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 $n=k+l$ .

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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:

$L_i=\epsilon_{ijk}x_jp_k$

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.

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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:

$H_0=\sum_ih_i$

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  is not like  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 $m_Lm_S$ 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 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 </math>T_2</math> will have thus to be diagonilized using eigenvectors $|J,m_J>$ common to operators $J_z$ and $J^2$ . each state is labelled by: