# 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

${\displaystyle V(r)=-{\frac {e}{r^{2}}}}$

It can be shown that eigenvalues of hamiltonian ${\displaystyle H}$ with central potential depend in general on two quantum numbers ${\displaystyle k}$ and ${\displaystyle l}$, but that for particular potential given by equation eqpotcenhy, eigenvalues depend only on sum ${\displaystyle n=k+l}$.

## 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:

${\displaystyle -[{\frac {\hbar ^{2}}{2\mu }}\Delta +V(r)]\phi (r)=E\phi (r).}$

Laplacian operator can be expressed as a function of ${\displaystyle L^{2}}$ operator.

Theorem: Laplacian operator ${\displaystyle \Delta }$ can be written as:

${\displaystyle \Delta =-{\frac {1}{r^{2}}}L^{2}+{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}}$

Proof: Here, tensorial notations are used (Einstein convention). By definition:

${\displaystyle L_{i}=\epsilon _{ijk}x_{j}p_{k}}$

So:

${\displaystyle {\begin{matrix}L_{i}L_{i}&=&\epsilon _{ijk}\epsilon _{ilm}x_{j}p_{k}x_{l}p_{m}\\&=&(\delta _{jl}\delta _{km}-\delta _{jm}\delta {kl})x_{j}p_{k}x_{l}p_{m}\\&=&x_{j}x_{j}p_{k}p_{k}-x_{j}p_{k}x_{k}p_{j}\end{matrix}}}$

The writing order of the operators is very important because operator do not commute. They obey following commutation relations:

${\displaystyle [x_{i},p_{j}]=i\hbar \delta _{ij}}$

${\displaystyle [x_{j},x_{k}]=0}$

${\displaystyle [p_{j},p_{k}]=0}$

From equation eqdefmomP, we have:

${\displaystyle p_{k}=-i\hbar {\frac {\partial }{\partial x_{k}}}}$

thus

${\displaystyle x_{j}x_{j}p_{k}p_{k}=-x^{2}\hbar ^{2}\Delta .}$

Now,

${\displaystyle {\begin{matrix}x_{j}p_{k}x_{k}p_{j}&=&[i\hbar \delta _{ik}+p_{k}x_{j}]x_{k}p_{j}\\&=&i\hbar x_{k}p_{k}+p_{k}x_{j}x_{k}p_{j}\\&=&i\hbar x_{k}p_{k}+p_{k}x_{k}x_{j}p_{j}\end{matrix}}}$

Introducing operator:

${\displaystyle {\tilde {D}}=x_{k}{\frac {\partial }{\partial x_{k}}}}$

we get the relation:

${\displaystyle {IMP/label|eql2pri}L^{2}=-x^{2}\Delta +{\tilde {D}}^{2}+{\tilde {D}}}$

Using spherical coordinates, we get:

${\displaystyle {\tilde {D}}=r{\frac {\partial }{\partial r}}}$

and

${\displaystyle {\tilde {D}}^{2}=(r{\frac {\partial }{\partial r}})(r{\frac {\partial }{\partial r}})=r^{2}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\partial }{\partial r}}}$

So, equation eql2pri becomes:

${\displaystyle \Delta =-{\frac {1}{r^{2}}}L^{2}+{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}}$

Let us use the problem's symmetries:

Since:

• ${\displaystyle L_{z}}$ commutes with operators acting on ${\displaystyle r}$
• ${\displaystyle L_{z}}$ commutes with ${\displaystyle L^{2}}$ operator ${\displaystyle L_{z}}$ commutes with ${\displaystyle H}$
• ${\displaystyle L^{2}}$ commutes with ${\displaystyle H}$

we look for a function ${\displaystyle \phi }$ that diagonalizes simultaneously ${\displaystyle H,L^{2},L_{z}}$ that is such that:

${\displaystyle {\begin{matrix}H\phi (r)&=&E\phi (r)\\L^{2}\phi (r)&=&l(l+1)\hbar ^{2}\phi (r)\\L_{z}\phi (r)&=&m\hbar \phi (r)\end{matrix}}}$

Spherical harmonics ${\displaystyle Y_{l}^{m}(\theta ,\phi )}$ can be introduced now:

Definition:

Spherical harmonics ${\displaystyle Y_{l}^{m}(\theta ,\phi )}$ are eigenfunctions common to operators ${\displaystyle L^{2}}$ and ${\displaystyle L_{z}}$. It can be shown that:

