Functional derivative technique
Ferdi Aryasetiawan Mathematical Physics, Lund University (Dated:)
1 I. THE SCHWINGER FUNCTIONAL DERIVATIVE TECHNIQUE
We will derive a closed set of equations which determine the self-energy by using the method of functional derivative, originally used by Schwinger in quantum electrodynamics and adopted into condensed matter by Hedin. The idea of the technique is to apply a time-dependent probing …eld to the system. The linear response of the one-particle Green’s function with respect to the probing …eld can be related to the two-particle Green’sfunction which in turn is related to the self-energy. Thus, the self-energy may be expressed as a functional derivative of the Green function with respect to the probing …eld. In the end, the …eld is set to zero since we only need to know the derivative calculated at zero …eld. Since a time-dependent probing …eld is applied, it is convenient to work in the interaction representation. In the interaction representation the Green function is de…ned as
^ ^ ^ T [S D(1) D(2)] iG(1; 2) = j j (1) D S^ E j j D E where
H^ = E0 ; (2) j i j i
S^ = U^D( ; ); (3) 1 1
t U^D(t; t0) = T exp[ i d()] (4) Zt0
^(t) = d3r^(r; t)'(r; t): (5) Z To simplify writing, we use the following notations:
r = (r; ); (6)
1 = (r1; t1): (7)
The …eld operators are in the interaction representation which is related to the …eld operators in the Heisenberg picture as follows (see Appendix):
2 ^ ^ ^ ^ H (t) = U D(0; t) D(t)U D(t; 0): (8)
The ground state is the interacting ground state without the probing …eld, i.e., it is the j i same as the Heisenberg ground state in the usual de…nition of the Green function:
^ ^ T [ H (1) Hy (2)] iG(1; 2) = j j : (9) D E h j i It is important to note that the …eld operators in the Dirac picture are independent of the applied …eld '(t) since
iHt^ iHt^ O^D(t) = e Oe^ (10)
and that G depends on the applied …eld only through S^. Let us relate the de…nition of G in (1) to the usual de…nition in (9). Using (8)
iG(1; 2) ^ ^ ^ ^ ^ ^ ^ ^ T [UD( ; 0)UD(0; t1) D(1)UD(t1; 0)UD(0; t2) Dy (2)UD(t2; 0)UD(0; ) = j 1 1 j D E U^D( ; 0)U^D(0; ) j 1 1 j D E D(0) T [ H (1) Hy (2)] D(0) = j j D D(0) D(0) E h j i H T [ H (1) Hy (2)] H = j j : (11) D H H E h j i To obtain the third line we have utilised the condition that the state at t = is the 1 interacting state without the probing …eld ' so that
D(0) = U^D(0; ) ; D(0) = U^D( ; 0): (12) j i 1 j i h j h j 1 The above condition implies that '( t) = '(t) up to a constant. The last line in (11) has been obtained from the fact that all pictures have their states coincide at t = 0. The above de…nition of the Green function is more general in that it includes the possibility of having a time-dependent probing …eld in the Hamiltonian. For ' = 0; D(0) is a state that 6 has evolved, due to the probing …eld '; from the interacting ground state at t = to a 1 state at t = 0. There is no assumption about adiabatic continuity. The probing …eld ' is
3 arbitrary which may be large and rapidly varying in time. For ' = 0 the above de…nition ^ ^ clearly reduces to (9) since S^ 1; D H ; and H becomes the interacting ground state ! ! . The Green function in (1) should be regarded as a functional of the probing …eld '. We will now take the functional derivative of G with respect to '. In order to do this it is only necessary to work out the functional derivative of S^ with respect to ' since the …eld operators in the Dirac picture do not depend on ': S^ = T exp[ i d4 ^(4)'(4) '(3) '(3) Z = iT [S^^(3)]: (13) We obtain ^ ^ ^ G(1; 2) T [S D(1) Dy (2)] i = j j '(3) '(3) D S^ E j j ^ D ^ ^E i T [S^(3) D(1) Dy (2)] = j j D S^ E j j ^ ^ D ^ E ^ i T [S D(1) Dy (2)] T [S^(3)] j j j j (14) D ED2 E S^ j j We de…ne the two-particle Green’sfunction asD E ^ ^ ^ ^ ^ T [S D(1) D(3) Dy (4) Dy (2) G(2)(1; 2; 3; 4) = (i)2 j j : (15) D S^ E j j We therefore obtain D E G(1; 2) = G(1; 2)(3) G(2)(1; 2; 3; 3+): (16) '(3) This relation is very important because it expresses the two-particle Green’sfunction, which corresponds to the Coulomb term in the equation of motion of the Green function, as a linear response of the one-particle Green’sfunction with respect to the probing …eld. It is readily veri…ed that ^ ^ ^ + ^ ^ T [S D(3) Dy (3 ) D(1) Dy (2)] G(2)(1; 2; 3; 3+) = j j : (17) D S^ E j j We now de…ne the mass operator as follows: D E
