Quantum field theory and Green s function

Transcription

1 Quantum field theory and Green s function Condensed matter physics studies systems with large numbers of identical particles (e.g. electrons, phonons, photons) at finite temperature. Quantum field theory arises naturally if we consider a quantum system composed by a large number of identical particles at finite temperature... Quantum statistical physics... Classical statistical physics Classical statistical physics tells us that in grand canonical ensemble, the expectation value of a thermal dynamic quantity (which will be called X here) can be written as: X = n X n exp-be n - m N n n exp-be n -mn n where b=t (here the Boltzmann constant k B is set to unity) and m is the chemical potential. Here, we sum over all possible states of the system ( n ). For the nth state, the total energy of the whole system is E n ; the particle number is N n and the value of the quantity X is X n. (.)..2. Quantum statistical physics For a quantum system (at finite temperature T), we can rewrite this formula in terms of quantum operators n n X` exp-bh` -mǹ n Tr X` exp-bh` -mǹ X = = n n exp-bh` -mǹ n Tr exp-bh` -mǹ (.2) where n sums over a complete basis of the Hilbert space n. H` is the Hamiltonian. Ǹ is the particle number operator and X` is the quantum operator of the quantity that we want to compute. The sum n n n is also known as a trace Tr. Here, we will not prove this operator formula, because it is not part of the main focus of this lecture. Instead, we will only demonstrate that by choosing a proper basis, this (quantum) operator formula recovers the classical formula shown above, and therefore, the operator formula is a natural quantum generalization of classical statistical physics. If H`, Ǹ = 0, we can choose the common eigenstates of H` and Ǹ as our basis. In this basis, we have H` n = E n n (.3) and Ǹ n = N n n (.4) where E n is the eigen-energy and N n is the eigen-value of the particle number operator. Using this basis, it is easy to check that exp-bh` -mǹ n = exp-be n -mn n n (.5)

2 2 Phys620.nb and therefore, n n X` exp-bh` -mǹ n n n X` n exp-be n -mn n X = = n n exp-bh` -mǹ n n n n exp-be n -mn n Here, we define X n = n X` n. = n X n exp-be n -mn n n exp-be n -mn n (.6) Note: the quantum field theory used in particle physics is the TÆ0 limit of this finite temperature theory. Q: What happens if H`, Ǹ π 0? A: If H`, Ǹ π 0, it implies that in this system, the particle number is not conserved. As will be discussed in later chapters, particle conservation law is related with the U() phase symmetry for charge neutral particles (or gauge symmetry for charged particles). The absences of particle conservation implies the breaking of phase/gauge symmetry, which means that the system is a superfluid/superconductor..2. Second quantization The terminology of second quantization is due to historical reasons. We are NOT quantizing something for a second time. We are just using a new basis to handle indistinguishable particles. Q: Why do we want to use second quantization? A: It is the most convenient way to handle a large number of indistinguishable particles..2.. wavefunctions for distinguishable particles It is easy to write down quantum wavefunctions for systems composed by distinguishable particles. For example, if we have two distinguishable particles, particle one in state y and particle two in y 2, the wavefunction is Yr, r 2 =y r y 2 r 2 (.7) If we have n distinguishable particles, the wavefunction can be written as Yr, r 2, r 3,..., r n =y r y 2 r 2 y n r n (.8) Here, the ith particle is in the quantum state y i for i =, 2, n wave functions for indistinguishable particles For indistinguishable particles, the wavefunction is very complicated, if the particle number is large. This is because we need to ensure that the wavefunction is symmetric (anti-symmetric) for bosons (fermions). If we have two indistinguishable particles, one particle in state y and the other in state y 2, the wavefunction is Yr, r 2 =y r y 2 r 2 y 2 r y r 2 (.9) Here, the + sign is for bosons and the - sign is for fermions. It is easy to check that the wavefunction is symmetric or anti-symmetric Yr, r 2 = Yr 2, r (.0) If we have 3 indistinguishable particles, the wavefunction contains 3!=6 terms Yr, r 2, r 3 = y r y 2 r 2 y 3 r 3 y r y 3 r 3 y 2 r 2 y 2 r 2 y r y 3 r 3 +y 3 r 3 y r y 2 r 2 +y 2 r 2 y 3 r 3 y r y 3 r 3 y 2 r 2 y r (.) If we have n indistinguishable particles, Yr, r 2, r 3,..., r n = P P y i r y i2 r 2 y in r n (.2) where P represents all permutations. For a system with n particles, there are n! different permutations and thus the r.h.s. has n! terms. For a system with a large number of indistinguishable particles, it is an extremely complicated to write down its wavefunction in this way. For example, for a system with just ten particles, n=0, there are 0!º2.6 million terms, which is impossible to write down. In condensed matter physics, a typical system has huge number of particles (~0 23 ) and thus we need a better way to write down our quantum theory.

3 Phys620.nb 3 Note: in particle physics, although one typically studies systems with a very small number of particles (e.g. two particles collides with each other in a collider), however, it is still necessary to consider a large number because we have virtue particles. For example, the E&M interactions between two electrons is realized by exchanging virtual photons. If one takes into account these virtual particles, the particle number is not very small, and thus we shall use second quantization Fock space The reason why it is hard to write down wavefunctions for indistinguishable particles is because when we write do the wavefunciton, we need to specify which particle is in which quantum state. For example, y i r j means the particle number j is in the quantum state y i. This procedure is natural for distinguishable particles, where we do know which particle is in a particular quantum state. However, for indistinguishable particles, we don t know which particle is in this state, because we cannot distinguish these particles. In other words, the traditional way to write down a wavefunction is designed for distinguishable particles. For indistinguishable particles, we need to first treat them as distinguishable particles, and then repeat the wavefunciton using all different permutations to make the wavefunction symmetric/anti-symmetric. This procedure is not natural and is very complicated. For indistinguishable particles, it is more natural to use the occupation number basis, which is known as the Fock space. In the Fock space, a many-body quantum state is written in terms of occupation numbers: Y = n, n 2, n 3 n N (.3) where n i is the number of particles in state y i (n i is known as the occupation number). Here we don t specify which particle is in the state y i. Instead, we just count the number of particles in this state. In this approach, the particle are indistinguishable automatically and thus all the complicates mentioned above are avoided Creation and annihilation operators: In the Fock space, physical operators can be written in terms of creation and annihilation operators. The creation operator increases the occupation number by and the annihilation operator reduces the occupation number by. For bosons, we have b i n, n 2, n 3 n N = n i + n, n 2,, n i +, n N (.4) b i n, n 2, n 3 n N = n i n, n 2,, n i -, n N (.5) and it is easy to check that these operators obey the following commutation relations: b i, b j = b i, b j = 0 b i, b j =d ij (.6) (.7) For fermions, due to the Pauli exclusive principle, each quantum state can at most have one particle (so n i = 0 or ). c i,, = 0 (.8) c i, 0, = - j= i- n i,, (.9) c i,, = 0 (.20) c i, 0, = - j= i- n i, 0, (.2) It is easy to check that fermion creation/annihilation operators obey anti-commutation relations: c i, c j = c i, c j = 0 c i, c j =d ij (.22) (.23)

