# Introduction to Mathematical Physics/N body problem in quantum mechanics/Atoms
## One nucleus, one electron

sechydrog
This case corresponds to the study of hydrogen atom.\index{atom} It is a particular case of particle in a central potential problem, so that we apply methods presented at section #secpotcent to treat this problem. Potential is here:
eqpotcenhy
$V(r)=-{\frac {e}{r^{2}}}$
It can be shown that eigenvalues of hamiltonian $H$ with central potential depend in general on two quantum numbers $k$ and $l$, but that for particular potential given by equation eqpotcenhy, eigenvalues depend only on sum $n=k+l$.

## Rotation invariance

secpotcent
We treat in this section the particle in a central potential problem ([#References|references]). The spectral problem to be solved is given by the following equation:
$-[{\frac {\hbar ^{2}}{2\mu }}\Delta +V(r)]\phi (r)=E\phi (r).$
Laplacian operator can be expressed as a function of $L^{2}$ operator.
Theorem: Laplacian operator $\Delta$ can be written as:
$\Delta =-{\frac {1}{r^{2}}}L^{2}+{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}$
Proof: Here, tensorial notations are used (Einstein convention). By definition:
$L_{i}=\epsilon _{ijk}x_{j}p_{k}$
So:
${\begin{matrix}L_{i}L_{i}&=&\epsilon _{ijk}\epsilon _{ilm}x_{j}p_{k}x_{l}p_{m}\\&=&(\delta _{jl}\delta _{km}-\delta _{jm}\delta {kl})x_{j}p_{k}x_{l}p_{m}\\&=&x_{j}x_{j}p_{k}p_{k}-x_{j}p_{k}x_{k}p_{j}\end{matrix}}$
The writing order of the operators is very important because operator do not commute. They obey following commutation relations:
$[x_{i},p_{j}]=i\hbar \delta _{ij}$
$[x_{j},x_{k}]=0$
$[p_{j},p_{k}]=0$
From equation eqdefmomP, we have:
$p_{k}=-i\hbar {\frac {\partial }{\partial x_{k}}}$
thus
$x_{j}x_{j}p_{k}p_{k}=-x^{2}\hbar ^{2}\Delta .$
Now,
${\begin{matrix}x_{j}p_{k}x_{k}p_{j}&=&[i\hbar \delta _{ik}+p_{k}x_{j}]x_{k}p_{j}\\&=&i\hbar x_{k}p_{k}+p_{k}x_{j}x_{k}p_{j}\\&=&i\hbar x_{k}p_{k}+p_{k}x_{k}x_{j}p_{j}\end{matrix}}$
Introducing operator:
${\tilde {D}}=x_{k}{\frac {\partial }{\partial x_{k}}}$
we get the relation:
${IMP/label|eql2pri}L^{2}=-x^{2}\Delta +{\tilde {D}}^{2}+{\tilde {D}}$
Using spherical coordinates, we get:
${\tilde {D}}=r{\frac {\partial }{\partial r}}$
and
${\tilde {D}}^{2}=(r{\frac {\partial }{\partial r}})(r{\frac {\partial }{\partial r}})=r^{2}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\partial }{\partial r}}$
So, equation eql2pri becomes:
$\Delta =-{\frac {1}{r^{2}}}L^{2}+{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}$
Let us use the problem's symmetries:
Since:
* $L_{z}$ commutes with operators acting on $r$
* $L_{z}$ commutes with $L^{2}$ operator $L_{z}$ commutes with $H$
* $L^{2}$ commutes with $H$
we look for a function $\phi$ that diagonalizes simultaneously $H,L^{2},L_{z}$ that is such that:
${\begin{matrix}H\phi (r)&=&E\phi (r)\\L^{2}\phi (r)&=&l(l+1)\hbar ^{2}\phi (r)\\L_{z}\phi (r)&=&m\hbar \phi (r)\end{matrix}}$
Spherical harmonics $Y_{l}^{m}(\theta ,\phi )$ can be introduced now:
Definition:
Spherical harmonics $Y_{l}^{m}(\theta ,\phi )$ are eigenfunctions common to operators $L^{2}$ and $L_{z}$. It can be shown that:
${\begin{matrix}L^{2}Y_{l}^{m}&=&l(l+1)Y_{l}^{m}\\L_{z}Y_{l}^{m}&=&mY_{l}^{m}\end{matrix}}$
Looking for a solution $\phi (r)$ that can\footnote{Group theory argument should be used to prove that solution actually are of this form.} be written (variable separation):
$\phi (r)=R(r)Y_{l}^{m}(\theta ,\phi )$
problem becomes one dimensional:
eqaonedimrr
$-[{\frac {\hbar ^{2}}{2\mu }}({\frac {d^{2}}{dr^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}})+{\frac {l(l+1)}{2\mu r^{2}}}\hbar ^{2}+V(r)]R_{l}(r)=E_{kl}R_{l}(r)$
where $R(r)$ is indexed by $l$ only. Using the following change of variable: $R_{l}(r)={\frac {1}{r}}u_{l}(r)$, one gets the following spectral equation:
$-[{\frac {\hbar ^{2}}{2\mu }}{\frac {d^{2}}{dr^{2}}}+V_{e}(r)]u_{kl}(r)=E_{kl}u_{kl}(r)$
where
$V_{e}(r)={\frac {l(l+1)}{2\mu r^{2}}}\hbar ^{2}+V(r)$
The problem is then reduced to the study of the movement of a particle in an effective potential $V_{e}(r)$. To go forward in the solving of this problem, the expression of potential $V(r)$ is needed. Particular case of hydrogen introduced at section sechydrog corresponds to a potential $V(r)$ proportional to $1/r$ and leads to an accidental degeneracy.

