# Assignment1 - Advanced Quantum Mechanics ```Assignment I
Diogo Ribeiro
Assignment I
Diogo Ribeiro (87311)
diogo.c.ribeiro@tecnico.ulisboa.pt
Problem 1
Show that a general two-particle operator (acting on a n-particle state) can be written as:
X
O=
hα, β| O(2) |γ, δi a†α a†β aδ aγ
α,β,γ,δ
where O(2) is the two-particle operator that acts on two-particle states.
Solution:
Lets start by looking at the simple case where the two-particle operator O(2) is given by:
O(2) = |α, βihγ, δ|
(1)
In this case, when we act on a two-particle state |a, bi we get:
O(2) |a, bi = |α, βihγ, δ| |a, bi = hγ|ai hδ|bi |α, βi
(2)
So, we are interested in finding the operator O that applies O(2) to every possible combination
of two single-particle states in a n-state function:
O |Ψi =
hγ, δ|ψ1 , ψ2 i |α, β, ψ3 , &middot; &middot; &middot; , ψn i
+ hγ, δ|ψ1 , ψ3 i |α, ψ2 , β, &middot; &middot; &middot; , ψn i
+ (&middot; &middot; &middot; )
+ hγ, δ|ψ1 , ψn i |α, ψ2 , ψ3 , &middot; &middot; &middot; , βi
+ hγ, δ|ψ2 , ψ1 i |β, α, ψ3 , &middot; &middot; &middot; , ψn i
+ hγ, δ|ψ2 , ψ3 i |ψ1 , α, β, &middot; &middot; &middot; , ψn i
+ (&middot; &middot; &middot; ) =
X
=
hγ, δ|ψi , ψj i |ψ1 , &middot; &middot; &middot; , ψi−1 , α, ψi+1 &middot; &middot; &middot; , ψj−1 , β, ψj+1 , &middot; &middot; &middot; , ψn i
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
i6=j
As we did in class, lets recall the definition of the number ζ:
(
1
Bosons
ζ=
(−1) Fermions
October 2019
1
(11)
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
With this we can account for both cases at the same time when we do a permutation in an
n-particle state. So, returning to our state, lets bring α to the front of the state:
X
hγ, δ|ψi , ψj i |ψ1 , &middot; &middot; &middot; , ψi−1 , α, ψi+1 &middot; &middot; &middot; , ψj−1 , β, ψj+1 , &middot; &middot; &middot; , ψn i =
(12)
i6=j
=
X
hγ, δ|ψi , ψj i ζ i−1 |α, ψ1 , &middot; &middot; &middot; ψi−1 , ψi+1 &middot; &middot; &middot; , ψj−1 , β, ψj+1 , &middot; &middot; &middot; , ψn i
(13)
i6=j
= a†α
X
hγ, δ|ψi , ψj i ζ i−1 |α, ψ1 , &middot; &middot; &middot; ψi−1 , ψi+1 &middot; &middot; &middot; , ψj−1 , β, ψj+1 , &middot; &middot; &middot; , ψn i
(14)
i6=j
In the process of permutating α with each left neighbor, the state beta might have changed
the position it took with respect to the beginning of state. If α was after β , then it remained
in position j. On the contrary, if α was before β, the state moved to position j − 1. To
account for this lets define the usefull variable:
(
ζ j−1
if j &gt; i
ηij =
(15)
j−2
ζ
if j &lt; i
With this we can bring β to the ”front line” by:
a†α
X
hγ, δ|ψi , ψj i ζ i−1 |ψ1 , &middot; &middot; &middot; , ψj−1 , β, ψj+1 , &middot; &middot; &middot; , ψn i =
(16)
hγ, δ|ψi , ψj i ζ i−1 ηij |β, ψ1 , &middot; &middot; &middot; (no ψi , ψj ) &middot; &middot; &middot; , ψn i =
(17)
i6=j
a†α
X
i6=j
a†α a†β
X
