Effective mass of nonrelativistic quantum electrodynamics

Effective mass of nonrelativistic quantum electrodynamics Fumio Hiroshima and K. R. Ito ecember 6, 5 Abstract The effective mass m eff of the nonrelativistic quantum electrodynamics with spin / is investigated. Let m eff /m = + a (Λ/me + a (Λ/me 4 + O(e 6, where m denotes the bare mass. a (Λ/m log(λ/m as Λ is well known. Also a (Λ/m Λ/m is established for a spinless case. It is shown that a (Λ/m (Λ/m in the case including spin /. Introduction. Quantum electrodynamics In this review we study an translation-invariant Hamiltonian minimally coupled to a quantized radiation field in the nonrelativistic quantum electrodynamics. Before going to discuss our problem, we informally derive our Hamiltonian from physical point of view. The conventional quantum electrodynamics is investigated through the Lagrangian density: L QE (x = ψ(x(iγ µ µ mψ(x 4 F µν(xf µν (x eψ(xγ µ ψ(xa µ (x, where x = (x, x R R 3, γ µ, µ =,,, 3, denotes 4 4 gamma matrices, ψ the spinor given by ψ = (ψ, ψ, ψ, ψ 3 T and ψ = ( ψ, ψ, ψ, ψ 3 γ, A µ (x a radiation field with F µν := µ A ν ν A µ, and m, e are the mass and the charge of an electron, respectively. The effective mass m eff is given through two point function R 4 (Ψ, T [ψ(xψ(]ψe i(x p x p dx and the effective charge through the two point function R 4 (Ψ, T [A µ (xa ν (]Ψe i(x p x p dx, where Ψ denotes a ground state of the Hamiltonian derived from L QE and T the time ordered product. In the perturbative quantum electrodynamics, Feynman diagrammatically, the e and e 4 terms of the effective mass is computed from the self-energy of electron, e.g., Figure : Electron self-energy Faculty of Mathematics, Kyushu University, Fukuoka 8-858, Japan. epartment of Math. and Phys., Setsunan University, 57-858, Osaka, Japan

and the effective charge from the self-energy of photon, e.g., Figure : Photon self-energy One can interpret the photon self-energy diagram as the emission of the pairs of virtual electrons and positrons. All these argument is successive from the physical point of view, but perturbative and implicit divergences are included.. Informal derivation of nonrelativistic quantum electrodynamics In this note we want to discuss the quantum electrodynamics nonperturbatically, but we assume that ( an electron is in low energy, ( we take the Coulomb gauge, (3 we introduce a form factor of an electron. ( implies that no emission of pairs of virtual electrons and positrons, i.e., there is no loop such as in Fig., then in our model the effective charge equals to the bare charge and the number of electrons is fixed. From ( the theory is not relatively covariant. From (3 it follows that the density of the electron charge is smoothly localised around the position of the electron and the ultraviolet divergence does not exist. Taking into account of (-(3, we modify the quantum electrodynamics as follows. Let E(t, x and B(t, x, (t, x R R 3, be an electric field and a magnetic field respectively, and q(t the position of an electron at time t R. The Maxwell equation with form factor is given by Ḃ = E, B =, Ė = B e( q(t q(t, E = e( q(t. Here Ẋ = dx/dt. Let (J(t, x, ρ(t, x = (e(x q(t q(t, e(x q(t. Then the Lagrangian density of the nonrelativistic quantum electrodynamics under consideration is given by L NRQE (t, x = m q(t + (E(t, x B(t, x + J(t, x A(t, x ρ(t, xφ(t, x, where A and φ are a vector potential and a scalar potential related to E and B such as E = Ȧ φ, B = A. Let L NRQE = L NRQE (t, xdx. Then the conjugate momenta are given p(t := L NRQE q = m q(t + e A(t, x(x q(tdx, Π(t, x := δl NRQE δȧ = Ȧ(t, x.

Then the Hamiltonian is given through the Legendre transformation as H NRQE = p(t q(t + Ȧ(t, xπ(t, xdx L NRQE = ( p(t e A(t, x(x q(tdx + V (q + {Ȧ(t, x + ( A(t, x } dx m where V is a smeared external potential given by V (q := (q y(q y e 4π y y dydy. In the next subsection we quantize H NRQE with spin / and total momentum p R 3, which is denoted by H(p and is called the Pauli-Fierz Hamiltonian, in the rigorous way from mathematical point of view..3 Non-relativistic quantum electrodynamics [ ] Let F be the boson Fock space given by F n s L (R 3 {, }, where n s denotes the n-fold symmetric tensor product with sl (R 3 {, } C. The Fock vacuum Ω F is defined by Ω {,,,..}. Let a(f be the creation operator and a (f the annihilation operator on F defined by n= (a (fψ (n+ n + S n+ (f Ψ (n, f L (R 3 {, }, and a(f = [a ( f], where S n denotes the symmetrizer. The scalar product on K is denoted by (f, g K which is linear in g and anti-linear in f. a and a satisfy canonical commutation relations: j=, [a(f, a (g] = ( f, g L (R 3 {,}, [a(f, a(g] =, [a (f, a (g] =. We write as a (k, jf(k, jdk for a (f with a formal kernel a (k, j. Let T be a selfadjoint operator on L (R 3. We define Γ(e itt a (f a (f n Ω a (e itt f a (e itt f n Ω. Thus Γ(e itt turns out to be a strongly continuous one-parameter unitary group in t, which implies that there exists a self-adjoint operator dγ(t on F such that Γ(e itt = e itdγ(t for t R. We define a Hilbert space H by H C F. The Pauli-Fierz Hamiltonian with total momentum p = (p, p, p 3 R 3 is given by a symmetric operator on H: H(p 3 σ m µ (p µ µ ea ˆµ + H f, µ= where m > and e R denote the mass and the charge( of an electron, respectively, ( i σ (σ, σ, σ 3 the Pauli-matrices given by σ, σ, i 3

