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\begin{document}





\begin{flushleft}
An Introduction to Second Quantization
\end{flushleft}


\begin{flushleft}
Sandeep Pathak∗ January 27, 2010
\end{flushleft}





1





\begin{flushleft}
Introduction
\end{flushleft}





\begin{flushleft}
We all know single particle quantum mechanics very well. The state of a system can be represented by a ket |$\psi$ and the space in which these states lie is called the Hilbert space. This Hilbert space contains states which can accomodate one particle. Thus, there is no room for more than one particles! What do we do to accomodate more guests? The answer is simple and that is to EXPAND! $\Rightarrow$ HN = H $\otimes$ · · · $\otimes$ H
\end{flushleft}


\begin{flushleft}
n$-$copies
\end{flushleft}





\begin{flushleft}
The idea is pretty similar to the idea of extension from 1-dimension to 2-dimensions. A point in 1-dimension is represented in R as x or y. If we take two copies of R, we get the space R $\otimes$ R in which we can represent a 2-dimensional point i.e. we go from x $\rightarrow$ (x, y). The operation $\otimes$ is known as direct product. Let us now examine how we can write a two particle state using two single particle states |1 , |2 . Suppose the particle 1 is in the state |1 and particle 2 is in the state |2 . Using the above idea, we can straight away write a state |1 $\otimes$ |2 in H $\otimes$ H. What is the meaning of this state? r1 , r2 | |1 $\otimes$ |2 = r1 |1 r2 |2 = $\phi$(r1 )$\phi$(r2 ) (1) If the particle 2 is in the state |1 and particle 1 is in the state |2 , the two particle state can be written as |2 $\otimes$ |1 and Thus, a general two particle state in H2 can be written as |$\psi$
\end{flushleft}


2





\begin{flushleft}
r1 , r2 | |2 $\otimes$ |1 = r1 |2 r2 |1 = $\phi$(r2 )$\phi$(r1 ) = $\alpha$|1 $\otimes$ |2 + $\beta$|2 $\otimes$ |1
\end{flushleft}





(2)





(3)





\begin{flushleft}
We know that the particles in quantum mechanics are indistinguishable and occur only in two possibilities Wavefunction Allowed Fermions Anti-symmetric |$\psi$ f = (|1 $\otimes$ |2 $-$ |2 $\otimes$ |1 ) 2 Bosons Symmetric |$\psi$ b = (|1 $\otimes$ |2 + |2 $\otimes$ |1 ) 2
\end{flushleft}





\begin{flushleft}
f 1 Consider the anti-symmetric state. We can drop $\otimes$ for brevity: |$\psi$ 2 = $\surd$2 (|1 |2 $-$ |2 |1 ). The real space representation is given by 1 r1 , r2 |$\psi$ f = $\surd$ ($\phi$1 (r1 )$\phi$2 (r2 ) $-$ $\phi$2 (r1 )$\phi$1 (r2 )) 2 2 1 $\phi$1 (r1 ) $\phi$1 (r2 ) = $\surd$ 2 $\phi$2 (r1 ) $\phi$2 (r2 ) $\equiv$ |1, 2| (4)
\end{flushleft}


∗





\begin{flushleft}
psandeep@physics.iisc.ernet.in
\end{flushleft}





1





\begin{flushleft}
\newpage
This form |1, 2| is known Slater determinant and is used to represent anti-symmetric wavefunctions. The generalization to the N -particle anti-symmetric wavefunction is trivial. If we have N states |n , (n = 1, 2, . . . , N ). Then the slater determinant |1, 2, . . . , N | is given by |$\psi$
\end{flushleft}


\begin{flushleft}
f N
\end{flushleft}





\begin{flushleft}
1 =$\surd$ N!
\end{flushleft}





\begin{flushleft}
$\phi$1 (r1 ) . . . $\phi$N (r1 )
\end{flushleft}





··· . . . ···





\begin{flushleft}
$\phi$1 (rN ) . . . $\phi$N (rN )
\end{flushleft}





(5)