4 4 Phys620.nb Note: the factor - j= i- n i is dropped in many textbooks, because + y and - y represent the same quantum state. However, this factor is curial to ensure the anti-commutation relation c i, c j =d ij. We will study this factor later when we examine D quantum systems Particle number operator For bosons, using the definition above, it is easy to check that b i b i, n i, = n i, n i, (.24) Therefore, the particle number operator for bosons in the quantum state y i is ǹ i = b i b i (.25) As a result, the total particle number operator is Ǹ = i ǹ i = i b i b i (.26) For fermions, it is easy to notice that c i c i, 0, = 0 c i c i,, = (.27) (.28) Therefore, the particle number operator for fermions in the quantum state y i is ǹ i = c i c i (.29) As a result, the total particle number operator is Ǹ = i ǹ i = i c i c i (.30) In the next a couple of sections, we will only consider fermions as our example (c and c ), but the same conclusions are applicable for bosons (b and b ).2.6. Quantum states In the Fock space, all quantum states can be written in terms of creation and annihilation operators. First, we need to define the vacuum (ground states in high energy physics) 0 by assuming that there is one and only one state in the Fock space that is annihilated by any annihilation operators. This state is our vacuum c i 0 = 0 for any c i (.3) It is easy to check that this quantum states has zero particle Ǹ 0 = i c i c i 0 = 0 (.32) Then, for any states with one particle (a one-particle state), we can write it as y= i a i c i 0 (.33) For two particle states, they can be written as y=a ij c j c i 0 (.34) For n particle states, they can be written as y=a i i n c in c i2 c i 0 (.35).2.7. quantum operators Any quantum operator can be written in terms of creation and annihilation operators c s and c s: X = f k,k 2 k m,q,q 2 q m' c k c k2... c km c q c q2 c qm' (.36)

5 Phys620.nb 5 If every term in this operator has m annihilation operators and m creation operators, this operator is known as a (m+m ) fermion operator Physical observables and correlation functions Since all quantum operators can be written in terms of creation and annihilation operators, the expectation value of any physical observable can be written in terms of the expectation values of creation and annihilation operators: X = X = f k,k 2 k m,q,q 2 q m' c k c k2... c km c q c q2 c qm' (.37) Therefore, in second quantization, anything we need to compute reduces to objects like this: c k c k2... c kn c q c q2 c qm. This type of objects are known as correlation functions. If there are N creation and annihilation operators, it is known as a N-point correlation function. Here, N = m + m'. Q: What are the simplest nontrivial correlation functions? Here, simplest means that we want the number of creation and annihilation operator to be as small as possible. nontrivial means the correlation function need to be nonzero. A: If the particle number is conserved (H`, Ǹ = 0, which is true for most of the cases we study), they are the two-point correlation functions. This is because, if H`, Ǹ = 0, we can prove that c k c k2... c km c q c q2 c qm' = 0 if m m. In other words, if we want to have a nontrivial correlation function, it must have the same number of creation and annihilation operators. Therefore, we only need to consider N-point correlation functions when N is even. The smallest positive even integer is 2, so the simplest nontrivial correlation functions are two-point correlation functions c k c q. The next one is four-point correlation c k c k2 c q c q2. Proof: As shown above, if H`, Ǹ = 0, we know that particle number is a conserved quantity and thus we can choose the common eigenstates of H` and Ǹ as our basis. In this basis, we have n n c k c k2... c km c q c q2 c qm' n exp-be n -mn n c k c k2... c km c q c q2 c qm' = n exp-be n -mn n (.38) Because n is the eigenstate of the total particle number operator Ǹ, we know that the quantum state n has N n particles. Define y = c k c k2... c km c q c q2 c qm' n (.39) because a creation operator increase the particle number by, while an annihilation operator reduce it by, it is easy to note that y has N n + m' - m particles. If m' m, N n + m' - m N n, which means that the quantum states n and y has different number of particles, i.e. they are both eigen states of Ǹ, but they have different eigenvalues. In quantum mechanics, we learned that two eigenstates of the same operator are orthogonal, if they have different eigenvalues, so we know immediately that n y = 0. And therefore, n c k c k2... c km c q c q2 c qm' n = 0. As a result, the correlation function c k c k2... c km c q c q2 c qm' = 0. Note: If H`, Ǹ π 0, we will have nonzero correlation functions with m π m'. For example, in superfluids, b π 0, and in superconductors c k c k2 π 0. These cases will be considered later Summary The terminology of second quantization is due to historical reasons. We are NOT quantizing something for a second time. We are just using a new basis to handle indistinguishable particles. In both high energy and condensed matter physics, quantum field theory utilize the second quantization construction. The reason is because quantum field theory deals with more than one indistinguishable palaces, and the second quantization formulas are the most natural way to describe this type of physics. In second quantization (same is true for quantum field theory), any physical quantities are reduced to computing various correlation functions. If particle number is conserved, only correlation functions with same number of creation and annihilation operators are nontrivial. The order matters: c k c q c q c k. Therefore, depending on the way to order the creation/annihilation operators, we can define different correlation functions. This topics will be address in the next section.

6 6 Phys620.nb.3. Hamiltonian in second quantization Let s consider a Hamiltonian with three terms: the kinetic energy H K, the potential energy H P and interactions between different particles H I. H = H K + H P + H I (.40).3.. Potential energy For particles in a potential Ur, the total potential energy (summed over all particles) is P.E. = r Ur rr (.4) where rr is the particle density at r. In second quantization, particle density is r`r =y r yr (.42) Here, y r creates an electron at r and y(r ) annihilate an electron at r. Therefore, the potential energy part of the Hamiltonian is H P = r Ur y r yr (.43).3.2. Kinetic energy Momentum space Total kinetic energy for a system with many particles is K.E. = k 2 p d ek nk (.44) Here, ek is the dispersion relation of a single particle and nk is the number of particles with momentum k. In the denominator, d is the spatial dimension of the system. In second quantization, the quantum operator for nk is ǹk =y k yk (.45) Here, y k creates an electron at k and y(k ) annihilate an electron at k. Therefore, H K = k 2 p d ek y k yk (.46) If we consider nonrelastivistic particles ek = k 2, H K = k 2 p d k 2 y k yk (.47) Q: What shall we do for relativistic particles? A: We cannot just set e(k)=c k or ek = c 2 k 2 + m 2 0 c 4. This is because x or x are not analytic functions (they are singular at k = 0). For fermions, it turns out that we need two species of fermions to have a relativistic dispersion. This topic will be discussed latter, when we study the Dirac theory. Real space The momentum space creation/annihilation operators and real space creation/annihilation operators are connected by the Fourier transformation. For the annihilation operator,

