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Functional Derivative Technique

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Functional Derivative Technique 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.
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