\begin{flushleft}
The subspace spanned by these anti-symmetric states is known as Fock's space for fermions. There is an ambiguity in the way we have deﬁned (4). There is nothing which prevents us from writing a slater determinant as r1 , r2 |$\psi$
\end{flushleft}


\begin{flushleft}
f 2
\end{flushleft}





\begin{flushleft}
1 $\phi$2 (r1 ) =$\surd$ 2 $\phi$1 (r1 )
\end{flushleft}





\begin{flushleft}
$\phi$2 (r2 ) $\phi$1 (r2 )
\end{flushleft}





(6)





\begin{flushleft}
Thus, we need a convention. If we have a two particle slater determinant |l, m|, then r1 , r2 |$\psi$
\end{flushleft}


\begin{flushleft}
f l,m
\end{flushleft}





\begin{flushleft}
1 $\phi$l (r1 ) = |l, m| = $\surd$ 2 $\phi$m (r1 )
\end{flushleft}





\begin{flushleft}
$\phi$l (r2 ) $\phi$m (r2 )
\end{flushleft}





(7)





\begin{flushleft}
for l $<$ m. For bosons, we get a space of symmetrized states which is the Fock's space for bosons. The bosonic wavefunctions are represented by permanents.
\end{flushleft}





2





\begin{flushleft}
Occupation number formalism
\end{flushleft}


\begin{flushleft}
f l,m ,
\end{flushleft}





\begin{flushleft}
Instead of writing Slater determinant state (7) as |$\psi$ |$\psi$
\end{flushleft}


\begin{flushleft}
f l,m l
\end{flushleft}





\begin{flushleft}
we can write it as (8)
\end{flushleft}





$\equiv$ |0, . . . , 0, 1 , 0, . . . , 0, 1 , 0, . . .


\begin{flushleft}
m
\end{flushleft}





\begin{flushleft}
i.e. we write '1's at positions l and m corresponding to participating single particle states in the slater determinant and '0's at all other positions. This is known as occupation number representation. Note that we have written (8) for a fermionic state. The same representation can be used to describe bosonic states as well. It is easy to see that, by construction, the state written in occupation number representation contains the information about the nature of particles (symmetry or anti-symmetry). Since, we are interested in situations where particle number is not ﬁxed, we want operators which can take us from one Fock space to another Fock space with diﬀerent particle number, say from Fn$-$1 to Fn . Thus, the states in a Fock space can be obtained by successively applying such operators on the vaccum state (the state with no particles).
\end{flushleft}





2.1





\begin{flushleft}
Fermionic Operators
\end{flushleft}





\begin{flushleft}
operator of c\dag{} and c\dag{} successively. m l |0 c\dag{} |0 m c\dag{} c\dag{} |0 l m
\end{flushleft}





\begin{flushleft}
\dag{} Let us deﬁne operator cn to be the operator which creates a particle in state |n . This is known as creation operator. There is a conjugate operator cn as well that destroys a particle from state |n and hence called an annihilation operator. Now we can create |$\psi$ f from the vacuum state by l,m
\end{flushleft}





= |0, . . . , 0, 0 , 0, . . . , 0, 0 , 0, . . .


\begin{flushleft}
l m
\end{flushleft}





= |0, . . . , 0, 0 , 0, . . . , 0, 1 , 0, . . .


\begin{flushleft}
l m
\end{flushleft}





\begin{flushleft}
= |0, . . . , 0, 1 , 0, . . . , 0, 1 , 0, . . . $\equiv$ |$\psi$
\end{flushleft}


\begin{flushleft}
l m
\end{flushleft}





\begin{flushleft}
f l,m
\end{flushleft}





(9)





2





\begin{flushleft}
\newpage
Now consider c\dag{} c\dag{} |0 = |$\psi$ m l Thus c\dag{} c\dag{} + c\dag{} c\dag{} |0 m l l m c\dag{} c\dag{} + c\dag{} c\dag{} m l l m = 0 = 0 (11)
\end{flushleft}


\begin{flushleft}
f m,l
\end{flushleft}





= $-$|0, . . . , 0, 1 , 0, . . . , 0, 1 , 0, . . .