7 Phys620.nb 7 yk = r - k ÿ r yr (.48) The inverse transformation is yr = k  kÿ r yk 2 p d (.49) The transformation for the creation operators can be obtained by taking the Hermitian conjugate y k = r  k ÿ r y r (.50) y r = k - kÿ r y k 2 p d (.5) Therefore, it is easy to check that H K = k 2 p d k 2 y k yk y r yr = r (.52) H k = y r yr r = r k - kÿ r y k 2 p d k '  k' ÿ r yk ' 2 p d = r k k 2 p d 2 p d k ' 2 p d k ' 2 p d k ÿ k ' y k yk ' expâ rk ' - kÿ r = k ÿ k ' y k yk ' r expâ rk ' - kÿ r (.53) = k 2 p d k ' 2 p d k ÿ k ' y k yk '2p d dk ÿ-k ' = k 2 p d k ÿ k y k yk = k 2 p d ek nk.3.3. Lattice systems In a lattice, if we ignore the interactions between electrons, the Hamiltonian contains two terms. y r yr H = H K + H P = r + Ur y r yr (.54) where the first term is the kinetic energy and the second term is the lattice potential, where Ur is a periodic potential. In principle, we could use this Hamiltonian. However, it is not the most convenient way to handle a lattice systems. A more convenient way utilizes Bloch waves and the band structure. In a lattice system, the kinetic energy of a single particle is described by the band structure e n k. Here n is the band index and k is a momentum point in the (reduced) Brillouin zone. Therefore, total energy of the system (if we ignore interactions is) k E = n e n k n n k BZ 2 p d (.55) Here, we sum over all bands n and the integral is over the first Brillouin zone BZ. n n k is the occupation number for the quantum state in band n at momentum k. In second quantization, the quantum operator for n n k is

8 8 Phys620.nb n n k =y n k y n k (.56) where y n k creates a particles in band n with momentum k. In other words, it is the creation operator for a Bloch wave. y n k is the annihilation operator for a Bloch wave. Therefore, the Hamiltonian for non-interacting electrons is k H k = n e n k y n k y n k BZ 2 p d For simplicity, we will not consider lattices in this chapter. Instead, we will only consider particles moving in free space with dispersion k 2. But please keep in mind that for a lattice system, we just need to substitute e=k 2 into the corresponding dispersion for Bloch waves. Both the kinetic energy part of the Hamiltonian and the potential energy part contain one creation and one annihilation operator. So they are both known as the quadratic terms, or two-fermion terms. If the Hamiltonian only contains these two terms, we call the system a non-interacting system, because there is no interaction between particles in this Hamiltonian..3.4.Interactions Let s consider Coulomb interaction between electrons. The total Coulomb energy is (.57) E Int = 2 r r' V r - r ' rr rr ' (.58) where Vr = e 2 r and rr is the particle density. In second quantization, we know that particle density is r`r =y r yr (.59) Therefore, the interaction part of the Hamiltonian is H I = 2 r r' V r - r ' y r yr y r ' yr ' (.60) The interaction term contains four creation/annihilation operators, and thus this term is called a quartic term or a four-fermi term. Typically, we reorder the operators using the anti-commutation relation between creation/annihilation operators. H I = 2 r r' V r - r ' y r yr y r' yr' = 2 r r' V = 2 r r' V = 2 r r' V r - r ' y r y r' yr' yr + 2 r r' V r - r ' dr - r' y r yr ' r - r ' y r y r' yr' yr + 2 rv0 y r yr r - r ' y r y V0 N r' yr' yr + 2 where N is the total number of particles in the system. The last term may looks problematic (it is singular), because for the Coulomb interaction, V 0 = 0 =, but this term will not cause any trouble. We can simply drop it, because this term just shifts the total energy by a constant. In a solid state material, this term is canceled by the potential energy from the nucleons. Very typically, when we write down the Hamiltonian, we put the operators in such an order that creation operators are put on the left and annihilation operators are on the right. This order is called normal order and the procedure to put operators in such an order is known as "normal ordering". Typically, one use two colons to represent normal ordering. If we put the product of some creation and annihilation operators operator between two colons, it means that we reorder these operators into normal order. For bosons, we just reorder the operators. For fermions, we need an extra factor + or -, depending on whether we have even or odd number of permutations to reorder the operators. For example, : y r yr' :=y r yr' (.62) (.6) : yr' y r :=-y r yr' (.63)

9 Phys620.nb 9 : yr' yr'' y r :=y r yr' yr'' (.64) For the interaction terms, instead of using the most straightforward formula, H I = 2 r r' V r - r ' y r yr y r' yr' (.65) we typically uses the normal ordered interaction term H I = 2 r r' V r - r ' : y r yr y r' yr' := 2 r r' V r - r ' y r y r' yr' yr These two definitions differ by a constant, which doesn t play any fundamental role. In other words, using normal order shifts the total energy by a constant. In fact, the reason we use normal ordering is because by doing so, the energy of the vacuum state is set to zero. In normal order, annihilation operator is put to the right side. By definition, the vacuum state is destroyed by any annihilation operator y 0 = 0 (.67) Therefore, (.66) : H : 0 = y 0 = 0 = 0 0 (.68) So the vacuum state is an eigenstate of : H : with zero eigen energy Summary In this chapter, we consider the simplest case H = y r yr r + 2 r r' V r - r ' y r y r' yr' yr (.69) The quadratic term: kinetic energy for non-relativistic particles e=k 2. The quartic term: normal ordered four-fermi interactions. However, it is important to keep in mind that the conclusions and methods can be generalized easily to other more complicated cases..4. Equation of motion for correlation functions In the Heisenberg picture, the equation of motion of an operator X is  X t = X t, Ht t Here, we set the Planck constant Ñ to unity for simplicity. (.70).4.. Equation of motion for annihilation operators For the Hamiltonian H = y r yr r + 2 r r' V r - r ' y r y r' yr' yr (.7) the equation of motion for the annihilation operator y is yr 0, t  t dr - r 0 yr = yr 0, H = r + 2 r r' dr - r0 V r - r' y r' yr' yr + 2 r r' V 2 yr 0, t =- + rv r 0 - r y r, t yr, t yr 0, t r - r' dr' - r0 y r yr' yr (.72)

