Physics 216 : Solution set 3 Problem 1 Solution 1
Physics 216 : Solution set 3
Problem 1 Define the retarded Green’s function through † † ⟨Ψα (t1 , r1 )Ψβ (t2 , r2 ) + Ψβ (t2 , r2 )Ψα (t1 , r1 )⟩ iGrαβ (t1 , r1 ; t2 , r2 ) = 0
if t1 − t2 > 0 (1) if t1 − t2 < 0.
(a) Calculate the Fourier-transformed Greens’ function Gr (ω, p) for the free Fermi gas at temperature T (i.e., the expectation value ⟨⟩ is taken with respect to the finite-temperature Fermi gas). (b) Specify where any poles occur. (c) Check that contour integration of your result indeed gives zero for negative t = t1 − t2 .
Solution 1 (a) Recall that Ψα (t, r) = e
iHt
ψα (r)e
−iHt
=e
iHt
(∑ k
) ∑ eikr ( ) eikr √ ckα e−iHt = √ eiHt ckα e−iHt V V k
(2)
∑ ∑ † † ′ ′ ′ Free Fermi gas has the Hamiltonian H = k′ α′ ϵk [ck′ α′ ck α , ckα ] = k,α ϵk ckα ckα , using [H, ckα ] = ( † ) ∑ † ′ ′ ′ ′ ′ = −ϵk ckα , we get k′ α′ ϵk ck′ α′ {ck α , ckα } − {ck′ α′ , ckα }ck α Ψα (t, r) =
∑ eikr √ e−iϵk t ckα . V k
(3)
Since there is a translational symmetry in the system, we can define iGrαβ (t, r) ≡ iGrαβ (t, r; 0, 0). Then, iGrαβ (t, r) = Θ(t)
1 ∑ i(kr−ϵk t) 1 ∑ ikr −iϵk t e ⟨e {ckα , c†k′ β }⟩ = Θ(t)δα,β e V V ′ k,k
(4)
k
The Fourier transformation of the Greens’ function is given as follows. ( ) ∫ ∫ 1 ∑ i(k·r−ϵk t) Gr (ω, p) = dt d3 rei(ωt−p·r) − iΘ(t)δα,β e V k ∫ ∞ = −iδα,β dte−i(ϵp −w)t 0
δα,β = , ω − ϵp + i0+
(5)
where 0+ is introduced to make the integral convergent. (b) The pole occurs at ϵp − i0+ , so slightly below the real axis. (c) Recall ∫ 3 d pdω i(p·r−ωt) δα,β r Gαβ (t, r) = e , (2π)4 ω − ϵp + i0+ 1
(6)
Physics 216 : Solution set 3 so when t < 0, e−iωt decays when Im ω > 0. Hence when we perform ω integral, we make this integral as a contour integral in (complex) ω by closing the contour in the upper plane. Since the pole exists below the real ω axis, the integral indeed becomes 0.
Problem 2 Consider two identical 1D non-relativistic bosons of mass m moving on a ring of Length L: x1 , x2 ∈ [0, L) ( ∂2 ) h ¯2 ∂2 and H = − 2m + ∂x 2 . ∂x21 2 (a) What are the energies of the ground state and first excited states? (b) Suppose a repulsive interaction of the form H ′ = cδ(r1 − r2 ) is added (c > 0). Construct the ground-state bosonic wavefunction by any means and find the ground state energy in the limit c → ∞.