( σ 3, and the free Hamiltonian H f, the momentum operator and quantum radiation field A ˆµ are given by H f dγ(ω, µ dγ(k µ, A ˆµ ˆ(k e µ (k, j(a (k, j + a(k, jdk, µ =,, 3. ω(k j=, Here e(k, j, j =,, denotes polarization vectors such that e(k, j =, e(k, e(k, =, k e(k, j = and e(k, e(k, = k/ k. We omit the tensor notation in what follows. Then H(p = m (p ea ˆ + H f e m σb ˆ, p R 3, where B ˆµ denotes the quantum magnetic field given by B ˆµ i j=, ˆ(k ω(k (k e(k, j µ (a (k, j a(k, jdk. Note that [A ˆµ, B ˆν ] = for µ, ν =,, 3. The spectrum of H(p is stuided in e.g., [, ] and see references in [5]. Mass renormalization. Main theorems Let where T m (e, p (p ea ˆm + H f e σb ˆ m, p R 3,, k < κ/m, ˆ m (k ˆ(mk = / (π 3, κ/m k Λ/m,, k > Λ/m. ( It is established in [4] that T m (e, p is self-adjoint on ( (H f for arbitrary Λ >, m >, p R 3, e R. Since A ˆ = ma ˆm, B ˆ = mb ˆm, H f = mhf and = mpf, where X = Y denotes the unitary equivalence, we have H m (e, p = mt m (e, ( p /mn z, where n z = (,,. Let :X : be the Wick product of X. We define H(e, ɛ :T m (e, ɛn z :, ɛ R. Set E(e, ɛ inf σ(h(e, ɛ. It is established in [5] that there exist constants e > and ɛ > such that for (e, ɛ a {(e, ɛ R e < e, ɛ < ɛ }, ( the dimension of Ker(H(e, ɛ E(e, ɛ is two, ( E(e, ɛ is an analytic function of e and ɛ on a, (3 there exists a strongly analytic ground state of H(e, ɛ. The effective mass m eff is defined by m m eff = ɛ E(e, ɛ ɛ=. 4

From this it immediately follows that effective mass m eff is an analytic function of e on {e R 3 e < e } with some e >. Set m m eff = a n (Λ/me n. n= The effective mass is investigated in e.g., [3, 5, 6, 8]. It is known and easily derived that a (Λ/m = 8 3π Our next issue is to study a (Λ/m. 4π ( Λ/m κ/m r + dr + Λ/m κ/m r (r + 3 dr. a (Λ/m Theorem. There exist positive constants c < c such that c lim Λ (Λ/m c.. Expansions We set A A ˆm and B B ˆm. Let us define H, E and g by H H(e, = H + eh I + e H II, E E(e, = n= e n n! E (n, g g (e, = n= e n n! (n, where H H f +, H I H ( I + H ( I, H ( I = A, H ( I σb, and H II : AA := A + A + + A + A + A A. Here we put A + j=, ˆm (k e(k, ja (k, jdk, A ω(k We can see that E ( = E (n+ =, n =,,, 3,..., and j=, ˆm (k ω(k e(k, ja(k, jdk. ( E ( = ( (, H ( I ( H = ( (, σb ( σb H ( H. ( Note that E ( (Λ/m as Λ. Moreover ( ( = Ω, ( = ( σb H (, ( = ( H II H ( + ( A ( H H + {( σb ( H H (3 = 3 H ( H II H ( σb σb ( σb ( E } ( (, ( + 3 H ( A H ( H II ( 5