10 0 Phys620.nb Similarly, for the conjugate operator y, we have y r 0, t - t 2 y r 0, t =- +y r 0 rv r 0 - r y r yr For noninteracting particles V = 0, this equation is very similar to the Schrodinger equation. But please keep in mind that y and y here are operators, instead of wavefunctions Equation of motion for correlation functions Define two point correlation functions (.73) G > r, t ; r 2 t2 =  yr, t y r 2, t2 (.74) Here, the factor  is introduced for historical reason. For systems with translational symmetry in space and time, G > r, t ; r 2 t2 only depends the time and position difference between and 2 G > r, t ; r 2 t2 = G > r - r2, t - t 2 = G > r, t (.75) Here, we define r = r - r2 and t = t - t 2 Q: What is the equations of motion for G > r, t? t G > r.t = t G > r.t - t 2 = t  yr, t y r 2, t2 = - t yr, t y r 2, t2 2 yr, t =-- + V r 0 - r y r, t yr, t yr, t y r 2, t 2 = r 2 yr, t y r 2, t 2 - V y y yy =  2 G > r, t - V <y y yy> If V = 0 (free systems), we have a closed partial differential equation for two fermion correlation functions. (.76)  t G > 0 r.t + 2 (.77) G 0 > r, t = 0 The sub-index 0 here implies that we are considering a non-interacting system without interaction. We can solve this partial differential > equation (with proper initial conditions and boundary conditions), and obtain the correlation function G 0 For interacting systems (V 0), the story is not as simple.  t G > r.t + 2 G> r, t =-ÂV <y y yy> So we have an inhomogeneous partial differential equation. (.78)  t G > r.t + 2 (.79) G> r, t = f r, t The terminology inhomogeneous equation means that the r.h.s. of the equation is nonzero. Inhomogeneous equation may look complicated, because for a different f r.t, it seems that we will need to solve a different equation. However, this is not the case. We just need to solve one question and then for any f r., t, we can get the solution directly. Q: How do we solve an inhomogeneous partial differential equation? A: Let s look at the E&M textbook. The Green s function method E&M: electric potential f(r) for charge distribution r(r) Gauss s law ÿe r =rr (.80)

11 Phys620.nb We know that E r =- fr (.8) so 2 fr =rr (.82) This is an inhomogeneous equation. How do we solve it? We first solve a different equation: 2 r Gr, r2 =dr2 (.83) Gr, r2 is the electric potential at point r induced by a point charge located at position r2. We know the solution of this equation, which is just the Coulomb s law Gr, r2 = 4 p r - r2 After we find Gr, r2, the solution for the inhomogeneous equation 2 fr =rr can be obtained easily as (.84) fr = r 0 Gr, r0 rr0 = r 0 4 p rr 0 r - r0 (.85) Mathematicians call Gr, r2 the Green s function. And this methods of solving inhomogeneous PDEs are known as the Green s function approach. In general, one first substitute the inhomogeneous part with a delta function. Then, one solve this new PDE, whose solution is the Green s function. Once the Green s function is obtained, one can write down the solution of the inhomogeneous PDE very easily using an integral. For the correlation function G >, the equation of motion is  t G > r.t + 2 G> r, t =-ÂV <y y yy> What we will need to do here is to substitute the r.h.s. by a delta function (.86)  t + 2 Gr, t = dr dt (.87) and Gr, t is our Green s function..4.4.time ordering: a trick to get delta functions Define time-ordered correlation functions (the Green s functions) Gr, t ; r 2 t 2 =  Tyr, t y r 2, t 2 (.88) Here, Ty r, t yr 2, t 2 is known as the time - ordered product. T yr, t y r 2, t 2 = yr, t y r 2, t 2 if t > t 2 y r 2, t 2 yr, t if t < t 2 (.89) For bosons, we use the + sign and for fermions we use the - sign. Another way to write down the same product is T yr, t y r 2, t 2 =yr, t y r 2, t 2 ht - t 2 y r 2, t 2 yr, t ht 2 - t (.90) where hx is the step function hx = for x > 0 and hx = 0 for x < 0. Now, let s consider the EOM for Gr, t. t Gr.t = t  T yr, t y r 2, t 2 =- t yr, t y r 2, t 2 ht - t 2 y r 2, t 2 yr, t ht 2 - t = - t yr, t y r 2, t 2 ht - t 2 y r 2, t 2 - t yr, t ht 2 - t -  yr, t y r 2, t 2 y r 2, t 2 yr, t dt - t 2 ( 9)

12 2 Phys620.nb = r 2 Tyr, t y r 2, t 2 - V T y y yy -Âdr - r 2 dt - t 2 =  2 Gr, t - V T y y yy - dr - r 2 dt - t 2 If V = 0 (non-interacting systems)  t G 0 r.t + (.92) 2 G 0 r, t = dr dt Bottom line, by time-ordering, we automatically get a delta function in the equation of motion, which makes G r, t a Green s function Green s function for free particles: G 0 r, t For the EOM,  t + 2 G 0r, t = dr dt it is easy to solve in the momentum space. r t exp-â kr+âwt  t + 2 G 0r, t = r t exp-â kr+âwtdrdt r t - t + 2 exp-â kr+âwtg 0r, t = w- k2 r t exp-â kr+âwt G 0 r, t = Define G 0 k, w = r t exp-â kr+âwtg 0 r, t (.93) (.94) (.95) (.96) (.97) It is easy to check that k w G 0 r, t = 2 p d 2 p expâ kr-âwtg 0k, w which is the inverse transformation. Using G 0 k, w, we get w- k2 G 0k, w = (.98) (.99) and therefore G 0 k, w = w- k2 More generally, for non-interacting particles with more complicated dispersion. G 0 k, w = w-ek For example, in a lattice system, if we ignore interactions, the Green s function for particles in band n is G 0 n k, w = w-e n k (.00) (.0) (.02)

13 Phys620.nb 3 Here, we assume that both the creation and annihilation operators are for Bloch waves in band n. If the two operators are for different bands, the correlation function is zero Note: there are many other reason to use T. () Path integral leads to T naturally. (2) The evolution operator Ut = Texp thtt (3) With T, bosons and fermions are unified together. Same theory with two different boundary conditions. Due to time-limitation, we will not address () and (2) in details in this course. But for theorists, it is very important to understand the path integral formula. The third point will be addressed in the next section..5. Boundary condition and connections between different Green s functions.5.. Other correlation functions and the boundary condition As mentioned in Sec., because the product between two operators is in general non-commutative AB BA, the order of operators matters a lot. Depending on the order of operators, we can define many different correlation functions. For example, the time-ordered correlation function defined above. In addition, we can define the following two correlation functions (and other correlation functions). G >, 2 =  y y 2 (.03) G <, 2 =  y 2 y (.04) Here, for simplicity, we use to represent r, t and 2 to represent r 2, t 2. Please notice that we cannot use the simple commutation/anticommutation relation y y 2 y 2 y d-2 (.05) This is because the commutation/anti-commutation relations we are familiar with are equal time commutation/anti-commutation relations. yr, t y r 2, t y r 2, t yr, t dr - r 2 (.06) Please notice that all the operators are at the same time (remember that we are using the Heisenberg picture, so that the operators are timedependent). What we considered here are operators at different time points y y 2 y 2 y =yr, t y r 2, t 2 y r 2, t 2 yr, t d - 2 (.07) Therefore, in general, it is not d(-2). And thus G >, 2 G <, 2 d - 2  (.08) In other words, we cannot simply relate these two Green s function using equal time commutation/anti-commutation relations. For statistical average, we have Boltzmann factor b H. For time-evolution, we have the evolution operator  H t. It seems that inverse temperature b is just the imaginary time. Let s try this idea by allowing time to be complex. For G > G >, 2 = Tr exp-bh` -mǹ yr, t y r 2, t 2  y y 2 =  Tr exp-bh` -mǹ (.09) = Tr exp-bh` -mǹ expâ H` t yr exp-â H` t expâ H` t 2 y r 2 exp-â H` t 2  Tr exp-bh` -mǹ If we use eigenenergy states to compute the sum