Solution 2 2
2
h ¯ ∂ (a) The eigenstate and eigenenergy of the single particle Hamiltonian H = − 2m ∂x2 in a ring is given 2 2 ( ) 2 2 h ¯ kn 1 h ¯ 2π ik x n by ψn (x) = √L e , En = 2m = 2m L n , where n is an integer. The ground state has both bosons in the ground state: EGS = 0 2 2 h ¯ The first excited states have one boson in n = 0 state and the other boson in n = ±1: E = 2π mL2 (b) Using our educated guess, the ground state can be written as ( ) Ψk1 ,k2 (x1 , x2 ) = sgn(x1 − x2 ) ei(k1 x1 +k2 x2 ) − ei(k2 x1 +k1 x2 ) , (7)
where k1 , k2 are constants which are determined later. Putting into the Schroedinger equation yields HΨk1 ,k2 = − =
) )( ¯2 ( h 2δ(x1 − x2 ) − 2δ(x1 − x2 ) + (−k12 − k22 )sgn(x1 − x2 ) ei(k1 x1 +k2 x2 ) − ei(k2 x1 +k1 x2 ) (8) 2m
¯2 2 h (k + k22 )Ψk1 ,k2 , 2m 1
(9) 2
h ¯ where we have used ∂x sgn(x) = 2δ(x) and we get eigenenergy 2m (k12 + k22 ). Using the periodic boundary 2π condition, k1 , k2 ∈ L Z. In order not to have a vanishing wave function k1 ̸= k2 must be satisfied. Hence the ground state is the case where k1 = 0 and k2 = 2π L : ( 2π ) 2π EGS (x1 , x2 ) = sgn ei L x1 − ei L x2 (10)
2
2
h L with the ground state energy EGS = 2¯ mL2 . (Alternatively, one can attack this problem by separating CM and rel motion degrees of freedom.)
Problem 3 Consider a two-dimensional electron moving in a constant perpendicular magnetic field. (a) Show that 2 2 in the rotationally symmetric gauge for A, the lowest Landau level eigenstates ψm ∼ z m e−|z| /4l with z ≡ (x + iy) and m a nonnegative integer are eigenstates for a particular choice of l (independent of m), and calculate the normalization Now suppose that the system of N noninteracting LLL electrons is put in a weak radial confining potential V (r) = αr2 , with α sufficiently small that no mixing occurs between Landau levels. The ground state of the N electrons is now a Slater determinant of the single-particle eigenstates m = 0, . . . , N − 1.
2
Physics 216 : Solution set 3 Now consider the low-energy edge excitations of this state. For example, moving the last electron out by angular momentum h ¯ gives an excited states. (b) How many excitations within the LLL are there of total momentum M , for M = 0, . . . , 5? (c) Give an estimate of the velocity of excitations at the edge.
Solution 3 Let’s use the following convention. The electron has negative charge, −e, in the magnetic field −B zˆ (B > 0) where x − y plane is the plane in which electrons are confined. Then the noninteracting Hamiltonian is given by N N 1 ∑( e )2 1 ∑ 2 H= pi + Ai = π , (11) 2m i=1 c 2m i=1 i where m is the mass of the electron and N is the total number of electrons. After a bit of algebra, one can compute√the commutation relation: [π x , π y ] = i¯heB/c. It is then convenient to define the magnetic length as lB ≡ ¯hc/eB. Let’s introduce the following creation/annihilation operator: i π x + iπ y a≡ √ 2 ¯h/lB −i π x − iπ y a† ≡ √ , 2 ¯h/lB
(12) (13) (14)