+6 ( A ( A ( σb H H H ( +6 ( A {( σb ( σb ( E } ( H H H ( +3 ( σb ( H II H ( + 6 ( σb ( A ( σb H H H H +6 ( σb {( σb ( σb ( E } ( H H H (. ( Although formula m m eff = 3 3 µ= (( + ea µ g, (H E ( + ea µ g H ( g, g H (3 is well known, it is not useful for our task, since expansion of (H E in e leads us to a complicated operator domain argument. Then, instead of (3, it is established in [6] that g s g (e, ɛ/ ɛ ɛ= satisfies that ( + ea 3 g ((H E with g = (H E ( + ea 3 g and H m = ( g, ( + ea 3 g. (4 m eff ( g, g H Using (4 in [5] it is proven that the effective mass is expanded as or where m m eff = 3 c (Λ/me 3 c (Λ/me 4 + O(e 6, (5 m eff m = + ( ( 3 c (Λ/me + 3 c (Λ/m + c (Λ/m e 4 + O(e 6, (6 3 Here 3 c (Λ/m (Ψ µ, H Ψ µ H, µ= 3 c (Λ/m µ= Ψ µ n { (Ψ µ, H Ψ µ H (Ψ µ, H Ψ µ H( (, ( H + R(Ψ µ, H Ψ µ H +(Ψ µ, H Ψ µ H + R(Ψ µ 3, H Ψ µ H (n! A µ (n + n! µ (n, n =,, 3, µ =,, 3, and H H, H H + H 3, H H 4 + H 5 + H 6 + H 7 + H 8, where we put H =, H = ( A, H 3 = ( σb, H H H H H }. (7 6

H 4 = ( H II, H 5 = ( A ( A, H H H H H H 6 = ( σb ( A, H 7 = H 6 = ( A ( σb H H H H H H 8 = {( σb ( σb ( E } (. H H H, H From above expressions of (, (, (3, it follows that for µ =,, 3, Ψ µ Φµ + Φµ, 6 Ψµ Φ µ i, Ψµ 3 6 Φ µ i, i= i=7 where Φ µ = A+ µ ( Φ µ = P f µ σb + H ( Φ µ 3 = A µ σb + H ( Φ µ 4 = P f µ A + A + H ( Φ µ 5 = P f µ ( A σb + H H ( Φ µ 6 = 4 P f µ σb + σb + H H ( Φ µ 7 = A µ A + A + H ( Φ µ 8 = A µ ( A σb + H H ( Φ µ 9 = 4 A µ σb + σb + H H ( Φ µ = 4 P f µ :AA: σb + H H ( Φ µ = P f µ ( A A + A + H H ( Φ µ = P f µ ( A ( A σb + H H H ( Φ µ 3 = 4 P f µ ( A σb + σb + H H H ( Φ µ 4 = 4 P f µ σb A + A + H H ( Φ µ 5 = 4 P f µ σb ( A σb + H H H ( Φ µ 6 = 8 P f µ σb σb + σb + H H H ( Substituting H,..., H 8 and Φ µ,..., Φµ 6 into (7, we see that { 3 8 c (Λ/m = ( Φ µ i, H l Φ µ i H + ( Φ µ i, H Φ µ i H µ= i= l=4 i= i= i= } 6 6 6 ( Φ µ i, (H + H 3 Φ µ i H + ( Φ µ i, H Φ µ i 6 H +( Φ µ i, H Φ µ i H. (8 i=3 i= i=3 i=3 i=7 i= From (8 it follows that c (Λ/m is decomposed into 76 terms. Fortunately it is, however, enough to consider terms containing even number of σb s, since the terms with odd number of σb vanishes by a symmetry. See Fig. 3-7. 7

No. Term σb H Order ( (Φ µ, H Φ µ ( (, ( H [log(λ/m] ( (Φ µ, H Φ µ ( (, ( H 3 [log(λ/m] Figure 3: ( i= Φ µ i, H i= Φ µ i H No. Term σb H Order (3 (Φ µ 3, H Φ µ 3 3 [log(λ/m] (4 (Φ µ 5, H Φ µ 3 4 [log(λ/m] (5 (Φ µ 3, H Φ µ 5 4 [log(λ/m] (6 (Φ µ 4, H Φ µ 4 3 + Λ/m (7 (Φ µ 6, H Φ µ 4 4 Λ/m (8 (Φ µ 4, H Φ µ 6 4 Λ/m (9 (Φ µ 5, H Φ µ 5 4 5 + Λ/m ( (Φ µ 6, H Φ µ 6 4 5 [log(λ/m] Figure 4: ( 6 i=3 Φ µ i, H 6i=3 Φ µ i H No. Term σb H Order ( (Φ µ, H 4Φ µ [log(λ/m] ( (Φ µ, H 4Φ µ 4 [log(λ/m] (3 (Φ µ, H 5Φ µ 3 log(λ/m (4 (Φ µ, H 5Φ µ 4 5 [log(λ/m] (5 (Φ µ, H 6Φ µ 4 [log(λ/m] (6 (Φ µ, H 6Φ µ 4 [log(λ/m] (7 (Φ µ, H 7Φ µ 4 [log(λ/m] (8 (Φ µ, H 7Φ µ 4 [log(λ/m] (9 (Φ µ, H 8Φ µ 3 [log(λ/m] ( (Φ µ, H 8Φ µ 4 5 [log(λ/m] Figure 5: ( i= Φ µ i, 8 l=4 H l i= Φ µ i H 8