14 4 Phys620.nb G >, 2 = Tr exp-bh` -mǹ expâ H` t yr exp-â H` t expâ H` t 2 y r 2 exp-â H` t 2  Tr exp-bh` -mǹ = expbmn expe n  t -b n yr exp-â H` t expâ H` t 2 y r 2 n exp-â E n t 2  Tr exp-bh` -mǹ n n expe n  t -Ât 2 -b n yr exp-â H` t expâ H` t 2 y r 2 n = expbmn  Tr exp-bh` -mǹ (.0) Because E n has no upper bound, as E n +, to keep the factor expe n  t -Ât 2 -b converge, we need to have Re t t 2 0, In other words, Im t - t 2 >-b. In the same time, using = m mm, we have G >, 2 = expb m N expe n  t -Ât 2 -b n yr n mm exp-â H` t expâ H` t 2 y r 2 nâ Tr exp-bh` -mǹ = m expbmn n,m expe n  t -Ât 2 -b n yr mm y r 2 n expâ E m t 2 - t  Tr exp-bh` -mǹ (.) To make sure that E m + shows no singularity, Imt - t 2 < 0. So, 0 > Im t - t 2 >-b For G <, we have 0 < Im t - t 2 <b Let s come back to G > G > r, t -Âb; r 2, t 2 = Tr exp-bh` -mǹ expb H` expâ H` t yr exp-â H` t exp-bh` expâ H` t 2 y r 2 exp-â H` t 2  Tr exp-bh` -mǹ = Tr expbmǹexpâ H` t yr exp-â H` t exp-bh` expâ H` t 2 y r 2 exp-â H` t 2  Tr exp-bh` -mǹ = Tr expâ H` t yr expbmǹ - exp-â H` t exp-bh` expâ H` t 2 y r 2 exp-â H` t 2  Tr exp-bh` -mǹ = e -bm Tr expâ H` t yr exp-â H` t exp-bh` -mǹ expâ H` t 2 y r 2 exp-â H` t 2  Tr exp-bh` -mǹ = (.2) Therefore, we find that e -bm Tr y exp-bh` -mǹ y 2  Tr exp-bh` -mǹ Tr exp-bh` -mǹ y 2 y = e -bm = e  Tr exp-bh` -bm G <, 2 -mǹ e bm G > r, t -Âb; r 2, t 2 = G < r, t ; r 2, t 2 (.3).5.2. BC for the time-ordered Green s function. The time-ordered Green s function can be defined along the imaginary axis also T yr, t y r 2, t 2 = yr, t y r 2, t 2 if Imt > Imt 2 y r 2, t 2 y r, t if Im t < Im t 2 (.4) So, T G = G> if Imt < Im t 2 G < if Im t > Im t 2 (.5) It is easy to check that for Gr, t, we have Gr, t = bm Gr, t -Âb (.6) This is a boundary condition for imaginary time. At m=0, Bosons have a periodic BC, and fermions have an anti-periodic BC the k,w space Define

15 Phys620.nb 5 G > k, w = r t - kr+â w t G > r, t = y k,w y k,w G < k, w =  r t - kr+â w t G < r, t = y k,w y k,w (.7) G < k, w =  r t - kr+â w t G < r, t =  e bm r t - kr+â w t G > r, t -Âb =Âe bm r t ' - kr+â w t'+â b G > r, t ' = -bw-m G > k, w (.8) Define the spectral function Ak, w = G > k, w G < k, w (.9) Because G < k, w = -bw-m G > k, w Ak, w = G > k, w G < k, w = G > k, w -bw-m G > k, w (.20) So bw-m Ak, w G > k, w = = Ak, w -bw-m bw-m = Ak, w = Ak, w f w bw-m Ak, w G < k, w = -bw-m G > k, w = -bw-m = Ak, w -bw-m bw-m Here f w is the boson/fermion distribution function. = Ak, w f w Note: Here, the only thing we assume is that the indistinguishable particles commute or anti-commute with each other. In Green s function, these two choices leads to two different boundary conditions, from which the Bose-Einstein and Fermi-Dirac distributions arise naturally G in the k,w space part I: the Matsubara frequencies Remember that Gr, t = bm Gr, t -Âb, which is (almost) a PBC (anti-pbc) for t. We know that boundary conditions implies quantization (discrete w). For example, for a periodic function f t = f t + T, we know that in the frequency space, f w is defined only on a discrete set of frequency points, w=2pnt. For anti-pbc, f t =-ft + T, w=2n + pt. For Gr, t = bm Gr, t -Âb, if we go to the k,w space and define (.2) (.22) Gr, t = k w  kr-â w t Gk, w (.23) The condition Gr, t = bm Gr, t -Âb implies that k w  kr-â w t Gk, w = bm -w b k w  kr-â w t Gk, w (.24) So, = bw-m = (.25) For bosons, bw-m = 2 n pâ w= 2 n p  +m b (.26) (.27) Following the historical convention, we write w=âw n w n = 2 n p -Âm b (.28) These discrete frequency points w n are known as the Matsubara frequencies

16 6 Phys620.nb For fermions, bw-m = 2 n + pâ 2 n + p  w= +m b w n = 2 n + p-âm b (.29) (.30) (.3).5.5. G in the k,w space part II: analytic continuation After we find Gk, Âw n, where w n takes discrete values, now let s define a new function Gk, z by simply replacing the discrete number Âw n into a complex number z that various continuously. For the function Gk, z, it is a function well-defined at every point on the complex z plane (there may be some singularity points). At the Matsubara frequencies, this new function Gk, z coincides with Gk, Âw n. This procedure is known as analytic continuation. This new function is very useful. Here, I will show you that by defining Gk, z, we can find a very easy way to get G > and G < from G. Later, we will use this Gk, z to compute Gk, Âw n. Gk, Âw n = 0 - b t -wn t Gk, t = - b t -w n t G > - b w k, t = t -w n t pâ - w t G > k, w = w - w -wn-â b - Ak, w -bw-m = 2 p w-âw n w - -w-âw n b Ak, w w -w-m b Ak, w w Ak, w 2 p w-âw n = -bw-m 2 p Âw n -w = -bw-m 2 p Âw n -w Therefore, Gk, z = w Ak, w 2 p z -w Now, we substitute z by w+âd Gk, w+âe = W Ak, W 2 p w+âe-w Gk, w+âe-gk, w-âe ImGk, w+âe = = W 2  2  2 p Ak, W w+âe-w - Ak, W w-âe-w = W Ak, W 2  2 p w-w 2 +e -2 Âe =- W e 2 2 p w-w 2 +e Ak, W =- 2 2 Wdw-WAk, W =- Ak, w 2 So, Ak, w =-2ImGk, w+âe (.32) (.33) (.34) (.35) (.36).5.6. Example: free particles  t G 0 r.t + 2 G 0 r, t = dr dt in the k,w space Âw n - k2 G 0k, Âw n = G 0 k, Âw n = Âw n - k2 More generic case: dispersion relation e(k) (.37) (.38) (.39)