i d3v(1 3)G(2)(1; 2; 3; 3+) = d3M(1; 3)G(3; 2): (18) Z Z 4 1 Multiplying both sides of the equation on the right by G (2; 4) and integrating over the variable 2 we obtain
(2) + 1 M(1; 4) = i d3d2v(1 3)G (1; 2; 3; 3 )G (2; 4): (19) Z Using the relation in (16) we obtain
G(1; 2) 1 M(1; 4) = i d3d2v(1 3) G(1; 2)(3) G (2; 4) '(3) Z G(1; 2) 1 = i d3v(1 3)(3)(1 4) + i d3d2v(1 3) G (2; 4) '(3) Z Z G(1; 2) 1 = VH (1)(1 4) + i d3d2v(1 3) G (2; 4): (20) '(3) Z Thus, by replacing the two-particle Green’s function using (16) the mass operator is now expressed entirely in terms of the single-particle Green’sfunction and its functional derivative with respect to the probing …eld. The self-energy is de…ned as the mass operator less the Hartree potential: G(1; 4) 1 (1; 2) = i d3d4v(1 3) G (4; 2): (21) '(3) Z Since 1 d4G(1; 4)G (4; 2) = (1 2) (22) Z it follows that 1 G(1; 4) 1 G (4; 2) d4 G (4; 2) + G(1; 4) = 0: (23) '(3) '(3) Z G(1; 2) G 1(4; 5) = d4d5 G(1; 4) G(5; 2) (24) '(3) '(3) Z We may write G 1(4; 2) (1; 2) = i d3d4v(1 3)G(1; 4) : (25) '(3) Z From the equation of motion we have (Exercise 1) @ i h(1) G(1; 2) d3(1; 3)G(3; 2) = (1 2); (26) @t 1 Z where h is the one-particle part of the Hamiltonian that includes the probing …eld ':
1 2 h = + Vext + VH + ': 2r Multiplying both sides on the right by the inverse of G yields
1 @ G (1; 2) = i h(1) (1 2) (1; 2) (27) @t 1 5 and therefore, using '(1)='(3) = (1 3), G 1(1; 2) V (1) (1; 2) = (1 3) + H (1 2) : (28) '(3) '(3) '(3) Using this expression in (25) gives V (4) (4; 2) (1; 2) = i d3d4v(1 3)G(1; 4) (4 3) + H (4 2) + '(3) '(3) Z V (2) = iv(1 2)G(1; 2) + i d3v(1 3)G(1; 2) H '(3) Z (4; 2) + i d3d4v(1 3)G(1; 4) : (29) '(3) Z This is an iterative equation for the self-energy which provides a procedure for making an expansion of the self-energy in powers of the Coulomb interaction. We recognised that the …rst term is just the exchange potential. It is a more general form of exchange with an interacting Green’sfunction instead of a non-interacting one. Let us work out the self-energy to second order in v and show that it is consistent with the conventional perturbation expansion result. In order to do this, we need to calculate
VH = and = to …rst order in v, as can be seen from (29). V (2) (4) H = d4 v(2 4) '(3) '(3) Z G(4; 4+) = i d4 v(2 4) '(3) Z G 1(5; 6) = i d4d5d6 v(2 4)G(4; 5) G(6; 4+): (30) '(3) Z where the identity in (23) has been used in the last line. To …rst order in v, V (2) H = i d4 v(2 4)G(4; 3)G(3; 4+): (31) '(3) Z To …rst order = ivG and using this we calculate (4; 2) G(4; 2) = iv(4 2) : (32) '(3) '(3) Using (24) (4; 2) G 1(5; 6) = i d5d6 v(4 2)G(4; 5) G(6; 2) '(3) '(3) Z = i d5d6 v(4 2)G(4; 5) Z V (5) (5; 6) (5 6) (5 3) + H + G(6; 2) '(3) '(3) = iv(4 2)G(4; 3)G(3; 2) + O(v2):
6 Using the result for VH = and = in (29) and keeping terms up to second order in v we obtain V (2) (1; 2) = iv(1 2)G(1; 2) + i d3v(1 3)G(1; 2) H '(3) Z (4; 2) + i d3d4v(1 3)G(1; 4) '(3) Z = iv(1 2)G(1; 2) + i d3v(1 3)G(1; 2)( i) d4 v(2 4)G(4; 3)G(3; 4+) Z Z + i d3d4v(1 3)G(1; 4)iv(4 2)G(4; 3)G(3; 2): (33) Z This is consistent with the result obtained using the traditional diagrammatic expansion. Since G is the interacting Green’sfunction, the self-energy consists only of skeleton diagrams. A skeleton diagram cannot be reduced further by absorbing interactions into renormalised Green function lines. The Schwinger functional derivative technique has several advantages over the conven- tional diagrammatic technique. The iterative equation for the self-energy in (29) is exact and it allows us to generate all the terms in the perturbation expansion without the need to enumerate topologically distinct connected diagrams and to employ the tedious Wick’sthe- orem. Moreover, the ground state in the de…nition of the Green function in the interaction picture in (1) is the exact interacting ground state. There is no assumption of adiabatic continuity, associated with switching on the Coulomb interaction slowly from t = to 1 t = 0, that is needed to make a one-to-one connection between the non-interacting ground state and the interacting one. The probing …eld ' in the functional derivative technique can be large and rapidly varying in time. Its role is to probe the response of the system, not to connect a non-interacting ground state to an interacting one. The Schwinger functional derivative technique provides a simple and elegant way of deriving the self-energy.