\begin{flushleft}
l m
\end{flushleft}





(10)





\begin{flushleft}
We get the anti-commutation relation as a result of the anti-symmetry of the state. Using this, we can deﬁne the operation of c\dag{} on any state |n1 , n2 , . . . l
\end{flushleft}


\begin{flushleft}
\dag{} cl |n1 , . . . , nl$-$1 , nl , . . . $\propto$ ($-$1)n1 +n2 +···nl$-$1 |n1 , . . . , nl$-$1 , nl + 1, . . .
\end{flushleft}





(12)





\begin{flushleft}
The proportionality constant is ﬁxed using normalization condition which give
\end{flushleft}


\begin{flushleft}
\dag{} cl |n1 , . . . , nl$-$1 , nl , . . . = ($-$1) Pl$-$1
\end{flushleft}


\begin{flushleft}
i=0
\end{flushleft}





\begin{flushleft}
ni
\end{flushleft}





$\surd$





\begin{flushleft}
nl + 1|n1 , . . . , nl$-$1 , nl + 1, . . .
\end{flushleft}





(13)





\begin{flushleft}
For fermions, nl + 1 has to be considered modulo 2 due to Pauli exclusion principle. Similarly, the conjugate of this operator is deﬁned as cl |n1 , . . . , nl$-$1 , nl , . . . = ($-$1)
\end{flushleft}


\begin{flushleft}
Pl$-$1
\end{flushleft}


\begin{flushleft}
i=0
\end{flushleft}





\begin{flushleft}
ni $\surd$
\end{flushleft}





\begin{flushleft}
nl |n1 , . . . , nl$-$1 , nl $-$ 1, . . .
\end{flushleft}





(14)





\begin{flushleft}
Using these relations, we can check that, for fermions, cl cm + cm cl = \{cl , cm \} = 0
\end{flushleft}





\begin{flushleft}
cl c\dag{} + c\dag{} cl = \{cl , c\dag{} \} = $\delta$lm m m m
\end{flushleft}





(15)





2.2





\begin{flushleft}
Bosonic Operators
\end{flushleft}





\begin{flushleft}
Bosonic creation and annhilation operators can be deﬁned as $\surd$ nl + 1|n1 , . . . , nl$-$1 , nl + 1, . . . b\dag{} |n1 , . . . , nl$-$1 , nl , . . . = l $\surd$ bl |n1 , . . . , nl$-$1 , nl , . . . = nl |n1 , . . . , nl$-$1 , nl $-$ 1, . . . Bosonic operators follow commutation relations bl bm $-$ bm bl = [bl , bm ] = 0
\end{flushleft}





(16)





\begin{flushleft}
b\dag{} b\dag{} $-$ b\dag{} b\dag{} = [b\dag{} , b\dag{} ] = 0 m l l m l m
\end{flushleft}





\begin{flushleft}
bl b\dag{} $-$ b\dag{} bl = [bl , b\dag{} ] = $\delta$lm m m m
\end{flushleft}





(17)





\begin{flushleft}
Irrespective of the nature of particles, one can check that \{|n1 , n2 , . . . \} forms a orthonormal basis for a N -particle system. n′ , n′ , . . . |n1 , n2 , . . . = $\delta$n′ ,n1 $\delta$n′ ,n2 · · · 1 2 1 2 Also, a general state |n1 , n2 , . . . can be written as (a\dag{} )n1 (a\dag{} )n1 $\surd$2 · · · |0 |n1 , n2 , . . . = $\surd$1 n1 ! n2 ! where a\dag{} 's can be either bosonic or fermionic creation operators. 3 (19) (18)
\end{flushleft}





\newpage
3





\begin{flushleft}
Observables
\end{flushleft}





\begin{flushleft}
Does it make sense to talk of the momentum pi of a single particle i? The answer is NO, since the particles are indistinguishable. We can only talk about sums such as i pi
\end{flushleft}





