Bradley j. nartowt; many body physics; dr. Muttalib; Wednesday, November 21, 2012; assignment 12 419 - pr 19-6 - luttinger liquid theory of fractional quantum Hall edge states: in this exercise we work with a simplistic version of the Luttinger liquid theory of fractional quantum Hall edge-states (Wen 1992, 2004). As mentioned mentioned on SSP 06 – [1.5] and Fig. 1.1, the eigenstate eigenstatess ψ nk of the free electrons in the xy-plane subjected kx to a strong perpendicular magnetic field B are given by: ψ nk ( x, y ) = e i φ n ( y − k ℓ 2 ) in which n is the integer-
Landau level level index, you you have the magnetic magnetic length length
ℓ=
ℏ
eB
, and k is the wavenumber wavenumber.. Note this gives the centercenter-
coordinate k ℓ 2 of the harmonic-oscillator-function φ n . Each landau level is degenerate degenerate,, with density density of states per per area given given by
dN dA
=
1 2π ℓ 2
. The filling per available
state, ν , , is called the filling fraction, fraction, [electron density per unit area ]; [filling fraction ] ≡ ν = ρ 0 ⋅ 2π ℓ 2 ; ρ0 =
(1.1)
In terms of this this filling fraction fraction (1.1): the classical Hall resistance is R H = h / (ν e2 ) . Near special filling fractions such such as ν = 1 / 3 , the Hall resistance resistance shows shows a plateau plateau as a function function of ν due to the properties of the many-body-state, but this is not the issue here. We focus entirely on the edges. Only states close to the edges are influenced by the confining potential, and there they acquire an energy dispersion: 1 ε nk ≈ ω C ( n + 12 ) + Vconf (k ℓ 2 ) + 12 mv mvD 2 ; Vconf ( y ) = [confining potential]; vD = eB Vc′onf ( y ) = [drift velocity]; (1.2) The confining potential is flat in the bulk of the sample, but bends upwards at the edge. The drift velocity is nonzero only for states near the edge. Following the analysis of section 19.5: our analysis of the fractional-Hall-state edges begins by studying the c †c . density-operator for electrons near the edge. Consider density, say, near the right edge, ρ R ( q ) = k > 0 k k + q
∑
(a) show that the commutator between density operators is: Cˆ R = [ ρ (q), ρ ( −q′)] ≈ 1 L sgn q ⋅ n ≈ q , q′
R
R
∑
2π
k
0< k < q
1 2π
nk = ck†ck ; sgn q ≡ Θ(q ) − Θ( −q );
L qν ;
(1.3)
Using [ AB, CD ] = A{C , B}D + {C , A}DB − AC{D , B}− C {D , A}B (c.f., QM 07 - 058 - Pr 01 - commutators and anticommutators), and {cκ† , cκ ′ } ≡ δ κκ ′ , and ck†ck + q− q′ | k>0 = ck†ck + q− q ′Θ( k + q ) and ck†+ q ′c k+ q |k >0 = ck†+ q ′ c k+q Θ( k − q′) , and consequently we get, R † † Cˆ = [ c c , c c
∑
q , q′
k
k +q
∑
k > 0
We then use
† k k +q −q′
(c c
∑
k >0
∑
† k k +q
† k +q′ k +q
×1 − c
c
− q ′> k > 0
∑
q> k >0
∑ (c {c
†
† k
, ck ′ck ′−q ′ ] =
k , k ′> 0
×1) =
( )Θ(k + χ ) = Θ ( χ )
q , q′
Proof: Θ( − q′)
∑ [c c k , k ′> 0
this makes (1.4) into, Cˆ R = Θ(q )
1
]=
k ′ k ′+ q ′
k ′> 0
k >0
=
∑
∑
∑
k > 0
χ > k > 0
† k k +q −q′
(c c
∑
∑
− q ′ + q′> k > 0 + q ′
† †
†
(1.4)
× Θ(k + q ) − c
k >0
†
, c k +q }ck ′−q ′ + 0ck ′−q ′ck +q − ck ck ′ 0 − ck ′ {ck ′−q ′ ,ck }ck +q ) † k +q ′ k +q
( ) , and Θ( − q ′)
ck†ck +q − q ′ − Θ( −q ′)
ck†+ q′ ck + q = Θ (− q′)
† k′
∑
c
− q ′> k > 0
ck†ck +q −q′ =
∑
× Θ( k − q′))
† ck + q′ ck + q =
q >k > 0
∑
1
k > 0
† ck ck + q− q ′ (proof ) and
(Θ(q ) − Θ( −q ′))ck†ck +q − q ′
ck†+q ′−q′ck + q − q′ = Θ( − q′)
∑
k > − q′
ck†ck + q− q ′ = 1⋅
see this just entails the dry practice of manipulating dummy/sum indices, so it is relegated to this footnote.