hγ, δ|ψi , ψj i ζ i−1 ηij |ψ1 , &middot; &middot; &middot; (no ψi , ψj ) &middot; &middot; &middot; , ψn i
(18)
i6=j
Lets now look back on the definition of the destruction operator. If we apply consecutively
two destruction operator to a n-particle state we get the following:
aα aβ |ψ1 , &middot; &middot; &middot; , ψn i =
n
X
= aα
ζ i−1 hβ|ψi i |ψ1 , &middot; &middot; &middot; (no ψi ) &middot; &middot; &middot; , ψn i
(19)
(20)
i
=
n
X
i
=
X
ζ i−1 hβ|ψi i
X
ηij hα|ψj i |ψ1 , &middot; &middot; &middot; (no ψi , ψj ) &middot; &middot; &middot; , ψn i
(21)
j6=i
hβ, α|ψi , ψj i ζ i−1 ηij |ψ1 , &middot; &middot; &middot; (no ψi , ψj ) &middot; &middot; &middot; , ψn i
(22)
j6=i
Comparing with the equation we had we see that our operator action becomes:
X
a†α a†β
hγ, δ|ψi , ψj i ζ i−1 ηij |ψ1 , &middot; &middot; &middot; (no ψi , ψj ) &middot; &middot; &middot; , ψn i = a†α a†β aδ aγ
(23)
i6=j
Considering now the most general operator O(2) we have:
X
X
O(2) =
|α, βihα, β| O(2) |γ, δihγ, δ| =
hα, β| O(2) |γ, δi |α, βihγ, δ|
α,β,γ,δ
October 2019
(24)
α,β,γ,δ
2
Prof. Guilherme Milhano
Assignment I
And this implies that its action on a n particle state is just:
X
O=
hα, β| O(2) |γ, δi a†α a†β aδ aγ
Diogo Ribeiro
(25)
α,β,γ,δ
October 2019
3
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
Problem 2
Take the expression from I or the equivalent one obtained in the lectures. Rewrite it in a
form relevant to be applied to a system of fermions with a ground state of G particles, that
is in terms of creation/annihilation operators for particles and holes. Interpret physically
each of terms obtained.
Solution:
We start by recalling the definition of the annihilation operator for holes bα and the associated
commutation relations. As we defined it, if we have a ground state, the destruction operator
for holes is defined as:
bα = a†α
(α ≤ G)
(26)
So that, destroying a hole is equivalent to creating a particle with the same index. As so,
we can write the commutation relations by means of the usual creation and annihilation
operators. If we assume that α &gt; G and β ≤ G we have that:
i
h
[aα , aα0 ]+ = [aα , bβ ]+ = aα , b†β = [bβ , bβ 0 ]+ = 0
(27)
+
i
h
†
bβ , bβ 0 = δβ,β 0
i
h
†
aα , aα0 = δα,α0
(28)
+
+
To rewrite the operator O in a useful form we need to separate the sums for all combinations
of α, β, γ, δ since these can separately be less than or grater than G, forming a total of 16
different terms. To keep track of all of them, it is useful to make a table. In the first 4
columns, a 1 denotes that the index is greater than G and a 0 that it is less than or equal
so that we can associate a binary number to each separate term. I shall refer to this binary
number as the signature s.
Also, we want to bring the operator into normal form (i.e. With all creation operators to
the left) and in the table, the steps to get there by means of the commutation relations are
presented.