17 Phys620.nb 7 G 0 k, Âw n = Âw n -ek Analytic continuation: G 0 k, z = z -ek the spectral function: Ak, w =-2ImGk, w+âe =-2Im w+âe-ek = 2 e w-e k 2 +e = 2 pdw-e k 2 (.40) (.4) (.42) particular number: n k t = y k t y k t = G < k, t - t = w 2 p G< k, w = w 2 p 2 pdw-e k f w = f e k w w n k t = y k t y k t = G > k, t - t = 2 p G> k, w = 2 p 2 pdw-e k f w = f e k (.43) (.44).5.7. Interacting particles. Once we found Gk, Âw n, we can get Ak, w using Ak, w =-2ImGk, w+âe (.45) Then we can get all other correlation functions like G < k, t - t and G > k, t - t using bw-m Ak, w G > k, w = = Ak, w -bw-m bw-m = Ak, w = Ak, w f w bw-m Ak, w G < k, w = -bw-m G > k, w = -bw-m (.47) = Ak, w = Ak, w f w -bw-m bw-m Bottom line: if we know one correlation function, we can get other. Therefore, among all the different correlation functions, we just need to focus on the one that is easiest to compute. The easiest one is the Green s function (time-ordered correlation functions), because it give us delta functions. (.46).6. Feynman diagram.6.. multi-particle Green s functions Single particle Green s function: G, ' = Â Ty y ' (.48) Two-particle Green s function: G 2, 2; ', 2' = Â 2 Ty y2 y 2' y ' (.49) three-particle Green s function: G 3, 2, 3; ', 2', 3' = Â 3 Ty y2 y3 y 3' y 2' y ' (.50) n-particle Green s function:

18 8 Phys620.nb G n, 2,, n;',2',,n' =  n Ty y2 yn y n' y 2' y ' (.5).6.2. the equations of motion of the Green s functions  yr 0, t = yr 0, H = r dr - r 0 yr + t 2 r r' dr - r 0 V r - r' y r' yr' yr + 2 r r' V r - r' dr' - r 0 y r yr' yr =- 2 yr 0, t + rv r 0 - r y r yr yr 0  t + 2 G, 2 = dr - r 2 dt - r 2 +  rv r 0 - r Ty r, t yr, t yr 0, t y r 2, t 2  t + 2 G, ' =d-' + rv r - r 2  d - ' + rv r - r 2 d - '  rv r - r 2  2 T y r 2, t yr 2, t yr, t y r ', t ' = 2 Ty r 2, t +d yr 2, t yr, t y r ', t ' =  2 Tyr, t yr 2, t y r 2, t +d y r ', t ' = (.52) (.53) (.54) d - '  rv r - r 2 G 2, 2; ', 2 + t =t 2 Here 2 + means that the time argument for 2 + is slightly larger than 2, to keep the operators in the right order t 2 + = t 2 +d.  t + 2 G, ' =d-'  r 2 V r - r 2 G 2, 2; ', 2 + t=t 2 (.55) This equation tells us that in order to get G, 2, we need to know G 2, 2; ', 2'. So we need to write down the EOM for G 2  t + 2 G 2, 2; ', 2' =d-' G2, 2' d-2' G2, '  r 2 V r - r 2 G 3, 2, 3; ', 2', 3 + t =t 3 (.56) This equation tells us that in order to get G 2, we need to know G 3. Repeat the same procedure, we find that if we want to know G n, we need to know G n+.  t + 2 G n, 2,, n;',2',,n' =d-' G n- 2,, n;2',,n' d-2' G n- 2,, n;',3',,n' +  r 2 V r - r 2 G n+, 2,, n + ; ', 2',, n + t =t n+ (.57) So we cannot get a close set of equations. In order words, the number of unknowns is always the number of equations+, and thus there is no way to solve these equations Q: How to solve this equation? A: the perturbation theory G, ' depends on interaction strength V. Let s expand G as a power series of V G, ' = G 0, + OV + OV 2 + OV 3 + (.58) Notice that G n- is related to V μg n. Therefore, to get G up to the order of OV n, we just need to keep G 2 to the order of OV n-, and G 3 to OV n-2 to OV 0 and set G n+2 = the zeroth-order approximation (free-particle approximation, or say non-interacting approximation) If we want to get G to the zeroth order, we need to set G 2 = 0

19 Phys620.nb 9  t + r 2 G 0, ' =d-' One equation, one unknown. It can be solved easily. G 0 k, Âw n = Âw n - k the first-order approximation (the Hartree-Fock approximation) If we want to get G to the first order, we need to set G 3 = 0 and keep G 2 to OV 0  t + r 2 G, ' =d-'  r 2 V r - r 2 G 2, 2; ', 2 + t=t 2  t + r 2 G 2, 2; ', 2' =d-' G2, 2' d-2' G2, ' + OV G 3 =d-' G 0 2, 2' d-2' G 0 2, ' two equations and two unknowns. The solution for the second equation is very simple: G 2, 2; ', 2' = G 0, ' G 0 2, 2' G 0, 2' G 0 2, ' (.59) (.60) (.6) (.62) (.63) Let s check it  t + r 2 G 0, ' G 0 2, 2' =d-' G 0 2, 2'  t + r 2 G 0, 2' G 0 2, ' =d, 2' G 0 2, ' So we can go back to the first equation to get G.  t + r 2 G, ' = (.64) (.65) (.66) d - '  r 2 V r - r 2 G 0, ' G 0 2, 2 + t=t 2 + r 2 V r - r 2 G 0, 2 + G 0 2, ' t=t 2  t + r 2 G, ' =d-' + f, ' This equation can be separated into two equations:  t + r 2 G 0, ' =d-'  t + r 2 G, ' = f, ' and G = G 0 + G. The first equation have been solved before. It is just the zeroth-order equation (the free theory)  t + r 2 G 0, ' =d-' For the second equation, the solution is straightforward using the Green s function technique G, ' = r 3 t 3 G 0, 3 f 3, ' (.67) (.68) (.69) (.70) (.7) Let s check this

