# Introduction to Mathematical Physics/N body problem in quantum mechanics/Atoms

## One nucleus, one electronEdit

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 invarianceEdit

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:

$L_i=\epsilon_{ijk}x_jp_k$

So:

$\begin{matrix} L_iL_i&=&\epsilon_{ijk}\epsilon_{ilm}x_{j}p_k x_l p_m\\ &=&(\delta_{jl}\delta_{km}-\delta_{jm}\delta{kl})x_{j}p_kx_l p_m\\ &=&x_jx_jp_kp_k-x_jp_kx_kp_j \end{matrix}$

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

$x_jx_jp_kp_k=-x^2\hbar^2\Delta.$

Now,

$\begin{matrix} x_jp_kx_kp_j&=&[i\hbar\delta_{ik}+p_kx_j]x_kp_j\\ &=&i\hbar x_kp_k+p_kx_jx_kp_j\\ &=&i\hbar x_kp_k+p_kx_kx_jp_j \end{matrix}$

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 $Y^m_l(\theta,\phi)$ can be introduced now:

Definition:

Spherical harmonics $Y^m_l(\theta,\phi)$ are eigenfunctions common to operators $L^2$ and $L_z$. It can be shown that:

$\begin{matrix} L^2Y^m_l&=&l(l+1)Y^m_l\\ L_zY^m_l&=&mY^m_l \end{matrix}$

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):

$\phi(r)=R(r)Y^m_l(\theta,\phi)$

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 electronsEdit

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 electronsEdit

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:

$H_0=\sum_ih_i$

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 termsEdit

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 $m_Lm_S$ 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 levelsEdit

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 [/itex]T_2[/itex] 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 $\vec L,\vec S,\vec J$.