# Contents

Few and Many Particle Systems
5.1  Bosons and Fermions
5.1.1  Boson Basis
5.1.2  Fermion Basis
5.2  The Helium Atom
5.2.1  Perturbation Theory
5.2.2  Variational Calculation
5.3  The H ${}_{2}$ Molecule
5.3.1  Hamilton Operator
5.3.2  Born-Oppenheimer Approximation
5.4  Brief Digression: Variational Approximations
5.5  Second quantization
Index

# List of Figures

1  Ground state wave function of He
2  Excited states of He
3  Pair correlation function
4  Level scheme of He
He ionization energies
$\left(Z-2\right)\tmspace+.1667\mathrm{em}\to \left(Z-1\right)$ ionization energies of several elements
7  Geometry and labels for the H ${}_{2}$ molecule

## 5Few and Many Particle Systems

In this section we first focus on two of the most important few particle quantum systems, namely an atom containing more than one electron and a simple molecule. Second, we set up the necessary tools to describe quantum systems with many particles.

### 5.1Bosons and Fermions

Let
 $\stackrel{~}{H}=\sum _{i}^{N}{h}_{i}$ $\left(1\right)$
be the Hamiltonian of $N$ particles acting in the Hilbert space $H$ and ${h}_{i}$ is a Hamiltonian acting on a single particle $i$, i.e. $\left[{h}_{i},{h}_{j}\right]=0, \forall i,j$. The matrix elements of ${h}_{i}$ are identical for each sub Hilbert space ${H}_{i}$ of particle $i$, i.e the single-particle Hamiltonians are 'identical'. Let $|\mu ,i⟩$be a orthonormal eigensystem (ONS) of ${h}_{i}$ in ${H}_{i}$
 $\begin{array}{ccc}\multicolumn{1}{c}{{h}_{i}|\mu ,i⟩}& =\hfill & {\epsilon }_{\mu }|\mu ,i⟩ ,\hfill \\ \multicolumn{1}{c}{⟨\mu ,i|\nu ,i⟩}& =\hfill & {\delta }_{\mu \nu }\hfill & \hfill \left(2\right)\end{array}$

then an ONS of $H$ in $H$ is
 $\begin{array}{ccc}\multicolumn{1}{c}{|\mu ⟩}& \equiv \hfill & |{\mu }_{1},1⟩\otimes \dots \otimes |{\mu }_{N},N⟩\equiv |{\mu }_{1},\dots {\mu }_{N}⟩\hfill \\ \multicolumn{1}{c}{\stackrel{~}{H}|\mu ⟩}& =\hfill & {E}_{\mu }|\mu ⟩\hfill \\ \multicolumn{1}{c}{{E}_{\mu }}& =\hfill & \sum _{i=1}^{N}{\epsilon }_{{\mu }_{1}}\hfill & \hfill \left(3\right)\end{array}$

Let $\left\{{A}_{l}\right\}$ be the complete set of commuting observables in each ${H}_{i}$. To be specific assume spin- $1/2$ particles with space coordinate ${\stackrel{^}{r}}_{i}$ and spin observable ${\stackrel{^}{\sigma }}_{i}=\left({\stackrel{^}{\sigma }}_{\mathrm{ix}},{\stackrel{^}{\sigma }}_{\mathrm{iy}},{\stackrel{^}{\sigma }}_{\mathrm{iz}}\right)$. I.e.
 $\begin{array}{ccc}\multicolumn{1}{c}{{\stackrel{^}{r}}_{i}|{r}_{i}⟩}& =\hfill & {r}_{i}|{r}_{i}⟩\hfill \\ \multicolumn{1}{c}{{\stackrel{^}{\sigma }}_{i}^{2}|\sigma ,{\sigma }_{\mathrm{iz}}⟩}& =\hfill & \sigma \left(\sigma +1\right)|\sigma ,{\sigma }_{\mathrm{iz}}⟩=\frac{3}{4}|\sigma ,{\sigma }_{\mathrm{iz}}⟩\hfill \\ \multicolumn{1}{c}{{\stackrel{^}{\sigma }}_{\mathrm{iz}}|\sigma ,{\sigma }_{\mathrm{iz}}⟩}& =\hfill & {\sigma }_{\mathrm{iz}}|\sigma ,{\sigma }_{\mathrm{iz}}⟩\hfill & \hfill \left(4\right)\end{array}$

Since the spin $=1/2$ for each particle we may drop the principal quantum number $\sigma$ for the remainder an we may simplify $|\sigma ,{\sigma }_{\mathrm{iz}}⟩\to |{\sigma }_{i}⟩$ with ${\sigma }_{i}=±1/2$. As usual set of eigenvalues $\left({r}_{i},{\sigma }_{i}\right)$ are called the coordinates of the particles. We abbreviate $\left({r}_{i},{\sigma }_{i}\right)\equiv 1$. As usual we may
represent the ONS in $H$ in these coordinates
 $\begin{array}{cccc}\multicolumn{1}{c}{{\Phi }_{\mu }\left(1,\dots N\right)}& =\hfill & ⟨1,\dots N|\mu ⟩=⟨1|{\mu }_{1},1⟩⟨2|{\mu }_{2},2⟩\dots ⟨N|{\mu }_{N},N⟩\hfill & \hfill \left(5\right)\\ \multicolumn{1}{c}{}& =\hfill & {\phi }_{{\mu }_{1}}\left(1\right){\phi }_{{\mu }_{2}}\left(2\right)\dots {\phi }_{{\mu }_{N}}\left(N\right)\equiv {\phi }_{1}\left(1\right){\phi }_{2}\left(2\right)\dots {\phi }_{2}\left(N\right)\hfill & \hfill \left(6\right)\end{array}$

Where ${\phi }_{{\mu }_{1}}\left(1\right)\equiv {\phi }_{1}\left(1\right)$ is the coordinate representation of the single particle state $|{\mu }_{1},1⟩$, or single particle wave function. Again, an abbreviation ${\mu }_{1}\to 1$ can be used wherever appropriate. Note, that in general with this abbreviation any combination ${\phi }_{l}\left(m\right)$ can occur where $l$ and $m$ are completely unrelated: the former labels the states, the latter the coordinates. I.e the particular state in eqn. (6) is read: "particle 1 with coordinates $1$ is in state $1$, a.s.o. ...".
Now, let $\left\{{O}_{l}\right\}$ be the set of all physical observables in $H$ and let $P$ be any permutation
 $\stackrel{^}{P}\left\{1,2,\dots N\right\}=\left\{{P}_{1},{P}_{2},\dots {P}_{N}\right\}.$ $\left(7\right)$
The $N$ particles are said to be
identical if
 $⟨{\mu }_{1},\dots {\mu }_{N}|{O}_{l}|{\mu }_{1},\dots {\mu }_{N}⟩=⟨{\mu }_{{P}_{1}},\dots {\mu }_{{P}_{N}}|{O}_{l}|{\mu }_{{P}_{1}},\dots {\mu }_{{P}_{N}}⟩$ $\left(8\right)$
for all ${O}_{l}$ and $P$. I.e., physical observations are invariant under particle interchange. For that it is sufficient (and necessary) that
 $|{\mu }_{{P}_{1}},\dots {\mu }_{{P}_{N}}⟩=c|{\mu }_{1},\dots {\mu }_{N}⟩ ,\mathrm{with}\mathrm{ }c={e}^{i\alpha }, \alpha \in \mathbbR$ $\left(9\right)$
In general the Hamiltonian $H$ of an $N$-particle system and the set of all physical observables $\left\{{O}_{l}\right\}$ in $H$ is not decomposable as in eqn. (1), however we may assume that
 $H, \left\{{O}_{l}\right\}=F\left(\left\{{A}_{1l}\right\},\left\{{A}_{2l}\right\},\dots \left\{{A}_{\mathrm{Nl}}\right\}\right)$ $\left(10\right)$
i.e. the Hamiltonian and the observables can be expressed as functions of the complete set of observables of each single particle Hilbert space. Then, for
identical particles we must have
 $H /\left\{{O}_{l}\right\} \left(\left\{{A}_{1l}\right\},\left\{{A}_{2l}\right\},\dots \left\{{A}_{\mathrm{Nl}}\right\}\right)=H /\left\{{O}_{l}\right\} \left(\left\{{A}_{{P}_{1}l}\right\},\left\{{A}_{{P}_{2}l}\right\},\dots \left\{{A}_{{P}_{N}l}\right\}\right)$ $\left(11\right)$
e.g. the Coulomb interaction or the kinetic energy
 $~\sum _{l\ne m}\frac{1}{|{r}_{l}-{r}_{m}|} ,\mathrm{ }\sum _{l}\frac{\hslash {}^{2}{p}_{l}^{2}}{2m}$ $\left(12\right)$
I.e. the Hamiltonian commutes with the
permutation operator
 $\left[H,\stackrel{^}{P}\right]=0$ $\left(13\right)$
i.e. instead of only eqn. (5), the eigenfunctions ${\Phi }_{\mu }\left(1,\dots N\right)$ can additionally be chosen to be eigenstates with eigenvalue $P$ for any permutation operator
 $\stackrel{^}{P}{\Phi }_{\mu }\left(1,\dots N\right)=P{\Phi }_{\mu }\left(1,\dots N\right)$ $\left(14\right)$
Each permutation is a product of
transpositions ${\stackrel{^}{\tau }}_{\mathrm{ij}}$ with ${\stackrel{^}{\tau }}_{\mathrm{ij}}{\Phi }_{\mu }\left(1..i..j..N\right)={\Phi }_{\mu }\left(1..j..i..N\right)$. Since ${\stackrel{^}{\tau }}_{\mathrm{ij}}^{2}=1$, the eigenvalues of ${\stackrel{^}{\tau }}_{\mathrm{ij}}$ are ${\tau }_{\mathrm{ij}}=±1$. Therefore $P=\Pi \left(±1\right)=±1$.
$P$ does not depend on the permutation and is either $+1$ or $-1$, the so-called parity under permutation, for any given ${\Phi }_{\mu }\left(1..i..j..N\right)$. In fact, assume for eg. 3 particles with ${\tau }_{12}=-1$ and ${\tau }_{\mathrm{ij}}=+1$ for all other $i,j$, then
 $\begin{array}{ccc}\multicolumn{1}{c}{-{\Phi }_{\mu }\left(2,1,3\right)}& =\hfill & {\Phi }_{\mu }\left(1,2,3\right)={\Phi }_{\mu }\left(3,2,1\right)={\Phi }_{\mu }\left(2,3,1\right)={\Phi }_{\mu }\left(2,1,3\right)\hfill \\ \multicolumn{1}{c}{⇒\mathrm{ }{\Phi }_{\mu }\left(1,2,3\right)}& =\hfill & 0\hfill & \hfill \left(15\right)\end{array}$