3.1





\begin{flushleft}
Single Particle Operators
\end{flushleft}





\begin{flushleft}
The operators which involve sum over only single particles are known as single particle operators ˆ F1 =
\end{flushleft}


\begin{flushleft}
i
\end{flushleft}





\begin{flushleft}
ˆ f1 (ri , pi )
\end{flushleft}





(20)





\begin{flushleft}
In second quantization, this operator can be written as ˆ F1 =
\end{flushleft}


\begin{flushleft}
l,l′
\end{flushleft}





\begin{flushleft}
ˆ l|f1 |l′ a\dag{} al′ l
\end{flushleft}





(21)





\begin{flushleft}
where, ˆ l|f1 |l′ = ˆ $\phi$∗ (r)f1 (r, p)$\phi$l′ (r) l (22)
\end{flushleft}





\begin{flushleft}
r
\end{flushleft}





3.2





\begin{flushleft}
Two Particle Operators
\end{flushleft}





\begin{flushleft}
The operators which involve sum over two particles are known as two particle operators 1 ˆ F2 = 2 ˆ f2 (ri , pi ; rj , pj )
\end{flushleft}


\begin{flushleft}
i=j
\end{flushleft}





(23)





\begin{flushleft}
In second quantization, this operator can be written as ˆ F2 =
\end{flushleft}


\begin{flushleft}
l1 ,l2 ,l3 ,l4
\end{flushleft}





\begin{flushleft}
ˆ l1 l2 |f2 |l4 l3 a\dag{}1 a\dag{}2 al3 al4 l l
\end{flushleft}





(24)





\begin{flushleft}
where, ˆ l1 l2 |f2 |l4 l3 = ˆ $\phi$∗1 (r1 )$\phi$∗2 (r2 )f2 (r1 , p1 ; r2 , p2 )$\phi$l4 (r1 )$\phi$l3 (r2 ) l l (25)
\end{flushleft}





\begin{flushleft}
r1 ,r2
\end{flushleft}





\begin{flushleft}
The good thing about this representation is that it is independent of the nature of the particles.
\end{flushleft}





4





\begin{flushleft}
Change of Basis
\end{flushleft}





\begin{flushleft}
Suppose we have creation operators a\dag{} corresponding to the basis \{|$\lambda$ \}. Now we want to go to a $\lambda$ ˜ diﬀerent basis \{|$\lambda$ \}. What are a\dag{} s in terms of a\dag{} s? ˜ $\lambda$ $\lambda$ ˜ Since \{|$\lambda$ \} is a basis, we can write a state |$\lambda$ as ˜ |$\lambda$ = Thus, a\dag{} |0 = ˜ $\lambda$ ˜ $\lambda$|$\lambda$ a\dag{} |0 $\lambda$ (27)
\end{flushleft}


\begin{flushleft}
$\lambda$
\end{flushleft}





\begin{flushleft}
˜ $\lambda$|$\lambda$ |$\lambda$
\end{flushleft}





(26)





\begin{flushleft}
$\lambda$
\end{flushleft}





4





\begin{flushleft}
\newpage
This gives a\dag{} = ˜ $\lambda$
\end{flushleft}


\begin{flushleft}
$\lambda$
\end{flushleft}





\begin{flushleft}
˜ $\lambda$|$\lambda$ a\dag{} $\lambda$
\end{flushleft}





(28)





\begin{flushleft}
Taking conjugate of the above equation, we get a$\lambda$ = ˜
\end{flushleft}


\begin{flushleft}
$\lambda$
\end{flushleft}





\begin{flushleft}
˜ $\lambda$|$\lambda$ a$\lambda$
\end{flushleft}





(29)





\begin{flushleft}
Example: Suppose a\dag{} creates a particle at site i on a lattice. What is operator a\dag{} in the momentum i k space? Using the above formula, we have a\dag{} = k
\end{flushleft}


\begin{flushleft}
i
\end{flushleft}





\begin{flushleft}
i|k a\dag{} i 1 $\surd$ N eik·Ri a\dag{} i
\end{flushleft}


\begin{flushleft}
i
\end{flushleft}





=





(30)