No. Term σb H Order ( (Φ µ 4, H Φ µ 3 log(λ/m ( (Φ µ 6, H Φ µ 4 [log(λ/m] (3 (Φ µ 3, H Φ µ 4 [log(λ/m] (4 (Φ µ 5, H Φ µ 4 5 Λ/m (5 (Φ µ 3, H 3Φ µ 4 [log(λ/m] (6 (Φ µ 5, H 3Φ µ 4 [log(λ/m] (7 (Φ µ 4, H 3Φ µ 4 Λ/m (8 (Φ µ 6, H 3Φ µ 4 5 Λ/m Figure 6: ( 6 i=3 Φ µ i, (H + H 3 ( i= Φ µ i H No. Term σb H Order (9 (Φ µ 7, H Φ µ [log(λ/m] (3 (Φ µ 9, H Φ µ 3 [log(λ/m] (3 (Φ µ, H Φ µ 3 = (3 (Φ µ 3, H Φ µ 4 = (33 (Φ µ 5, H Φ µ 4 = (34 (Φ µ 8, H Φ µ 4 [log(λ/m] (35 (Φ µ, H Φ µ 4 Λ/m (36 (Φ µ, H Φ µ 4 5 [log(λ/m] (37 (Φ µ 4, H Φ µ 4 [log(λ/m] (38 (Φ µ 6, H Φ µ 4 5 (Λ/m Figure 7: ( 6 i=7 Φµ i, H i= Φ µ i H 9

.3 Feynman diagrams spin = σb + = σ(k, f = σ (k e(k, j, polarization = A µ = e µ (k, j, propagator( = (π 3 ω(k, propagator( = = H ω(k i + k i /, E i propagator(3 = = H ω(k + ω(k + k + k /, E propagator(4 = ( = ( H ω(k + k /, ( A + = k e(p, j, µ = k µ.. Figure 8: Items of diagrams The 38 terms in Fig. 3-7 can be represented by Feynman diagrams. The items of diagrams are in Fig. 8. Example. We compute (Φ µ 5, H Φ µ 5 as an example. Since (Φ µ 5, H Φ µ 5 = 4 (P f µ ( A + σb + H H (, H µ ( A + σb + H H (, its diagram is given as in Fig. 9. Figure 9: (Φ µ 5, H Φ µ 5.

Then (Φ µ 5, H Φ µ 5 = d 3 k d 3 k 4 4(π 6 ω ω σ (k e (k + k µ (k + k µ E E E E d 3 k d 3 k k + k = 4(π 6 ω ω E 3 ( σ σ (k e (k e E ( (k + k e σ + (k + k e σ E E + σ σ (k e (k e E E Here and in what follows we set e = e(k, j, e = ( e(k, j (, σ = σ(k, j, σ = σ(k, j and X denotes the expectation value of X: X = (, X C. The diagrams consists of three kinds of diagrams; AA type, Aσ type and σσ type. particular AA type corresponds to the spinless model discussed in [6]. In (Φ µ 4, H Φ µ 4 = O( Λ/m (Φ µ, H 4Φ µ = O([log(Λ/m] (Φ µ, H 5Φ µ = O(log Λ/m (Φ µ 4, H Φ µ = O(log Λ/m (Φ µ 7, H Φ µ = O([log(Λ/m] (Φ µ, H Φ µ = Figure : AA type Feynman diagrams

(Φ µ 3, H Φ µ 3 = O([log(Λ/m] (Φ µ 5, H Φ µ 3 = O([log(Λ/m] (Φ µ 3, H Φ µ 5 = O([log(Λ/m] (Φ µ 5, H Φ µ 5 = O( Λ/m (Φ µ, H 4Φ µ = O([log(Λ/m] (Φ µ, H 5Φ µ = O([log(Λ/m] (Φ µ, H 7Φ µ = O([log(Λ/m] (Φ µ, H 7Φ µ = O([log(Λ/m] (Φ µ, H 8Φ µ = O([log(Λ/m] Figure : Aσ type Feynman diagrams

(Φ µ 6, H Φ µ = O([log(Λ/m] (Φ µ 3, H Φ µ = O([log(Λ/m] (Φ µ 5, H Φ µ = O( Λ/m (Φ µ 3, H 3Φ µ = O([log(Λ/m] (Φ µ 5, H 3Φ µ = O([log(Λ/m] (Φ µ 4, H 3Φ µ = O( Λ/m (Φ µ 9, H Φ µ = O([log(Λ/m] (Φ µ 3, H Φ µ = (Φ µ 5, H Φ µ = Figure : Aσ type Feynman diagrams 3

(Φ µ 8, H Φ µ = O([log(Λ/m] (Φ µ, H Φ µ = O( Λ/m (Φ µ, H Φ µ = O([log(Λ/m] (Φ µ 4, H Φ µ = O([log(Λ/m] (Φ µ 6, H Φ µ 4 = O( Λ/m (Φ µ 4, H Φ µ 6 = O( Λ/m (Φ µ, H 6Φ µ = O([log(Λ/m] (Φ µ, H 6Φ µ = O([log(Λ/m] (Φ µ, H Φ µ = O([log(Λ/m] Figure 3: Aσ type Feynman diagrams 4