An analogous procedure applies to any number of particles.
For identical particles eqn. (9) is a conditio sine qua non, however eqn. (14) is not. One of the main points is, that as of today almost all known quantum systems of identical particles obey the more restrictive condition eqn. (14). The corresponding particles are called
 $P=\left\{\begin{array}{c}\hfill 1\mathrm{ }\mathrm{Bosons}\hfill \\ \hfill -1\mathrm{ }\mathrm{Fermions}\hfill \end{array}$ $\left(16\right)$
A famous example of eqn. (9) but not eqn. (14) are the excitations in the lowest Landau level of a 2D electron gases in strong magnetic fields
at filling fractions $\nu =1/m, m\in \mathbbN$ (fractional quantum Hall effect). They obey $\alpha =\pi /m$. Such particles have a plethora of names: anyons, fractons, semions,... Another famous example are electrons in one-dimension. Here it turns out, that there is no difference between Fermions and Bosons and the concept of Fermi liquids to describe many-electron systems is not valid anymore. It is replaced by the concept of a Luttinger liquid.
Exercise 1 A model for particles with fractional statistics.
a) Consider a boson with mass $m$ and a 'strange' charge $q$ circling around a thin solenoid in $z$-direction.The charge $q$ is assumed to be proportional to the flux $\Phi$
 $q=C\Phi \mathrm{ }C\in \mathbbR$ $\left(17\right)$
Imagine the flux to occur by increasing the magnetic field $B$ through the solenoid, starting at $B\left(t=0\right)=0$ where the angular momentum ${l}_{z}$ of the particle is zero. Show that
 ${\stackrel{·}{l}}_{z}=-\frac{C\Phi \stackrel{·}{\Phi }}{2}\mathrm{ }i.e.\mathrm{ }{l}_{z}=-\frac{q\Phi }{4\pi }$ $\left(18\right)$
Now, view the boson and its solenoid as a '
composite particle'.
b) Consider two identical composite particles as in a). Show that in terms of the relative coordinate $r={r}_{1}-{r}_{2}$, momentum $p=\left({p}_{1}-{p}_{2}\right)/2$ and the center of mass momentum $P={p}_{1}+{p}_{2}$ the Hamiltonian of the two-particle system is
 $H=\frac{{P}^{2}}{4m}+\frac{\left(p-q{A}_{{r}_{1}-{r}_{2}}{\right)}^{2}}{m}$ $\left(19\right)$
[Hint: The vector potential of a thin solenoid is $A=±\Phi \left({e}_{z}×r/{r}^{2}\right)/2\pi$]

c) The total wave function for two bosons is $\Psi \left(R,r\right)=\phi \left(R\right)\psi \left(r,\Theta \right)$, where $R=\left({r}_{1}+{r}_{2}\right)/2$ is the center of mass and $\left(r,\Theta \right)$ is the relative coordinate in cylindrical coordinates w.r.t. the solenoid direction. What is the relation
 $\psi \left(r,\Theta +\pi \right)\overset?↔\psi \left(r,\Theta \right)$ $\left(20\right)$
d) Using the singular (why?) gauge transformation $\stackrel{~}{A}=A-\nabla \chi$, with $\chi =\Phi \Theta /2\pi$, show that
 $\begin{array}{ccc}\multicolumn{1}{c}{\stackrel{~}{H}}& =\hfill & \frac{{P}^{2}}{4m}+\frac{p{}^{2}}{m}\hfill \\ \multicolumn{1}{c}{\stackrel{~}{\psi }\left(r,\Theta +\pi \right)}& =\hfill & {e}^{-\mathrm{iq}\Phi /2}\stackrel{~}{\psi }\left(r,\Theta \right)\hfill & \hfill \left(21\right)\end{array}$

I.e. the composite particles
interacting via ${A}_{{r}_{1}-{r}_{2}}$ can be transformed into non-interacting particles with fractional statistics $\alpha =q\Phi /2$.

#### 5.1.1Boson Basis

For $P=1$ we need ${\Phi }_{\mu }^{B}\left(1,\dots N\right)$ to be totally symmetric

 ${\Phi }_{\mu }^{B}\left(1,\dots N\right)=C\sum _{P\text{'}}{\phi }_{P\text{'}1}\left(1\right){\phi }_{P\text{'}2}\left(2\right)\dots {\phi }_{P\text{'}N}\left(N\right)$ $\left(22\right)$
where $P\text{'}$ are all
distinct permutations. Normalization
 $C=\sqrt{\frac{{n}_{1}!{n}_{2}!...{n}_{N}!}{N!}}$ $\left(23\right)$
where ${n}_{i}$ is the
occupation number of state $i$.
Exercise 2 Prove eqn. (23)
Exmpl.: ${\Phi }_{\mu }\left(1,2,3\right)={\phi }_{\alpha }\left(1\right){\phi }_{\alpha }\left(2\right){\phi }_{\beta }\left(3\right)$ and $\sqrt{2!1!/3!}=1/\sqrt{3}$
 ${\Phi }_{\mu }^{B}\left(1,2,3\right)=\frac{1}{\sqrt{3}}\left[{\phi }_{\alpha }\left(1\right){\phi }_{\alpha }\left(2\right){\phi }_{\beta }\left(3\right)+{\phi }_{\alpha }\left(3\right){\phi }_{\alpha }\left(2\right){\phi }_{\beta }\left(1\right)+{\phi }_{\alpha }\left(1\right){\phi }_{\alpha }\left(3\right){\phi }_{\beta }\left(2\right)\right]$ $\left(24\right)$

#### 5.1.2Fermion Basis

For $P=-1$ we need ${\Phi }_{\mu }^{F}\left(1,\dots N\right)$ to be totally antisymmetric
 $\begin{array}{ccc}\multicolumn{1}{c}{{\Phi }_{\mu }^{F}\left(1,\dots N\right)}& =\hfill & \frac{1}{\sqrt{N}}\sum _{P}\mathrm{sign}\left(P\right) {\phi }_{P1}\left(1\right){\phi }_{P\text{'}2}\left(2\right)\dots {\phi }_{\mathrm{PN}}\left(N\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & \frac{1}{\sqrt{N}}det\left[\begin{array}{ccc}\hfill {\phi }_{1}\left(1\right)\hfill & \hfill \dots \hfill & \hfill {\phi }_{1}\left(N\right)\hfill \\ \hfill :\hfill & \hfill \ddots \hfill & \hfill :\hfill \\ \hfill {\phi }_{N}\left(1\right)\hfill & \hfill \dots \hfill & \hfill {\phi }_{N}\left(N\right)\hfill \end{array}\right]\hfill & \hfill \left(25\right)\end{array}$

Exercise 3 Prove the normalization $1/\sqrt{N!}$ in eqn. (25)
This wave function is called a Slater-determinant. A consequence is Pauli's exclusion principle: no two Fermions can be in an identical one-particle state. Otherwise the determinant eqn. (25) vanishes.
Exmpl.: ${\Phi }_{\mu }\left(1,2\right)={\phi }_{\alpha }\left(1\right){\phi }_{\beta }\left(2\right)$
 ${\Phi }_{\mu }^{F}\left(1,2\right)=\frac{1}{\sqrt{2}}\left[{\phi }_{\alpha }\left(1\right){\phi }_{\beta }\left(2\right)-{\phi }_{\alpha }\left(2\right){\phi }_{\beta }\left(1\right)\right]$ $\left(26\right)$
Obviously this is zero if $\alpha =\beta$.