∑
k>0
(1.5)
ck† ck + q −q ′ ; you
ˆ R is diagonal in q, q′ so Using the approximation k F >> q, q′ = O ( 2mk BT ) → q ≈ q′ , we see that C q , q′ R ˆ R δ . On introducing this Kronecker function, we gain occurrence of the sgn q function as in (1.3), Cˆ q, q′ = C q , q′ q ,q ′
ˆR Cˆ R q , q ′ ≈ Cq , q′δ qq′ = δ qq ′
∑ (Θ(q) − Θ(−q′))c c
† k k +q −q ′
q >k > 0
= sgn q ⋅
∑
† k k
q >k > 0
nk = sgn q ⋅
q >k > 0
∑ (Θ(q) − Θ( −q))c c
=
q>k>0
k
≈ sgn q ⋅
q>k> 0
∑ ν = ν sgn q ⋅ ∑ 1 ≈ ν sgn q ∫
q>k> 0
∑ sgn q ⋅ n
=
q
L ⋅ dk
0
2π
= sgn q
L qν
∑
nk
q>k > 0
→
(1.6)
ˆR Cˆ R q ,q ′ ≈ C q , q ′δ qq ′
2π
=
1 2π
(sgn q )(L qν )
Observation: The special nature of the fractional-quantum-Hall (FQH) groundstate only enters in the last approximation: it would not make sense to replace the occupation nk by unity, but it should rather be the occupation of the FQH bulk-state; i.e., nk = ν . From this observation now follow some interesting consequences. (b) in analogy with the treatment in section 19.5.3: introduce the fields φ ( x ) and P( x) through Fourier1
transformations of the density operators:
π
∂ xφ =
ρ L ( x ) + ρ R ( x ) and ν
1
π
P( x ) = ρ L ( x ) − ρ R ( x ) , and show that
in order to preserve the canonical commutation-relations, an additional factor of 1/ ν must be included in the definition of P, as here indicated. The canonical commutation relations amidst the density operators are given by [19.34], [ ρ R (q ), ρ R (− q′)] = + 21π Lqνδ qq′ ; [ ρ L ( q ), ρ L (− q′ )] = − 21π L qνδ qq′ ; [ ρ R (q ), ρ L (− q′ )] = 0; (1.7)
The commutators amidst the ∂ qφ and α P( −q′) , for α an unknown constant, then appear as, Cˆ −φ q, P′q = [∂ qφ , α P( −q′)] = α =α
π ν
( −δ
qq ′
π π π [ ρ L ,q , ρ L ,− q′ ] − [ ρ L , q , ρ R ,−q′ ] [ ρ L, q + ρ R, q , ρ L ,− q′ − ρ R ,− q′ ] = α ν ν +[ ρ R ,q , ρ L ,− q′ ] − [ ρ R ,q , ρ R ,− q′ ]
− 0 + 0 − δ qq′
)
(1.8)
q L 2π
ν = −αδ qq′ qL → α = 1 = ν 0
qL ˆ φ ,φ = [∂ φ , ∂ φ ] = π [ ρ + ρ , ρ + ρR ,− q ′ ] = π (− δqq ′ + 0 + 0 + δ qq ′ ) 2π = 0 ; C q L ,q R,q L ,− q ′ − q ′q − q′ qL P P Cˆ − q, ′q = [α Pq , α P− q′ ] = α 2 νπ [ ρ L ,q − ρ R ,q , ρ L ,−q ′ − ρR ,− q′ ] = πα 2 ( − δ qq ′ + 0 + 0 − − δ qq ′ ) 2π = 0 ;
(1.9)
(c) the second basic assumption is that the energy of edge-excitations is quadratic in charge-density. In other words: we assume that there is no gap for charge-fluctuations near the edge, and then we expand the energy in density-fluctuations. Show, in analogy with section 19.5, that this results in a bosonic-Hamiltonian with a structure similar to that of the Luttinger-liquid Hamiltonian, given by [E.19.24], some velocity of the edge-states, π v′ v′ [ E.19.24] → H = D ρ (q ) ρ (− q) = D ν (ν [P ( x )]2 + ν 1 [∂ xφ ]2 ) dx ; vD′ = ; (1.10) related to the drift-velocity, v 2 L q D
∑
∫
Justifying the Hamiltonian: Notice that H =
′ π v D L
∑ ρ (q) ρ ( −q) contains no reference to ν , so it follows from q
2
using SSP 06 – 359 – [19.38], except near one of the two edges , L or R, say, the left edge, as,
2
In (1.11): we have the constant C ( N ) = C ( N L , N R ) =
π vF L
( NL 2 + NR2 ) =
π vF 2 L
(N L 2 + J 2 ) .