II. THE HEDIN EQUATIONS
The self-energy expression presented in the previous section was derived by Hedin. It was already realised at that time that a …nite expansion in powers of the bare Coulomb interaction v leads to unphysical results in solids. The reason is that in solids screening e¤ects are of paramount importance. At low energies, the e¤ective Coulomb interaction is
7 much smaller than the bare value, especially for metals. It was then proposed by Hedin that one should make an expansion of the self-energy in powers of the screened interaction. Since the screened interaction is much weaker than than the bare one the self-energy may be expected to converge much faster. This turns out to be not the case, due to the presence of energy dependence in the screened interaction. Nevertheless the Hedin equations, which are a set of self-consistent equations relating the self-energy to the screened interaction, provide a powerful tool for studying the electronic structure of matter from …rst-principles. Instead of ' it is permissible to work with the total …eld
V = ' + VH : (34)
In place of (28) we have
G 1(1; 2) (1; 2) = (1 2)(1 3) : (35) V (3) V (3) Using the chain rule the self-energy in (25) becomes
G 1(4; 2) V (5) (1; 2) = i d3d4d5 v(1 3)G(1; 4) : (36) V (5) '(3) Z V=' is simply the de…nition of the inverse dielectric function: it is the ratio between the total and the applied …elds. 1 V (1) (1; 2) : (37) '(2) The Coulomb interaction is then screened by the dielectric function.
V (5) 1 d3 v(1 3) = d3 (5; 3)v(3 1) W (5; 1): '(3) Z Z Thus, (1; 2) = i d4d5 W (5; 1)G(1; 4)(4; 2; 5); (38) Z where we have de…ned the vertex function as
G 1(1; 2) (1; 2; 3) V (3) (1; 2) = (1 2)(1 3) + : (39) V (3)
8 Using the chain rule and the identity in (23) we can set up an integral equation for the vertex.
(1; 2) (1; 2; 3) = (1 2)(1 3) + V (3) (1; 2) G(4; 5) = (1 2)(1 3) + d4d5 G(4; 5) V (3) Z (1; 2) G 1(6; 7) = (1 2)(1 3) d4d5 d6d7G(4; 6) G(7; 5) G(4; 5) V (3) Z Z (1; 2) = (1 2)(1 3) + d4d5d6d7 G(4; 6)G(7; 5)(6; 7; 3): (40) G(4; 5) Z
We now need an integral equation that determines the screened interaction W . From (37) 1 V (1) W (1; 2) = d3 (1; 3)v(3 2) = d3 v(3 2) (41) '(3) Z Z and using (34)
V (1) W (1; 2) = d3 (1 3) + H v(3 2) '(3) Z (4) = v(1 2) + d3d4 v(1 4) v(3 2) '(3) Z (4) V (5) = v(1 2) + d3d4d5 v(1 4) v(3 2) V (5) '(3) Z (4) 1 = v(1 2) + d3d4 d5v(1 4) (5; 3)v(3 2) V (5) Z = v(1 2) + d4d5 v(1 4)P (4; 5)W (5; 2) (42) Z where we have de…ned the polarisation function
(1) P (1; 2) : (43) V (2) The linear density response function, hereafter referred to as response function, is de…ned as (1) R(1; 2) : (44) '(2) The polarisation function is the response function with respect to the total …eld. We can
9 readily write down the integral equation for the response function from its de…nition.
(1) R(1; 2) = '(2) (1) V (3) = d3 V (3) '(2) Z (4) = d3P (1; 3) (3 2) + d4v(3 4) '(2) Z Z = P (1; 2) + d3d4P (1; 3)v(3 4)R(4; 2); (45) Z which is identical in form to the integral equation for the screened interaction. The polarisation function can be expressed in terms of the Green function.
