Chiral fermions and chemical potential Some thoughts Rajamani Narayanan Department of Physics Florida International University NCSU, July 21
Introduction of chemical potential on the lattice for chiral fermions Overlap Dirac operator at nonzero chemical potential and random matrix theory; Jacques Bloch, Tilo Wettig; hep-lat/0604020 Domain-wall and overlap fermions at nonzero quark chemical potential; Jacques Bloch, Tilo Wettig; arxiv:0709.4630 [hep-lat] Energy density for chiral lattice fermions with chemical potential; Christof Gattringer, Ludovit Liptak; arxiv:0704.0092 [hep-lat] Thermodynamics of the ideal overlap quarks on the lattice; Debasish Banerjee, R.V. Gavai, Sayantan Sharma; arxiv:0803.3925 [hep-lat] Basic idea in hep-lat/0604020 Follow Hasenfratz and Karsch (Phys. Lett. B125:308, 1983) Replace U 4 by e µ U 4 and U 4 by e µ U 4 everywhere in the Wilson-Dirac kernel. Extend the definition of the sign function. Rajamani Narayanan 2
Massless overlap Dirac operator Massless overlap Dirac operator in even dimensions is D o = 1 2 (1 + γ d+1ɛ[h w (U, m)]) H W is the hermitian Wilson Dirac operator. U is the background gauge field. m is the negative Wilson mass taken to be in the range [0, 2]. ɛ function on a Hermitian matrix is defined as follows: If H w = V ΛV where V is the unitary matrix that diagonalizes H w and Λ ij = λ i δ ij with real λ i being the eigenvalues of H w ; then ɛ(h w ) = V Λ Λ V where Λ ij = λ i δ ij. Rajamani Narayanan 3
Wilson Dirac operator with a chemical potential In the presence of a chemical potential, µ, H w is not hermitian. It takes the form [C L ] xαi,yβj = 1 2 d 1 + 1 2 σαβ d H w = k=1 σαβ k ( ) B CR C L B [δ y,x+ˆk (U k(x)) ij δ x,y+ˆk(u k (y)) ij] [δ y,x+ ˆd eµ (U d (x)) ij δ x,y+ ˆde µ (U d (y)) ij] [C R ] xαi,yβj = 1 d 1 2 k=1 [σ k ]αβ [δ y,x+ˆk(u k (x)) ij δ x,y+ˆk(u k (y)) ij] 1 2 [σ d ]αβ [δ y,x+ ˆde µ (U d (x)) ij δ x,y+ ˆde µ (U d (y)) ij] 1 [B] xαi,yβj = 2 δ d 1 αβ k=1 [2δ xyδ ij δ y,x+ˆk(u k (x)) ij δ x,y+ˆk(u k (y)) ij] + 1 2 δ αβ[2δ xy δ ij δ y,x+ ˆde µ (U d (x)) ij δ x,y+ ˆde µ (U d (y)) ij] mδ xαi,yβj C L (µ) = C R( µ) B (µ) = B( µ) Rajamani Narayanan 4
Bloch and Wettig, hep-lat/0604020 The definition of the ɛ function for a hermitian matrix is extended to a general complex matrix as follows: If H w = V ΛV 1 where V is a complex matrix that diagonalizes H w and Λ ij = λ i δ ij with complex λ i being the eigenvalues of H w ; then where [ e L sλ ] ij = el sλ j δ ij. ɛ(h w ) = V ɛ(λ)v 1 = V [ lim L s e LsΛ ] 1 V 1 e L sλ + 1 If then lim L s λ j = R j + ii j e L sλ j 1 e L sλ j + 1 V 1 = R j R j = Re λ j Re λ j Rajamani Narayanan 5
A domain-wall justification The domain wall action for massless fermions can be wrriten as (H. Neuberger, hep-lat/9710089) D = S = 2L s s=1 Φ s (DΦ) s ( Φ1 Φ2 Φ2Ls 1 Φ 2Ls ) = ( χ R 1 χ L 1 χ R L s 1 χl L s ) C R B + 1 0 0 0 0...... 0 0 B + 1 C L 1 0 0 0...... 0 0 0 1 C R B + 1 0 0...... 0 0 0 0 B + 1 C L 1 0...... 0 0 0 0 0 1 C R B + 1...... 0 0 0 0 0 0 B + 1 C L.... 0 0.............................................................. 0 0 0 0 0 0...... B + 1 C L The physical fermion is ψ = ( χ R 1 χ L L s ) Rajamani Narayanan 6
Pseudofermions The contribution from all the unphysical fermions are subtracted by the pseudofermion action D pf = C R B + 1 0 0 0 0...... 0 1 B + 1 C L 1 0 0 0...... 0 0 0 1 C R B + 1 0 0...... 0 0 0 0 B + 1 C L 1 0...... 0 0 0 0 0 1 C R B + 1...... 0 0 0 0 0 0 B + 1 C L.... 0 0.............................................................. 1 0 0 0 0 0...... B + 1 C L Rajamani Narayanan 7
Fermion determinant Neuberger, hep-lat/9710089 [ 1 T det D = (det B) k k det 2 1 + T k ] γ 5 2 det D pf = (det B) k det [ ( 1 + T k) γ 5 ] T = 1 B+1 1 B+1 C L C R 1 B+1 C R 1 B+1 C L + B + 1 det D det D pf = det 1 2 [ 1 + γ 5 tanh T L s ] 1 T L s + 1 Rajamani Narayanan 8
Using tanh to define ɛ T is a general complex matrix in the presence of a chemical potential and we are interested in T L s 1 T L s + 1 Let T = V EV 1 where V is the general complex matrix that diagonalizes T and E ij = e i δ ij is the diagonal matrix made up of the complex eigenvalues, e i, of T. Then T L s 1 = V E Ls 1 V 1 T L s + 1 E L s + 1 lim L s e L s i 1 e L s i + 1 = { 1 if e i < 1 1 if e i > 1 Rajamani Narayanan 9