which has the canonical commutation relation [a, a† ] = 1. The Hamiltonian can be written as ) N ( ∑ 1 H = ¯hωc a†i ai + , 2 i=1
(15)
eB is the cyclotron frequency. One immediately sees that this is nothing but a harmonic oscillator where ωc = mc ( ) with eigenenergies ¯hωc n + 12 for n ∈ {0, 1, 2, · · · }. Our task here is to construct the ground states, which are annihilated by a operator. (a) The symmetric gauge is given as A = B2 (y, −x). Now, identify the 2D plane as a complex plane via z ≡ x + iy. This amounts to giving: ( ) 1 z ∂ a= √ + 2lB ∗ (16) ∂z 2 2lB ( ∗ ) 1 z ∂ a† = √ − 2lB ∗ (17) ∂z 2 2lB
Solving the eigenequation: aψ(z, z ∗ ) = 0, the eigenstates are given by 2 1 1 ψm (r) = √ (z/lB )m e− 4 |r| . 2 m 2π2 m!lB ∫ ∗ One can check that the normalization condition holds: d2 rψm (r)ψn (r) = δm,n In this gauge, the angular momentum operator becomes
(18)
Lz = xpy − ypx = ¯h(z∂z − z ∗ ∂z∗ ), so ψm has angular momentum ¯hm. (b) If we add a confining potential V (r) = αr2 , the ground state of N N −1 z1 z2N −1 N −2 z2N −2 [ ] 1 z1 Ψ = A ψ0 , ψ1 · · · , ψN −1 ∼ √ .. .. N . . 1 1 3
(19) electrons is: ··· ··· .. . ···
N −1 zN N −2 zN .. . 1
,
(20)
Physics 216 : Solution set 3 which is the unique M = 0 excitation. Other excitations are as follows: [ ] M =1: A ψ0 , · · · , ψN −2 , ψN [ ] { A ψ0 , · · · , ψN −2 , ψN +1 [ ] M =2: A ψ0 , · · · , ψN −3 , ψN −1 , ψN [ ] A ψ0 , · · · , ψN −2 , ψN +2 [ ] M =3: A ψ0 , · · · , ψN −3 , ψN −1 , ψN +1 ] [ A ψ0 , · · · , ψN −4 , ψN −2 , ψN −1 , ψN ] [ A ψ0 , · · · , ψN −2 , ψN +3 [ ] A ψ0 , · · · , ψN −3 , ψN −1 , ψN +2 [ ] M =4: A ψ0 , · · · , ψN −3 , ψN , ψN +1 [ ] A ψ0 , · · · , ψN −4 , ψN −2 , ψN −1 , ψN +1 [ ] A ψ0 , · · · , ψN −5 , ψN −3 ψN −2 , ψN −1 , ψN ] [ A ψ0 , · · · , ψN −2 , ψN +4 [ ] A ψ0 , · · · , ψN −3 , ψN −1 , ψN +3 [ ] A ψ0 , · · · , ψN −3 , ψN , ψN +2 [ ] M =5: A ψ0 , · · · , ψN −4 , ψN −2 , ψN −1 , ψN +2 ] [ A ψ0 , · · · , ψN −4 , ψN −2 , ψN , ψN +1 [ ] A ψ0 , · · · , ψN −5 , ψN −3 ψN −2 , ψN −1 , ψN +1 ] [ A ψ0 , · · · , ψN −6 , ψN −4 , ψN −3 ψN −2 , ψN −1 , ψN
(21) (22)
(23)
(24)
(25)
2 (c) The edge has angular momentum Lz = N ¯h ∼ rmv. Using ⟨r2 ⟩ = N 2lB , the velocity of excitations at the edge can be estimated as √ N ¯h ¯h 1 N √ v∼ = (26) 2 m mlB 2 N 2lB
Problem 4 (a) Verify, directly from the definition: ˆ = R(ξ, θ, ϕ)|S, S⟩ |Ω⟩ =e
iS z ϕ iS y θ S z ξ
e
e
|S, S⟩,
(27) (28)
ˆ ≡ (sin θ cos ϕ, sin θ sin ϕ, cos θ), that |Ω⟩ ˆ is an eigenstate of the spin component in where the unit vector Ω ˆ direction: the Ω ˆ · S|Ω⟩ ˆ = S|Ω⟩ ˆ Ω (29) ∑ ˆ = ∏ |Ω ˆ i⟩ (b) Show that the expectation value of the Heisenberg model H = i,j Si ·Sj in a coherent state |Ω⟩ i is 2 ∑ ˆ ≡ ⟨Ω|H| ˆ ˆ =S ˆi · Ω ˆj. H[Ω] Ω⟩ Ji,j Ω (30) 2 i,j
Solution 4 (a) Let R be the rotation given by Euler angles (ξ, θ, ϕ). (Ω = Rˆ z .) Then R−1 SR = RS 4
(31)
Physics 216 : Solution set 3 holds because S is a vector operator. Using Sz |S, S⟩ = S|S, S⟩, the following equation holds: ( ) ( ) ˆ · S|Ω⟩ ˆ = zˆ R−1 S |Ω⟩ ˆ = zˆ RSR−1 R|S, S⟩ = Rˆ Ω z · S|S, S⟩ ˆ = RSz |S, S⟩ = RS|S, S⟩ = SR|S, S⟩ = S|Ω⟩.