(1) P (1; 2) = V (2) G(1; 1+) = i V (2) G 1(3; 4) = i d3d4G(1; 3) G(4; 1+) V (2) Z = i d3d4G(1; 3)(3; 4; 2)G(4; 1+): (46) Z From the de…nition of the screened interaction
1 W (1; 2) = d3 (1; 3)v(3 2) Z V (1) = d3 v(3 2) '(3) Z V (1) = d3 (1 3) + H v(3 2) '(3) Z = v(1 2) + d3d4 v(1 4)R(4; 3)v(3 2) (47) Z
In summary, the closed set of self-consistent Hedin’sequations is given by
(1; 2) = i d4d5 W (5; 1)G(1; 4)(4; 2; 5): (48) Z
(1; 2; 3) = (1 2)(1 3) (49) (1; 2) + d4d5d6d7 G(4; 6)G(7; 5)(6; 7; 3): (50) G(4; 5) Z
10 W (1; 2) = v(1 2) + d3d4 v(1 3)P (3; 4)W (4; 2): (51) Z P (1; 2) = i d3d4G(1; 3)(3; 4; 2)G(4; 1+): (52) Z G(1; 2) = G0(1; 2) + d3d4 G0(1; 3)(3; 4)G(4; 2): (53) Z The non-interacting Green function G0 is de…ned to be the Hartree Green function.
III. PHYSICAL SIGNIFICANCE OF W
The screened interaction is given by
W (1; 2) = v(1 2) + d3d4v(1 3)R(3; 4)v(4 2) (54) Z We can think of v(1 2) as the Coulomb potential at 1 of a test charge located at 2, which may be chosen to be at the origin of the coordinate system:
W (1) = v(1) + d3d4v(1 3)R(3; 4)v(4) Z Consider
(3) = d4R(3; 4)v(4) Z in the second term on the right hand side. This is the induced charge density of the system at point 3 due to the presence of the test charge at 2, which has been chosen to be at the origin. This induced charge density gives rise to a change in the Hartree potential at point 1:
VH (1) = d3v(1 3)(3); Z which acts as a screening potential for the test charge so that the total potential of the test charge is given by the screened interaction
W (1) = v(1) + VH (1):
Thus, W (1; 2) is a screened potential at point 1 of a test charge located at point 2.
11 IV. LINEAR RESPONSE THEORY: THE KUBO FORMULA
We will derive the Kubo formula in two ways, following the original derivation of Kubo and using the Schwinger functional derivative method. As will be seen, the latter technique provides a much simpler way of deriving the formula. Consider applying an external …eld '(r; t); which may be time dependent, to a system in its ground state. We wish to work out a formula which allows us to calculate the change in a physical quantity represented by an operator O^ to …rst order in the applied …eld. Since we apply an external …eld, it is convenient to work in the interaction picture. The equation of motion of the state vector in the interaction representation is given by
@ ^ i D(t) = D(t) D(t) (55) @t j i j i where ^ iHt^ ^ iHt^ D(t) = e (t)e : (56)
The formal solution to the time-development of the state vector is given by
D(t) = U^D(t; 0) D(0) (57) j i j i where t ^ i dt D(t ) U^(t; 0) = T e 0 0 0 : (58) R The expectation value of an operator O^ at time t is then
O(t) = D(t) O^D(t) D(t) j j D E = D(0) U^(0; t)O^D(t)U^(t; 0) D(0) (59) j j D E where iHt^ iHt^ O^D(t) = e Oe^ : (60)
To …rst order in the applied …eld the time-development operator is given by
t ^ U^D(t; 0) = 1 i dt0 D(t0); (61) Z0 t ^ ^ ^ UD(0; t) = UDy (t; 0) = 1 + i dt0 D(t0): (62) Z0
12 Substituting these in the expression for O(t) yields
O(t) = D(t) O^D(t) D(t) D(0) O^D(t) D(0) j j j j D t E D E ^ = i dt0 D(0) [D(t0); O^D(t)] D(0) : (63) 0 j j Z D E Note that
iHt^ iHt^ D(0) O^D(t) D(0) = e Oe^ j j j j
D E D iE0t iE0t E = e Oe^ j j D E = O^ : j j D E We recall that the Dirac operators are the same as the Heisenberg operators without the appled …eld and all representations have the same state at t = 0 so that D(0) is just the interacting ground state without the applied …eld. Let us consider an external …eld that is coupled to the charge density, i.e.,
^ ^ ^ D(t) = dr Dy (rt)'(rt) D(rt) = dr '(rt)^D(rt); (64) Z Z and let us calculate the density response to this applied …eld. Using O^D = ^D in (63) we …nd
t ^ (rt) = i dt0 D(0) [D(t0); ^D(rt)] D(0) 0 j j Z D t E = i dr0 dt0'(r0t0) D(0) [^D(r0t0); ^D(rt)] D(0) : (65) h j j i Z Z0 r The retarded linear density response function R (rt; r0t0) is de…ned by
1 r (rt) = dr0 dt0 R (rt; r0t0)'(r0t0) (66) Z Z0 so that
r R (rt; r0t0) = i D(0) [^D(r0t0); ^D(rt)] D(0) (t t0) (67) h j j i = i [^H (r0t0); ^H (rt)] (t t0) h j j i is the retarded response function. The second line ie obtained by setting the probing …eld ' to zero.