(Φ µ 6, H Φ µ 6 = O([log(Λ/m] (Φ µ, H Φ µ = O([log(Λ/m] (Φ µ 6, H 3Φ µ = O(Λ/m (Φ µ 6, H Φ µ = O((Λ/m (Φ µ, H 8Φ µ = O([log(Λ/m] Figure 4: σσ type Feynman diagrams.4 Proof of Theorem. We shall prove Theorem.. Note that the following formulas are useful. Lemma.3 It follows that e i e i =, i =,, (9 (e e (e e = + (ˆk ˆk, ( (k e (k e = k ( (ˆk ˆk, ( (k e (e e (e k = (k, k ( (ˆk ˆk, ( σ i σ i = k i, i =,, (3 R σ σ (e k (k e = k k ((ˆk ˆk, (4 5

Proof: Note that R σ σ (e e = k k (ˆk ˆk, (5 σ σ σ σ = 4 k k, (6 R σ σ σ σ = k k ( (ˆk ˆk. (7 e µ (k, je ν (k, j = (δ µν k µk ν k, (k e(k, j µe ν (k, j = ɛ µνα k α, R σ µ σ ν = δ µν. Here ɛ αβγ denotes the antisymmetric tensor with ɛ 3 = +. In what follows, the summation over repeated index is understood. We see that ( e i e i = e µ (k i, je µ (k i, j =, i =,, ( (e e (e e = e µ (k, je µ (k, j e ν (k, je ν (k, j = (δ µν k µk ν k (δ µν k µk ν k = (3 + (ˆk ˆk k k = + (ˆk ˆk, (3 (k e (k e = k µ e µ (k, jk ν e ν (k, j = k µ k ν (δ µν k µk ν k = k ( (ˆk ˆk, (4 (k e (e e (e k = k µ e µ (k, je λ (k, je λ (k, j k ν e ν (k, j = k µ (δ µλ k µk λ k k ν(δ λν k λk ν k = (k, k ( (ˆk ˆk, (5 R σ σ = R σ µ σ ν (k e(k, j µ (k e(k, j ν = δ µν (k e(k, j µ (k e(k, j ν = (k e(k, j µ (k e(k, j µ = k ( e(k, + e(k, = k, (6 R σ σ (e k (k e = R σ µ σ ν (k e(k, j µ (k e(k, j ν e α (k, jk α e β (k, j k β = (k e(k, j µ (k e(k, j µ e α (k, jk α e β (k, j k β = ( ɛ µαγ k γ ( ɛ µβδ k δ k α k β = k k = k k ( (ˆk ˆk, (7 R σ σ (e e = R σ µ σ ν (k e(k, j µ (k e(k, j ν e α (k, je α (k, j = ( ɛ µαβ k β ( ɛ µαγ k γ = (k k, (8 σ σ σ σ = σ σ µ σ ν σ (k e(k, j µ (k e(k, j ν = σ σ (k e(k, j µ (k e(k, j µ = (k e(k, j (k e(k, j = k k = 4 k k, (9 R σ σ σ σ = σ σ σ σ + (k e(k, j µ (k e(k, j µ R σ σ = 4 k k + (k e(k, j µ (k e(k, j µ (k e(k, j ν (k e(k, j ν = 4 k k + (δ µν k µk ν k (δ µν k µk ν k = 4 k k + k k ( + (ˆk ˆk = k k ( (ˆk ˆk. Using the diagrams presented in Fig.-4 and (9-(7, we can easily expressed 38 terms as integrals on = {(k, k R 3 R 3 κ/m k Λ/m, κ/m k Λ/m}. 6

Note that imaginary part of σ a σ b does not contribute integrals. We show the results:. (Φ µ, H Φ µ ( (, ( = 4 (A+ µ (, A + µ H ( ( σb + H (, σb + H ( = d 3 k d 3 k 4 4(π 6 e e σ σ ω ω E E E = d 3 k d 3 k 4 k 4 4(π 6 ω ω E E. (Φ µ, H Φ µ ( (, ( = 4 (P f µ σb + H (, H µ σb + H ( ( σb + H (, σb + H ( = d 3 k d 3 k 4 4(π 6 k σ σ σ σ ω ω E E E E E = d 3 k d 3 k 4 k 4 k 4 4(π 6 ω ω E 3 E 3. (Φ µ 3, H Φ µ 3 = 4 (A+ µ = 4 = 4 H σb + (, H A + µ H σb + ( d 3 k d 3 k 4(π 6 ( σ σ (e e + σ σ (e e ω ω E E E E d 3 k d 3 k 4(π 6 ( 4 k + k k (ˆk ˆk ω ω E E E E 4. (Φ µ 5, H Φ µ 3 = 4 (P f µ ( A + σb + H H (, = d 3 k d 3 k 4 4(π 6 ω ω E = d 3 k d 3 k 4 4(π 6 ω ω E H A + µ H σb + ( E ((k e (k e σ σ E + (k e (k e σ σ E ( k 4 ( (ˆk ˆk + k k ( (ˆk ˆk E E E 5. (Φ µ 3, H Φ µ 5 = (Φµ 5, H Φ µ 3 7