5


5.1





\begin{flushleft}
Applications of Second Quantization
\end{flushleft}


\begin{flushleft}
1 Single spin- 2 operator
\end{flushleft}





\begin{flushleft}
A spin- 1 can be represented as 2 1 ˆ \{S i \}$\alpha$,$\alpha$′ = \{ˆ i \}$\alpha$,$\alpha$′ $\sigma$ 2 where, $\sigma$ i s are Pauli matrices ˆ $\sigma$x = 0 1 1 0 , $\sigma$y = 0 i $-$i 0 , $\sigma$z = 1 0 0 $-$1 (32) (31)
\end{flushleft}





\begin{flushleft}
The basis states here are eigen states of S z i.e. | $\uparrow$ ad | $\downarrow$ . This operator in second quantized language can be written as ˆ Si =
\end{flushleft}


\begin{flushleft}
$\alpha$,$\alpha$′
\end{flushleft}





\begin{flushleft}
ˆi c\dag{} S$\alpha$,$\alpha$′ c$\alpha$ $\alpha$ 1 2
\end{flushleft}


\begin{flushleft}
i c\dag{} $\sigma$$\alpha$,$\alpha$′ c$\alpha$ $\alpha$ˆ $\alpha$,$\alpha$′
\end{flushleft}





\begin{flushleft}
= which gives Sx = Sy = Sz =
\end{flushleft}





(33)





\begin{flushleft}
1 \dag{} (c c$\downarrow$ + c\dag{} c$\uparrow$ ) $\downarrow$ 2 $\uparrow$ 1 \dag{} \dag{} (c c$\downarrow$ $-$ c$\downarrow$ c$\uparrow$ ) 2i $\uparrow$ 1 \dag{} (c$\uparrow$ c$\uparrow$ $-$ c\dag{} c$\downarrow$ ) $\downarrow$ 2
\end{flushleft}





(34)





5.2





\begin{flushleft}
Local Density Operator
\end{flushleft}





\begin{flushleft}
Local density $\rho$(r) is deﬁned as $\rho$(r) =
\end{flushleft}


\begin{flushleft}
i
\end{flushleft}





\begin{flushleft}
$\delta$(r $-$ ri ) 5
\end{flushleft}





(35)





\begin{flushleft}
\newpage
In second quantized language, it is written as $\rho$(r) = dd r′ a\dag{} (r′ )$\delta$(r $-$ r′ )a(r′ ) (36)
\end{flushleft}





\begin{flushleft}
= a\dag{} (r)a(r) Total number of particles are given by ˆ N= dd ra\dag{} (r)a(r)
\end{flushleft}





(37)





5.3





\begin{flushleft}
Free Electrons
\end{flushleft}





\begin{flushleft}
If the potential is constant, then we have free particles. The free particle Hamiltonian is given by H0 = k2 \dag{} c ck$\sigma$ 2m k$\sigma$ (38)
\end{flushleft}





\begin{flushleft}
k$\sigma$
\end{flushleft}





\begin{flushleft}
The ground state of such system can be easily determined. For a system of size Lx × Ly × Lz , the allowed momentum states k have components ki = 2$\pi$ni /Li where ni ǫ Z. Each such state because of Pauli exclusion principle can accomodate two electrons ($\uparrow$, $\downarrow$). Thus, if we have N electrons, all states upto an energy EF , called Fermi energy, are occupied. The momentum kF corresponding to 2 kF Fermi energy ( 2m ) is called Fermi momentum. All the k points which have smaller momentum than Fermi momentum forms the Fermi surface. In second quantized language, the ground state can be written as |F S = c\dag{} |0 k$\sigma$ (39)
\end{flushleft}





\begin{flushleft}
|k|$<$kF ,$\sigma$
\end{flushleft}





\begin{flushleft}
In most of the problems, we are interested in excitations about this state. For example, destroying a electron inside Fermi surface and creating one outside. Can we deﬁne operators such that this state is the ''physical vacuum'' of the theory? c\dag{} = ˜k$\sigma$ c\dag{} k$\sigma$ ck$\sigma$ if k $>$ kF if k $\leq$ kF ck$\sigma$ = ˜ ck$\sigma$ c\dag{} k$\sigma$ if k $>$ kF if k $\leq$ kF (40)
\end{flushleft}





\begin{flushleft}
We can see that if we apply ck$\sigma$ on |F S , we will get zero. ˜
\end{flushleft}