13 As another demonstration of the usefulness of the Schwinger functional derivative tech- nique, we can readily derive the Kubo formula from (14). Since (1) = iG(1; 1+) the linear density response R = = is
G(1; 1+) R(1; 2) = i '(2) ^ T [S^D(1)] = j j '(2) D S^ E j j D E i T [S^^D(2)^D(1)] = j j D S^ E j j D E i T [S^^D(1)] T [S^^D(2)] + j j j j (68) D ED 2 E S^ j j D E +) After taking the functional derivative of the density D(1) = iG(1; 1 with respect to the applied …eld ', we set ' = 0. This implies that S^ = 1 and the Dirac …eld operator becomes the Heisenberg …eld operator. We obtain without much e¤ort the time-ordered linear density response function
R(1; 2) = i ^H (2)^H (1) (t2 t1) i ^H (1)^H (2) (t1 t2) (69) h j j i h j j i where
^H = ^H (70) is the density ‡uctuation operator measuring the density ‡uctuation from the ground-state
density (1) = ^H (1) : h j j i To obtain the response function in the frequency representation we …rst insert a complete
14 set of eigenstates of H^ in between the density operators yielding
^H (2)^H (1) (t2 t1) + ^H (1)^H (2) (t1 t2) h j j i h j j i = ^H (2) n n ^H (1) (t2 t1) n h j j i h j j i X + ^H (1) n n ^H (2) (t1 t2) n h j j i h j j i X iHt^ 2 iHt^ 2 iHt^ 1 iHt^ 1 = e ^(r2)e n n e ^(r1)e (t2 t1) n j j i h j j X D E iHt^ 1 iHt^ 1 iHt^ 2 iHt^ 2 + e ^(r1)e n n e ^(r2)e (t1 t2) n j j i h j j X D E i(En E0)(t2 t1)( = ^(r2) n n ^(r1) e (t2 t1) n h j j i h j j i X i(En E0)(t1 t2)( + ^(r1) n n ^(r2) e (t1 t2) (71) n h j j i h j j i X Performing the Fourier transform as in the Lehman representation of the Green function yields the famous Kubo’sformula
^(r0) n n ^(r) ^(r) n n ^(r0) R(r; r0; !) = h j j i h j j i h j j i h j j i (72) ! E + E + i ! + E E i n n 0 n 0 X The response function gives information about the excitation spectrum of the system.
Im R(!) exhibits peaks when ! = En E0, corresponding to the N-particle excitation energies. If there is no magnetic …eld, i.e., if time-reversal symmetry is obeyed,
^(r0) n n ^(r) = ^(r) n n ^(r0) (73) h j j i h j j i h j j i h j j i is real so that R satis…es
R(r; r0; !) = R(r; r0; !); (74)
R(r; r0; !) = R(r0; r; !): (75)
The response function R is the time-ordered response which di¤ers from the retarded response Rr: The two are related as follows:
Re R(!) = Re Rr(!); (76)
Im R(!)sgn(!) = Im Rr(!); sgn(!) != ! (77) j j
15 valid for real !. Compared with the original derivation of Kubo, the Schwinger functional derivative tech- nique provides a simple way of deriving the response functions. We have derived the Kubo formula speci…cally for linear density response function. However, the method is applicable to a more general response function since any expectation value of a single-particle operator in the ground state is expressible in terms of the Green function. The response function can be written in terms of its spectral representation.