It is assumed that, at every step we are summing over all indexes with the conditions imposed
by the binary code. Acknowledging this we make use of simple fact regarding the operator
definition. Restating it in terms of general operators ci we have:
X
X αβ
O=
hα, β| O(2) |γ, δi cα cβ cγ cδ ≡
Oγδ cα cβ cγ cδ
(29)
α,β,γ,δ
α,β,γ,δ
Since we are dealing with fermions we have that permuting the matrix elements states
implies:
X αβ
X βα
X βα
X αβ
Oδγ cα cβ cγ cδ (30)
Oγδ cα cβ cγ cδ = −
Oγδ cα cβ cγ cδ =
Oδγ cα cβ cγ cδ = −
α,β,γ,δ
α,β,γ,δ
α,β,γ,δ
α,β,γ,δ
αβ
We can now relabel indexes so that we make all Oijkl terms equal to Oγδ
(this will allow us
to sum similar terms):
X αβ
X αβ
X αβ
X αβ
Oγδ cα cβ cγ cδ = −
Oγδ cβ cα cγ cδ =
Oγδ cβ cα cδ cγ = −
Oγδ cα cβ cδ cγ (31)
α,β,γ,δ
October 2019
α,β,γ,δ
α,β,γ,δ
4
α,β,γ,δ
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
As so, we can change the sign of a summation term just by trading labels in the operators:
by changing α → β we gain a minus sign and the same goes for γ → δ. An important thing
to note is that, after the change, the indexes do not necessarily respect the same conditions.
As an example, consider the case of α, δ, γ &gt; G and β ≤ G (s = 1011).The operators in the
original definition become:
(s = 1011)
a†α bβ aδ aγ = −a†β bα aδ aγ
(s = 0111)
(32)
So that, by trading indexes we changed the signature of the term. The boxed equations in
the table represent the simplified terms after indexes are swapped (to joint similar terms)
and next to it the signature is presented. The colored boxes indicate similar terms that can
be joined after the appropriate change in the indexes. Here’s the table:
α β γ
δ
a†α a†β aδ aγ
bα bβ b†δ b†γ = bα (δβδ − b†δ bβ )b†γ
= δβδ bα b†γ − bα b†δ bβ b†γ
= δβδ (δαγ − b†γ bα ) − (δαδ − b†δ bα )bβ b†γ
= δβδ δαγ − δβδ b†γ bα − δαδ bβ b†γ + b†δ bα bβ b†γ
0
0
0
0
= δβδ δαγ − δβδ b†γ bα − δαδ (δβγ − b†γ bβ ) + b†δ bα (δβγ − b†γ bβ )
= δβδ δαγ − δβδ b†γ bα − δαδ δβγ + δαδ b†γ bβ + δβγ b†δ bα − b†δ bα b†γ bβ
= δβδ δαγ − δβδ b†γ bα − δαδ δβγ + δαδ b†γ bβ + δβγ b†δ bα − b†δ (δαγ − b†γ bα )bβ
= δβδ δαγ − δβδ b†γ bα − δαδ δβγ + δαδ b†γ bβ + δβγ b†δ bα − δαγ b†δ bβ + b†δ b†γ bα bβ
= (δβδ δαγ − δαδ δβγ ) + (−δβδ b†γ bα + δαδ b†γ bβ + δβγ b†δ bα − δαγ b†δ bβ ) + b†δ b†γ bα bβ
= 2δβδ δαγ + 4δαδ b†γ bβ + b†δ b†γ bα bβ
(0000)
bα bβ aδ b†γ = − bα bβ b†γ aδ
= − bα (δβγ − b†γ bβ )aδ
= − δβγ bα aδ + bα b†γ bβ aδ
0
0
0
1
= − δβγ bα aδ + (δαγ − b†γ bα )bβ aδ
= − δβγ bα aδ + δαγ bβ aδ − b†γ bα bβ aδ
= (δαγ bβ aδ − δβγ bα aδ ) − b†γ bα bβ aδ = 2δαγ bβ aδ + b†γ bβ bα aδ
October 2019
5
(0001)
Prof. Guilherme Milhano
Assignment I
α β γ
δ
Diogo Ribeiro
a†α a†β aδ aγ
bα bβ b†δ aγ = bα (δβδ − b†δ bβ )aγ