5.4





\begin{flushleft}
Free electrons in a magnetic ﬁeld
\end{flushleft}





\begin{flushleft}
Suppose we have N electrons in a magnetic ﬁeld. The Hamiltonian for such a system is given by H= 1 2m eA(ri ) pi $-$ c
\end{flushleft}


2





\begin{flushleft}
i
\end{flushleft}





\begin{flushleft}
$-$$\gamma$
\end{flushleft}





\begin{flushleft}
i
\end{flushleft}





\begin{flushleft}
$\sigma$i · B(ri )
\end{flushleft}





(41)





\begin{flushleft}
The Hamiltonian in the second quantization language can be written as H= eA(r) 1 \dag{} c$\sigma$ (r) $-$i$\nabla$r $-$ d r 2m c
\end{flushleft}


\begin{flushleft}
d 2
\end{flushleft}





\begin{flushleft}
$\sigma$
\end{flushleft}





\begin{flushleft}
c$\sigma$ (r) $-$ $\gamma$
\end{flushleft}





\begin{flushleft}
$\alpha$,$\alpha$′
\end{flushleft}





\begin{flushleft}
dd r c\dag{} (r)$\sigma$$\alpha$,$\alpha$′ c$\alpha$′ (r) · B(r) $\alpha$
\end{flushleft}





(42)





6





\newpage
5.5





\begin{flushleft}
Non-interacting particles
\end{flushleft}


\begin{flushleft}
pi ˆ2 + V (ri ) 2m
\end{flushleft}





\begin{flushleft}
Consider spinless non-interacting particles. The Hamiltonian is given by H= (43)
\end{flushleft}





\begin{flushleft}
i
\end{flushleft}





\begin{flushleft}
In real space basis, the second quantized Hamiltonian can be written as H= ˆ where p = $-$i$\nabla$r dd ra\dag{} (r) p2 ˆ + V (r) a(r) 2m (44)
\end{flushleft}





5.6





\begin{flushleft}
Electrons in a periodic potential
\end{flushleft}





\begin{flushleft}
Eigen-states of a periodic Hamiltonian are the Bloch states which are represented as $\psi$kn (r) = eik·r ukn (r) (45)
\end{flushleft}





\begin{flushleft}
where ukn (r) has the same periodicity as V (r). k is the crystal momentum which takes values inside the Brillouin zone and n is the band index. Let us assume that the bands are well separated and we are interested in the lowest band only. The Hamiltonian in the momentum space can be written as H= ǫk c\dag{} ck$\sigma$ k$\sigma$
\end{flushleft}


\begin{flushleft}
k$\sigma$
\end{flushleft}





(46)





5.7





\begin{flushleft}
Coulomb Interaction
\end{flushleft}


\begin{flushleft}
e2 |ri $-$ rj |
\end{flushleft}





\begin{flushleft}
The term corresponding to Coulomb interaction is given by Hee = In momentum basis, this can be written as Hee = 1 2 k1 $\sigma$1 , k2 $\sigma$2 | e2 \dag{} |k4 $\sigma$4 , k3 $\sigma$3 ck1 $\sigma$1 c\dag{} 2 $\sigma$2 ck3 $\sigma$3 ck4 $\sigma$4 k |ri $-$ rj | (48) 1 2 (47)
\end{flushleft}





\begin{flushleft}
i=j
\end{flushleft}





\begin{flushleft}
\{ki \},\{$\sigma$i \}
\end{flushleft}





\begin{flushleft}
Then, the overlap matrix element is given by k1 $\sigma$1 , k2 $\sigma$2 | e2 |k4 $\sigma$4 , k3 $\sigma$3 |ri $-$ rj | = = =
\end{flushleft}