## One nucleus, N electrons

This case corresponds to the study of atoms different from hydrogenoids atoms. The Hamiltonian describing the problem is:
$H=\sum _{i}-{\frac {\hbar ^{2}}{2m}}\Delta _{i}-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r_{i}}}+\sum _{j{\mathrel {>}}i}{\frac {1}{4\pi \epsilon _{0}}}{\frac {e^{2}}{r_{ij}}}+T_{2}$
where $T_{2}$ represents a spin-orbit interaction term that will be treated later. Here are some possible approximations:

### N independent electrons

This approximation consists in considering each electron as moving in a mean central potential and in neglecting spin--orbit interaction. It is a ``mean field approximation. The electrostatic interaction term
$-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r_{i}}}+\sum _{j{\mathrel {>}}i}{\frac {1}{4\pi \epsilon _{0}}}{\frac {e^{2}}{r_{ij}}}$
is modelized by the sum $\sum W(r_{i})$, where $W(r_{i})$ is the mean potential acting on particle $i$. The hamiltonian can thus be written:
$H_{0}=\sum _{i}h_{i}$
where $h_{i}=-{\frac {\hbar ^{2}}{2m}}\Delta _{i}+W(r_{i})$.
Remark:
More precisely, $h_{i}$ is the linear operator acting in the tensorial product space $\otimes _{i=1}^{N}E_{i}$ and defined by its action on function that are tensorial products:
$[1_{1}\otimes \dots \otimes 1_{i-1}\otimes h_{i}\otimes 1_{i+1}\dots \otimes 1_{N}](\phi _{1}\otimes \dots \phi _{N})=h_{i}(\phi _{1})\otimes \dots \phi _{N}$
It is then sufficient to solve the spectral problem in a space $E_{i}$ for operator $h_{i}$. Physical kets are then constructed by anti symmetrisation (see example exmppauli of chapter chapmq) in order to satisfy Pauli principle.\index{Pauli} The problem is a central potential problem (see section #secpotcent). However, potential $W(r_{i})$ is not like $1/r$ as in the hydrogen atom case and thus the accidental degeneracy is not observed here. The energy depends on two quantum numbers $l$ (relative to kinetic moment) and $n$ (rising from the radial equation eqaonedimrr). Eigenstates in this approximation are called electronic configurations.
Example:
For the helium atom, the fundamental level corresponds to an electronic configuration noted $1s^{2}$. A physical ket is obtained by anti symetrisation of vector:
$|1:n,l,m_{l},m_{s}>\otimes |2:n,l,m_{l},m_{s}>$