left edge 2π v = L ∑ q >0 [0 + ρ L ( q) ρ L ( −q)] left edge = ... (1.11) ∑ q >0 ( ρ R (−q) ρ R (q) + ρ L (q) ρ L ( −q) ) = right edge 2π v = [ ρ R ( −q) ρ R ( q) + 0] L ∑ q > 0 F
H O = H 0 − C ( N ) =
2π vF L
F
Now: putting in the coordinates 2 π ρ L ( q ) = ∂ qφ + ν P (q ) into (1.11) (the extra factor of ν included in P( q) as * from (1.8) to make the transform canonical), and using P(−q ) = P ( q) with Im P (q ) = 0 , and Σ = ∫ L ⋅ dx ,we get, 2 2 ∂ qφ +ν P (q) ∂ − qφ +ν P ( −q) 2π vF v 2 = F ∂ qφ ⋅ ∂ − q φ +ν ∂ qφ P( −q) +ν P (q )∂ − qφ +ν P (q ) P ( −q )) ) H O = ( 2L q> 0 L q >0 2 π 2 π
∑
=
∑
vF
∑ ( −(∂ φ ) 2L
2
q
−ν∂ qφ ⋅ 2 Im P( q) −ν
2
P
q >0
H O =
−vFν
2L
∑(
1
ν
(∂ qφ ) 2 +ν P 2 ) =
v′D
2
q >0
2
)
(1.12)
∫ (ν [ P( x)]
2
ν
; ) dx
2
+ ν 1 [ ∂ xφ ]
vD′ = −v F ;
(d) in order to probe this interesting structure of the edge-states, tunneling experiments were suggested by Wen and have been performed recently. To calculate the tunneling density of states: we need a single-electron operator Ψ( x) , and as in section 19.6 we must demand that [ ρ ( x), Ψ( x′)] = Ψ ( x) ⋅ δ ( x − x′) , in which ρ = ρ L . Show that, because of the (unusual) commutation-relation (1.3), that Ψ ( x ) ~ exp( −2π i ⋅
x
∫
−∞
[ ρ ( x′) / ν ]dx′) , with
an additional ν appearing. Preliminary (1): we should notice about our ansatz that, for a vanishing d Ρ = ρ ( x) / ν and Ρ( −∞) ≡ 0 ) we have the property, antiderivative at −∞ , e.g., ( ∃Ρ = Ρ( x) : dx Ψ ( x + y ) ~ exp( −2π i ⋅ =e
Thus: −2π i ⋅
x
∫
−∞
−2π iΡ ( x )
e
x + y
∫
−∞
[ ρ ( x′) / ν ]dx′) = exp( −2π i ⋅ ( Ρ( x) + Ρ( y) − 2 Ρ( −∞) )) = exp( −2 π i ⋅ ( Ρ( x) + Ρ( y ) ))
−2π iΡ ( y )
=e
−2π i ( Ρ ( x ) + Ρ ( y ) )
x
∫
= exp( −2π i ⋅
−∞
[ ρ( x′) / ν ]dx ′) exp( −2π i ⋅
∫
y
−∞
(1.13) [ ρ( x ′) / ν ]dx ′)
[ ρ ( x′) / ν ]dx′ = B is a linear function of the boson-operator. It is shown in the previous exercise B
B
19.1 that if B is a linear function of a boson operator, then [ O, e ] = [ O, B]e . This means we have the property, [ ρ ( x′), e
−2π iΡ ( y )
] = [ ρ ( x′), ( −2π iΡ( y ))]e
−2π iΡ ( y )