0 S(r; r0; !0) 1 S(r; r0; !0) R(r; r0; !) = d!0 + d!0 (78) ! !0 i 0 ! !0 + i Z 1 Z
S(r; r0; !0 0) = ^(r0) n n ^(r) (!0 En + E0) (79) n h j j i h j j i X To ensure the symmetry of R we de…ne
S(r; r0; !) = S(r; r0; !): (80) It can be readily veri…ed that the spectral representation of R is equivalent to (72). The
spectral function S(r; r0; !) is real for systems with time reversal symmetry. Using the symmetry property of the spectral function the spectral representation of R can be rewritten as 1 1 1 R(r; r0; !) = d!0S(r; r0; !0) : (81) ! ! + i ! + ! i Z0 0 0
A. The Random-Phase Approximation (Time-Dependent Hartree Approxima- tion)
When we perturb a system of electrons by applying an external …eld, the density will change and the change in the density can be calculated if we know the response function of the system. In many cases it is su¢ cient to calculate the linear response function as de…ned in (66). However, as the perturbation …eld becomes larger, non-linear response functions may become important. The random-phase approximation (RPA) is an approximation for the linear response function. It was originally derived for the electron gas by Bohm and Pines using the equation of motion method. Gell-Mann and Brueckner subsequently derived the RPA using the many-body diagrammatic technique. The derivation is very involved and
16 requires the assumption that the electron density is high, much higher than the average electron density found in the alkali metals and in most materials. In fact, the RPA is equivalent to the time-dependen Hartree approximation (TDHA), which does not require the assumption of high electron density. Here, we derive the RPA using the functional derivative method, yet another demonstration of the usefulness of the method. The linear response function is given by
(1) R(1; 2) = : (82) '(2)
In terms of the Green function the density is given by
(1) = iG(1; 1+): (83) Thus taking the functional derivative we obtain
G(1; 1+) R(1; 2) = i '(2) G 1(3; 4) = i d3d4 G(1; 3) G(4; 1+) '(2) Z V (4) (3; 4) = i d3d4 G(1; 3) (3 4)(3 2) + (3 4) H + G(4; 1+): '(2) '(2) Z (84)
The second step has been obtained using the identity in (23) and the third step from (28). The RPA corresponds to neglecting the response of the self-energy to the applied …eld ', keeping only the change in the Hartree potential with respect to the applied …eld. This is why we may regard the RPA as the time-dependent Hartree approximation. In this approximation we have
V (3) R(1; 2) = i d3 G(1; 3)G(3; 1+) (3 2) + H '(2) Z V (3) = d3 P (1; 3) (3 2) + H '(2) Z (4) = P (1; 2) + d3d4 P (1; 3)v(3 4) : (85) '(2) Z Thus we obtain the well-known RPA equation
R(1; 2) = P (1; 2) + d3d4 P (1; 3)v(3 4)R(4; 2); (86) Z
17 with P (1; 2) = iG(1; 2)G(2; 1+); (87) which is equivalent to approximating the vertex in (46) by
(1; 2; 3) = (1 2)(1 3): (88) We note that the form of the RPA equation is the same as the exact equation for the response function in (45). The di¤erence lies in P , where the vertex function is approximated by (88). If we solve (86) iteratively, we obtain
R = P + P vP + P vP vP + ::: (89)
If we represent the right hand side by diagrams, we obtain the so-called ring diagrams, where each ring diagram represents the polarisation P = iGG and the two lines forming the ring represent the Green functions G. Let us work out the polarisation function explicitly using some non-interacting G0. In practice this is chosen to be the LDA Green function. The Fourier expansion of the Green function is
d! i!(t1 t2) G(1; 2) = e G(r ; r ; !): (90) 2 1 2 Z
d!1 i!1(t1 t2) d!2 i!2(t2 t1) P (1; 2) = i e G(r1; r2; !1) e G(r2; r1; !2): (91) 2 2 Z Z Taking the Fourier transform of P yields, with = t1 t2
i! P (r1; r2; !) = d e P (r1; r2; ) Z d!1 d!2 i(! !1+!2) = i d e G(r1; r2; !1)G(r2; r1; !2) 2 2 Z d!1 d!2 = i 2(! !1 + !2)G(r1; r2; !1)G(r2; r1; !2) 2 2 Z d!2 = i G(r1; r2; ! + !2)G(r2; r1; !2): (92) 2 Z which also follows directly from the convolution theorem. Using a non-interacting Green’s function occ unocc 0 kn(r)kn(r0) kn(r)kn(r0) G (r; r0; !) = + : (93) ! "kn i ! "kn + i kn kn X X