### 5.2The Helium Atom

We now apply some of the concepts of the last section to a very simple many body system. Namely, the two-electron elements or ions ${\mathrm{Li}}^{+}$, $\mathrm{He}$, and ${H}^{-}$ with nuclear charge $Z=3$, 2, and 1. Consider only Coulomb interaction (i.e. no relativistic phenomena like $\mathrm{LS}$-coupling)
 $H=\frac{1}{2m}\left({p}_{1}^{2}+{p}_{2}^{2}\right)-{\mathrm{Ze}}^{2}\left(\frac{1}{|{r}_{1}|}+\frac{1}{|{r}_{2}|}\right)+\frac{{e}^{2}}{{r}_{12}}$ $\left(27\right)$
with ${r}_{12}=|{r}_{1}-{r}_{2}|$. The eigenstates are
 $\begin{array}{ccc}\multicolumn{1}{c}{|q,\chi ⟩}& =\hfill & |q⟩\otimes |\chi ⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & |\mathrm{real} \mathrm{space}⟩\otimes |\mathrm{spin} \mathrm{space}⟩\hfill & \hfill \left(28\right)\end{array}$

with $H|q⟩={\epsilon }_{q}|q⟩$, and $|\chi ⟩$
any state acted upon by the spin operators ${\stackrel{^}{\sigma }}_{1,2}$ of the two electrons. Coordinate representation of $|q,\chi ⟩$
 $\Psi \left(1,2\right)=⟨1,2|q,\chi ⟩=⟨{r}_{1}{r}_{2}|q⟩⟨{\sigma }_{1}{\sigma }_{2}|\chi ⟩=\Phi \left({r}_{1},{r}_{2}\right)\chi \left({\sigma }_{1},{\sigma }_{2}\right)$ $\left(29\right)$
Because of the Pauli principle $\Psi \left(1,2\right)=-\Psi \left(2,1\right)$, we have two options
 $\begin{array}{ccc}\hfill \Phi \left({r}_{1},{r}_{2}\right)=-\Phi \left({r}_{1},{r}_{2}\right)\hfill & \hfill \hfill & \hfill \chi \left({\sigma }_{1},{\sigma }_{2}\right)=+\chi \left({\sigma }_{1},{\sigma }_{2}\right)\hfill \\ \hfill \hfill \\ \hfill " =+ "\hfill & \hfill \hfill & \hfill " =- "\hfill \end{array}$ $\left(30\right)$
or using S(A)=(anti)symmetric
 $\Psi =\left\{\begin{array}{c}\hfill {\Phi }^{A}{\chi }^{S}\hfill \\ \hfill {\Phi }^{S}{\chi }^{A}\hfill \end{array}$ $\left(31\right)$
Spin space wave-functions
The single-particle spin-1/2 spinors (spin-wave functions) form a two-dimensional space $|↑⟩$ and $|↓⟩$. Their $\sigma$-representation is given by
 $⟨\sigma |↑⟩\equiv \alpha \left(\sigma \right)=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]\mathrm{ }⟨\sigma |↓⟩\equiv \beta \left(\sigma \right)=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]$ $\left(32\right)$
The
two-particle spin-space has dimension ${2}^{2}=4$. It can be spanned by the 1 antisymmetric and 3 symmetric two-particle wave-functions which can be built from eqn. (32)
 $\begin{array}{cccccccc}\hfill \mathrm{State}\hfill & \hfill \hfill & \hfill \mathrm{Coordinate} \mathrm{reps}.\hfill & \hfill \hfill & \hfill \mathrm{Ket}\hfill & \hfill \hfill & \hfill {S}^{2}\hfill & \hfill {S}_{z}\hfill \\ \hfill \hfill \\ \hfill {\chi }^{A}=\hfill & \hfill \hfill & \hfill \frac{1}{\sqrt{2}}\left[\alpha \left({\sigma }_{1}\right)\beta \left({\sigma }_{2}\right)-\alpha \left({\sigma }_{2}\right)\beta \left({\sigma }_{1}\right)\right]\hfill & \hfill \hfill & \hfill \frac{1}{\sqrt{2}}\left[|↑↓⟩-|↓↑⟩\right]\hfill & \hfill \hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill {\chi }^{S}=\hfill & \hfill \hfill & \hfill \alpha \left({\sigma }_{1}\right)\alpha \left({\sigma }_{2}\right)\hfill & \hfill \hfill & \hfill |↑↑⟩\hfill & \hfill \hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill {\chi }^{S}=\hfill & \hfill \hfill & \hfill \frac{1}{\sqrt{2}}\left[\alpha \left({\sigma }_{1}\right)\beta \left({\sigma }_{2}\right)+\alpha \left({\sigma }_{2}\right)\beta \left({\sigma }_{1}\right)\right]\hfill & \hfill \hfill & \hfill \frac{1}{\sqrt{2}}\left[|↑↓⟩+|↓↑⟩\right]\hfill & \hfill \hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill {\chi }^{S}=\hfill & \hfill \hfill & \hfill \beta \left({\sigma }_{1}\right)\beta \left({\sigma }_{2}\right)\hfill & \hfill \hfill & \hfill |↓↓⟩\hfill & \hfill \hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}$ $\left(33\right)$
The last two columns refer to the eigenvalues of the states $\chi$ for w.r.t. the
total spin and total spin-z operators
 $\begin{array}{ccc}\multicolumn{1}{c}{\stackrel{^}{S}}& =\hfill & {\stackrel{^}{\sigma }}_{1}+{\stackrel{^}{\sigma }}_{2}\hfill \\ \multicolumn{1}{c}{{\stackrel{^}{S}}_{z}}& =\hfill & {\stackrel{^}{\sigma }}_{1z}+{\stackrel{^}{\sigma }}_{2z}\hfill & \hfill \left(34\right)\end{array}$

I.e., luckily enough, the $\chi$s are eigenkets of the total spin by construction.
i) ${\Phi }^{S}{\chi }^{A}$ = spin singlet = named para-Helium
i) ${\Phi }^{A}{\chi }^{S}$ = spin triplet = named ortho-Helium
ii) Since $⟨{\chi }^{A}|{\chi }^{S}⟩=0$, transition matrix elements of type $⟨{\Phi }^{S}{\chi }^{A}|W|{\Phi }^{A}{\chi }^{S}⟩$ are $\ne 0$ only if $W=\mathrm{fct}\left({\stackrel{^}{\sigma }}_{1},{\stackrel{^}{\sigma }}_{2}\right)$. Therefore dipole-transitions, where $W~\left({r}_{1}+{r}_{2}\right)$, will not occur. Transition which involve spin-orbit coupling $~c \stackrel{^}{L}·\stackrel{^}{S}$ can occur. However the coupling constant $c$ is usually a small relativistic corrections with a scale of $v/c$. Therefore the life-time for transitions between ortho- and para-Helium can be days to weeks.
iii) Spin-orbit coupling will lift the degeneracy of ortho-Helium spin-multiplets. Para-Helium = single line in spectrum, ortho-Helium = three lines.

#### 5.2.1Perturbation Theory

Now we approximate the real space wave function by time-independent perturbation theory.
 $H=\sum _{i=1}^{2}{H}_{0i}+W ,\mathrm{ }W=\frac{{e}^{2}}{{r}_{12}}$ $\left(35\right)$
The single particle eigenfunctions are ${\phi }_{q{}_{i}}\left({r}_{i}\right)$, where $q=\left(\mathrm{nlm}\right)$ refers to the hydrogen atoms principal and angular momentum quantum numbers for nuclear charge $Z$. The unperturbed energies are
 $\begin{array}{ccc}\multicolumn{1}{c}{{E}_{Q}}& =\hfill & {\epsilon }_{{q}_{1}}+{\epsilon }_{{q}_{2}} ,\mathrm{ }{\epsilon }_{q}=-{Z}^{2}\mathrm{Ry}\frac{1}{{n}^{2}}\hfill \\ \multicolumn{1}{c}{{E}_{0}}& =\hfill & -4\mathrm{Ry}-4\mathrm{Ry}=-8\mathrm{Ry}\mathrm{ }\mathrm{for} {\mathrm{He}}_{2}, i.e. Z=2\hfill & \hfill \left(36\right)\end{array}$

This implies that the unperturbed He ${}_{2}$ has an
ionization energy of ${E}_{i}=4\mathrm{Ry}$. The eigenfunctions of $H$ can be expressed as
 $\Phi \left({r}_{1},{r}_{2}\right)=\sum _{{q}_{1},{q}_{2}}{c}_{{q}_{1},{q}_{2}}{\phi }_{{q}_{1}}\left({r}_{1}\right){\phi }_{{q}_{2}}\left({r}_{2}\right)$ $\left(37\right)$
and must be symmetrized according to 30.

