# Electronic Properties of Materials/Quantum Mechanics for Engineers/Perturbation Methods

This is the ninth chapter of the first section of the book Electronic Properties of Materials.
**INCOMPLETE**
Most operators (Hamiltonians) are not simple. Fortunately, with a bit of effort, we can sometimes rewrite the operator (${\hat {H}}$) as ${\hat {H}}={\hat {H}}_{o}+\lambda H'$, where ${\hat {H}}$ is a Hamiltonian for which we know the solution.
${\hat {H}}_{o}\ \psi _{\eta }^{(o)}=E_{\eta }^{(o)}\ \psi _{\eta }^{(o)}$
Here, $\psi _{\eta }^{(o)}$ are non degeneratory orthogonal eigenfunctions, and $H'$ is a small perturbation to the $H_{o}$. Additionally, $\lambda$ is a real arbitrary parameter, and when $\lambda =0$, we have:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikibooks.org/v1/":): {\displaystyle \begin{array}{lcl} H = H_o & \ \ when \ \lambda=1 \\ H = H_o + H' & \quad \end{array}}
The problem we want to solve is: $H\psi _{\eta }=E_{\eta }\psi _{\eta }$
The perturbation is small and in the limit $\lambda$ goes to zero.
$E_{\eta }=E_{\eta }^{(o)}+\psi _{\eta }=\psi _{\eta }^{(o)}$
We will assume that $E_{\eta }$ and $\psi _{\eta }$ can be written as powers of $\lambda$.
${\begin{aligned}E_{\eta }&=\sum _{j=0}^{\infty }\lambda ^{j}E_{\eta }^{(j)}=E_{\eta }^{(o)}+\lambda E_{\eta }^{(1)}+\lambda ^{2}E_{\eta }^{(2)}+\cdots +\lambda ^{n}E_{\eta }^{(n)}\\\psi _{\eta }&=\sum _{j=0}^{\infty }\lambda ^{k}\psi _{\eta }^{(j)}\end{aligned}}$
Substituting:
$\left(H_{o}+\lambda H'\right)\left(\psi _{\eta }^{(o)}+\lambda \psi _{\eta }^{(1)}+\lambda ^{2}\psi _{\eta }^{(o)}+\cdots +\lambda ^{n}\psi _{\eta }^{(n)}\right)=\left(E_{\eta }^{(o)}+\lambda E_{\eta }^{(1)}+\lambda ^{2}E_{\eta }^{(o)}+\cdots +\lambda ^{n}E_{\eta }^{(n)}\right)\left(\psi _{\eta }^{(o)}+\lambda \psi _{\eta }^{(1)}+\lambda ^{2}\psi _{\eta }^{(o)}+\cdots +\lambda ^{n}\psi _{\eta }^{(n)}\right)$
Multiply through & collect common properties to form equations for each power of $\lambda$:${\begin{aligned}\lambda ^{o}:&\quad H_{o}\psi _{n}^{(o)}=E_{n}^{(o)}\psi _{n}^{(o)}\\\lambda ^{1}:&\quad H_{o}\psi _{n}^{(1)}+H'\psi ^{(o)}=E_{n}^{(o)}\psi _{n}^{(1)}+E_{n}^{(1)}\psi _{n}^{(o)}\\\lambda ^{2}:&\quad H_{o}\psi _{n}^{(2)}+H'\psi ^{(1)}=E_{n}^{(0)}\psi _{n}^{(2)}+E_{n}^{(1)}\psi _{n}^{(1)}+E_{n}^{(2)}\psi _{n}^{(o)}\end{aligned}}$
The powers of $\lambda ^{o}$ are just our unperturbed $H_{o}$. We will begin by looking at powers of $\lambda ^{1}$.