${\displaystyle {\begin{matrix}L^{2}Y_{l}^{m}&=&l(l+1)Y_{l}^{m}\\L_{z}Y_{l}^{m}&=&mY_{l}^{m}\end{matrix}}}$

Looking for a solution ${\displaystyle \phi (r)}$ that can\footnote{Group theory argument should be used to prove that solution actually are of this form.} be written (variable separation):

${\displaystyle \phi (r)=R(r)Y_{l}^{m}(\theta ,\phi )}$

problem becomes one dimensional:

eqaonedimrr

${\displaystyle -[{\frac {\hbar ^{2}}{2\mu }}({\frac {d^{2}}{dr^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}})+{\frac {l(l+1)}{2\mu r^{2}}}\hbar ^{2}+V(r)]R_{l}(r)=E_{kl}R_{l}(r)}$

where ${\displaystyle R(r)}$ is indexed by ${\displaystyle l}$ only. Using the following change of variable: ${\displaystyle R_{l}(r)={\frac {1}{r}}u_{l}(r)}$, one gets the following spectral equation:

${\displaystyle -[{\frac {\hbar ^{2}}{2\mu }}{\frac {d^{2}}{dr^{2}}}+V_{e}(r)]u_{kl}(r)=E_{kl}u_{kl}(r)}$

where

${\displaystyle V_{e}(r)={\frac {l(l+1)}{2\mu r^{2}}}\hbar ^{2}+V(r)}$

The problem is then reduced to the study of the movement of a particle in an effective potential ${\displaystyle V_{e}(r)}$. To go forward in the solving of this problem, the expression of potential ${\displaystyle V(r)}$ is needed. Particular case of hydrogen introduced at section sechydrog corresponds to a potential ${\displaystyle V(r)}$ proportional to ${\displaystyle 1/r}$ 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:

${\displaystyle H=\sum _{i}-{\frac {\hbar ^{2}}{2m}}\Delta _{i}-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r_{i}}}+\sum _{j{\mathrel {>}}i}{\frac {1}{4\pi \epsilon _{0}}}{\frac {e^{2}}{r_{ij}}}+T_{2}}$

where ${\displaystyle T_{2}}$ 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

${\displaystyle -{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r_{i}}}+\sum _{j{\mathrel {>}}i}{\frac {1}{4\pi \epsilon _{0}}}{\frac {e^{2}}{r_{ij}}}}$

is modelized by the sum ${\displaystyle \sum W(r_{i})}$, where ${\displaystyle W(r_{i})}$ is the mean potential acting on particle ${\displaystyle i}$. The hamiltonian can thus be written:

${\displaystyle H_{0}=\sum _{i}h_{i}}$

where ${\displaystyle h_{i}=-{\frac {\hbar ^{2}}{2m}}\Delta _{i}+W(r_{i})}$.

Remark:

More precisely, ${\displaystyle h_{i}}$ is the linear operator acting in the tensorial product space ${\displaystyle \otimes _{i=1}^{N}E_{i}}$ and defined by its action on function that are tensorial products:

${\displaystyle [1_{1}\otimes \dots \otimes 1_{i-1}\otimes h_{i}\otimes 1_{i+1}\dots \otimes 1_{N}](\phi _{1}\otimes \dots \phi _{N})=h_{i}(\phi _{1})\otimes \dots \phi _{N}}$

It is then sufficient to solve the spectral problem in a space ${\displaystyle E_{i}}$ for operator ${\displaystyle h_{i}}$. 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 ${\displaystyle W(r_{i})}$ is not like ${\displaystyle 1/r}$ as in the


hydrogen atom case and thus the accidental degeneracy is not observed here. The energy depends on two quantum numbers ${\displaystyle l}$ (relative to kinetic moment) and ${\displaystyle n}$ (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 ${\displaystyle 1s^{2}}$. A physical ket is obtained by anti symetrisation of vector:

${\displaystyle |1:n,l,m_{l},m_{s}>\otimes |2:n,l,m_{l},m_{s}>}$

### Spectral terms

Let us write exact hamiltonian ${\displaystyle H}$ as:

${\displaystyle H=H_{0}+T_{1}+T_{2}}$

where ${\displaystyle T_{1}}$ represents a correction to ${\displaystyle H_{0}}$ due to the interactions between electrons. Solving of spectral problem associated to ${\displaystyle H_{1}=H_{0}+T_{1}}$ using perturbative method is now presented.