1) Bare two-particle excitations
 ${E}_{\left(100,100\right)}=-4\mathrm{Ry}\left(\frac{1}{4}+\frac{1}{4}\right)=-2\mathrm{Ry}>{E}_{i}$ $\left(38\right)$
above the ionization threshold and will be discarded in a construction of the interacting eigenstates eqn. (37).
2) Symmetrized unperturbed real-space basis
From 1) the construction is simple
 $\begin{array}{ccc}\multicolumn{1}{c}{{\varphi }^{S}\left({r}_{1},{r}_{2}\right)={\varphi }_{\mathrm{nlm}}^{S}\left({r}_{1},{r}_{2}\right)}& =\hfill & \frac{1}{\sqrt{2}}\left({\phi }_{\mathrm{nlm}}\left({r}_{1}\right){\phi }_{100}\left({r}_{2}\right)+{\phi }_{100}\left({r}_{1}\right){\phi }_{\mathrm{nlm}}\left({r}_{2}\right)\right) ,\mathrm{ }n=2,3,...\hfill \\ \multicolumn{1}{c}{{\varphi }_{100}^{S}\left({r}_{1},{r}_{2}\right)}& =\hfill & {\phi }_{100}\left({r}_{1}\right){\phi }_{100}\left({r}_{2}\right) ,\mathrm{ }n=1\hfill \\ \multicolumn{1}{c}{}\\ \multicolumn{1}{c}{{\varphi }^{A}\left({r}_{1},{r}_{2}\right)={\varphi }_{\mathrm{nlm}}^{A}\left({r}_{1},{r}_{2}\right)}& =\hfill & \frac{1}{\sqrt{2}}\left({\phi }_{\mathrm{nlm}}\left({r}_{1}\right){\phi }_{100}\left({r}_{2}\right)-{\phi }_{100}\left({r}_{1}\right){\phi }_{\mathrm{nlm}}\left({r}_{2}\right)\right) ,\mathrm{ }n=2,3,...\hfill \\ \multicolumn{1}{c}{{\varphi }_{100}^{A}\left({r}_{1},{r}_{2}\right)}& =\hfill & 0 ,\mathrm{ }n=1\hfill & \hfill \left(39\right)\end{array}$

As can be check directly, these states are already eigenstates of the total angular momentum
 $l={l}_{1}+{l}_{2}$ $\left(40\right)$
 State ${l}^{2}$ ${l}_{z}$ ${P}_{12}$ ${\varphi }_{\mathrm{nlm}}^{S}\left({r}_{1},{r}_{2}\right)$ $\hslash l\left(l+1\right)$ $m$ + ${\varphi }_{\mathrm{nlm}}^{A}\left({r}_{1},{r}_{2}\right)$ " " -
Now replace ${c}_{{q}_{1},{q}_{2}}{\phi }_{{q}_{1}}\left({r}_{1}\right){\phi }_{{q}_{2}}\left({r}_{2}\right)$ in eqn. (37) by ${c}_{\mathrm{nlm}}{\varphi }_{\mathrm{nlm}}\left({r}_{1},{r}_{2}\right)$. Since $\left[{H}_{0}\left(W\right),{l}^{2}\right]=0$, $\left[{H}_{0}\left(W\right),{l}_{z}\right]=0$, and $\left[{H}_{0}\left(W\right),{\stackrel{^}{P}}_{12}\right]=0$ we have
 $⟨{\varphi }_{\mathrm{nlm}}^{p}|W|{\varphi }_{\mathrm{nl}\text{'}m\text{'}}^{p\text{'}}⟩\propto {\delta }_{\mathrm{pp}\text{'}}{\delta }_{\mathrm{ll}\text{'}}{\delta }_{\mathrm{mm}\text{'}}$ $\left(41\right)$
(Brief reminder: ${A}^{+}=A$, $A|{a}_{j}⟩={a}_{i}|{a}_{j}⟩$, and $\left[A,B\right]=0$ then $0=⟨{a}_{i}|\left[A,B\right]|{a}_{j}⟩=\left({a}_{i}-{a}_{j}\right)⟨{a}_{i}|B|{a}_{j}⟩$. I.e. $⟨{a}_{i}|B|{a}_{j}⟩$ can be non-zero only for ${a}_{i}={a}_{j}$.) Moreover $⟨{\varphi }_{\mathrm{nlm}}^{p}|{H}_{0}|{\varphi }_{\mathrm{mlm}}^{p}⟩\propto {\delta }_{\mathrm{nm}}$. Therefore, we may use
non-degenerate perturbation theory to evaluate te 1 ${}^{\mathrm{st}}$-order corrections in $W$ to the energies
 $\begin{array}{ccc}\multicolumn{1}{c}{\delta {E}_{\mathrm{nlm}}^{S}}& =\hfill & ⟨{\varphi }_{\mathrm{nlm}}^{S}|W|{\varphi }_{\mathrm{nlm}}^{S}⟩\ne f\left(m\right)\hfill \\ \multicolumn{1}{c}{\delta {E}_{\mathrm{nlm}}^{A}}& =\hfill & ⟨{\varphi }_{\mathrm{nlm}}^{A}|W|{\varphi }_{\mathrm{nlm}}^{A}⟩\ne f\left(m\right)\hfill & \hfill \left(42\right)\end{array}$

where the independence of $m$ is physically conceivable (but will not be proven here).

1) $\nexists$ ortho-He in the $100$-state because of Pauli's principle. I.e. the ground state of He is pure para He. This is a simple example proving Hund's 1 ${}^{\mathrm{st}}$ and 2 ${}^{\mathrm{nd}}$ rule: maximize S (rule 1) and $l$ (rule 2) which obey Pauli's principle.
2) Figure 1: The ground state of He is 'spherically symmetric' in real and spin space because of $l=0$ and $S=0$.
Figure 1: Ground state wave function of He

3) Figure 2: For all excited states $n\ne 1,\mathrm{lm}$, para-He has a higher energy than ortho-He, i.e. ${E}_{\mathrm{nlm}}^{S}>{E}_{\mathrm{nlm}}^{A}$. This is due to the Coulomb repulsion.
Figure 2: Excited states of He
Such phenomena are called electron correlation and exchange phenomena. (Note: para has ground state erg. but all excited states are higher in erg. than ortho)
4) Item 3) is at the heart of Hund's rule coupling or direct exchange: to minimize the Coulomb energy of electrons in orthogonal orbitals (here $\mathrm{lmn}=200$ and $100$), spin will align ferromagnetically. The magnetic exchange energy related to this must be of the order of $\mathrm{eV}$, i.e. many orders of magnitude larger than classical magnetic dipole coupling.
Excited state energies
According the eqn. (42), the shift of the energy to $O\left(W\right)$ for $n>1$ (!!) is
 $\begin{array}{cccc}\multicolumn{3}{c}{\delta {E}_{\mathrm{nl}}^{S/A}={e}^{2}\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{{\varphi }_{\mathrm{nlm}}^{S/A☆}\left({r}_{1},{r}_{2}\right){\varphi }_{\mathrm{nlm}}^{S/A}\left({r}_{1},{r}_{2}\right)}{|{r}_{1}-{r}_{2}|}}& \hfill \left(43\right)\\ \multicolumn{1}{c}{}& =\hfill & \frac{{e}^{2}}{2}\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{\left(\mathrm{dupl}. r. \mathrm{bracket}{\right)}^{☆}\left({\phi }_{\mathrm{nlm}}\left({r}_{1}\right){\phi }_{100}\left({r}_{2}\right)±{\phi }_{100}\left({r}_{1}\right){\phi }_{\mathrm{nlm}}\left({r}_{2}\right)\right)}{|{r}_{1}-{r}_{2}|}\hfill \\ \multicolumn{1}{c}{}& =\hfill & \int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{e|{\phi }_{\mathrm{nlm}}\left({r}_{1}\right){|}^{2}e|{\phi }_{100}\left({r}_{2}\right){|}^{2}±\left(e{\phi }_{\mathrm{nlm}}^{☆}\left({r}_{1}\right){\phi }_{100}\left({r}_{1}\right)\right)\left(e{\phi }_{\mathrm{nlm}}\left({r}_{2}\right){\phi }_{100}^{☆}\left({r}_{2}\right)\right)}{|{r}_{1}-{r}_{2}|}\hfill \\ \multicolumn{1}{c}{}& \equiv \hfill & \int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{{\rho }_{\mathrm{nl}}\left({r}_{1}\right){\rho }_{10}\left({r}_{2}\right)±{\rho }_{\mathrm{nl},10}^{X☆}\left({r}_{1}\right){\rho }_{\mathrm{nl},10}^{X}\left({r}_{2}\right)}{|{r}_{1}-{r}_{2}|}\hfill & \hfill \left(44\right)\\ \multicolumn{1}{c}{}& \equiv \hfill & {C}_{\mathrm{nl}}±{X}_{\mathrm{nl}}\hfill & \hfill \left(45\right)\end{array}$