6. 7. (Φ µ 4, H Φ µ 4 = [6, (3.3] = 4 (P f µ (A + A + H (, H µ (A + A + H ( = d 3 k d 3 k 4 4(π 6 ω ω E 3 k + k (e e (e e + e e = d 3 k d 3 k 4 4(π 6 k + k ( + (ˆk ˆk ω ω E 3 (Φ µ 6, H Φ µ 4 = 8 (P f µ σb + σb + H H (, H µ (A + A + H ( = d 3 k d 3 k 8 4(π 6 ω ω E 3 k + k ( σ σ e e + σ σ e e E E = d 3 k d 3 k 8 4(π 6 k + k k k (ˆk ˆk ( + ω ω E E E 3 8. (Φ µ 4, H Φ µ 6 = (Φµ 6, H Φ µ 4 9... (Φ µ 5, H Φ µ 5 = 4 (P f µ ( A + σb + H H (, = d 3 k d 3 k 4 4(π 6 ω ω E 3 = d 3 k d 3 k 4 4(π 6 ω ω E 3 H µ ( A + σb + H H ( k + k (k e ((k e σ σ + (k e σ σ E E E k + k ( k 4 ( (ˆk ˆk + k k ( (ˆk ˆk E E E (Φ µ 6, H Φ µ 6 = E = 6 (P f µ σb + σb + H H (, = d 3 k d 3 k 6 4(π 6 ω ω E 3 = d 3 k d 3 k 6 4(π 6 ω ω E 3 H µ σb + σb + H H ( E k + k ( σ σ σ σ E + σ σ σ σ E k + k ( 4 k k + k k ( (ˆk ˆk E E E (Φ µ, H 4Φ µ = [6, (3.34] = (A ν = = d 3 k d 3 k 4(π 6 (e e (e e ω ω E E d 3 k d 3 k 4(π 6 ( + (ˆk ˆk ω ω E E H A + µ (, A ν H A + µ ( 8

. 3. 4. 5. 6. (Φ µ, H 4Φ µ = 8 (P f µ σb + H (, (A + A + + A + A + A A H H µ σb + H ( = 4 (A ν H µ σb + H (, A ν H µ σb + H ( = d 3 k d 3 k 4 4(π 6 σ σ (k k (e e ω ω E E E E = d 3 k d 3 k k k (ˆk ˆk 4 4(π 6 ω ω E E (Φ µ, H 5Φ µ = [6, (3.36] = (A+ µ (, = = H ( A + H ( A + H A + µ ( d 3 k d 3 k 4(π 6 (k e ((k e (e e + (k e (e e ω ω E E E E d 3 k d 3 k 4(π 6 ( k ( (ˆk ˆk + (k k ( (ˆk ˆk ω ω E E E E (Φ µ, H 5Φ µ = 4 ( ( A + H H µ σb + H (, ( A + H µ σb + H ( = d 3 k d 3 k 4 4(π 6 ω ω E E (k e ((k e k σ σ E + (k e (k k σ σ E = d 3 k d 3 k 4 4(π 6 ω ω E E ( k 6 ( (ˆk ˆk E + (k k ( k k ( (ˆk ˆk E (Φ µ, H 6Φ µ = 4 ( σb H H µ σb + H (, ( A + A + µ H ( = d 3 k d 3 k 4 4(π 6 (k e ((k e σ σ + (k e σ σ ω ω E E E E E E = d 3 k d 3 k k k ( (ˆk ˆk 4 4(π 6 ω ω E E E (Φ µ, H 6Φ µ = 4 (σb+ A + µ H (, ( A + H H µ σb + H ( = d 3 k d 3 k 4 4(π 6 ((k e (k e σ σ + (k e (k e σ σ ω ω E E E E E = d 3 k d 3 k k 4 ( (ˆk ˆk 4 4(π 6 ω ω E E E 9

7. 8. 9.. (Φ µ, H 7Φ µ = 4 ( ( A + H H µ σb + H (, σb + A + µ H ( = d 3 k d 3 k 4 4(π 6 ((k e (k e σ σ + (k e (k e σ σ ω ω E E E E E E = d 3 k d 3 k k 4 ( (ˆk ˆk 4 4(π 6 ω ω E E E (Φ µ, H 7Φ µ = 4 (σb+ ( A + A + µ H H (, H µ σb + H ( = d 3 k d 3 k 4 4(π 6 σ σ ((k e (k e + (k e (k e ω ω E E E E E = d 3 k d 3 k k k ( (ˆk ˆk 4 4(π 6 ω ω E E E (Φ µ, H 8Φ µ = E = 4 (σb+ H A + µ (, 4 (σb+ (, H σb + ( (A + µ (, E H σb + H A + µ ( A + µ H H ( = d 3 k d 3 k 4 4(π 6 ((e e σ σ + (e e σ σ ω ω E E E E d 3 k d 3 k 4 4(π 6 σ σ (e e ω ω E E E = d 3 k d 3 k 4 4(π 6 ( 4 k + k k (ˆk ˆk d 3 k d 3 k ω ω E E E 4 4(π 6 4 k ω ω E (Φ µ, H 8Φ µ = E 3 = 6 (σb+ H µ σb + H (, σb + H H µ σb + H ( 6 (σb+ (, σb + H ( (σb + (, H H µ H µ σb + H ( = d 3 k d 3 k 6 4(π 6 ω ω E E d 3 k d 3 k 6 4(π 6 k σ σ σ σ ω ω E E E E E = d 3 k d 3 k 6 4(π 6 ω ω E E ( 4 k k 4 E d 3 k d 3 k 6 4(π 6 4 k k 4 ω ω E E 4 E ( σ σ σ σ k E E + (k k σ σ σ σ E E + (k k ( k k ( (ˆk ˆk E