Rearrange:
$(H_{o}-E_{n}^{(o)})\psi _{n}^{(1)}+(H'-E_{n}^{(1)})\psi _{n}^{(o)}=0$
multiply left by $\psi _{n}^{*(o)}$ and integrate
$\int \psi _{n}^{*(o)}(H_{o}-E_{n}^{(o)})\psi _{n}^{(1)}+\int \psi _{n}^{*(o)}(H'-E_{n}^{(1)})\psi _{n}^{(o)}=0$
Begin with left term. These operators are Hermitian. They have special properties, namely that they obey the postulates of quantum mechanics, including a few revations that are useful for proofs. One such property is:
$\int \Psi ^{*}H\Phi =\left(\int \Phi ^{*}H\Psi \right)^{*}$
Which we will use here:
${\begin{aligned}\int \psi _{n}^{(o)^{*}}H_{o}\psi _{n}^{(1)}=&\left(\int \psi _{n}^{(1)^{*}}H\psi _{n}^{(o)}\right)^{*}\\=&\left(\int \psi _{n}^{(1)^{*}}E_{n}^{(o)}\psi _{n}^{(o)}\right)^{*}\\=&E_{n}{(o)}\left(\int \psi _{n}^{(1)^{*}}\psi _{n}^{(o)}\right)^{*}\\=&E_{n}^{(o)}\int \psi _{n}^{(o)^{*}}\psi _{n}^{(1)}\end{aligned}}$
Thus, our entire term equals zero. As a result:
$\int \psi _{n}^{(o)}H'\psi _{n}^{(o)}=E_{n}^{(1)}\int \psi _{n}^{(o)^{*}}\psi _{n}^{(o)}=E_{n}^{(1)}$
Therefore, the first order correction to the eigenvalue is:
$E_{n}^{(1)}=\int \psi _{n}^{(o)^{*}}H'\psi _{n}^{(o)}$
Following the same steps we can find the higher order perturbations:
${\begin{aligned}E_{n}^{(2)}&=\int \psi _{n}^{(o)^{*}}\left(H'-E_{n}^{(1)}\right)\psi _{n}^{(1)}\\E_{n}^{(3)}&=\int \psi _{n}^{(1)^{*}}\left(H'-E_{n}^{(1)}\right)\psi _{n}^{(1)}-2E_{n}^{(2)}\int \psi _{n}^{(o)^{*}}\psi _{n}^{(1)}\end{aligned}}$
Most simple theories do not require these higher order corrections, but how do we get the wavefunctions in the first place? Lets assume that ψ$_n$$^{(1)}$=∑$_k$a$_{nk}$$^{(1)}$ψ$_k$$^{(o)}${\textstyle \psi _{n}^{(1)}=\sum _{k}{a_{nk}}^{(1)}{\psi _{k}}^{(o)}} where $a_{nk}$ coefficient is the projection of ψ$_n$$^{(1)}${\textstyle \psi _{n}^{(1)}} onto ψ$_k$$^{(o)}${\textstyle \psi _{k}^{(o)}}. Returning to our original $\lambda '$ term, gather:
$H_{o}\psi _{n}^{(1)}+H'\psi _{n}^{(o)}=E_{n}^{(o)}\psi _{n}^{(1)}+E_{n}^{(1)}\psi _{n}^{(o)}$
Rearranging and substituting gives us:
$0=\left(H_{o}-E_{n}^{(o)}\right)\sum a_{n}^{(1)}\psi _{k}^{(o)}+\left(H'-E_{n}^{(1)}\right)\psi _{n}^{(o)}$
Multiplying the right side by $\psi _{\ell }^{(o)^{*}}$and integrating gets us:
${\begin{aligned}0&=\int \psi _{\ell }^{(o)^{*}}E_{n}^{(o)}\sum a_{nk}^{(1)}\psi _{k}^{(o)}-E_{n}^{(o)}\int \psi _{\ell }^{(o)^{*}}\sum _{k}a_{nk}^{(1)}\psi _{k}^{(o)}+\int \psi _{\ell }^{(o)^{*}}H'\psi _{n}^{(o)}-E_{n}^{(1)}\int \psi _{\ell }^{(o)^{*}}\psi _{n}^{(o)}\\&=E_{\ell }^{(o)}a_{n\ell }^{(1)}-E_{n}^{(o)}a_{n\ell }^{(1)}+\int \psi _{\ell }^{(o)^{*}}H'\psi _{n}^{(o)}-E_{n}^{(1)}\delta _{n\ell }\end{aligned}}$
When n=ℓ{\textstyle n=\ell }, we loose all a$_{nℓ}$$^{(1)}${\textstyle a_{n\ell }^{(1)}} terms, giving us: $E_{n}^{(1)}=\int \psi _{\ell }^{(o)^{*}}H'\psi _{n}^{(o)}$
However, when n≠ℓ{\textstyle n\neq \ell } we get:
${\begin{aligned}0&=\left(E_{\ell }^{(o)}-E_{n}^{(1)}\right)a_{n\ell }^{(1)}+\int \psi _{\ell }^{(o)^{*}}H'\psi _{n}^{(o)}\\a_{n\ell }&={-\int \psi _{\ell }^{(o)^{*}}H'\psi _{n}^{(o)} \over E_{\ell }^{(o)}-E_{n}^{(o)}}\end{aligned}}$