(32)
ˆ i |Si |Ω ˆ i ⟩ = i ⟨S, S|R−1 Si Ri |S, S⟩i = Ri i ⟨S, S|Si |S, S⟩i = Ri S zˆ = SRi zˆ = S Ω ˆi ⟨Ω i
(34)
(33)
(b) Observe that:
Then the following holds: ˆ ˆ = ⟨Ω|H| Ω⟩
S2 ∑ ˆi · Ω ˆj. Ji,j Ω 2 i,j
(35)
Problem 5 (a) Write integrals for the first two terms (“Hartree” and “exchange”) for G(ω, k) in momentum-space perturbation theory in the interaction strength V . (b) Use bubble diagrams (“RPA”) to calculate simple Thomas-Fermi screening of the electron-electron interaction (that is, ω = 0 and small k/kF ). The screened 0 interaction is V˜ = q24πe , k02 = 4πe2 ∂n ∂µ . +k2 0
(c) Consider the screening of the Coulomb interaction in a metal. The Thomas-Fermi approximation in quantum mechanics consists of computing the electron density at each point in space in a potential V (r) by using the macroscopic density for a constant chemical potential. Write a Poisson equation of the form −∇2 V = 4πeδ(r) + (density responses) and reproduce the screening of a Coulomb potential.
Solution 5 (a) Let’s follow the notation of Mattuck. The Hartree term is given by:
5
Physics 216 : Solution set 3
6
Physics 216 : Solution set 3 and the exchange term is given by:
(b) RPA contribution can be computed as:
7
Physics 216 : Solution set 3
which yields the Thomas-Fermi screening of the electron-electron interaction (c) The Thomas-Fermi amounts to having: µ → µ(r) = µ0 − V (r) = µ0 − eϕ(r). This gives the den-
8
Physics 216 : Solution set 3 sity as ∫
∫
µ(r)
n(r) =
D(ϵ)dϵ = 0
∫
µ0 0
≈ n0 − eϕ(r)
µ0 −eϕ(r)
D(ϵ)dϵ +
D(ϵ)dϵ
(36)
µ0
dn0 dµ
(37)
So the electric potential satisfies −∇2 V (r) = 4πeδ(r) − 4πe2 V (r)
dn0 dµ
(38)
0 Let k02 = 4πe2 dn dµ , then we get:
(−∇2 + k02 )V (r) = 4πeδ(r) ∫ d3 k ir·q which can be solved in the Fourier space, V (r) = (2π) V (q), the above equation becomes 3e ∫
] d3 k ir·q [ 2 e (q + k02 )V (q) = 3 (2π)
∫
] d3 k ir·q [ e 4πe . 3 (2π)
(39)
(40)
Which gives the screened interaction: V˜ (q) =
4πe , q2 + k02
k02 = 4πe2
9
∂n0 ∂µ
(41)
Problem 1 Define the retarded Green’s function through † † ⟨Ψα (t1 , r1 )Ψβ (t2 , r2 ) + Ψβ (t2 , r2 )Ψα (t1 , r1 )⟩ iGrαβ (t1 , r1 ; t2 , r2 ) = 0
if t1 − t2 > 0 (1) if t1 − t2 < 0.
(a) Calculate the Fourier-transformed Greens’ function Gr (ω, p) for the free Fermi gas at temperature T (i.e., the expectation value ⟨⟩ is taken with respect to the finite-temperature Fermi gas). (b) Specify where any poles occur. (c) Check that contour integration of your result indeed gives zero for negative t = t1 − t2 .
Solution 1 (a) Recall that Ψα (t, r) = e
iHt
ψα (r)e
−iHt
=e
iHt
(∑ k
) ∑ eikr ( ) eikr √ ckα e−iHt = √ eiHt ckα e−iHt V V k
(2)
∑ ∑ † † ′ ′ ′ Free Fermi gas has the Hamiltonian H = k′ α′ ϵk [ck′ α′ ck α , ckα ] = k,α ϵk ckα ckα , using [H, ckα ] = ( † ) ∑ † ′ ′ ′ ′ ′ = −ϵk ckα , we get k′ α′ ϵk ck′ α′ {ck α , ckα } − {ck′ α′ , ckα }ck α Ψα (t, r) =
∑ eikr √ e−iϵk t ckα . V k
(3)
Since there is a translational symmetry in the system, we can define iGrαβ (t, r) ≡ iGrαβ (t, r; 0, 0). Then, iGrαβ (t, r) = Θ(t)
1 ∑ i(kr−ϵk t) 1 ∑ ikr −iϵk t e ⟨e {ckα , c†k′ β }⟩ = Θ(t)δα,β e V V ′ k,k
(4)
k
The Fourier transformation of the Greens’ function is given as follows. ( ) ∫ ∫ 1 ∑ i(k·r−ϵk t) Gr (ω, p) = dt d3 rei(ωt−p·r) − iΘ(t)δα,β e V k ∫ ∞ = −iδα,β dte−i(ϵp −w)t 0
δα,β = , ω − ϵp + i0+
(5)
where 0+ is introduced to make the integral convergent. (b) The pole occurs at ϵp − i0+ , so slightly below the real axis. (c) Recall ∫ 3 d pdω i(p·r−ωt) δα,β r Gαβ (t, r) = e , (2π)4 ω − ϵp + i0+ 1
(6)
Physics 216 : Solution set 3 so when t < 0, e−iωt decays when Im ω > 0. Hence when we perform ω integral, we make this integral as a contour integral in (complex) ω by closing the contour in the upper plane. Since the pole exists below the real ω axis, the integral indeed becomes 0.