20 20 Phys620.nb So  t + 2 r r 3 t 3 G 0, 3 f 3, ' = r 3 t 3 d - 3 f 3, ' = f, ' (.72) G, ' = r 3 t 3 G 0, 3 f 3, ' =  r 3 t 3 r 2 G 0, 3 V r 3 - r 2 G 0 3, ' G 0 2, 2 + t3 =t 2 + r 3 t 3 r 2 G 0, 3 V r 3 - r 2 G 0 3, 2 + G 0 2, ' t3=t 2 (.73) For G, ' G, ' = G 0, '  r 3 t 3 r 2 G 0, 3 V r 3 - r 2 dt 3 - t 2 G 0 3, ' G 0 2, (.74)  r 3 t 3 r 2 G 0, 3 V r 3 - r 2 dt 3 - t 2 G 0 3, 2 + G 0 2, ' t3 =t 2 The first term is the free propagator, the second therm is known as the Hartree term and the last term is the Fock term second order: If we want to get G to the order of OV 2, we set G 4 = 0, keep G 3 to OV 0 and G 2 to OV  t + r 2 G, ' =d-'  r 2 V r - r 2 G 2, 2; ', 2 + t=t 2  t + 2 G 2, 2; ', 2' =d-' G2, 2' +d-2' G2, '  r 2 V r - r 2 G 3, 2, 3; ', 2', 3 + t =t 3  t + r 2 G 3, 2, 3; ', 2', 3' =d-' G 2 2, 3; 2', 3' d-2' G 2 2, 3; ', 3' +d-3' G 2 2, 3; ', 2' The last equation give us (keep only OV 0 terms) G 3, 2, 3; ', 2', 3' = G 0, ' G 0 2, 2' G 0 3, 3' G 0, ' G 0 2, 3' G 0 2, 3' G 0, 2' G 0 2, ' G 0 3, 3' + G 0, 2' G 0 2, 3' G 0 3, ' + G 0, 3' G 0 2, ' G 0 3, 2' G 0, 3' G 0 2, 2' G 0 3, ' (.75) (.76) (.77) (.78) So we can get G 2 and then G Too complicated! Hard to compute when n is large. There is a simple way to directly get the finally answer we want, thanks the very smart technique designed by Feynman Feynman diagrams and Feynman rules Each integration coordinate is represented by a point; A propagator, G 0, ', is represented by a solid line: A creation operator is represented by a solid line attached to the point with an arrow from the point; An annihilation operator is represented by a solid line attached to the point with an arrow to the point; The interaction V - 2 is represented by a dashed line connecting two points (r and r 2 ). And each ending point has a creation operator and an annihilation operator: Connect the lines, and keep the direction of the arrows Find all the diagrams obey the rules described above, each of them represent a term in the power series expansion of G. Zeroth order terms contains no V, first order terms contains just one V, nth order term contains n V. Here we consider two point correlation functions G, ' as an example. Because G, ' = Ty  y ' has one creation and one annihilation term, we have one ending point and one starting point for solid lines. At the zeroth order, we don t have V, and these two points are the only thing we have in our diagrams. As a result the only diagram we have is a solid line connecting these two points.

21 Phys620.nb 2 The first order terms has one V, and it is easy to note that there are only two ways to connect the diagrams Therefore, up to the first order G, ' = G 0, ' Â r 3 t 3 r 2 G 0, 3 V r 3 - r 2 dt 3 - t 2 G 0 3, ' G 0 2, (.79) Â r 3 t 3 r 2 G 0, 3 V r 3 - r 2 dt 3 - t 2 G 0 3, 2 + G 0 2, ' t3 =t 2 If we want higher order terms, we just add more Vs to the diagrams. For example, if we have two Vs (second order terms), we have We just need to find all the plots and then write them into integrals (and the compute the integrals) k,w-space If the system have momentum and energy conservation laws, it is typically much easier to compute Green s functions in the momentum-energy space. In the k,w-space, we use the same diagrams and in addition, we also: Assign momentum and frequency to each line and keep the momentum and energy conservation law at each ending point or crossing points. For each loop, there is a pair of unknown q and ÂW n and we need to integrate/sum over them. For each fermionic loop, we get an extra factor of (-), which comes the commutation relation. For mth order diagram, we get a factor - m For example, using the same diagram above, we can compute the two point correlation functions as Gk, Âw n = G 0 k, Âw n q 2 p 3 b G 0 k, Âw n G 0 k, Âw n V0, 0 G 0 k, ÂW n - ÂW n q 2 p 3 b G 0 k, Âw n G 0 k, Âw n G 0 k + q, Âw n +ÂW n Vq, ÂW n +... ÂW n (.80).7. the Dyson s equation If we look at the diagrams, there are many repeating structures. These repeating structures can be utilized to simplify the calculation..7.. The sum of a geometric series How do we compute the sum of a geometric series: X = a + aq+ aq 2 + aq 3 + (.8) First, we notice that

22 22 Phys620.nb qx= aq+ aq 2 + aq 3 + aq 4 + (.82) then, we rewrite X as X = a + aq+ aq 2 + aq 3 + = a + qx (.83) So, X - qx= a X = a - q (.84) (.85) We can use the same trick to sum many Feynman diagrams example : the Hartree approximation For diagrams, we can use the same trick. For example, the following diagrams can be summed together Here we sum over these diagrams and ignore others. This approximation is known as the Hartree approximation. Gk, Âw n º G 0 k, Âw n + G 0 k, Âw n S H k, Âw n G 0 k, Âw n + G 0 k, Âw n S H k, Âw n G 0 k, Âw n S H k, Âw n G 0 k, Âw n + G 0 k, Âw n S H k, Âw n G 0 k, Âw n S H k, Âw n G 0 k, Âw n S H k, Âw n G 0 k, Âw n + (.86) where S H k, Âw n = = q 2 p 3 b V0, 0 G 0 q, ÂW n ÂW n (.87) This term is known as the self-energy correction. It is the same diagram we considered above but with external legs removed. Using the same trick mentioned above, we find G H k, Âw n = G 0 k, Âw n -S H k, Âw n G 0 k, Âw n - = G 0 k, Âw n - -S H k, Âw n - = w-ek-s H k, Âw n This technique and this formula is known as the Dyson s equation. Here, we find that the Green s function for interacting particles is very similar to the free Green s function G 0. The only thing interactions does is to change the single particle energy ek into ek +S H k, Âw n. In other words, the interactions changes the energy of the particle by S H k, Âw n. This is the reason why this term is called a self-energy correction. However, it is important to keep in mind that this term is NOT really a shift in ek, because it is also a function of frequency example 2: the Hartree-Fock approximation We can get more accurate results by adding more diagrams into our calculation. For example, (.88) 2 The approximation that sums over these diagrams (and ignore others) are known as the Hartree-Fock approximation. Gk, Âw n º G 0 k, Âw n + G 0 k, Âw n S HF k, Âw n G 0 k, Âw n + G 0 k, Âw n S HF k, Âw n G 0 k, Âw n S HF k, Âw n G 0 k, Âw n + G 0 k, Âw n S HF k, Âw n G 0 k, Âw n S HF k, Âw n G 0 k, Âw n S HF k, Âw n G 0 k, Âw n + (.89) here the Hartree-Fock self-energy correction S HF k, Âw n is