In eqn. (44) we have defined
charge density
 ${\rho }_{\mathrm{nl}}\left(r\right)=e|{\phi }_{\mathrm{nlm}}\left(r\right){|}^{2}\mathrm{ }{\rho }_{10}\left(r\right)=e|{\phi }_{100}\left(r\right){|}^{2}$ $\left(46\right)$
and the
exchange charge density
 ${\rho }_{\mathrm{nl},10}^{X}\left(r\right)=e{\phi }_{\mathrm{nlm}}\left(r\right){\phi }_{100}^{☆}\left(r\right)$ $\left(47\right)$
both of which do not depend on $m$ (see. statement after eqn. (42)). In eqn. (45) we have defined the Coulomb integral
 ${C}_{\mathrm{nl}}=\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{{\rho }_{\mathrm{nl}}\left({r}_{1}\right){\rho }_{10}\left({r}_{2}\right)}{|{r}_{1}-{r}_{2}|}>0$ $\left(48\right)$
and the exchange integral
 ${X}_{\mathrm{nl}}=\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{{\rho }_{\mathrm{nl},10}^{X☆}\left({r}_{1}\right){\rho }_{\mathrm{nl},10}^{X}\left({r}_{2}\right)}{|{r}_{1}-{r}_{2}|}>0$ $\left(49\right)$
In particular ${E}^{\mathrm{para}}-{E}^{\mathrm{ortho}}=\delta {E}_{\mathrm{nl}}^{S}-\delta {E}_{\mathrm{nl}}^{A}=2{X}_{\mathrm{nl}}>0$.The Coulomb integral meets the classical expectation of an energy shift of the electrons due to the Coulomb force between them. The exchange integral has no classical analog (the classical descriptions of the electron gas in terms of a plasma does not know about this phenomenon!). Meaning: consider $\delta {E}_{\mathrm{nl}}^{A}$, i.e. $↑↑$. Define the pair correlation-function
 ${e}^{2}{g}_{↑↑}\left({r}_{1},{r}_{2}\right)={\rho }_{\mathrm{nl}}\left({r}_{1}\right){\rho }_{10}\left({r}_{2}\right)-{\rho }_{\mathrm{nl},10}^{X☆}\left({r}_{1}\right){\rho }_{\mathrm{nl},10}^{X}\left({r}_{2}\right)$ $\left(50\right)$
From its definition via the denominator of eqn. (43) this is the probability to find a $↑$-spin electron at ${r}_{1}$, if there is already a $↑$-spin electron at ${r}_{2}$. This is sketched in figure 3
Figure 3: Pair correlation function
where
 ${g}_{↑↑}\left(r,r\right)=0$ $\left(51\right)$
due to Pauli's principle. The suppression of probability is called the
exchange hole. Fig. 3 also shows ${g}_{↑↓}\left({r}_{1},{r}_{2}\right)$. The latter need not be $0$ at ${r}_{1}={r}_{2}$. Therefore a difference arises in the Coulomb repulsion energy of the two electrons due to different spin alignment in ${\varphi }_{\mathrm{nlm}}^{A}\left({r}_{1},{r}_{2}\right)$ and ${\varphi }_{\mathrm{nlm}}^{S}\left({r}_{1},{r}_{2}\right)$ - this is taken into account by $±{X}_{\mathrm{nl}}$.
Exercise 4 Pair correlation function in a Fermi sea. Consider the fermion basis from en. (25) for a state $\mu =\left({m}_{1},\dots {m}_{N}\right)$ of the $N$ particles.
a) The propabiltiy for finding a fermion at coordinates $1$ if another fermion is at coordinate $2$ is $\rho \left(1,2\right)=\int d3d4\dots dN{\Phi }_{\mu }^{F ☆}\left(1,\dots N\right){\Phi }_{\mu }^{F}\left(1,\dots N\right)$, where $1=\left({r}_{1},{\sigma }_{1}\right)$ and $\int di$ refers to integration over coordinates and summation over spin. Show that
 $\begin{array}{ccc}\multicolumn{1}{c}{\rho \left(1;2\right)}& =\hfill & \frac{1}{N!}\sum _{P}\left[{\phi }_{{m}_{P1}}^{☆}\left(1\right){\phi }_{{m}_{P2}}^{☆}\left(2\right)-{\phi }_{{m}_{P2}}^{☆}\left(1\right){\phi }_{{m}_{P1}}^{☆}\left(2\right)\right]{\phi }_{{m}_{P1}}\left(1\right){\phi }_{{m}_{P2}}\left(2\right)\hfill \\ \multicolumn{1}{c}{}& \hfill & \frac{1}{N\left(N-1\right)}\sum _{\mathrm{ab}}\left[{\phi }_{a}^{☆}\left(1\right){\phi }_{b}^{☆}\left(2\right)-{\phi }_{b}^{☆}\left(1\right){\phi }_{a}^{☆}\left(2\right)\right]{\phi }_{a}\left(1\right){\phi }_{b}\left(2\right)\hfill & \hfill \left(52\right)\end{array}$

b) Assume the one particle states to be momentum and spin-z eigenfunctions with $a\equiv \left(k,\alpha \right)$, where $k$ is the wave vector and $\alpha =±1$ depending on spin-up or down, i.e.
 ${\phi }_{a}\left(1\right)={\phi }_{k\alpha }\left(r,\sigma \right)=\frac{1}{\sqrt{V}}{e}^{ikr}{\delta }_{\alpha \sigma }$ $\left(53\right)$
Furthermore assume that ${\Phi }_{\mu }^{F}\left(1,\dots N\right)$ consists of all one-particle states with both spin directions up to the
Fermi wave vector $|k|\le {k}_{F}$. This is a $\mathrm{Fermi}$ sea. Show that
 $\rho \left(r,\sigma ;r\text{'},\sigma \text{'}\right)=\frac{1}{N\left(N-1\right){V}^{2}}\sum _{k\ne k\text{'};k,k\text{'}<{k}_{F}}\left\{\begin{array}{cc}\hfill \left[1-{e}^{i\left(k-k\text{'}\right)\left(r-r\text{'}\right)}\right]\hfill & \hfill , \sigma =\sigma \text{'}\hfill \\ \hfill 1\hfill & \hfill , \sigma \ne \sigma \text{'}\hfill \end{array}$ $\left(54\right)$
c) What is the value for $\rho \left(r,\sigma ;r\text{'},-\sigma \right)$? Try to show that
 $\rho \left(r,\sigma ;r\text{'},\sigma \right)\frac{N-2}{{N}^{2}\left(N-1\right)4{V}^{2}}\left[1-9{\left(\frac{\mathrm{sin}\left({k}_{F}d\right)-{k}_{F}r\mathrm{cos}\left({k}_{F}d\right)}{\left({k}_{F}d{\right)}^{3}}\right)}^{2}\right]$ $\left(55\right)$
where $d=|r-r\text{'}|$. Plot $\rho \left(r,\sigma ;r\text{'},\sigma \text{'}\right)$ as a function of ${k}_{F}d$. Interpret the result.
For an actual calculation of $\delta {E}_{\mathrm{nl}}^{A/S}$ perturbation theory to $O\left(W\right)$ is not good enough. Higher order corrections, i.e. screening of the states ${\phi }_{\mathrm{nlm}}\left(r\right)$ with $n>1$ from the nuclear charge due to the presence of the $100$ electron, plays an important role. One way to approximate this is by setting
 $\begin{array}{ccc}\multicolumn{1}{c}{Z=2}& \mathrm{ }\mathrm{for}\hfill & {\phi }_{100}\left(r\right)\hfill \\ \multicolumn{1}{c}{Z=1}& "\hfill & {\phi }_{n>1\mathrm{lm}}\left(r\right)\hfill & \hfill \left(56\right)\end{array}$