. (Φ µ 4, H Φ µ = [6, (3.33] = (P f µ A + A + H (, ( A + A + µ H H ( = d 3 k d 3 k 4(π 6 ω ω E (k e (k e ((e e + (e e E = d 3 k d 3 k 4(π 6 ( (k k ( (ˆk ˆk ω ω E E. (Φ µ 6, H Φ µ = 4 (P f µ σb + σb + H H (, = d 3 k d 3 k 4 4(π 6 ω ω E = d 3 k d 3 k 4 4(π 6 ω ω E H ( A + H A + µ ( E ((k e (k e σ σ E + (k e (k e σ σ E E ( k k ( (ˆk ˆk ( E + E 3. (Φ µ 3, H Φ µ = 4 (A+ µ σb + H (, ( A + H H µ σb + H ( = d 3 k d 3 k 4 4(π 6 ω ω E E ((k e (k e σ σ + (k e (k e σ σ E E = d 3 k d 3 k 4 4(π 6 k 4 ( (ˆk ˆk ω ω E E 3 4. (Φ µ 5, H Φ µ = 4 (P f µ ( A + σb + H H (, = d 3 k d 3 k 4 4(π 6 ω ω E = d 3 k d 3 k 4 4(π 6 ω ω E E ( A + H H µ σb + H ( (k e (k + k k ((k e σ σ + (k e σ σ E E E E k (k + k ( k 4 ( (ˆk ˆk E + k k ( (ˆk ˆk E 5. (Φ µ 3, H 3Φ µ = 4 (A+ µ σb + H (, σb + A + µ H H ( = d 3 k d 3 k 4 4(π 6 ( σ σ (e e + σ σ (e e ω ω E E E E = 4 d 3 k d 3 k 4(π 6 ω ω E ( k k (ˆk ˆk + 4 k E E E

6. (Φ µ 5, H 3Φ µ = 4 (P f µ ( A + σb + H H (, = d 3 k d 3 k 4 4(π 6 ω ω E = d 3 k d 3 k 4 4(π 6 ω ω E σb + A + µ H H ( (k e ( σ σ (k e + σ σ (k e E E E ( k k ( (ˆk ˆk + k 4 ( (ˆk ˆk E E E 7. (Φ µ 4, H 3Φ µ = 8 (P f µ A + A + H (, σb H H µ σb + H ( = d 3 k d 3 k 8 4(π 6 ω ω E σ σ (k + k k (e e + e e E E = d 3 k d 3 k 8 4(π 6 4k (k + k k k (ˆk ˆk ω ω E E 8. (Φ µ 6, H 3Φ µ = 6 (P f µ σb + σb + H H (, = d 3 k d 3 k 6 4(π 6 ω ω E = d 3 k d 3 k 6 4(π 6 (k + k k ( 4 k k ω ω σb H H µ σb + H ( (k + k k ( σ σ σ σ + σ σ σ σ E E E E E E E + k k ( (ˆk ˆk E 9. (Φ µ 7, H Φ µ = [6, (3.3] = ( H A + A + (, A + µ = = d 3 k d 3 k 4(π 6 (e e ((e e + (e e ω ω d 3 k d 3 k 4(π 6 ( + (ˆk ˆk ω ω E E H A + µ ( 3. (Φ µ 9, H Φ µ = 4 ( σb + σb + H H (, A + µ A + µ H ( = d 3 k d 3 k 4 4(π 6 (e e ( σ σ + σ σ ω ω E E E E = d 3 k d 3 k 4 4(π 6 k k (ˆk ˆk ( + ω ω E E E E