Problem 2 Consider two identical 1D non-relativistic bosons of mass m moving on a ring of Length L: x1 , x2 ∈ [0, L) ( ∂2 ) h ¯2 ∂2 and H = − 2m + ∂x 2 . ∂x21 2 (a) What are the energies of the ground state and first excited states? (b) Suppose a repulsive interaction of the form H ′ = cδ(r1 − r2 ) is added (c > 0). Construct the ground-state bosonic wavefunction by any means and find the ground state energy in the limit c → ∞.
Solution 2 2
2
h ¯ ∂ (a) The eigenstate and eigenenergy of the single particle Hamiltonian H = − 2m ∂x2 in a ring is given 2 2 ( ) 2 2 h ¯ kn 1 h ¯ 2π ik x n by ψn (x) = √L e , En = 2m = 2m L n , where n is an integer. The ground state has both bosons in the ground state: EGS = 0 2 2 h ¯ The first excited states have one boson in n = 0 state and the other boson in n = ±1: E = 2π mL2 (b) Using our educated guess, the ground state can be written as ( ) Ψk1 ,k2 (x1 , x2 ) = sgn(x1 − x2 ) ei(k1 x1 +k2 x2 ) − ei(k2 x1 +k1 x2 ) , (7)
where k1 , k2 are constants which are determined later. Putting into the Schroedinger equation yields HΨk1 ,k2 = − =
) )( ¯2 ( h 2δ(x1 − x2 ) − 2δ(x1 − x2 ) + (−k12 − k22 )sgn(x1 − x2 ) ei(k1 x1 +k2 x2 ) − ei(k2 x1 +k1 x2 ) (8) 2m
¯2 2 h (k + k22 )Ψk1 ,k2 , 2m 1
(9) 2
h ¯ where we have used ∂x sgn(x) = 2δ(x) and we get eigenenergy 2m (k12 + k22 ). Using the periodic boundary 2π condition, k1 , k2 ∈ L Z. In order not to have a vanishing wave function k1 ̸= k2 must be satisfied. Hence the ground state is the case where k1 = 0 and k2 = 2π L : ( 2π ) 2π EGS (x1 , x2 ) = sgn ei L x1 − ei L x2 (10)
2
2
h L with the ground state energy EGS = 2¯ mL2 . (Alternatively, one can attack this problem by separating CM and rel motion degrees of freedom.)
Problem 3 Consider a two-dimensional electron moving in a constant perpendicular magnetic field. (a) Show that 2 2 in the rotationally symmetric gauge for A, the lowest Landau level eigenstates ψm ∼ z m e−|z| /4l with z ≡ (x + iy) and m a nonnegative integer are eigenstates for a particular choice of l (independent of m), and calculate the normalization Now suppose that the system of N noninteracting LLL electrons is put in a weak radial confining potential V (r) = αr2 , with α sufficiently small that no mixing occurs between Landau levels. The ground state of the N electrons is now a Slater determinant of the single-particle eigenstates m = 0, . . . , N − 1.
2
Physics 216 : Solution set 3 Now consider the low-energy edge excitations of this state. For example, moving the last electron out by angular momentum h ¯ gives an excited states. (b) How many excitations within the LLL are there of total momentum M , for M = 0, . . . , 5? (c) Give an estimate of the velocity of excitations at the edge.