### Spectral terms

Let us write exact hamiltonian $H$ as:
$H=H_{0}+T_{1}+T_{2}$
where $T_{1}$ represents a correction to $H_{0}$ due to the interactions between electrons. Solving of spectral problem associated to $H_{1}=H_{0}+T_{1}$ using perturbative method is now presented.
Remark:
It is here assumed that $T_{2}<<T_{1}$. This assumption is called $L$--$S$ coupling approximation.
To diagonalize $T_{1}$ in the space spanned by the eigenvectors of $H_{0}$, it is worth to consider problem's symmetries in order to simplify the spectral problem. It can be shown that operators $L^{2}$, $L_{z}$, $S^{2}$ and $S_{z}$ form a complete set of observables that commute.
Example:
Consider again the helium atom ([ph:mecaq:Cohen73]). From the symmetries of the problem, the basis chosen is:
$|1:n_{1},l_{1};2:n_{2},l_{2};L,m_{L}>\otimes |S,m_{S}>$
where $L$ is the quantum number associated to the total kinetic moment\index{kinetic moment}:
$L\in \{l_{1}+l_{2},l_{1}+l_{2}-1,\dots ,|l_{1}-l_{2}|\}$
and $S$ is the quantum number associated to total spin of the system\index{spin}:
$S\in \{0,1\}$
Moreover, one has:
$m_{L}=m_{l_{1}}+m_{l_{2}}$
and
$m_{S}=m_{s_{1}}+m_{s_{2}}$
Table Tab. tabpauli represents in each box the value of $m_{L}m_{S}$ for all possible values of $m_{L}$ and $m_{S}$. \begin{table}[hbt]
tabpauli
Theorem:
theopair
For an atom with two electrons, states such that $L+S$ is odd are excluded.
Proof:
We will proof this result using symmetries. We have:
${\begin{matrix}|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}\\&&=\sum _{m}\sum _{m'}{\mathrel {<}}l,l',m,m'|L,M_{L}{\mathrel {>}}|1:n,l,m;2:n',l',m'{\mathrel {>}}\end{matrix}}$
Coefficients ${\mathrel {<}}l,l',m,m'|L,M_{L}{\mathrel {>}}$ are called Glebsh-Gordan\index{Glesh-Gordan coefficients} coefficients. If $l=l'$, it can be shown (see ([ph:mecaq:Cohen73]) that:
${\mathrel {<}}l,l,m,m'|L,M_{L}{\mathrel {>}}=(-1)^{L}{\mathrel {<}}l,l,m',m|L,M_{L}{\mathrel {>}}.$
Action of $P_{21}$ on $|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}$ can thus be written:
$P_{21}|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}=(-1)^{L}|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}$
Physical ket obtained is:
${\begin{matrix}|n,l,n,l;L,M_{L};S,M_{S}{\mathrel {>}}\\&&=\left\{{\begin{array}{ll}0&{\mbox{ if }}L+S{\mbox{ is odd }}\\|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}\otimes |S,M_{S}{\mathrel {>}}&{\mbox{ if }}L+S{\mbox{ is even }}\end{array}}\right.\end{matrix}}$

### Fine structure levels

Finally spectral problem associated to
$H=H_{0}+T_{1}+T_{2}$
can be solved considering $T_{2}$ as a perturbation of $H_{1}=H_{0}+T_{1}$. It can be shown ([ph:atomi:Cagnac71]) that operator $T_{2}$ can be written $T_{2}=\xi (r_{i}){\vec {l}}_{i}{\vec {s}}_{i}$. It can also be shown that operator ${\vec {J}}={\vec {L}}+{\vec {S}}$ commutes with $T_{2}$. Operator $T_{2}$ will have thus to be diagonilized using eigenvectors $|J,m_{J}>$ common to operators $J_{z}$ and $J^{2}$. each state is labelled by:
$^{2S+1}L_{J}$
where $L,S,J$ are azimuthal quantum numbers associated with operators ${\vec {L}},{\vec {S}},{\vec {J}}$.