3. 3. 33. 34. 35. 36. (Φ µ, H Φ µ = [6, RHS of(3.3] = ( A + A + H (, ( A + H µ A + µ H ( = d 3 k d 3 k 4(π 6 (e k (e k (e e + e e = ω ω E E E (Φ µ 3, H Φ µ = 4 (P f µ ( A + σb + σb + H H H (, A + µ H ( = d 3 k d 3 k 4 4(π 6 (e k (k e ( σ σ + σ σ = ω ω E E E E E (Φ µ 5, H Φ µ = 4 (P f µ σb + ( A + σb + H H H (, A + µ H ( d 3 k d 3 k = 4(π 6 (e k ((k e σ σ + (k e σ σ = ω ω E E E E E (Φ µ 8, H Φ µ = 4 ( ( A + σb + H H (, A + µ H µ σb + H ( = d 3 k d 3 k 4 4(π 6 (k e ((k e σ σ + (k e σ σ ω ω E E E E E = d 3 k d 3 k 4 4(π 6 ( k 4 ( (ˆk ˆk + k k ( (ˆk ˆk ω ω E E E E (Φ µ, H Φ µ = 8 (P f µ A + A σb + H H (, H µ σb + H ( = 4 (A µ σb + H (, A µ H ν H ν σb + H ( = d 3 k d 3 k 4 4(π 6 k σ σ (e e ω ω E E E E = d 3 k d 3 k 4 4(π 6 k k 3 (ˆk ˆk ω ω E E 3 (Φ µ, H Φ µ = 4 ( ( A + σb + H H (, ( A + H µ H µ σb + H ( = d 3 k d 3 k 4 4(π 6 (k e k ((k e σ σ + (k e σ σ ω ω E E E E E E = d 3 k d 3 k 4 4(π 6 k ( k 4 ( (ˆk ˆk + k k ( (ˆk ˆk ω ω E E E E 3 3

37. 38. (Φ µ 4, H Φ µ = 8 ( A + A + H (, σb + H µ H µ σb + H ( = d 3 k d 3 k 8 4(π 6 k σ σ (e e + e e ω ω E E E E = d 3 k d 3 k 4 k k 3 (ˆk ˆk 8 4(π 6 ω ω E E 3 (Φ µ 6, H Φ µ = E 4 = 6 ( σb + σb + H H (, σb + H µ H µ σb + H ( = d 3 k d 3 k 6 4(π 6 k ( σ σ σ σ + σ σ σ σ ω ω E E E E E E = d 3 k d 3 k 6 4(π 6 k ( k k ( (ˆk ˆk + 4 k k ω ω E E E E 3 As is seen above, integrands in each terms are functions of k, k, (ˆk ˆk. Changing variables k, k, (ˆk ˆk to r, r, X = cos θ, θ < π, respectively, it is seen that We see that F (Λ/m = d d(λ/m F (Λ/m = = π (π 6 d 3 k d 3 k 4(π 6 f( k, k, (ˆk ˆk ω ω dx Λ/m κ/m Λ/m dr dr r r f(r, r, X. κ/m π Λ/m (π 6 dx drr(λ/m[f(λ/m, r, X + f(r, Λ/m, X]. (8 κ/m To see the asymptotic behavior of F (Λ/m, we estimate the right hand side of (8. Then we can see that (Φ µ i lim, H jφ µ k <, µ =,, 3, Λ Λ/m where (i, j, k (, 8,, (, 8,, (6,,. In the case of (i, j, k = (, 8,, (, 8,, each term is divided into two integrals. Although each integral diverges as (Λ/m as Λ, cancelation happens. See Fig. 5 and 6. Then the terms (i, j, k = (, 8,, (, 8, diverge as [log(λ/m]. Finally we can directly see that the term (Φ µ 6, H Φ µ diverges as (Λ/m. Then the proof is complete. 4

Remark.4 Informally we can see that the diagram behaves as (Λ/m. Figure 5: (Φ µ, H 8Φ µ Figure 6: (Φ µ, H 8Φ µ Figure 7: (Φ µ 6, H Φ µ Terms (Φ µ, H 8Φ µ, (Φµ, H 8Φ µ and (Φµ 6, H Φ µ include diagram. Although, as is seen in Fig. 5, 6 and 7, (Φ µ, H 8Φ µ and (Φµ, H 8Φ µ have counter terms, (Φµ 6, H Φ µ does not. 5

Acknowledgements. We thank Tetsuya Hattori and Takashi Hara for helpful comments. K. R. I. thanks Grant-in-Aid for Science Research (C 554 from MEXT and F. H. thanks Grant-in-Aid for Science Research (C 5549 from MEXT. References [] T. Chen, Operator-theoretic infrared renormalization and construction of dressed -particle states in non-relativistic QE, mp-arc -3, preprint,. [] J. Fröhlich, Existence of dressed one electron states in a class of persistent models, Fortschritte der Physik (974, 59 98. [3] C. Hainzl and R. Seiringer, Mass renormalization and energy level shift in non-relativistic QE, Adv. Theor. Math. Phys. 6 (, 847-87. [4] F. Hiroshima, Fiber Hamiltonians in nonrelativistic quantum electrodynamics, in preparation. [5] F. Hiroshima and K. R. Ito, Mass renormalization in non-relativistic quantum electrodynamics with spin /, mp-arc 4-46, preprint 4. [6] F. Hiroshima and H. Spohn, Mass renormalization in nonrelativistic QE, J. Math. Phys. 46 (5 43. [7] E. Lieb and M. Loss, A bound on binding energies and mass renormalization in models of quantum electrodynamics, J. Stat. Phys. 8, ( 57 69. [8] H. Spohn, Effective mass of the polaron: A functional integral approach, Ann. Phys. 75 (987, 78 38. 6