Ground state energy
For $n=1$ eqn. (43-45) simplify because the numerator only contains ${\rho }_{10}\left(r\right)$ and no exchange contributions
 ${E}_{100}^{S}\equiv {E}_{0}^{S}={E}_{0}+{C}_{10}=-2{Z}^{2}\mathrm{Ry}+{e}^{2}\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}\frac{|{\phi }_{100}\left({r}_{1}\right){|}^{2}|{\phi }_{100}\left({r}_{2}\right){|}^{2}}{|{r}_{1}-{r}_{2}|}$ $\left(57\right)$
with a 'hydrogen'-like ground state wave function for a nuclear charge $Z$
 ${\phi }_{100}\left(r\right)=\frac{1}{\sqrt{\pi {a}^{3}}}{e}^{-r/a},\mathrm{ }a=\frac{{a}_{0}}{Z},\mathrm{ }{a}_{0}=\frac{\hslash {}^{2}}{{\mathrm{me}}^{2}}$ $\left(58\right)$
This yields
 ${C}_{10}=\frac{5}{8}\frac{{e}^{2}}{a}=Z\frac{5}{4}\frac{{e}^{2}}{2{a}_{0}}=Z\frac{5}{4}\mathrm{Ry}=\frac{5}{2}\mathrm{Ry} ,\mathrm{ }\mathrm{for} Z=2$ $\left(59\right)$
Exercise 5 Prove eqn. (59). Do not use eqn. (57) directly. Instead
a) Consider the Poisson equation for the charge density set by eqn. (46) and enq. (58) in spherical coordinates. Show that the electrostatic potential is
 $\nu \left(r\right)=\frac{e}{r}\left\{1-\left(1+\frac{r}{a}\right){e}^{-\frac{2r}{a}}\right\}$ $\left(60\right)$

b) Use eqn. (60) to obtain eqn. (59).
From this the ionization energy of He is
 ${E}_{i}={E}_{\mathrm{continuum}}-{E}_{0}^{S}=-4\mathrm{Ry}-\left(-8\mathrm{Ry}+\frac{5}{2}\mathrm{Ry}\right)=\left(\frac{8}{2}-\frac{5}{2}\right)\mathrm{Ry}=\frac{3}{2}\mathrm{Ry}$ $\left(61\right)$
This is strongly reduced from the unperturbed ionization energy $4\mathrm{Ry}$. The experimental value is ${E}_{i}^{\mathrm{exp}}\approx 1.82\mathrm{Ry}$. The level scheme of He is depicted in fig. 4
Figure 4: Level scheme of He

#### 5.2.2Variational Calculation

Variational calculations may outperform perturbative approaches, depending on the quality of the variational Ansatz. Here we pursue the Ansatz to evaluate the ground state energy with a 'hydrogen-like' wave functions with however a modified nuclear charge $Z\text{'}$. Next we consider two points of views of such an Ansatz.
First point of view:
Idea: let us modify $W\to \stackrel{~}{W}$ such as to have an 'optimally small' interaction
 $\begin{array}{ccc}\multicolumn{1}{c}{H}& =\hfill & \frac{1}{2m}\left({p}_{1}^{2}+{p}_{2}^{2}\right)-Z\text{'}{e}^{2}\left(\frac{1}{|{r}_{1}|}+\frac{1}{|{r}_{2}|}\right)\hfill \\ \multicolumn{1}{c}{}& \hfill & +{e}^{2}\left(\frac{1}{{r}_{12}}+\left(Z\text{'}-Z\right){e}^{2}\left(\frac{1}{|{r}_{1}|}+\frac{1}{|{r}_{2}|}\right)\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & {\stackrel{~}{H}}_{0}\left(Z\text{'}\right)+\stackrel{~}{W}\left(Z\text{'}\right)\hfill & \hfill \left(62\right)\end{array}$

Procedure: Choose $Z\text{'}$ such that $\delta {E}_{\mathrm{nl}}^{S/A}\left(Z\text{'}\right)$ is minimal. For this no change in the original set of basis functions has to be made because ${\stackrel{~}{H}}_{0}\left(Z\text{'}\right)$ is of identical form as ${H}_{0}$.
Physics: $e\Delta Z=e\left(Z-Z\text{'}\right)$ is a screening charge of the nuclear charge for each electron due to the presence of the other electron.
The calculation proceeds exactly as before, however with
 $\begin{array}{ccc}\multicolumn{1}{c}{{E}_{0}^{\text{'}}}& =\hfill & -2{Z}^{\text{'}2}\mathrm{Ry}=-{Z}^{\text{'}2}\frac{{e}^{2}}{{a}_{0}}\hfill \\ \multicolumn{1}{c}{{C}_{10}}& =\hfill & {Z}^{\text{'}}\frac{5}{8}\frac{{e}^{2}}{{a}_{0}}\hfill \\ \multicolumn{1}{c}{⟨{\varphi }_{100}^{S}|\left(Z\text{'}-Z\right){e}^{2}\left(\frac{1}{|{r}_{1}|}+\frac{1}{|{r}_{2}|}\right)|{\varphi }_{100}^{S}⟩}& =\hfill & 2\left(Z\text{'}-Z\right)⟨{\phi }_{100}|\frac{{e}^{2}}{r}|{\phi }_{100}⟩\hfill \\ \multicolumn{1}{c}{}& \hfill & \genfrac{}{}{0}{}{\mathrm{Virial} \mathrm{theorem}}{=}2\left(Z\text{'}-Z\right)\frac{{Z}^{\text{'}}{e}^{2}}{{a}_{0}}\hfill & \hfill \left(63\right)\end{array}$

The last equation uses a well-known expectation value of $1/r$ for 'hydrogen-like' wave functions from basic QM. Altogether
 ${\stackrel{~}{E}}_{0}^{S}=-\frac{{e}^{2}}{{a}_{0}}\left[{Z}^{\text{'}2}-{Z}^{\text{'}}\frac{5}{8}+2{Z}^{\text{'}}\left(Z-{Z}^{\text{'}}\right)\right]$ $\left(64\right)$
Now, the interaction's contribution to eqn. (64) is
 $⟨\stackrel{~}{W}⟩={Z}^{\text{'}}\frac{{e}^{2}}{{a}_{0}}\left[\frac{5}{8}-2\left(Z-{Z}^{\text{'}}\right)\right]$ $\left(65\right)$
This can be made 'optimally small', namely $\equiv 0$ for one non-trivial case, i.e. ${Z}^{\text{'}}=Z-5/16$. This was our goal. It implies a
screening charge of $5e/16$. Another way of viewing this is to consider the total energy which can be minimized to yield
 $\frac{\partial {\stackrel{~}{E}}_{0}^{S}}{\partial {Z}^{\text{'}}}=0\mathrm{ }⇒\mathrm{ }\frac{5}{8}-2\left(Z-{Z}^{\text{'}}\right)=0\mathrm{ }⇒\mathrm{ }{Z}^{\text{'}}=Z-\frac{5}{16}$ $\left(66\right)$
The ionization energy now is
 ${E}_{i}={E}_{\mathrm{continuum}}-{\stackrel{~}{E}}_{0}^{S}=\frac{217}{128}\mathrm{Ry}\approx 1.70\mathrm{Ry}$ $\left(67\right)$
This is an improvement w.r.t. eqn. (61) and pretty close to ${E}_{i}^{\mathrm{exp}}\approx 1.82\mathrm{Ry}$.
Second point of view:
Idea: Let us claim an optimum ground state wave function
 ${\stackrel{~}{\phi }}_{100}\left(r\right)=\frac{1}{\sqrt{\pi {a}^{\text{'}3}}}{e}^{-r/{a}^{\text{'}}} ,\mathrm{ }{a}^{\text{'}}=\frac{{a}_{0}}{{Z}^{\text{'}}} ,\mathrm{ }{a}_{0}=\frac{\hslash {}^{2}}{{\mathrm{me}}^{2}}$ $\left(68\right)$
which experiences a modified nuclear charge ${Z}^{\text{'}}$. The expectation value of the energy is
 $\begin{array}{ccc}\multicolumn{1}{c}{⟨H⟩=⟨{\stackrel{~}{\varphi }}_{100}^{S}|H|{\stackrel{~}{\varphi }}_{100}^{S}⟩}& =\hfill & ⟨{\stackrel{~}{\varphi }}_{100}^{S}|\left[{\stackrel{~}{H}}_{0}\left(Z\text{'}\right)+\stackrel{~}{W}\left(Z\text{'}\right)\right]|{\stackrel{~}{\varphi }}_{100}^{S}⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{{e}^{2}}{{a}_{0}}\left[{Z}^{\text{'}2}-{Z}^{\text{'}}\frac{5}{8}+2{Z}^{\text{'}}\left(Z-{Z}^{\text{'}}\right)\right]\hfill & \hfill \left(69\right)\end{array}$

Ritz's variational principle requires $\partial ⟨H⟩/\partial {Z}^{\text{'}}=0$, i.e eqn. (66). The various results from perturbation theory, variational approximation and experiment are depicted in fig. 5.
Figure 5: He ionization energies
1) From the variational principle we know, that $⟨H⟩$ is always higher than the exact ground state energy. I.e. the variational ionization energy is always less than the exact one. State-of-the-art variational calculations use several hundred variational parameters and include nuclear coordinates, (hype-)fine-structure, QED, ...
2) Applying eqn. (64) and eqn. (66) to the general case of the ionization energy ${E}_{I}$ of an ion of charge $Z-2$ into one with charge $Z-1$ we get
 ${E}_{I}=\left({Z}^{2}-\frac{5}{4}Z+2{\left(\frac{5}{16}\right)}^{2}\right)\mathrm{Ry}$ $\left(70\right)$
This can be compared with experimental values from Landolt-Börnstein as shown in fig. 6
Figure 6: $\left(Z-2\right) \to \left(Z-1\right)$ ionization energies of several elements

### 5.3The H ${}_{2}$ Molecule

In this section we will consider two prime mechanisms for the formation of solids, namely the Born-Oppenheimer approximation and the theory covalent chemical chemical bonding.

