HW1-6 - phonon interacting with localized, uncharged fermion: A fermion: A system of phonons phonons with with state-energy state-energy ε d interacting with a localized level which can contain a single (spinless) fermionic particle is described by, H = ( ε d +
∑
M k (ak + ak ) ) c c + †
k
†
∑
k
ωk ak† ak
†
†
≡ ε d c c + c cH M + H ω ;
(1.1)
"phonon on--k" opera operator tor] ; c = ["loc "local aliz ized ed-s -sta tate te fermi fermion" on" oper operat ator or ] ; ak = ["phon
Note straight away: we have the interaction M k written in terms of a k-space representation,
∫
∫
3 3 * * M k = [ k-space fermion/phonon coupling ] ≡ M (r )e i⋅k•r d r = ( M (r )e− i⋅k •r d r ) = M − k = M − k ∈ ℝ ; (1.2)
Find the energy spectrum of this Hamiltonian. Do so by following these steps: a) Subject the Hamiltonian (1.1) M k 1 to canonical transformation: H → H ′ = U † HU = e S He − S , where S ≡ c † c ( a † − a ) ≡ c † cAˆ = Aˆ c †c .
∑
k ω k
−k
k
S Preliminaries: baker-Hausdorff expansion, and proof that the operator e is unitary. Also, cite fermion vs. phonon commutation relations, noting they occupy their own operator-spaces,
]] + 16 [ A, [ A, [ A, B ]] ]]] + .. ... ≡ B + [1 A , B ] + 12 [2 A, B ] + 16 [3 A , B ] + .. . .. = e A Be − A = B + [ A, B] + 12 [ A, [ A, B ]] † k′
†
† k′
† k
†
†
†
∑
n →∞ n =0
1 n n!
[ A , B ]; ];
†
(1.3)
[ ak , a ] = δ kk′ ; [ ak , ak′ ] = 0 = 0 = [a , a ]; {c, c } = 1; 1; {c , c} = 0 = 0 = {c , c };
Another thing we’ll need: the product-rule for (anti)commutators (both indicated), A[ B, C ]− + [ A, C ]− B = A[ B, C] + [ A, C] B [ AB, C ]± = ABC ± CAB = A[B, C ]+ − [ A, C ]+ B = A{B, C} − { A, C}B
(1.4)
Transforming: The baker-Hausdorff expansion (1.3) used to effect the (unitary) congruency transform will “hit” the Hamiltonian (1.1), and ak ↔ ak† operators in the decomposition H = ε d c †c + c† cH M + H ω of (1.1) will
“move through” the c ↔ c † operators, and vice versa. Also, c† ↔ c only occurs as c†c , and any operator commutes with itself, n ∞ [ c†cAˆ , H ] 1 † ˆ 1 2 † ˆ 1 3 † ˆ H ′ = H + [ c cA, H ] + 2 [ c cA, H ] + 6 [ c cA, H ] + ... = H + n! n =1
∑
∞
= H +
∑
[ c cAˆ ,ε d c c + c cH M + H ω ] n
†
†
†
n!
n =1 ∞
H ′ = H +
∑
∞
= H +
∑
ε d Aˆ[n c †c , c †c ] + [ n c †cAˆ ,c †cH M ] + (c †c )n [n Aˆ , H ω ] n!
n =1
ε d Aˆ ⋅ 0 + (c c) (c c ) [ Aˆ , H M ] + (c c ) n [ n Aˆ , H ω ] n
†
†
n n
†
n!
n =1
∞
= H +
∑
(1.5)
n n † 2n n † (c c) [ Aˆ , H M ] + (c c) [ Aˆ , H ω ]
n!
n =1
Commutators (only the n = 0, n = 1 ones survive): The only commutators left to compute appear as, M k M k ′ n † M k M k′ n † † [n Oˆ , H M ] = [ a − a− k , ak ′ + ak′ ] = ([ ak , ak′ ] + [n ak† , ak†′ ] − [a− k , ak′ ] − [a− k , ak†′ ]); k kk ′ kk ′
∑
[ Oˆ , H ω ] = n
∑
ωk
∑
M k ωk ′
† k
† k′
[ a − a− k , a ak′ ] =
∑
M kω k′
kk ′ ωk ω k And subsequently: the (1.6) needs to be further broken down,
1
kk ′
n
ω k
(1.6) n
† k
† k′
n
† k′
([ a , a ak′ ] − [ a− k , a ak′ ]);
Recall: (1) transform of a product is the product of a transform, and (2) the eigenvalues of Hermitian operators are invariant.