18 the integral over the frequency !2 consists of four terms of the form 1 1 d!2 ; (94) ! + !2 "kn i !2 "k n i Z occ 0 0 occ 1 1 d!2 ; (95) ! + !2 "kn i !2 "k n + i Z occ 0 0 unocc 1 1 d!2 ; (96) ! + !2 "kn + i !2 "k n i Z unocc 0 0 occ 1 1 d!2 : (97) ! + !2 "kn + i !2 "k n + i Z unocc 0 0 unocc The integrand of the …rst integral has both of its poles in the upper-half plane so that the integral is zero because the contour can be chosen to be along a semicircle in the lower-half plane. The integrand of the fourth integral has both of its poles in the lower-half plane and by choosing the contour to be along a semicircle in the upper-half plane the integral is also zero. Only the second and third integrals are non-zero. Performing the second contour integral by closing the contour in the lower-half semicircle yields 1 1 d!2 ! + !2 "kn i !2 "k n + i Z occ 0 0 unocc 2i = : (98) ! + "unocc "occ i k0n0 kn Similarly 1 1 d!2 ! + !2 "kn + i !2 "k n i Z unocc 0 0 occ 2i = : (99) ! + "occ "unocc + i k0n0 kn Using these results in (92) gives
occ unocc 0 kn(r)kn(r0)k0n0 (r0)k n (r) P (r; r0; !) = 0 0 ! + "k0n0 "kn i kn k0n0 unoccXoccX kn(r) (r0)k n (r0) (r) + kn 0 0 k0n0 (100) ! + "k n "kn + i kn k n 0 0 X X0 0 Interchanging the dummy indices kn k0n0 in the second term on the right hand side $ occ unocc 0 kn(r)k0n0 (r)k n (r0)kn(r0) P (r; r0; !) = 0 0 ! "k n + "kn + i kn k n 0 0 X X0 0 kn(r) (r)k n (r0) (r0) k0n0 0 0 kn : (101) ! + "k n "kn i 0 0
19 Since P 0 is just the response function of a non-interacting system, we could have used (72), where the many-electron wave functions are single Slater determinants, to obtain the above equation directly. As can be seen the polarisation function satis…es the following symmetry:
P (r0; r; !) = P (r; r0; !): (102) To take into account the spin degree of freedom, we have to multiply the above expression by a factor of two for paramagnetic system. For spin-polarised systems, the above expression is valid for each spin channel and the total P 0 is then the sum of the contribution from each spin channel. We may rewrite the expression for P 0 as follows.
(r)k n (r) (r0)kn(r0) 0 kn 0 0 k0n0 P (r; r0; !) = nkn(1 nk0n0 ) ! "k n + "kn + i kn k n 0 0 X X0 0 kn(r) (r)k n (r0) (r0) k0n0 0 0 kn nkn(1 nk0n0 ) : (103) ! + "k n "kn i 0 0 where
nkn = 1 if kn is occupied
= 0 if kn is unoccupied. (104)
Interchanging kn k0n0 in the second term gives $ 0 P (r; r0; !) = (r) (r) (r0) (r0) kn k0n0 k0n0 kn kn k n X X0 0 nkn(1 nk n ) nk n (1 nkn) 0 0 0 0 : (105) ! "k n + "kn + i ! "k n + "kn i 0 0 0 0 For systems with time reversal symmetry we have for every kn a corresponding kn = kn with the same eigenvalue. Therefore
0 P (r; r0; !) 1 = [kn(r)k0n0 (r)k n (r0)kn(r0) + kn(r) k0n0 (r) k n (r0) kn(r0)] 2 0 0 0 0 kn k n X X0 0 nkn(1 nk n ) nk n (1 nkn) 0 0 0 0 ! " + " + i ! " + " i k0n0 kn k0n0 kn 1 = [ (r)k n (r) (r0)kn(r0) + c.c.] 2 kn 0 0 k0n0 kn k n X X0 0 nkn(1 nk n ) nk n (1 nkn) 0 0 0 0 : (106) ! "k n + "kn + i ! "k n + "kn i 0 0 0 0
20 This means we only need to consider the real part of the product of the four wave functions 0 0 and P is symmetric with respect to an interchange of r r0. Thus P can be written as $
Re[ (r)k n (r) (r0)kn(r0)] 0 kn 0 0 k0n0 Re P (r; r0; !) = (nkn nk0n0 ) (107) ! "k n + "kn kn k n 0 0 X X0 0 occ unocc 0 Im P (r; r0; !) = 2 Re[ (r)k n (r) (r0)kn(r0)]( ! "k n + "kn) (108) kn 0 0 k0n0 j j 0 0 kn k n X X0 0 showing that the spectral function is real for systems with time-reversal symmetry. P 0 obeys the following symmetry 0 0 P (r; r0; !) = P (r; r0; !); (109) 0 0 P (r0; r;!) = P (r; r0; !); (110) valid for systems with time-reversal symmetry.
V. PHYSICAL INTERPRETATION OF THE RPA
The underlying physical assumption in the RPA is that the response of an interacting system to a external perturbation is given by the response to the total …eld (external per- turbation plus induced Hartree potential) as if the system is non-interaccting. In other words
0 = R' = P (' + VH ):
Since the change in the Hartree potential is given by
VH = v = vR' we …nd
R' = P 0(' + vR') = (P 0 + P 0vR)':
Since the perturbing …eld is arbitrary we conclude that
R = P 0 + P 0vR which is the RPA equation.