#### 5.3.1Hamilton Operator

Figure 7: Geometry and labels for the H ${}_{2}$ molecule
The geometry of the H ${}_{2}$ moelcule is shown in fig. 7. From this we read of the Hamiltonian
 $H=\frac{1}{2M}\left({P}_{a}^{2}+{P}_{b}^{2}\right)+\frac{1}{2m}\left({p}_{1}^{2}+{p}_{2}^{2}\right)+{e}^{2}\left(\frac{1}{{R}_{\mathrm{ab}}}+\frac{1}{{r}_{12}}+\frac{1}{{r}_{a1}}+\frac{1}{{r}_{b2}}+\frac{1}{{r}_{b1}}+\frac{1}{{r}_{a2}}\right)$ $\left(71\right)$
For the remainder of this chapter we neglect relativistic effects, in particular spin-orbit coupling and nuclear-spin electron-spin coupling. I.e. the wave function factorizes in terms orbit and spin space
 $\Phi \left({R}_{a},{R}_{b},{r}_{1},{r}_{2},{\sigma }_{a},{\sigma }_{b},{\sigma }_{1},{\sigma }_{2}\right)=\Psi \left({R}_{a},{R}_{b},{r}_{1},{r}_{2}\right)\chi \left({\sigma }_{a},{\sigma }_{b}\right)\chi \left({\sigma }_{1},{\sigma }_{2}\right)$ $\left(72\right)$
The nucleons (electrons) are identical particles, fermions (!), separately and the Hamiltonian commutes with their permutation operator ${P}_{\mathrm{ab}}$ ( ${P}_{12}$) respectively. I.e. Pauli's principle has to applied to the nucleons (electrons) separately. In turn, the wave-function eqn. (72) has to be antisymmetric w.r.t. the permutation of the nucleon (electron) coordinates separately. Eg.

#### 5.3.2Born-Oppenheimer Approximation

Based on the small parameter $m/M\approx 1/2000$ we make the approximation that the nucleons move very slow as compared to the electrons. This suggests a two stage solution of the Schrödinger equation for eqn. (71) by writing
 $\Psi \left({R}_{a},{R}_{b},{r}_{1},{r}_{2}\right)=\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)g\left({R}_{a},{R}_{b}\right)$ $\left(74\right)$
where $\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)$ solves the
electronic Schrödinger equation (ESG) for eqn. (71) keeping the nuclear coordinates fixed. Along with this we normalize $\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)$ for each ${R}_{a},{R}_{b}$
 $\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}{\varphi }^{☆}\left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)\equiv 1 ,$ $\left(75\right)$
i.e., $|\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right){|}^{2}$ can be interpreted as the probabiltiy to observe two electrons in a volume element ${\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}$ at ${r}_{1},{r}_{2}$ holding the nucleons a some fixed locations ${R}_{a},{R}_{b}$. To abreviate the notation we will discard the coordinates in $\varphi ,g$ whenever possible.
The probabiltiy to observe two nucleons in a volume element ${\mathrm{dR}}_{a}^{3}{\mathrm{dR}}_{b}^{3}$ at ${R}_{a},{R}_{b}$ follows from integrating over the electron coordinates
 ${\mathrm{dR}}_{a}^{3}{\mathrm{dR}}_{b}^{3} \int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}{\Psi }^{☆}\left({R}_{a},{R}_{b},{r}_{1},{r}_{2}\right)\Psi \left({R}_{a},{R}_{b},{r}_{1},{r}_{2}\right)={\mathrm{dR}}_{a}^{3}{\mathrm{dR}}_{b }^{3}|g\left({R}_{a},{R}_{b}\right){|}^{2}$ $\left(76\right)$
Inserting $\varphi$ into the ESG we write
 $\left[\frac{1}{2m}\left({p}_{1}^{2}+{p}_{2}^{2}\right)+{e}^{2}\left(\frac{1}{{r}_{12}}+\frac{1}{{r}_{a1}}+\frac{1}{{r}_{b2}}+\frac{1}{{r}_{b1}}+\frac{1}{{r}_{a2}}\right)\right]\varphi ={E}_{\mathrm{el}}\left({R}_{\mathrm{ab}}\right)\varphi ,$ $\left(77\right)$
where we use that the electronic energy ${E}_{\mathrm{el}}\left({R}_{\mathrm{ab}}\right)$ at the fixed nuclear coordinates ${R}_{a},{R}_{b}$ depends only on the inter-nuclear
distance. Inserting $\Psi =\varphi g$ into the full SG reads
 $\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}{\varphi }^{☆}\left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)×|\mathrm{ }\left[\frac{1}{2M}\left({P}_{a}^{2}+{P}_{b}^{2}\right)+\frac{{e}^{2}}{{R}_{\mathrm{ab}}}+{E}_{\mathrm{el}}\left({R}_{\mathrm{ab}}\right)\right]\varphi g=E\varphi g ,$ $\left(78\right)$
where we do a left scalar-multiply with ${\varphi }^{☆}$. At this point we need to consider the dependence of $\varphi \left(\dots |{R}_{a},{R}_{b}\right)$ on the nuclear coordinates. Consider the action of ${P}_{a}^{2}$ first

Using

and eqn. (75) and a similar set of arguments for ${P}_{b}^{2}$ we get from eqn. (78)

Now, the main point is, that $\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)$ is essentially a function of the differences ${r}_{1,2}-{R}_{a,b}$. Therefore we may approximate
 $\begin{array}{ccc}\multicolumn{3}{c}{\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}{\varphi }^{☆}\left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)\frac{{P}_{a}^{2}+{P}_{b}^{2}}{2M}\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)}\\ \multicolumn{1}{c}{}& \hfill & ~\frac{m}{M} O\left[\int {\mathrm{dr}}_{1}^{3}{\mathrm{dr}}_{2}^{3}{\varphi }^{☆}\left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)\frac{{p}_{1}^{2}+{p}_{2}^{2}}{2m}\varphi \left({r}_{1},{r}_{2}|{R}_{a},{R}_{b}\right)\right]~\frac{m}{M} {E}_{\mathrm{el}}\left({R}_{\mathrm{ab}}\right) .\hfill & \hfill \left(82\right)\end{array}$

This allows to neglect the 2nd term on the 1st line of eqn. (81) to $O\left(m/M\right)$.
This is the content of Born's approximation: the effect of the electrons on the nuclei is to generate an effective potential ${E}_{\mathrm{el}}\left({R}_{\mathrm{ab}}\right)$ - a kind of electronic glue - in which the nucleons move in addition to their own Coulomb repulsion
TO BE CONTINUED

### 5.4Brief Digression: Variational Approximations

For the He atom and the H ${}_{2}$ molecule we have used the variational method to determine approximate eigenvectors of a quantum system. Here we briefly summarize the basis of this method: let $H$ be the Hilbert space and $|\psi ⟩\in H$. Consider the energy functional
 $E\left(\psi \right)=\frac{⟨\psi |H|\psi ⟩}{⟨\psi |\psi ⟩}$ $\left(83\right)$
then, any state which makes this functional stationary, i.e. $E\left(\delta \psi \right)\equiv 0$, is an eigenstate of $H$. Namely
 $\begin{array}{ccc}\multicolumn{1}{c}{\delta E}& =\hfill & \frac{⟨\delta \psi |H|\psi ⟩+⟨\psi |H|\delta \psi ⟩}{⟨\psi |\psi ⟩}-\frac{⟨\psi |H|\psi ⟩\left(⟨\delta \psi |\psi ⟩+⟨\psi |\delta \psi ⟩\right)}{⟨\psi |\psi {⟩}^{2}}\equiv 0\hfill \\ \multicolumn{1}{c}{⟨\psi |\psi ⟩\delta E}& =\hfill & ⟨\delta \psi |\left(H-E\right)|\psi ⟩+⟨\psi |\left(H-E\right)|\delta \psi ⟩\equiv 0\hfill & \hfill \left(84\right)\end{array}$