= −2π i[ ρ ( x ′), Ρ( y)] e
− 2π iΡ ( y )
Preliminary (2): The Kac-Moody algebra for [ ρ L ( x), ρ L ( x′′)] , using (1.3), appears as, C xx′′ = [ ρ L ( x ), ρ L (x′′)] = =
C xx′′ =
(∑
d ν 2π L d ( ix ) d ν 2π iL dx
∫
∞
0+
q >0
e
1 2
L
∑
iq ( x − x ′′ )
i qx − q ′′x ′′ )
qq ′′
−
'e (
∑
q<0
e
[ ρR ,q , ρR ,− q′′ ] =
iq ( x − x′′ )
2i sin q( x − x′′) ⋅ L ⋅ dq =
−ν
2π i
)=
d ν 2π L d ( ix )
L 2
L
∑
∑
q> 0
d ⋅ dx δ ( x − x′′) ⋅ α 0 ;
q
'e
(e
iq ( x − x′′ ) 1 2π
iq ( x − x′′ )
−e
(sgn q )( qν ) =
− iq ( x − x ′′ )
)=
∑ 'sgn q e ∑ 2i sin q( x − x′′) ν
2π L
d ν 2π i L dx
(1.14)
q
d d ( ix )
iq ( x − x′′ )
q> 0
(1.15)
α 0 = [arbitrary constant];
Seeing if the ansatz satisfies the required commutator: Thus: computing the commutator and using the “unusual” (1.3) throughout, we get, C xx′ ~ [ ρ ( x ), Ψ ( x′)] = [ ρ ( x), e = α 0e
−2π i ⋅Ρ ( x ′ )
−2π i ⋅Ρ ( x ′ )
]=e
−2π i⋅Ρ ( x′ )
−2π i
ν
δ ( x − x′) ~ α 0Ψ ( x′)δ ( x − x′) → α 0
x ′
∫ [ ρ ( x), ρ ( x′′)]dx′′ = e
−∞
= 1
− 2π i⋅Ρ ( x ′ )
−2π iν
x′
∫ −2π iν
−∞
d dx
δ ( x − x ′′) ⋅ α 0dx′′
(1.16)
iπ / ν
(e) show that the anti-commutator of Ψ( x) and Ψ ( x′) is: Ψ ( x) Ψ ( x′) = Ψ( x′) Ψ ( x)e . It is easier to first i calculate the commutator between the two field operators, Cˆ xη′x = [Ψη ( x), Ψη ( x′)] = Ψη ( x′)Ψη ( x)( e π / ν − 1) ;
proving this goes as, ′ ′ Cˆ η ~ [e −2π iΡ ( x ) , e −2π iΡ ( x ) ] = −(−2π i) 2[ Ρ( x′), Ρ( x )]e− 2π iΡ ( x )e − 2π iΡ ( x )
x′x
= ( 2νπ i
x ′
x
∫ ∫ )( ∫ δ ( x′ − x′′′) dx′′′)e 2
= ( 2νπ ) (
−∞
−∞
[ ρ L ( x′′), ρ L ( x′′′)]dx′′dx′′′)e
x
−2π iΡ ( x )
−∞
e
−2π iΡ ( x )
−2π iΡ ( x ′ )
e
− 2π iΡ ( x′ )
2
= ( 2νπ ) (
= ( 2ν π i )δ ( x′ − x )e
????
iπ / ν η −1); Cˆ x′x ~ ( 2ν π i )δ ( x′ − x ) Ψ L ( x ) Ψ L ( x′) = Ψη ( x′) Ψη ( x)(e
x′
∫ ∫ −∞
−2π iΡ ( x )
e
x
−∞
−ν
2π i
− 2π iΡ ( x′ )
⋅ dxd ′′ δ ( x ′′ − x ′′′) dx ′′dx ′′′)e
− 2π iΡ ( x )
e
−2π iΡ ( x′ )
[ couldn't figure it out...]
(f) determine which values of the filling-fraction ν in the former question that leads to proper Fermi-statistics.
We have
iπ
ν
= iπ + 2π in ↔
1
ν
= 2n + 1 (that is to say:
ν =
1 2 n +1
).
(1.17)