21 VI. THE GW APPROXIMATION
In practice, it is virtually impossible to solve the Hedin equations exactly for real systems. As always, we have to resort to approximations in order to make progress. An approxima- tion to the self-energy due to Hedin which has been found to be very fruitful is the GW approximation. This approximation may be regarded as an expansion of the self-energy to …rst order in the screened interaction W obtained by approximating the vertex in (49) by
(1; 2; 3) = (1 2)(1 3); (111) neglecting the self-energy response to the total …eld =V: With this approximation the self-energy becomes (1; 2) = iG(1; 2)W (2; 1); (112) hence the name GW approximation. For systems with time reversal symmetry (without magnetic …eld) W (1; 2) = W (2; 1). The polarisation function becomes
P (1; 2) = iG(1; 2)G(2; 1+): (113)
The screened interaction is calculated from (51) with P approximated by P 0.
W (1; 2) = v(1 2) + d3d4 v(1 3)P 0(3; 4)W (4; 2): (114) Z In frequency space the self-energy in the GWA is given by
d!0 (r; r0; !) = i G(r; r0; ! + !0)W (r0; r; !0) (115) 2 Z which follows from the convolution theorem of the Fourier transform. The self-energy con- sists of the exchange and correlation parts. The exchange part is given by
x d!0 i! (r; r0; !) = i G(r; r0; ! + !0)v(r r0)e 0 : (116) 2 Z
The converging factor exp(i!0) arises from the rule that equal time Green’sfunction should + be taken as G(rt; r0t ). Using the spectral representation of G in (117) we obtain
A(r; r0; !0) 1 A(r; r0; !0) G(r; r0; !) = d!0 + d!0 : (117) ! !0 i ! !0 + i Z 1 Z
22 x d!0 (r; r0; !) = iv(r r0) 2 Z A(r; r ; ! ) 1 A(r; r ; ! ) 0 1 0 1 i!0 d!1 + d!1 e ! + !0 !1 i ! + !0 !1 + i Z 1 Z = v(r r0) d!1A(r; r0; !1): (118) Z 1 Due to the converging factor exp(i!0) the contour can be closed in the upper-half circle and only the …rst term in the bracket contributes. If we replace the spectral fucntion A by a non-interacting one 0 A (r; r0; !) = n(r)n (r0)(! "n) (119) n X we …nd the familiar expression of the Fock exchange.
occ x (r; r0) = v(r r0) n(r)n (r0) (120) n X
As for the response function, the correlation part of the screened interaction de…ned by
W c = W v (121) can be written in terms of its spectral representation because from (47)
W c = vRv (122) so that the frequency dependence of W c is determined by the response function R. Thus,
0 c D(r; r0; !0) 1 D(r; r0; !0) W (r; r0; !) = d!0 + d!0 ! !0 i 0 ! !0 + i Z 1 Z 1 1 1 = d!0D(r; r0; !0) : (123) ! ! + i ! + ! i Z0 0 0 Like the response function R the symmetry property of W c is given by
c c W (r0; r; !) = W (r; r0; !) (124) and the spectral function has the following symmetry
D(r; r0; !) = D(r; r0; !): (125) For systems with time-reversal symmetry
D(r0; r; !) = D(r; r0; !): (126)
23 Using the spectral representation of G in (117) and the above spectral representation for W c the self-energy consists of four integrals but only two survive. c d!0 1 A(r; r0; !1) D(r0; r; !2) (r; r0; !) = i d!1 d!2 2 0 ! + !0 !1 i !0 !2 + i Z Z 1 Z d!0 1 1 A(r; r0; !1) D(r0; r; !2) i d!1 d!2 2 0 ! + !0 !1 + i !0 + !2 i Z Z Z 1 A(r; r0; !1)D(r0; r; !2) = d!1 d!2 0 ! + !2 !1 i Z 1 Z 1 1 A(r; r0; !1)D(r0; r; !2) + d!1 d!2 ! !2 !1 + i Z Z0 1 1 A(r; r0; !1)D(r0; r; !2) = d!1( !1) d!2(!2) ! + !2 !1 i Z 1 Z 1 1 1 A(r; r0; !1)D(r0; r; !2) + d!1(!1 ) d!2(!2) : (127) ! !2 !1 + i Z 1 Z 1 The correlation part of the self-energy can also be written as a spectral representation. c (r; r0; !1) 1 (r; r0; !1) (r; r0; !) = d!1 + d!1 (128) ! !1 i ! !1 + i Z 1 Z where 1 c (r; r0; !) = sgn(! ) Im (r; r0; !): (129) Approximating the spectral function A by its non-interacting value
0 A (r; r0; !) = n(r)n (r0)(! "n) (130) n X we obtain
c 1 n(r)n (r0)D(r0; r; !2) (r; r0; !) = ( " ) d! (! ) n 2 2 ! + ! " i n 2 n X Z 1 1 (r) (r )D(r ; r; ! ) + (" ) d! (! ) n n 0 0 2 (131) n 2 2 ! ! " + i n 2 n X Z 1 so that occ