturbation theory or calculus of variations.. Do I need to look at the small change in the LHS as well ($$E + \delta E$$) and express that as a power series? I'm still not sure how that would help.

• I don’t think $\delta$ is a scalar (number) so for starters I can’t imagine pulling it out of the integral. – ZeroTheHero Apr 11 '19 at 12:02
• Oh okay, I thought it would have been some small dimensionless parameter – Tapedeck Apr 11 '19 at 12:15

Here's a crash course in calculus of variations. $$E$$ depends on the function $$\Psi$$, so if you substitute in $$\Psi\rightarrow \Psi+\delta\Psi$$, where $$\delta\Psi$$ is a function that is small everywhere, that will change $$E\rightarrow E+\delta E$$. To find $$\delta E$$, make this substitution and keep only 1st order terms in $$\delta\Psi$$. For terms that end up being derivatives of $$\delta\Psi$$ you will need to do integration by parts.
• Thanks octonion, so when I put this substitution in I get $\int d^3r [\frac{\hbar^2}{2m}\nabla(\Psi + \delta\Psi)\cdot\nabla(\Psi + \delta\Psi) + V(\vec r)(\Psi + \delta\Psi)^2]$ Expanding the dot product I would put $\int d^3 \nabla\Psi\cdot\nabla\Psi + 2\nabla\Psi\cdot\nabla(\delta\Psi) + H.O.T$ I'm not sure how I would integrate the $2\nabla\Psi\cdot\nabla(\delta\Psi)$ by parts- would I need an identity like $\nabla\Psi\cdot\nabla\Psi = -\Psi\nabla^2\Psi + \nabla(\Psi\nabla\Psi)$ ? – Tapedeck Apr 12 '19 at 9:42
• @Tapedeck. You're on the right track, but you want to integrate the $2\nabla\Psi\cdot \nabla(\delta\Psi)$ term by parts, we don't care about the $\nabla\Psi\cdot\nabla\Psi$ term because it is telling us $E$ not $\delta E$. Also that is a true identity you wrote, but you need to consider the last term to be a divergence $\nabla\cdot(\Psi\nabla\Psi)$. Think about what happens when you integrate a total divergence like that over all space. You're almost there! – octonion Apr 12 '19 at 10:01
• Hi octonion, Sorry yes that was meant to be a divergence. So that term is fine, you can use the divergence theorem to change it to a surface integral where the boundary goes to infinity, and $\Psi$ should vanish at infinity. I think the part I'm struggling with is integrating the dot product $2\nabla\Psi\cdot\nabla(\delta\Psi)$ by parts, I'm not clear on how that can be done. Thanks again! – Tapedeck Apr 12 '19 at 10:18
• @Tapedeck, You integrate it by parts using the very same identity you wrote. That is just the product rule for divergence $\nabla\cdot(A\vec{B})=\nabla A \cdot \vec{B}+A\nabla\cdot \vec{B}$ – octonion Apr 12 '19 at 10:35
• I did notice that when I was playing around that it looks like $<\delta\Psi|\hat H|\Psi>$ Oh can we use $\hat H | \Psi > = E | \Psi>$ and then orthonormality on $\Psi$ & $\delta \Psi$ ? – Tapedeck Apr 25 '19 at 4:27