Remark:

It is here assumed that ${\displaystyle T_{2}<. This assumption is called ${\displaystyle L}$--${\displaystyle S}$ coupling approximation.

To diagonalize ${\displaystyle T_{1}}$ in the space spanned by the eigenvectors of ${\displaystyle H_{0}}$, it is worth to consider problem's symmetries in order to simplify the spectral problem. It can be shown that operators ${\displaystyle L^{2}}$, ${\displaystyle L_{z}}$, ${\displaystyle S^{2}}$ and ${\displaystyle S_{z}}$ 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:

${\displaystyle |1:n_{1},l_{1};2:n_{2},l_{2};L,m_{L}>\otimes |S,m_{S}>}$

where ${\displaystyle L}$ is the quantum number associated to the total kinetic moment\index{kinetic moment}:

${\displaystyle L\in \{l_{1}+l_{2},l_{1}+l_{2}-1,\dots ,|l_{1}-l_{2}|\}}$

and ${\displaystyle S}$ is the quantum number associated to total spin of the system\index{spin}:

${\displaystyle S\in \{0,1\}}$

Moreover, one has:

${\displaystyle m_{L}=m_{l_{1}}+m_{l_{2}}}$

and

${\displaystyle m_{S}=m_{s_{1}}+m_{s_{2}}}$

Table Tab. tabpauli represents in each box the value of ${\displaystyle m_{L}m_{S}}$ for all possible values of ${\displaystyle m_{L}}$ and ${\displaystyle m_{S}}$. \begin{table}[hbt]

tabpauli

Theorem:

theopair

For an atom with two electrons, states such that ${\displaystyle L+S}$ is odd are excluded.

Proof:

We will proof this result using symmetries. We have:

${\displaystyle {\begin{matrix}|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}\\&&=\sum _{m}\sum _{m'}{\mathrel {<}}l,l',m,m'|L,M_{L}{\mathrel {>}}|1:n,l,m;2:n',l',m'{\mathrel {>}}\end{matrix}}}$

Coefficients ${\displaystyle {\mathrel {<}}l,l',m,m'|L,M_{L}{\mathrel {>}}}$ are called Glebsh-Gordan\index{Glesh-Gordan coefficients} coefficients. If ${\displaystyle l=l'}$, it can be shown (see ([ph:mecaq:Cohen73]) that:

${\displaystyle {\mathrel {<}}l,l,m,m'|L,M_{L}{\mathrel {>}}=(-1)^{L}{\mathrel {<}}l,l,m',m|L,M_{L}{\mathrel {>}}.}$

Action of ${\displaystyle P_{21}}$ on ${\displaystyle |1:n,l;2:n,l';L,M_{L}{\mathrel {>}}}$ can thus be written:

${\displaystyle P_{21}|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}=(-1)^{L}|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}}$

Physical ket obtained is:

${\displaystyle {\begin{matrix}|n,l,n,l;L,M_{L};S,M_{S}{\mathrel {>}}\\&&=\left\{{\begin{array}{ll}0&{\mbox{ if }}L+S{\mbox{ is odd }}\\|1:n,l;2:n,l';L,M_{L}{\mathrel {>}}\otimes |S,M_{S}{\mathrel {>}}&{\mbox{ if }}L+S{\mbox{ is even }}\end{array}}\right.\end{matrix}}}$

### Fine structure levels

Finally spectral problem associated to

${\displaystyle H=H_{0}+T_{1}+T_{2}}$

can be solved considering ${\displaystyle T_{2}}$ as a perturbation of ${\displaystyle H_{1}=H_{0}+T_{1}}$. It can be shown ([ph:atomi:Cagnac71]) that operator ${\displaystyle T_{2}}$ can be written ${\displaystyle T_{2}=\xi (r_{i}){\vec {l}}_{i}{\vec {s}}_{i}}$. It can also be shown that operator ${\displaystyle {\vec {J}}={\vec {L}}+{\vec {S}}}$ commutes with ${\displaystyle T_{2}}$. Operator [/itex]T_2[/itex] will have thus to be diagonilized using eigenvectors ${\displaystyle |J,m_{J}>}$ common to operators ${\displaystyle J_{z}}$ and ${\displaystyle J^{2}}$. each state is labelled by:

${\displaystyle ^{2S+1}L_{J}}$

where ${\displaystyle L,S,J}$ are azimuthal quantum numbers associated with operators ${\displaystyle {\vec {L}},{\vec {S}},{\vec {J}}}$.