23 Phys620.nb 23 S HF k, Âw n = + = q 2 p 3 b V0, 0 G 0 q, ÂW n - ÂW n q 2 p 3 b G 0 k + q, Âw n +ÂW n Vq, ÂW n ÂW n (.90) Using the same trick, (the Dyson s equation), we find that G HF k, Âw n = G 0 k, Âw n -S HF k, Âw n G 0 k, Âw n - = G 0 k, Âw n - -S HF k, Âw n - = w-ek-s HF k, Âw n (.9) The final result is almost the same as the Hartree approximation. We just need to change S H k, Âw n into S HF k, Âw n example 3: include more diagrams If we include more diagrams, the Green s function still takes the same structure Gk, Âw n = G 0 k, Âw n + G 0 k, Âw n Sk, Âw n G 0 k, Âw n + G 0 k, Âw n Sk, Âw n G 0 k, Âw n Sk, Âw n G 0 k, Âw n + G 0 k, Âw n Sk, Âw n G 0 k, Âw n Sk, Âw n G 0 k, Âw n Sk, Âw n G 0 k, Âw n + (.92) Here, the self-energy correction contains more diagrams Sk, Âw n = (.93) Using Dyson s equation, we find that Gk, Âw n = G 0 k, Âw n -Sk, Âw n G 0 k, Âw n = -Sk, Âw G 0 k,âw n n = Âw n -ek-sk, Âw n (.94).7.5. One-particle irreducible diagrams If we use the Dyson s equation, the key is to avoid double counting. For example, the second order diagram has been included in the Hartree approximation when we include in the self-energy. Therefore, when we compute second order self-energy corrections, we should not include this diagram again. The rule to avoid double counting is to use -particle irreducible diagrams in the selfenergy S. One-particle irreducible diagrams means that if we cut one internal link (solid line), the diagram is still connected. For example, the following diagrams are -particle irreducible diagrams, and thus they should be include din the self-energy: This diagram is not -particle irreducible, and thus we should not include them in the self-energy correction (it has already been taken care of by the first order term ).7.6. Summary Draw all possible -particle-irreducible diagrams. Remove external legs to get the self energy (note: each fermion loop contributes a factor -. For nth order diagram, we need an factor - n in the momentum space formula). Use Dyson s equation to get the full Green s function. GÂw n = Âw n -ek-sk,âw n Other particles can be treated using the same approach (photons, phonons, etc.)

24 24 Phys620.nb.8. Physical meaning The physical meaning of the spectral function Ak, w is: if we have a particle with momentum k, Ak, w2 p is the probability for this particle to have energy w..8.. Ak, w 0 The proof can be found in the book of Mahan (page 5). We will not show it here w Ak, w= 2 p Ak, w = G > k, w G < k, w (.95) Integrate over all w w - 2 p Ak, w = w w - 2 p G> k, w - 2 p G< k, w Notice that w - 2 p Âwt G > k, w =ÂG > k, t = yk, t 0 + t y k, t 0 If we set t = 0, we find that by integrate over all w, we get the equal-time correlation function w - 2 p G> k, w = yk, t 0 y k, t 0 Similarly, w - 2 p G< k, w = y k, t 0 yk, t 0 Therefore, w - 2 p Ak, w = w w - 2 p G> k, w - 2 p G< k, w = yk, t 0 y k, t 0 y k, t 0 yk, t 0 = = In other words, - w 2 p Ak, w measures the commutator/anit-commutator between y and y, which is unity Free particles For particles without interactions, the Green s function is G 0 k, Âw n = Âw n -ek The spectral function Ak, w is obtained by Ak, w =-2lim e0 ImGk, w+âe =-2lim e0 Im w+âe-ek = lim 2 e e0 w-e k 2 +e = 2 pdw-e k 2 (.96) (.97) (.98) (.99) (.200) (.20) (.202) For a fixed k, if we plot Ak, w as a function of w, we find a delta peak at w=e k. This delta peak means that for a free particle with momentum k, the energy of this particle can only take one value: w=e k. And it is easy to check that w Ak, w = - 2 p.8.4. Interacting particles When interactions are taken into account, (.203)

25 Phys620.nb 25 Gk, Âw n = Âw n -ek-sk, Âw n = Âw n -ek- ReSk, Âw n -ÂImSk, Âw n (.204) Sometimes, the real part of S is labeled as S, where the imaginary part is S 2 Sk, z =S k, z +ÂS 2 k, z (.205) Therefore, the spectral function is Ak, w = -2 lim e0 ImGk, w+âe = Âw n -ek-sk, w+âe =-2lim e0 Im w+âe-ek-s k, w+âe-â S 2 k, w+âe = lim e0 2 S 2 k, w+âe w-ek -S k, w+âe 2 + S 2 k, w+âe 2 = 2 S 2 k, w w-ek -S k, w 2 +S 2 k, w 2 (.206).8.5. The real part of self-energy correction Here, we consider the real part of Sk, w. For simplicity, we consider the limit that ImSk, w 0. When S 2 (k,w)0 2 S 2 k, w Ak, w = w-ek-s k, w 2 +S 2 k, w =dw-ek-s k, w 2 (.207) If S k, w only depends on k and is independent of w Ak, w =dw-ek -S k (.208) Here, the spectrum function is also a delta function, which implies that for a particle with momentum k the energy of this particle is ek +S k. In other words, interactions between electrons change the energy of a particle from ek into ek +S k. The physical meaning here is that for free particles, the energy of a particle is just the kinetic energy ek. But for interacting particles, the energy of a particle contains both kinetic energy ek and contributions from interactions S k. In reality, S k, w also depends on w. Here, for a fixed k, we can solve the following equation Ak, w =dw-ek -S k, w = - w S k, w w=e è k dw-e è k (.209) where e è k is the solution of the equation w-ek -S k, w = 0 (we assume that there is only one solution for simplicity). Here, we used the fact that d f x = i dx - x i f ' x i (.20) Notice that at fixed k, Ak, w = Z k dw-e è k - ws k,w w=e è k is a function of k, which we will call Z k. (.2) Here, we learned that the real part of the self-energy can do two things: () renormalizing the dispersion relation and (2) changing the coefficient in front of the delta function. But remember that w Ak, w = - 2 p while w (.23) - 2 p Z k dw-e è k = Z k This means that if Z k, ImSk, w cannot be zero for all frequency. So we must consider ImSk, w to fully understand interaction effects. (.22)