# 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}< . 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 ${\vec {L}},{\vec {S}},{\vec {J}}$ .