Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Ubungen zur Theoretischen Physik F¨ SS 15
Prof. Dr. A. Mirlin Blatt 11
Dr. Una Karahasanovic, Dr. Ivan Protopopov Besprechung 10.07.2015
1. Nearest-neighbour hopping
(a) Let γk† = PN
n=1Uknc†n. Then it follows that γk = PN
n=1Ukn∗ cn. In order for γs to satisfy fermionic commutation relations, we must have{γk, γp†}=δkp. We have that
{γk, γp†} = {
N
X
n=1
Ukn∗ cn,
N
X
m=1
Upmc†m}
=
N
X
m,n=1
Upm(U†)nk{cn, c†m}
| {z }
δnm
=
N
X
n=1
UpnUnk† = (U U†)pk (1)
Therefore, we need to have (U U†)pk =δpk, i.e. U needs to be an unitary matrix.
(b) We need to show that this choice of U is unitary. We just multiply out UpnUnk† = 1
N X
n
e2πipnN e−2πiknN = 1 N
N
X
n=1
e2πin(p−k)N
| {z }
N δpk
=δpk (2)
which proves that this choice of U is unitary.
(c) Energy spectrum. We need to expresscs in terms ofγs. We get thatc†n=PN
p=1Unp† γp†, and that cn =PN
k=1Uknγk. The Hamiltonian then becomes H = −t
N
X
p,k,n=1
(Up,n+1∗ Ukn+Upn∗ Uk,n+1)γp†γk
= −t 1 N
X
p,k,n
(e2πi(k−p)nN e−2πipN +e2πi(k−p)nN e−2πikN )γp†γk
= −tX
p
(e−2πipN +e2πipN )γp†γp =X
p
Epγp†γp (3)
withEp =−2tcos 2πpN
, and where in the second line we have used that N1 P
ne2πi(k−p)nN = δkp.
2. Two level system
(a) The commutator [H, B+F]. We calculate
[H, B+F] = [b†b+ωf†f+λ(b†f+f†b), b†b+f†f]
= λ[b†f +f†b, b†b+f†f]
= λ
f b†[b†, b]
| {z }
−1
+[b†f, f†f] +f†[b, b†]
| {z }
1
b+ [f†b, f†f]
(4) Further we calculate
[b†f, f†f] =b†f f†f
|{z}
1−f†f
−f†f b†f =b†f (5)
since f2 = 0. Similarly, using the same identities we get that
[f†b, f†f] =−f†b. (6)
Then it follows from (4), and (5) and (6) that
[H, B+F] =λ(−f b†+b†f +f†b−f†b) = 0. (7) (b) Eigenstates of H. In the previous part of the question we have shown that H and B +F commute, which means that they have the same eigenstates. Hence need to prove that |ψi is an eigenstate of F +B (and therefore it’s automatically an eigenstate of H). Therefore we examine
(B+F)|ψi = (f†f+b†b)(un(b†)n−1f†+vn(b†)n)|0i
= un(b†)n−1(f f†+ 1)f†|0i −vn(b†)nf†f|0i
|{z}
0
+ unf†b†b(b†)n−1|0i+vnb†b(b†)n|0i (8) Further one can use that
[b,(b†)n] =n(b†)n−1 (9)
Then it follows that
(B+F)|ψi = un(b†)n−1f†|0i+unf†b†((n−1)(b†)n−2+ (b†)n−1b)|0i+vnb†n(b†)n−1|0i
= un(b†)n−1f†|0i+un(n−1)(b†)n−1f†|0i+nvn(b†)n|0i
= n|ψi (10)
(c) Finding the eigenvalues of H. To find the eigenvalues of H, we need to to solve H|ψi=M|ψi, whereM is the energy eigenvalue.We have that
H|ψi = (b†b+ωf†f +λ(b†f +f†b))
un(b†)n−1f†+vn(b†)n
|0i
=
vn(n−1)(b†)n−1f†|0i+nun(b†)n|0i
+ωvn(b†)n−1f†|0i+λb(b†)n−1(f†)2
| {z }
0
|0i + λunb(b†)nf†|0i
| {z }
n(b†)n−1f†|0i
+λvn(b†)n f f†|0i
| {z }
(1−f†f)|0i
+λun(b†)n+1f|0i
|{z}
0
= (b†)n−1f†|0i(vn(n−1) +ωvn+λnun) + (b†)n|0i(nun+λvn)
= M vn(b†)n−1f†|0i+M un(b†)n|0i (11)
where the last line is the obtained from the energy eigenvalue equation. We compare the ultimate and penultimate line of the above. We must have that