Solution 3 Let’s use the following convention. The electron has negative charge, −e, in the magnetic field −B zˆ (B > 0) where x − y plane is the plane in which electrons are confined. Then the noninteracting Hamiltonian is given by N N 1 ∑( e )2 1 ∑ 2 H= pi + Ai = π , (11) 2m i=1 c 2m i=1 i where m is the mass of the electron and N is the total number of electrons. After a bit of algebra, one can compute√the commutation relation: [π x , π y ] = i¯heB/c. It is then convenient to define the magnetic length as lB ≡ ¯hc/eB. Let’s introduce the following creation/annihilation operator: i π x + iπ y a≡ √ 2 ¯h/lB −i π x − iπ y a† ≡ √ , 2 ¯h/lB
(12) (13) (14)
which has the canonical commutation relation [a, a† ] = 1. The Hamiltonian can be written as ) N ( ∑ 1 H = ¯hωc a†i ai + , 2 i=1
(15)
eB is the cyclotron frequency. One immediately sees that this is nothing but a harmonic oscillator where ωc = mc ( ) with eigenenergies ¯hωc n + 12 for n ∈ {0, 1, 2, · · · }. Our task here is to construct the ground states, which are annihilated by a operator. (a) The symmetric gauge is given as A = B2 (y, −x). Now, identify the 2D plane as a complex plane via z ≡ x + iy. This amounts to giving: ( ) 1 z ∂ a= √ + 2lB ∗ (16) ∂z 2 2lB ( ∗ ) 1 z ∂ a† = √ − 2lB ∗ (17) ∂z 2 2lB
Solving the eigenequation: aψ(z, z ∗ ) = 0, the eigenstates are given by 2 1 1 ψm (r) = √ (z/lB )m e− 4 |r| . 2 m 2π2 m!lB ∫ ∗ One can check that the normalization condition holds: d2 rψm (r)ψn (r) = δm,n In this gauge, the angular momentum operator becomes
(18)
Lz = xpy − ypx = ¯h(z∂z − z ∗ ∂z∗ ), so ψm has angular momentum ¯hm. (b) If we add a confining potential V (r) = αr2 , the ground state of N N −1 z1 z2N −1 N −2 z2N −2 [ ] 1 z1 Ψ = A ψ0 , ψ1 · · · , ψN −1 ∼ √ .. .. N . . 1 1 3
(19) electrons is: ··· ··· .. . ···
N −1 zN N −2 zN .. . 1
,
(20)
Physics 216 : Solution set 3 which is the unique M = 0 excitation. Other excitations are as follows: [ ] M =1: A ψ0 , · · · , ψN −2 , ψN [ ] { A ψ0 , · · · , ψN −2 , ψN +1 [ ] M =2: A ψ0 , · · · , ψN −3 , ψN −1 , ψN [ ] A ψ0 , · · · , ψN −2 , ψN +2 [ ] M =3: A ψ0 , · · · , ψN −3 , ψN −1 , ψN +1 ] [ A ψ0 , · · · , ψN −4 , ψN −2 , ψN −1 , ψN ] [ A ψ0 , · · · , ψN −2 , ψN +3 [ ] A ψ0 , · · · , ψN −3 , ψN −1 , ψN +2 [ ] M =4: A ψ0 , · · · , ψN −3 , ψN , ψN +1 [ ] A ψ0 , · · · , ψN −4 , ψN −2 , ψN −1 , ψN +1 [ ] A ψ0 , · · · , ψN −5 , ψN −3 ψN −2 , ψN −1 , ψN ] [ A ψ0 , · · · , ψN −2 , ψN +4 [ ] A ψ0 , · · · , ψN −3 , ψN −1 , ψN +3 [ ] A ψ0 , · · · , ψN −3 , ψN , ψN +2 [ ] M =5: A ψ0 , · · · , ψN −4 , ψN −2 , ψN −1 , ψN +2 ] [ A ψ0 , · · · , ψN −4 , ψN −2 , ψN , ψN +1 [ ] A ψ0 , · · · , ψN −5 , ψN −3 ψN −2 , ψN −1 , ψN +1 ] [ A ψ0 , · · · , ψN −6 , ψN −4 , ψN −3 ψN −2 , ψN −1 , ψN
(21) (22)
(23)
(24)
(25)
2 (c) The edge has angular momentum Lz = N ¯h ∼ rmv. Using ⟨r2 ⟩ = N 2lB , the velocity of excitations at the edge can be estimated as √ N ¯h ¯h 1 N √ v∼ = (26) 2 m mlB 2 N 2lB