= δβδ bα aγ − bα b†δ bβ aγ
0
0
1
0
= δβδ bα aγ − (δαδ − b†δ bα )bβ aγ
= (δβδ bα aγ − δαδ bβ aγ ) + b†δ bα bβ aγ = 2δβδ bα aγ + b†δ bα bβ aγ
0
0
1
1
bα bβ aδ aγ
(0010)
(0011)
bα a†β b†δ b†γ = − a†β bα b†δ b†γ
= − a†β (δαδ − b†δ bα )b†γ
= − δαδ a†β b†γ + a†β b†δ bα b†γ
0
1
0
0
= − δαδ a†β b†γ + a†β b†δ (δαγ − b†γ bα )
= − δαδ a†β b†γ + δαγ a†β b†δ − a†β b†δ b†γ bα
= (δαγ a†β b†δ − δαδ a†β b†γ ) − a†β b†δ b†γ bα = 2δαγ a†β b†δ + a†β b†γ b†δ bα
(0100)
bα a†β aδ b†γ = − a†β bα aδ b†γ
= a†β bα b†γ aδ
0
1
0
1
= a†β (δαγ − b†γ bα )aδ
= δαγ a†β aδ − a†β b†γ bα aδ =
δαγ a†β aδ − a†β b†γ bα aδ
(0101)
bα a†β b†γ aγ = − a†β bα b†γ aγ
0
1
1
0
= − a†β (δαδ − b†δ bα )aγ
= − δαδ a†β aγ + a†β b†δ bα aγ ⇒ δαγ a†β aδ − a†β b†γ bα aδ
0
1
1
1
October 2019
bα a†β aδ aγ = − a†β bα aδ aγ = a†β bα aγ aδ
6
(0101)
(0111)
Prof. Guilherme Milhano
Assignment I
α β γ
δ
Diogo Ribeiro
a†α a†β aδ aγ
a†α bβ b†δ b†γ = a†α (δβδ − b†δ bβ )b†γ
= δβδ a†α b†γ − a†α b†δ bβ b†γ
1
0
0
0
= δβδ a†α b†γ − a†α b†δ (δβγ − b†γ bβ )
= (δβδ a†α b†γ − δβγ a†α b†δ ) + a†α b†δ b†γ bβ = 2δβδ a†α b†γ + a†α b†δ b†γ bβ
(1000)
a†α bβ aδ b†γ = − a†α bβ b†γ aδ
1
0
0
= − a†α (δβγ − b†γ bβ )aδ
1
= − δβγ a†α aδ + a†α b†γ bβ aδ ⇒ δβδ a†α aγ − a†α b†δ bβ aγ
(1010)
a†α bβ b†δ aγ = a†α (δβδ − b†δ bβ )aγ
1
0
1
0
= δβδ a†α aγ − a†α b†δ bβ aγ = δβδ a†α aγ − a†α b†δ bβ aγ
a†α bβ aδ aγ
1
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
a†α a†β b†δ aγ
1
1
1
1
a†α a†β aδ aγ
October 2019
a†α a†β b†δ b†γ
(1010)
(1011)
(1100)
a†α a†β aδ b†γ = −a†α a†β b†γ aδ = a†β a†α b†γ aδ
(1110)
(1110)
(1111)
7
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
Joining now the similar terms we can write all the terms of the operator in yet another
table. The column on the right gives the intuition to what the term does. Lets begin with
the scalar ones:
0-Operator Terms
The terms with no operators are just scalar terms. We have one single term like this. It
arises from the change of reference from the vacuum state to the new ground state.
2-Operator Terms
This terms refer to processes involving single particles just like we saw with the general
particle operator in class.
4-Operator Terms
Finally, we have the four operator terms. These refer to interactions and double processes
that are not completely disconnected. For example, we have terms that might be interpreted
as stimulated deexitation and excitation. We present the terms in the next page. It is
interesting to note that the holes behave just like particles and can interact with each other.
Also the sign of the term is not important since we can always exchange indexes (and by
doing so the order by which the particles/holes are created/destroyed).