\begin{flushleft}
eik4 ·r eik3 ·r e$-$ik2 ·r e2 · $\surd$ · $\surd$ $\delta$ $\sigma$1 $\sigma$4 $\delta$ $\sigma$2 $\sigma$3 d r d r $\surd$ · $\surd$ · |r $-$ r′ | V V V V e2 $\delta$ $\sigma$1 $\sigma$4 $\delta$ $\sigma$2 $\sigma$3 ′ ′ d3 r d3 r′ e$-$i(k1 +k2 $-$k3 $-$k4 )·r e$-$i(k1 $-$k4 )·(r$-$r ) V2 |r $-$ r′ | e2 $\delta$ $\sigma$1 $\sigma$4 $\delta$ $\sigma$2 $\sigma$3 ′ d3 $\rho$ e$-$i(k1 $-$k4 )·$\rho$ d3 r′ e$-$i(k1 +k2 $-$k3 $-$k4 )·r V2 $\rho$
\end{flushleft}


\begin{flushleft}
3 $-$ik1 ·r 3 ′e
\end{flushleft}





′





′





= =





\begin{flushleft}
$\delta$ $\sigma$1 $\sigma$4 $\delta$ $\sigma$2 $\sigma$3 V2 $\delta$ $\sigma$1 $\sigma$4 $\delta$ $\sigma$2 $\sigma$3 $\delta$(k1 + k2 $-$ k3 $-$ k4 ) V 7
\end{flushleft}





\begin{flushleft}
Change of variable: r$-$r′ =$\rho$ e2 × V $\delta$(k1 + k2 $-$ k3 $-$ k4 ) × (k1 $-$ k4 )2
\end{flushleft}


\begin{flushleft}
V $\delta$(k1 +k2 $-$k3 $-$k4 )
\end{flushleft}





(49)





\begin{flushleft}
\newpage
Figure 1: Feynman diagram depicting the interaction of two particles via Coulomb interaction. We see that the Coulomb interaction conserves the total momentum. Thus, Hee = e2 2
\end{flushleft}


\begin{flushleft}
\dag{} c\dag{} (k+q)$\sigma$ c(k′ $-$q)$\sigma$ ′
\end{flushleft}





\begin{flushleft}
k,k′ ,q,$\sigma$,$\sigma$ ′
\end{flushleft}





\begin{flushleft}
1 q2
\end{flushleft}





\begin{flushleft}
ck′ $\sigma$′ ck$\sigma$
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(50)





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This interaction of two particles via Coulomb interaction is shown in the Feynman diagram in ﬁg. 1. Note that for q = 0, the matrix element diverges but this term in a lattice is cancelled by the uniform background of positive ionic charge. This model is known as Jellium model.
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5.8


5.8.1





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Tight Binding Model
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Wannier States
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Suppose we have atomic orbitals localized at atomic sites in a lattice. Let's assume that the ''physical extent'' of such orbitals is smaller than the inter-atomic spacing. In such cases, we say that the orbitals are tightly bound to the lattice centers. It is thus, convenient to work in the local basis that corresponds to the atomic orbital states of the isolated ion. Wannier state localized at site Ri is deﬁned as 1 |$\psi$i = $\surd$ N e$-$ik·Ri |$\psi$k (51)
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k
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where |$\psi$k 's are the Bloch states. It is easy to see that this set of states forms an orthogonal basis. Thus, |r = This gives us c\dag{} (r) = $\sigma$
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i ∗ $\psi$i (r)c\dag{} i$\sigma$
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i
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$\psi$i |r |$\psi$i
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(52)





(53)





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Similarly, invertng (51) gives us 1 c\dag{} = $\surd$ k$\sigma$ N eik·Ri c\dag{} i$\sigma$
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i
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(54)





8





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\newpage
Thus, Bloch Hamiltonian (46) in this basis becomes H= where tij = site j.
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1 N k \dag{} ǫk ck$\sigma$ ck$\sigma$ = k$\sigma$
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1 N
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ij
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k$\sigma$
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eik·(Ri $-$Rj ) ǫk c\dag{} cj$\sigma$ $\equiv$ i$\sigma$
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tij c\dag{} cj$\sigma$ i$\sigma$
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ij$\sigma$
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(55)





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eik·(Ri $-$Rj ) ǫk is known as hopping amplitude for the electron to go from site i to
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9





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