Statistical Physics & Condensed Matter Theory I:
Exercise
The resonant level model: Solution
a)
The total number of fermions will be conserved under time evolution provided [HRLM , Nf ] = 0.
We can compute this term-by-term. Besides the obviously vanishing terms [c†k ck , c†k0 ck0 ] = 0,
[c†k ck , d† d] = 0, [d† d, d† d] = 0, we have the nontrivial commutators
h
i
c†k d, c†k0 ck0 = c†k dc†k0 ck0 − c†k0 ck0 c†k d = dc†k0 {c†k , ck0 } = −δk,k0 c†k d,
h
i
d† ck , c†k0 ck0 = d† ck c†k0 ck0 − c†k0 ck0 d† ck = d† {ck , c†k0 }ck0 = δk,k0 d† ck ,
and (remembering that d2 = 0 and (d† )2 = 0),
h
i
c†k d, d† d = c†k dd† d − d† dc†k d = c†k (1 − d† d)d − 0 = c†k d,
†
d ck , d† d = d† ck d† d − d† dd† ck = 0 − d† (1 − d† d)ck = −d† ck .
Therefore,
[HRLM , Nf ] = t
i
i
Xh †
Xh †
ck d + d† ck , c†k0 ck0 + t
ck d + d† ck , d† d
k,k0
k
=t
X
−c†k d + d† ck + c†k d − d† ck = 0.
k
b)
We need to explicitly calculate (dropping obviously zero terms proportional to [c†k ck , d† ] = 0 and
[d† d, ck ] = 0 and the like)


h
i
i X
h
i
Xh †
X
†
†
†
†
†
†
†
†
HRLM , fn =
εk Mn,k0 ck ck , ck0 + Ln εd d d, d + t Ln
ck d, d +
Mn,k0 d ck , ck0 
k,k0
=
X
k
εk Mn,k c†k + Ln εd d† + tLn
k,k0
k
X
k
c†k + t
X
X
X
Mn,k d† =
(εk Mn,k + tLn )c†k + (εd Ln + t
Mn,k )d†
k
k
k
(using e.g. [c†k d, d† ] = c†k dd† − d† c†k d = c†k {d, d† } = c†k and similar). Equating this to En fn† yields
the conditions
X
En Mn,k = εk Mn,k + tLn ,
En Ln = εd Ln + t
Mn,k
k
which are the sought-after coupled equations for Mn,k and Ln .
1
c)
Using Mn,k = tLn /(En − εk ), we get (En − εd )Ln = t2 Ln
En − εd =
X
k
P
1
k En −εk
and therefore
t2
.
En − εk
d)
We have
X
X
Mn,k0 c†k0 + Ln d)|0i
h0|fn fn† |0i = h0|( (Mn,k )∗ ck + L∗n d)(
k0
k
X
X
(Mn,k )∗ Mn,k0 h0|ck c†k0 |0i + |Ln |2 h0|dd† |0i =
=
|Mn,k |2 + |Ln |2
k,k0
k
and thus (substituting for Mn,k from above)
|Ln | =
1+
X
k
t2
(En − εk )2
!−1/2
.
e)
For your enlightenment, let’s first prove the identity given in the question:
X
X
1
2 X
1
1
=
=
2E
n
En − εk
En − ∆(k − 1/2)
∆
− (2k − 1)
k∈Z
k∈Z ∆
k∈Z
!
∞
∞
2 X
π
πEn
1
4En /∆
2 X
1
=
+ 2En
=
.
= − tan
2En
n 2
2
∆
∆
∆
∆
( 2E
∆ − (2k − 1)
∆ + (2k − 1)
∆ ) − (2k − 1)
k=1
k=1
Thus,
X
k∈Z
1
∂
=−
(En − εk )2
∂En
π
− tan
∆
πEn
∆
=
π2
1
n
∆2 cos2 πE
∆
But we also have
X
En − εd
1
π
=
= − tan
t2
E n − εk
∆
k
πEn
∆
so
(En − εd )2
π2
=
t4
∆2
!
1
πEn
∆
cos2
−1
→
1
cos2
πEn
∆
=1+
∆2
(En − εd )2 .
π 2 t4
This allows us to write
−2
|Ln |
π 2 t2
=1+ 2 +
∆
En − εd
t
2
so finally
|Ln |2 =
(En − εd
)2
t2
,
+ t2 (1 + π 2 t2 /∆2 )
and this has Lorentzian form I/((En − α)2 + γ 2 ) with I = t2 , α = εd and inverse lifetime
γ = t(1 + π 2 t2 /∆2 )1/2 .
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