Overlap Dirac operator with chemical potential lim L s T L s 1 T L s + 1 = ɛ [ ln T ] lim L s det D det D pf = det D o D o = 1 2 [1 + γ 5ɛ ( ln T )] ln T H w as the lattice spacing in the (d + 1) direction goes to zero. The definition of the overlap Dirac operator with a chemical potential in hep-lat/0604020 seems justified. Rajamani Narayanan 10
Isospin chemical potential H w (µ) = ( ) B(µ) CR (µ) C L (µ) B(µ) C L (µ) = C R( µ) B (µ) = B( µ) H w(µ) = H w ( µ) ɛ (H w ( µ)) = [ɛ (H w (µ))] det D o (µ) = [det D o ( µ)] Rajamani Narayanan 11
A reminder of the derivation of the overlap Dirac operator Overlap fermions provides a solution to the problem of putting chiral fermions on the lattice. Assume there is no chemical potential. Form two many body operators: Then H = a γ 5 a H + = a H w a det C L = b b+ det C R = t t+ where b± > are the normalized lowest energy states of H ± and t± > are the normalized highest energy states of H ±. Phases of these states have to be fixed such that det C L = det C R Rajamani Narayanan 12
The computation Let ( ) α β V = γ δ ( α γ) be the unitary matrix that diagonalizes H w with and ( β δ) spanning the positive and negative eigenvalues of H w respectively. ( 1 0) and ( 0 1) span the positive and negative eigenvalues of γ 5 respectively. Therefore, det C L = δ and det C R = α up to a phase. Since V is unitary, one can show that Since det V det V = 1, it follows that and therefore det C L det C L = det C R det C R choice. det V = det α det δ det α det α = det δ det δ are the same and independent of the phase Rajamani Narayanan 13
Derivation of the overlap Dirac operator ɛ(h w )V = γ 5 ɛ(h w )V = D o V = ( ) α β γ δ ( α ) β γ δ ( ) α 0 0 δ det α det D o = det α det δ det δ det D o = det δ det δ Rajamani Narayanan 14
Addition of the chemical potential V = ( ) α β γ δ ɛ(h w )V = ( ) α β γ δ γ 5 ɛ(h w )V = D o V = ( α ) β γ δ ( ) α 0 0 δ det D o det V = det α det δ det D o = det V 1 det α det δ Rajamani Narayanan 15
Remarks H w is not hermitian. H + is not a hermitian many body operator. a should really be replaced by a 1. It carries the same meaning. a 1 is the creation operator and is the inverse of a, the annihilation operator. If (a, a 1 ) obey canonical anticommutation relations, and if b = V 1 a, then (b, b 1 ) also obey canonical anticommutation relations. det C L det C R is not real and positive. There should be no ambiguity in the definition of det C L det C R. Under V DV where D is an arbitary complex diagonal matrix, det V 1 det α det δ and therefore det D o is invariant. The propagator G o = D 1 o 1 = is clearly chiral and is invariant under V DV. ( ) 0 βδ 1 γα 1 0 Rajamani Narayanan 16
Eigenvalues of S = γ 5 ɛ ɛ 2 = 1 Let Sψ = sψ Then ɛψ = sγ 5 ψ ψ = sɛγ 5 ψ 1 s [γ 5ψ] = S [γ 5 ψ] There is a pairing of eigenvalues of the form, (s, 1/s). s = ±1 are not paired. s = 1 corresponds to a zero mode of D o. If ɛ is hermitian, S is unitary and all eigenvalues lie on the unit circle. In the presence of µ. eigenvalues inside the unit circle have partners outside the unit circle. Rajamani Narayanan 17
det Do(µ) Assume we are in the zero topological sector. When µ = 0, let s j = e iφ j with 0 φ j < π be half the eigenvalues of S. Then, det D o (0) = j cos 2 φ j 2 When µ 0, let us assume that s j < 1 be half the eigenvalues of S. Then, det D o (0) = 1 [ ] 2 + sj + s 1 j 4 j Whether µ = 0 or µ 0, s j close to 1 cause a suppression and this is just the role of almost zero modes. What if all the s j with s j < 1 get close to zero? (A poosible scenario as µ is increased) Then, the determinant gets very large (opposite of suppression). In addition, if the phase of s j gets uniformly distributed on the unit circle, then the determinant will remain large. Rajamani Narayanan 18
Phase of det Do(µ) Phase of the fermion determinant results in the sign problem. What happens in large N c QCD with finite number of fermions flavors? Can we work in the quenched approximation even with a chemical potential in the large N c limit? The fermion determinant should still be one power of N c less than the vacuum polarization. But the fermion determinant will have a factor of N c and this implies that the phase of the determinant can be anywhere on the unit circle. If so, phase averaging is a problem. What is the phase distribution of the eigenvalues s j with s j > 1? Rajamani Narayanan 19