Since for $\delta E=0$ eqn. (84) must be fullfilled for
any $|\delta \psi ⟩$, consider the one with $|\delta \psi ⟩\to i|\delta \psi ⟩$, i.e. $⟨\delta \psi |\to -i⟨\delta \psi |$
 $⟨\psi |\psi ⟩\delta E=-i⟨\delta \psi |\left(H-E\right)|\psi ⟩+i⟨\psi |\left(H-E\right)|\delta \psi ⟩\equiv 0$ $\left(85\right)$
multiplying eqn. (84) by $i$ an combining we get
 $⟨\delta \psi |\left(H-E\right)|\psi ⟩\equiv 0\mathrm{ }⟨\psi |\left(H-E\right)|\delta \psi ⟩\equiv 0$ $\left(86\right)$
or
 $⟨\psi |E=⟨\psi |H\mathrm{ }H|\psi ⟩=E|\psi ⟩$ $\left(87\right)$
where the latter two equations are identical, since ${H}^{+}=H$ and ${E}^{☆}=E$, where the 2nd follows from eqn. (83).
As a consequence: any state $|\phi ⟩\in H$, which is not necessarily an eigenstate of the Schrödinger equation has an expectation value of its energy which is larger than the ground state energy. Consider a CONS $|n⟩$ of $H$ with $|0⟩$ being the ground state, i.e. $E\left(0\right)={E}_{0}$ is stationary and the absolute minimum of $E\left(\psi \right)$
 $\begin{array}{ccc}\multicolumn{1}{c}{E\left(\phi \right)-{E}_{0}}& =\hfill & \frac{⟨\phi |H-{E}_{0}|\phi ⟩}{⟨\phi |\phi ⟩}=\sum _{n\ne 0}\left({E}_{n}-{E}_{0}\right)\frac{⟨\phi |n⟩⟨n|\phi ⟩}{⟨\phi |\phi ⟩}\hfill \\ \multicolumn{1}{c}{}& =\hfill & \sum _{n\ne 0}\left(\mathrm{positive}×\mathrm{positive}\right)\geqslant0\hfill & \hfill \left(88\right)\end{array}$

If eqn. (83) is minimized in a reduced space of variations $|\delta \psi ⟩\subseteq H$, then the solution is called a variational approximation.
Particular case: subspace-diagonalization. Let $|\psi ⟩=\sum _{i=1}^{N}{c}_{i}|{\mu }_{i}⟩\in \mathrm{Span}\left\{|{\mu }_{1}⟩,|{\mu }_{2}⟩\dots |{\mu }_{N}⟩\right\}\subseteq H$, where the set $\left\{|{\mu }_{1}⟩,|{\mu }_{2}⟩\dots |{\mu }_{N}⟩\right\}$ does not necessarily have to be orthogonal or normalized, then eqn. (86) reads
 $\sum _{i,j=1}^{N}\delta {c}_{i}^{☆}⟨{\mu }_{i}|\left(H-E\right)|{\mu }_{j}⟩{c}_{j}=0$ $\left(89\right)$
Since $\delta {c}_{i}^{☆}$ can be arbitrary, we may chose $\delta {c}_{i}^{☆}={\delta }_{\mathrm{li}}$ for fixed $l=1,\dots N$, which leads to
 $\sum _{j=1}^{N}{H}_{\mathrm{ij}}{c}_{j}=E\sum _{j=1}^{N}{S}_{\mathrm{ij}}{c}_{j}$ $\left(90\right)$
which is a so-called
generalizd eigenvalue problem.If the overlap matrix ${S}_{\mathrm{ij}}=⟨{\mu }_{i}|{\mu }_{j}⟩={\delta }_{\mathrm{ij}}$, i.e. if the $|{\mu }_{i}⟩$'s are orthonormal, then eqn. (90) simply corresponds to solving Schrödiger's equation in a reduced subspace.

### 5.5Second quantization

TO DO
Exercise 6 Tight-binding: a simple approach to band structure in solids. Assume a one-dimensional Lattice of $N$ sites. Assume a one-particle Hilbert space of states $|i,\alpha ⟩$, with $i=0,1,\dots N-1$ and $\alpha =1,2$. Assume that the particles a fermions and neglect the spin for simplicity. These states model electronic states on a one-dimensional lattice of ficticious two-level atoms which are located at the sites $i$. Assume that the Hamiltonian is a one-particle operator of the form
 $H=\sum _{i=0}^{N-1}{h}_{i} ,$ $\left(91\right)$
where
 $⟨m,\beta |{h}_{i}|n,\alpha ⟩={\delta }_{m,n}{\delta }_{\alpha ,\beta }{\epsilon }_{\alpha }+\frac{1}{2}\left({\delta }_{m,n+1}+{\delta }_{m,n-1}\right){t}_{\alpha \beta } ,$ $\left(92\right)$
with ${\epsilon }_{\alpha },{t}_{\alpha \beta }\in \mathrm{Re}$ and ${t}_{\alpha \beta }={t}_{\beta \alpha }$, and and we use so-called periodic boundary conditions, namely that the site $m=N$
is equivalent to the site $m=0$.
a) What is the physical meaning of ${\epsilon }_{\alpha }$ and ${t}_{\alpha \beta }$?
b) Let ${c}_{i,\alpha }^{\left(+\right)}$ be the (creation)destruction operators of electrons on site $i$ in level $\alpha$. Show that the Hamilton operator in its 2nd quantized form is
 $H=\sum _{i=0}^{N-1}\left[{\epsilon }_{\alpha }{c}_{i,\alpha }^{+}{c}_{i,\alpha }+\frac{1}{2}{t}_{\alpha \beta }\left({c}_{i+1,\alpha }^{+}{c}_{i,\beta }+{c}_{i-1,\alpha }^{+}{c}_{i,\beta }\right)\right] .$ $\left(93\right)$
Prove that $H$ is hermitean. What is the physical meaning of the two addends in $H$.

c) Perform a basis change in eqn. (93) from the site-representation $i$ onto (lattice-)momentum $k$ by
 ${a}_{k,\alpha }^{+}=\frac{1}{\sqrt{N}}\sum _{n=0}^{N-1}{e}^{\mathrm{ikn}}{c}_{i,\alpha }^{+} ,$ $\left(94\right)$
where $k=2\pi m/N$ with $m=0,1,\dots N-1$. [Supplementary question: why is $k$ chosen like this?]. Show that
 $H=\sum _{k}\left[{a}_{k,1}^{+} {a}_{k,2}^{+}\right]\left[\begin{array}{cc}\hfill {\epsilon }_{1}+{t}_{11}\mathrm{cos}\left(k\right)\hfill & \hfill {t}_{12}\mathrm{cos}\left(k\right)\hfill \\ \hfill {t}_{21}\mathrm{cos}\left(k\right)\hfill & \hfill {\epsilon }_{2}+{t}_{22}\mathrm{cos}\left(k\right)\hfill \end{array}\right]\left[\begin{array}{c}\hfill {a}_{k,1}\hfill \\ \hfill {a}_{k,2}\hfill \end{array}\right] .$ $\left(95\right)$
Using this show that
 $H=\sum _{k,\alpha }{E}_{k\alpha }{b}_{k,\alpha }^{+}{b}_{k,\alpha } ,$ $\left(96\right)$
where ${b}_{k,\alpha }^{\left(+\right)}$ are also fermions. Evaluate ${E}_{k\alpha }$. What is the physical meaning of ${E}_{k\alpha }$, $\alpha$, and ${b}_{k,\alpha }^{\left(+\right)}$.

d) Discuss ${E}_{k\alpha }$ for $\alpha =1,2$ and ${\epsilon }_{1}=0$, ${\epsilon }_{2}=1$, ${t}_{11}={t}_{22}=1$ as a function of $k$ and ${t}_{12}$. At ${t}_{12}\ne 0$: what is the ground state of $N/2$, of $N$, and of $3N/2$ fermions (assuming $N$ is even) expressed in terms of the ${b}_{k,\alpha }^{+}$ and the vacuum $|⟩$?

## Index (showing section)

 anyons, 5.1 charge densities, 5.2 correlation, 5.2 Coulomb integral, 5.2 direct exchange, 5.2 energy functional, 5.4 exchange, 5.2 exchange charge density, 5.2 exchange hole, 5.2 exchange integral, 5.2 fractional quantum Hall effect, 5.1 generalizd eigenvalue problem, 5.4 Hund's rule, 5.2 Hund's rule coupling, 5.2 identical particles, 5.1 ionization energy, 5.2 Luttinger liquid, 5.1 ortho-Helium, 5.2 overlap matrix, 5.4 pair correlation-function, 5.2 para-Helium, 5.2 Pauli's exclusion principle, 5.1 permutation, 5.1 plasma, 5.2 screening, 5.2 Slater-determinant, 5.1 subspace-diagonalization, 5.4 transpositions, 5.1

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On 7 Jan 2010, 15:28.