Scalar Terms
X αβ
X αβ
Oβα
2
Oγδ δβδ δαγ = 2
α,β≤G
γ,δ≤G
α,β≤G
Two-Operator Terms
X αβ
X X αβ †
4
Oγδ δαδ b†γ bβ = −4
Oαδ bδ bβ
α≤G β≤G
δ≤G
α,β≤G
γ,δ≤G
X
4
αβ
Oγδ
δαγ bβ aδ = 4
X
αβ
Oγδ
δαγ a†β b†δ = 4
α,γ,δ≤G
β&gt;G
4
X
α,γ≤G
β,δ&gt;G
October 2019
αβ
Oαδ
bβ aδ
α≤G β≤G
δ&gt;G
α,β,γ≤G
δ&gt;G
4
XX
αβ
Oγδ
δαγ a†β aδ = 4
XX
αβ † †
Oαδ
aβ b δ
α≤G β&gt;G
δ≤G
XX
αβ †
Oαδ
aβ aδ
α≤G β&gt;G
δ&gt;G
8
Interpretation
Ground State Energy Shift
Interpretation
Hole Movement
Particle-Hole pair
destruction
Particle-Hole pair
creation
Particle Movement
Prof. Guilherme Milhano
Assignment I
Four-Operator Terms
X αβ †
Oγδ bδ b†γ bα bβ
Interpretation
Hole-Hole interaction
α,β≤G
γ,δ≤G
X
2
αβ †
Oγδ
b γ b β b α aδ
Hole movement
Particle Deexitation
α,β,γ≤G
δ&gt;G
X
αβ
b α b β aδ aγ
Oγδ
Two Particle
Deexitation
α,β≤G
γ,δ&gt;G
X
2
αβ † † †
aβ b γ b δ b α
Oγδ
Hole movement
Particle Exitation
α,γ,δ≤G
β&gt;G
−4
X
αβ † †
Oγδ
aβ b γ b α aδ
Hole movement
Particle movement
α,γ≤G
β,δ&gt;G
2
X
αβ †
Oγδ
aα bβ aδ aγ
Particle Movement
Particle Deexites
α,γ,δ&gt;G
β≤G
X
αβ † † † †
Oγδ
aα aβ b δ b γ
Two Particle
Exitation
α,β&gt;G
γ,δ≤G
2
X
αβ † † †
Oγδ
aβ aα bγ aδ
Particle Movement
Particle Exites
α,β,δ&gt;G
γ≤G
X
αβ † †
Oγδ
aα aβ aδ aγ
Particle Paticle
interaction
α,β&gt;G
γ,δ&gt;G
October 2019
Diogo Ribeiro
9
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
Problem 3
Consider an interacting electron gas (the two-body interaction is given by a Coulomb potencial). Search for ’Wannier orbitals’ which are the eigenstates of this systems. Write the
second quantized Hamiltonian.
Solution:
For the problem, I will consider the generic Hamiltonian for the electron Gas:
H =T +U +V
(33)
where T is the kinectic energy operator, U is some generic potential and V is the coulomb
interaction between electrons. Denoting by u(r) the external potential felt by an electron at
position r and v(r, r0 ) the coulomb interaction:
e2
1
v(ri , rj ) = v(ri − rj ) =
4π0 |ri − rj |
(34)
We can write each operator separately as:
N
T =−
~2 X 2
∇
2m i=1 i
U=
N
X
u(ri )
V =
i=1
1X
v(ri , rj )
2 i6=j
(35)
To write the Hamiltonian in second quantization formalism we need to choose a basis. The
momentum eigenstates form a complete and orthonormal set of functions and simplify our
life when calculating boring integrals. As so we use as our basis the eigenfunctions:
1
eik&middot;r
(36)
(2π)3/2
The momentum operator is a single particle operator so its second quantized form is:
φk (r) =
T =
X
hk0 |
k,k0
p2
|ki c†k0 ck
2m
(37)
Calculating the matrix element:
p2
hk |
|ki =
2m
0
Z
3
dr
~2 k2
=
2m
1
0
e−ik &middot;r
3/2
(2π)
1
(2π)3
Z
~2 2
1
ik&middot;r
e
−
∇
2m
(2π)3/2
i(k−k0 )&middot;r
3
dre
{z
|
=
~2 k2
δk 0 k
2m
(38)
(39)
}
δk 0 k
As we see, the matrix elements for the operator T are diagonal, as one would expect as the
eigenfunctions of p (our chosen basis) are also eigenfunctions of p2 /2m. Plugging the result
back in the quantized form of T we have:
T =
X ~2 k2
k
October 2019
2m
10
c†k ck
(40)
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
Since the external potential is also a one particle operator, its second quantized form is, just
like for T :
X
U=
hk0 | u |ki c†k0 ck
(41)
k,k0
And similarly:
Z
0
3
hk | u |ki =
dr
=
1
(2π)3
|
1
1
−ik0 &middot;r
ik&middot;r
u(r)
e
e
(2π)3/2
(2π)3/2
Z
−i(k0 −k)&middot;r
3
d r u(r) e
{z
(42)
= uk0 −k
(43)
}
uk0 −k
Where uk0 −k denotes the Fourier transform of the potential. Changing the name of variables
so that q = k0 − k we can rewrite the potential operator as:
X
X
U=
uk−k0 c†k0 ck =
uq c†k+q ck
(44)
k,k0
k,q
In this last form we see that we are describing a scattering process from a state with momentum k to a state with momentum k + q.
