0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don’t think $\delta$ is a scalar (number) so for starters I can’t imagine pulling it out of the integral. $\endgroup$ – ZeroTheHero Apr 11 at 12:02
  • $\begingroup$ Oh okay, I thought it would have been some small dimensionless parameter $\endgroup$ – Tapedeck Apr 11 at 12:15
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\nabla\Psi\cdot\nabla\Psi = -\Psi\nabla^2\Psi + \nabla(\Psi\nabla\Psi)$ ? $\endgroup$ – Tapedeck Apr 12 at 9:42
  • $\begingroup$ @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! $\endgroup$ – octonion Apr 12 at 10:01
  • $\begingroup$ 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! $\endgroup$ – Tapedeck Apr 12 at 10:18
  • $\begingroup$ @Tapedeck, You integrate it by parts using the very same identity you wrote. That is just the product rule for divergence $\nabla\cdot(A\vec{B})=\nabla A \cdot \vec{B}+A\nabla\cdot \vec{B}$ $\endgroup$ – octonion Apr 12 at 10:35
  • 1
    $\begingroup$ 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 $\Psi$ & $\delta \Psi$ ? $\endgroup$ – Tapedeck Apr 25 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.