Since $a_{nn}^{(1)}$ does not seem to be determined from these equations, there is an uncomfortable degree of arbitrariness in selecting $a_{nn}^{(1)}$. Require normalization:
${\begin{aligned}1&=\int \Psi ^{*}\psi \\1+0\left(\lambda ^{j+1}\right)&=\int \left({\psi _{n}^{*}}^{(o)}+\lambda {\psi _{n}^{*}}^{(1)}+\lambda ^{2}{\psi _{n}^{*}}^{2}+\cdots \right)\left(\psi _{n}^{(o)}+\lambda \psi _{n}^{(1)}+\lambda ^{2}\psi _{n}^{(2)}+\cdots \right)\end{aligned}}$
Where: ${\begin{aligned}\lambda ^{o}:\ 1&=\int {\psi _{n}^{(o)}}^{*}\psi _{n}^{(o)}\\\lambda ^{1}:\ 0&=\int {\psi _{n}^{(0)}}^{*}\psi _{n}^{(1)}+\int {\psi _{n}^{(1)}}^{*}\psi _{n}^{(o)}\\\lambda ^{3}:\ 0&=\int {\psi _{n}^{(0)}}^{*}\psi _{n}^{(2)}+\int {\psi _{n}^{(1)}}^{*}\psi _{n}^{(1)}+\int {\psi _{n}^{(2)}}^{*}\psi _{n}^{(o)}\end{aligned}}$
Thus:
$0=\underbrace {\int {\psi _{n}^{(o)}}^{*}\psi _{n}^{(1)}} _{=a_{nn}}+\underbrace {\int {\psi _{n}^{(1)}}^{*}\psi _{n}^{(o)}} _{{=a_{nn{^{(1)}}}}^{*}}$
which is a projection of $\psi _{n}^{(1)}$ onto $\psi _{n}^{(o)}$.
$a_{nn}^{(1)}+{a_{nn}^{(1)}}^{*}=0$ is a complex number.
Complex number formula: ${\begin{aligned}0&=a+bi+a-bi\\0&=a\end{aligned}}$
What is $b$? Here, we choose $b=0$.
<FIGURE> "Title" (Description)
In quantum mechanics usually, but not always, $\Psi$ can have arbitrary phase $\phi$, so long as magnitude of $\Psi$ is correct. Here we choose ϕ=0{\textstyle \phi =0}. Therefore:
$a_{nn}^{(1)}=0$
This dictates that all of $\psi _{n}^{(1)}$ is orthogonal to $\psi _{n}^{(o)}$.
As an example, consider adding a correction to the hydrogen atom, what is actually a fairly common occurrence.
${\begin{aligned}H&=H_{o}+H'\\H_{o}&={-\hbar ^{2} \over 2m}\Delta ^{2}-{e^{2} \over r}\\H'&=-G{mM_{p} \over r}\end{aligned}}$
This last equation is the influence of gravitational attraction between the positive ion and the negative ion. This is a first order energy correction.
${\begin{aligned}E_{1s}^{(1)}&=\int {\psi _{1s}^{(o)}}^{*}\left({-GmM_{p} \over r}\right)\psi _{1s}^{(o)}\operatorname {d} \!right\\\psi _{1s}^{(o)}&={1 \over {\sqrt {\pi }}}{1 \over a_{o}^{3 \over 2}}\exp \left[{-2r \over a_{o}}\right]{-GmM \over r}\\\\E_{1s}^{(1)}&=\int _{0}^{\pi }\sin \!\theta \operatorname {d} \!\theta \int _{0}^{2\pi }\operatorname {d} \!\phi \int _{0}^{\infty }r^{2}\operatorname {d} \!r{1 \over \pi }{1 \over a_{o}}\exp \left[{-2r \over a_{o}}\right]{-GmM \over a_{o}^{2}}\left({a_{o}^{2} \over 4}\right)\\&=4\pi {-GmM \over \pi a_{o}^{3}}\int _{0}^{\infty }\operatorname {d} \!rr\exp \left[{-2r \over a_{o}}\right]\\&=-4{GmM \over a_{o}^{3}}\left({a_{o}^{2} \over 4}\right)\end{aligned}}$
When working with degenerate wavefunctions, the problem becomes slightly more complicated because the interactions amongst the degenerate wavefunctions must be carefully accounted for. That said, this is just bookkeeping. The general procedure for Rayleigh-Schrodinger perturbation theory is as outlined here.