Finally, we have the two particle operator related to the Coulomb interaction. As we saw in
the first problem, in second quantization formalism, it is written as:
X
V =
= hk1 , k2 | v |k4 , k3 i c†k1 c†k2 ck3 ck4
(45)
k1 ,k2 ,k3 ,k4
The matrix elements are a bit more complicated than the single particle operators but,
thanks to the choice of momentum basis, relatively easy:
ik4 &middot;r ik3 &middot;r0 −ik1 &middot;r −ik2 &middot;r0 Z
Z
e
e
e
e
0
3
3 0
v(r − r )
hk1 , k2 | v |k4 , k3 i = d r d r
3/2
3/2
3/2
(2π)
(2π)
(2π)
(2π)3/2
1
=
(2π)6
Z
3
dr
Z
3 0
0
−i(k1 −k4 )&middot;r
d r v(r − r ) e
e
−i(k2 −k3 )&middot;r0
(46)
If we now change one of the variables of integration, namely r to R = r − r0 we have:
Z
Z
1
1
3
−i(k1 −k4 )&middot;R
3 0
−i(k2 −k3 +k1 −k4 )&middot;r0
d
R
v(R)
e
d
r
e
= vk1 −k4 δk1 ,k3 +k4 −k2
(2π)3
(2π)3
|
{z
} |
{z
}
vk1 −k4
δk1 ,k3 +k4 −k2
(47)
denotes the fourier transform of v. Plugging it back
Where, as in the potential term, vk1 −k4
in the quantized operator we have:
X
X
vk3 −k2 c†k3 +k4 −k2 c†k2 c†k3 ck4
V =
= vk1 −k4 δk1 ,k3 +k4 −k2 c†k1 c†k2 ck3 ck4 =
(48)
k2 ,k3 ,k4
k1 ,k2 ,k3 ,k4
Renaming the labels as k4 = k , k3 = k0 and k3 − k2 = q we have:
October 2019
11
Prof. Guilherme Milhano
Assignment I
X
V =
=
X
Diogo Ribeiro
vq c†k+q c†k0 −q ck0 ck
(49)
k,k0 ,q
k1 ,k2 ,k3 ,k4
Similarly to the one particle potential operator, by relabeling the indexes we get a much
bigger insight of the problem. The operator describes the scattering of two particles by
exchange of momentum q with amplitude vq . To calculate explicitly this amplitude we have
to solve the integral:
Z
Z
1
1
e2
−i q&middot;R
3
−i q&middot;R
3
vq =
e
(50)
d
R
v(R)
e
=
d
R
(2π)3
(2π)3
4π0 R
To do so, we choose spherical coordinates with the z-axis pointing in the direction of q:
Z π Z 2π
Z
Z ∞
e−i qR cos θ 2
e−i q&middot;R
3
dφ
dθ
dR
dR
=
R sin θ
(51)
R
R
0
0
0
The φ integral is straigthforward and letting w = cos θ we have that:
Z
e−i q&middot;R
= −2π
dR
R
Z
3
∞
Z
1
−i qRw
dR
dw wR e
−1
0
2πi
=−
q
Z
∞
dR e−i qR − ei qR
(52)
0
To solve this last integral we need to use a simple trick so that we get a finite result. Lets
multiply the integrand by a real decreasing exponential:
Z
∞
dR e
0
−λR
−i qR
e
−e
i qR
∞
e−(i q+λ)R e(i q−λ)R
=
+
iq + λ
iq − λ 0
∞
−(i q+λ)R
(iq − λ)e
(iq + λ)e(i q−λ)R
=
+
q 2 + λ2
q 2 + λ2
0
2iq
λ→0 2i
−−→
= 2
2
q +λ
q
If we let λ → 0 we get the integral we were seeking and obtain finally:
Z
e−i q&middot;R
4π
e2
d3 R
= 2
=⇒ vq =
R
q
(2π)3 0 q 2
(53)
(54)
(55)