vn(n−1) +ωvn+λnun−M vn= 0
nun+λvn−M un = 0 (12)
In order for this system of equation to have a non-trivial solutions inun and vn, its determinant should be zero, from where it follows that
(n−M)2+ (ω−)(n−M)−λ2n= 0. (13) From here we get that
M =n+ω−
2 ±
s
ω− 2
2
+nλ2. (14)
3. Diagonalizing quadratic Hamiltonians
(a) Similarly as was done in the lecture, we introduce Bogoliubov operators c† = ua†+vb
d† = ub†+va (15)
In order forc, dto obey bosonic commutation relations we have shown (see lecture notes) that one must have that|u|2− |v|2 = 1, which then lead to a parametrization u= coshθ and v = sinhθ. Similarly, the Hamiltonian
H =Aa†a+Bb†b+λ(a†b†+ba) (16) can be written in the matrix notation like
H = 1/2(a†bb†a)
a λ 0 0 λ b 0 0 0 0 b λ 0 0 λ a
a b† b a†
(17)
Let us consider the upper block of the matrix above. We have that Hu =H = (a†b)
a λ λ b
a b†
(18) We can rewrite the Bogouliubov transformation in the matrix form
a b†
=
u −v
−v u
c d†
(19) If we plug (19) into (18) and perform the matrix multiplication we get that
Hu =H = (c†d)
au2+bv2−2λuv −(a+b)uv+λ(u2+v2)
−(a+b)uv+λ(u2+v2) bu2+av2−2λuv
c d†
(20)
In order for off diagonal terms to dissapear one must haveλcosh 2θ+12sinh (2θ)(a+ b) = 0. If we square this equation we get that cosh2(2θ) = ((a+b)2
a+b)2−4λ2, and then sinh (2θ) = −2λ
a+b cosh (2θ). The Hamiltonian is then
H =ωcc†c+ωdd†d (21) with
ωc = v2a+u2b+ 2λuv
ωd = v2b +u2a+ 2λuv (22)
From here it follows that ωC = 1
2 n
(A+B)2−4λ21/2
−(B−A)o , ωD = 1
2 n
(A+B)2−4λ21/2
+ (B−A)o
. (23)
The expectation value of hc†ci = 0 since the ground state is the vacuum of c operators. However, this ground state is not a vacuum of a operators and we get that
ha†ai=h0|(uc†−vd)(uc−vd†)|0i=v2, (24) where we used the fact that c|0i= 0 and d|0i= 0, and the usual bosonic commu- tation relations.
(b) Holstein-Primakoff exoansion.
This Hamiltonian is exactly of the form of the Hamiltonian in the previous part of the question. So we know that the dispersions are given by Eq (23). That is
2ωD/C(k) =
J2z2(Sa+sb)2−16J2SaSb(coskxa+ coskya+ coskza)21/2
±J z(Sa−S−b)(25) where the cos terms came from the Pz
j=1exp (ik.ej). For d = 3 the coordination number isz = 6. If one expands the above dispersions for (SA−SB)SA one gets that
2ωD/C(k) = J z(Sa+Sb)
1−4 SaSb
(Sa+Sb)2 + 4 SaSb (Sa+Sb)2
k2a2 3
1/2
±6J(Sa−Sb) (26) The square root in the equation above can be re-written as
[...]1/2 = Sa−Sb Sa+Sb 1 +
k2a2 3 (Sa+Sb)2
4SaSb −1
!1/2
(27) Then one sees that there are two regimes. In the first regime
k2a2 3 (Sa+Sb)2
4SaSb −1 1 (28)
ork k0 where k0 =√
3√S4Sa−Sb
aSba. In this regime the dispersion is a linear function of momentum, i.e. we get that
2ωD/C(k) = 6J(Sa−Sb)k
k0 ±6J(Sa−Sb) (29) that isα = 1. In the other regime wherekak0a, the dispersion is given by
2ωD/C(k) = 6J(Sa−Sb)
1 + k2 2k02
±6J(Sa−Sb), (30) that is – in this regimeα = 2.