Problem 4 (a) Verify, directly from the definition: ˆ = R(ξ, θ, ϕ)|S, S⟩ |Ω⟩ =e
iS z ϕ iS y θ S z ξ
e
e
|S, S⟩,
(27) (28)
ˆ ≡ (sin θ cos ϕ, sin θ sin ϕ, cos θ), that |Ω⟩ ˆ is an eigenstate of the spin component in where the unit vector Ω ˆ direction: the Ω ˆ · S|Ω⟩ ˆ = S|Ω⟩ ˆ Ω (29) ∑ ˆ = ∏ |Ω ˆ i⟩ (b) Show that the expectation value of the Heisenberg model H = i,j Si ·Sj in a coherent state |Ω⟩ i is 2 ∑ ˆ ≡ ⟨Ω|H| ˆ ˆ =S ˆi · Ω ˆj. H[Ω] Ω⟩ Ji,j Ω (30) 2 i,j
Solution 4 (a) Let R be the rotation given by Euler angles (ξ, θ, ϕ). (Ω = Rˆ z .) Then R−1 SR = RS 4
(31)
Physics 216 : Solution set 3 holds because S is a vector operator. Using Sz |S, S⟩ = S|S, S⟩, the following equation holds: ( ) ( ) ˆ · S|Ω⟩ ˆ = zˆ R−1 S |Ω⟩ ˆ = zˆ RSR−1 R|S, S⟩ = Rˆ Ω z · S|S, S⟩ ˆ = RSz |S, S⟩ = RS|S, S⟩ = SR|S, S⟩ = S|Ω⟩.
(32)
ˆ i |Si |Ω ˆ i ⟩ = i ⟨S, S|R−1 Si Ri |S, S⟩i = Ri i ⟨S, S|Si |S, S⟩i = Ri S zˆ = SRi zˆ = S Ω ˆi ⟨Ω i
(34)
(33)
(b) Observe that:
Then the following holds: ˆ ˆ = ⟨Ω|H| Ω⟩
S2 ∑ ˆi · Ω ˆj. Ji,j Ω 2 i,j
(35)
Problem 5 (a) Write integrals for the first two terms (“Hartree” and “exchange”) for G(ω, k) in momentum-space perturbation theory in the interaction strength V . (b) Use bubble diagrams (“RPA”) to calculate simple Thomas-Fermi screening of the electron-electron interaction (that is, ω = 0 and small k/kF ). The screened 0 interaction is V˜ = q24πe , k02 = 4πe2 ∂n ∂µ . +k2 0
(c) Consider the screening of the Coulomb interaction in a metal. The Thomas-Fermi approximation in quantum mechanics consists of computing the electron density at each point in space in a potential V (r) by using the macroscopic density for a constant chemical potential. Write a Poisson equation of the form −∇2 V = 4πeδ(r) + (density responses) and reproduce the screening of a Coulomb potential.
Solution 5 (a) Let’s follow the notation of Mattuck. The Hartree term is given by:
5
Physics 216 : Solution set 3
6
Physics 216 : Solution set 3 and the exchange term is given by:
(b) RPA contribution can be computed as:
7
Physics 216 : Solution set 3
which yields the Thomas-Fermi screening of the electron-electron interaction (c) The Thomas-Fermi amounts to having: µ → µ(r) = µ0 − V (r) = µ0 − eϕ(r). This gives the den-
8
Physics 216 : Solution set 3 sity as ∫
∫
µ(r)
n(r) =
D(ϵ)dϵ = 0
∫
µ0 0
≈ n0 − eϕ(r)
µ0 −eϕ(r)
D(ϵ)dϵ +
D(ϵ)dϵ
(36)
µ0
dn0 dµ
(37)
So the electric potential satisfies −∇2 V (r) = 4πeδ(r) − 4πe2 V (r)
dn0 dµ
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0 Let k02 = 4πe2 dn dµ , then we get:
(−∇2 + k02 )V (r) = 4πeδ(r) ∫ d3 k ir·q which can be solved in the Fourier space, V (r) = (2π) V (q), the above equation becomes 3e ∫
] d3 k ir·q [ 2 e (q + k02 )V (q) = 3 (2π)
∫
] d3 k ir·q [ e 4πe . 3 (2π)
(39)
(40)
Which gives the screened interaction: V˜ (q) =
4πe , q2 + k02
k02 = 4πe2
9
∂n0 ∂µ
(41)