(56)
It is important to note that q cannot not be zero. This makes sense since there is no
interaction if the exchange of momentum is 0. Putting everything together we arrive at the
second quantization Hamiltonian:
X
X X ~ 2 k2 †
e2
†
H=
ck ck +
uq ck+q ck +
c†k+q c†k0 −q ck0 ck
(57)
3 q2
2m
(2π)
0
k
k,q
k,k0 ,q
A Detour into Solid State Physics
Consider now that this electron gas is in a perfect crystal. In this case, we have a latticeperiodic potential such that:
UL (r) = UL (r + Rn )
October 2019
12
(58)
Prof. Guilherme Milhano
Assignment I
Diogo Ribeiro
Rn being any lattice vector. According to Bloch’s theorem, the eigenstates of such system
can be written as:
Ψk (r) = eik&middot;r uk (r)
(59)
where uk (r) has the periodicity of the potential UL . These are just the momentum eigenfunctions we used before modulated by a periodic function uk . We can also think the other
way around and say that the uk functions are being modulated by an envelope eik&middot;r . This
envelope makes the wave function very broad in space. We can try to get a more localized
wavefunction. To do so, we can combine many Bloch functions over a large interval of k
values, but, since we are working in a crystal and all electron waves are described by vectors
inside the Brillouin zone, the best we can do is to sum up all Bloch states with k inside
the Brillouin zone. Mathematicaly we can write this as a Fourier transform of the Bloch’s
functions:
Z
V
d3 k e−ik&middot;R Ψk (r)
(60)
w(r − R) ≡ wR (r) =
3
(2π) BZ
These are the so called Wannier functions we were asked to find. We see that, for each
lattice vector R we have a Wannier state |Ri with the respective creation and annihilation
operators c†R and cR .Note also that we cannot write the Wannier functions explicitly without
knowing the underlying potential. Nevertheless, we can sketch how one would proceed to
write the Hamiltonian in this basis. The operators would take the form:
X Z
p2
~2 2
†
3
∗
|Ri cR0 cR =
∇ wR (r) c†R0 cR
d r wR0 (r) −
T =
hR |
2m
2m
R,R0
R,R0
U=
X
0
X
0
hR
| u |Ri c†R0 cR
R,R0
V =
X
=
X Z
3
dr
∗
wR
0 (r)
(61)
u(r) wR (r) c†R0 cR
(62)
R,R0
hR1 , R2 | v |R4 , R3 i c†R1 c†R2 cR3 cR4 =
(63)
R1 ,R2
R3 ,R4
=
X Z
3
dr
Z
3 0
dr
∗
∗
wR
(r)wR
(r)
1
2
v(r − r ) wR4 (r)wR3 (r) c†R1 c†R2 cR3 cR4
0
R1 ,R2
R3 ,R4
and so, by knowing the Wannier functions, the procedure would be to calculate the integrals
just like we did for the momentum basis and arrive at a similar expression to (57).
October 2019
13
Prof. Guilherme Milhano
```