[ n ak† , ak ′ ] = [ ak† ,...[ak† ,[ak† , ak′ ]]] = [ak† ,...[a k† , −δ kk′ ]] = −[n −1 a †k , δ kk′ ] = −δ kk′δ n1 ; n
†
†
n
†
n
†
[ ak , ak ′ ] = 0 = 0 = −[ a−k , ak′ ]; [ a − k , ak′ ] = [
n −1
a− k , δ − kk′ ] = δ − kk′δ n1 ;
(1.7)
[n ak† , ak†′ ak′ ] = −[ n −1 ak† (ak†′ [ak′ , ak† ] + [ak†′ , a k† ]a k′ )] = −[ n−1 ak† (ak†′δ k′k + 0ak′ )] = −ak†′δ k′kδ n1 ; n
†
[ a− k , ak′ ak′ ] = −[
n −1
†
†
a− k (ak′ [ak′ , a− k ] + [ak′ , a − k ]a k′ )] = −[
n −1
†
a − k (a k′ 0 − δ k′ , − ka k′ )] = +δ k′ , − ka k′δ n1;
Victory lap: plugging in the simple commutators: By (1.7), we see that only n = 0 and n = 1 terms survive in the expansion (1.5); this then appears as, ∞
Hɶ =
H +
∑
(c † c ) 2 n
∑
M k M k ′ kk ′
ωk
( −δ
1 1!
M k ω k ′
ω k
kk′
( −ak†′δ k′kδ n1 − δ k′ , − k ak′δ n1 )
n!
n =1
= H −
∑
δ n1 + 0 − 0 − δ − k ,k′δ n1 ) + (c †c) n
kk ′
(c†c )1
∑
2
M k † † † + M k ( ak + a− k )) = ε d c c − c c (2 k ωk
b) put H ′ in the form H ′ = [ε d − Σ]c † c +
obvious that Σ = − − c †c
∑
∑
k
∑
2M k k
(1.8)
2
+
ω k
∑
k
ω k ak† a k
ω k ak† ak , and determine the “self-energy”
Σ.
Glancing at (1.8), it is
2 (2 / ω k ) . Question for grader: What is a good qualitative description of “selfM k k
energy”? I have heard it come up in my research a few times when we want to turn a physical junction at which quantum transport occurs from an infinite-dimensional Hamiltonian to a 4x4 one. It seems to come up whenever we have infinite-dimensional (albeit: simple) matrices. and finally c) recognizing the Hamiltonian as a sum of independent quadratic forms, and thereby reading off the spectrum. We have (1.8) acting upon a state containing phonons with energy ε s (k ) = ℏω s (k ) and fermions
with energy ε f as,
ε
ɶ = εɶ = (ε − Σ) c c + = H = H d †
=
εd
−Σ
β ( ε Fer − µ )
e
+
+1
∑
ℏωk k
β ℏωk
e
ℏωk
k
ε d − Σ
≈
−1
∑
e
β (ε Fer − µ )
ak† ak +
+ 1
ℏω k d
V
(2π )3
∫e
β ℏω k
3
(1.9)
k
−1
Just for fun, we could plot this vs. temperature using a Debye dispersion, ω s( k) = ck ⋅ Θ( kD − k) , which would turn (1.9) into,
ε (T )
(ε d − Σ) / V
=
V
e
( ε Fer − µ )/( k BT )
+1
(ε d − Σ) / V
=
e
( ε Fer − µ )/( k BT )
+
+1
3
(2π ) β ( β ℏc) +
k D
1 (k BT ) 4 π 4 2π 2 ( ℏc) 3 15
3
∫ 0
β ℏck ⋅ 4π ( β ℏck ) 2 d ( β ℏck ) (ε d β ℏck
e
(ε d − Σ) / V
=
e
(ε Fer − µ )/ (k BT )
+1
=
−1
+
e
− Σ) / V
(ε Fer − µ ) /(k BT )
+1
+
∞
( k BT ) 4 2
2π ( ℏc )
3
x 3dx
∫ e
x
0
−1
( k BT ) 4 π 2
(1.10)
(ℏc)3 30
There are five (5) different energy scales here: {ε d , Σ, ε Fer, µ , k BT } , so it’s difficult to judge which energy scale should be compared to which scale, as far as plotting is concerned. Plotting the function
ε ( x ) V
~
1 e1/ x +1
+ 13 x
4
just
results in a monotonically-increasing function, which isn’t too interesting. After doing so much fancy math, I want to plot something interesting to impart relevance. Can you make a suggestion?
