The Fermion Bag Approach Anyi Li Duke University In collaboration with Shailesh Chandrasekharan 1
Motivation Monte Carlo simulation Sign problem Fermion sign problem Solutions to the sign problem Fermion bag approach Massless Thirring model Conclusion 2
QCD phase diagram Lattice QCD at zero density Competing interactions First principles simulation at nonzero density is still unavailable 3
Phase diagram of High T c Material Competing interactions The first principle within Good model calculation is still missing 4
Reason? Sign problem! 5
Monte Carlo simulation Quantum fluctuation is important Strong interactions Many degree of freedom Analytic first principle calculations are difficult Monte Carlo method is the only known first principle approach to solve strongly correlated problems 6
Monte Carlo simulation Problems of interest are converted to a classical statistical mechanics problem Z = W s s Where W s > 0 is the Boltzmann weight of the configuration [s] The Monte Carlo method allows one to generate important configurations [s], thus compute expectation values efficiently 7
But for quantum partition functions this is not always possible! Z = Tr exp ( H T ) Where H is a Hermitian operator Z = s 0 e εh s M 1 s M 1 e εh s M 2 s 2 e εh s 1 s 1 e εh s 0 s 1,s 2,,s M 1 Although Z is guaranteed to be real and positive, each term in the sum can in general be complex! Let us define Ω s 0, s 1,, s M 2, s M 1 = s 0 e εh s M 1 s M 1 e εh s M 2 s 2 e εh s 1 s 1 e εh s 0 clearly Ω s 0, s 1,, s M 2, s M 1 = Ω s 0, s M 1,, s 2, s 1 8
So if we define W s = Real Ω s 0, s 1,, s M 1, s M Sign s = Sign Real Ω s 0, s 1,, s M 1, s M We can only guarantee that we can write Z = [s] Sign( [s])w([s]) If the partition function has the above form with configurations [s] such that Sign[s] = -1, we say this particular representation of Z has a sign problem. 9
Can we find representation to make sign always to be 1? Solution to sign problem 10
Need fresh new ideas to solve sign problem 11
The sign problem Suppose we wish to measure an observable O O = 1 Z [s] O s Sign s W( s ) Z = Sign s W( s ) [s] The brute force method produces [s] according to the distribution P s = W( s ) W( s ) [s] In a sense we have defined a new classical partition function for [s] Z c = W( s ) [s] Z 12
We can see that We can define Sign c = Thus we get O = Z c Z [s] 1 Z c O s Sign s W( s ) [s] Sign s W( s ) [s] W( s ) O = O Sign c Sign c = Z Z c Generically since Z and Zc are partition functions in the thermodynamic limit we expect Sign c = Z Z c = exp f V T Usually <Sign>is exponentially small at large V and low T! Thus, the signal for <O>will be exponentially noisy. 13
A fresh look at the fermion path integral 14
Fermion path integral What does the Grassmann path integral (on the lattice) mean? Z = Tr e H/T = dψ dψ e S(ψ,ψ) There are statements in the literature which say... there is no way to represent Grassmann variables on a computer so we integrate them away!... But we will argue that Grassmann variables help enumerate fermionic world-line configurations 15
Here ψ and ψ are Grassmann valued fields on a Hypercubic lattice Grassmann integration dψ = 0 dψ ψ = 1 ψ 2 = 0 ψ 1 ψ 2 = ψ 2 ψ 1 Example : Free staggered fermion S ψ, ψ = η ij ψ i ψ j + η ji ψ j ψ i m ψ i ψ i <ij> i What does the partition function of this theory mean? 16
The partition function is made of products of terms on each bond. Each of this term can be one of the following e η ijψ i ψ j = 1 + η ij ψ i ψ j = + ψ i j i j ψ e mψ iψ i = 1 + mψ i ψ i = i + i Grassmann nature: Each site can have one incoming and one outgoing line There are sign factors that come from Grassmann ordering. Grassmann variables help enumerate world-lines which are self avoiding loops! 17
Fermion World-line configuration from Grassmann Variabale Fermion loop Monomers = mass terms 18
Fermion sign problem Thus, the fermionic partition function can be written as Z = Sign C W( C ) C fermion loops The sign function depends on the loop and the model. Signs factors come from local phases. Every fermion loop has a negative sign. Can we solve fermion sign problems? Research over the past decade shows that the fermion sign problem can be solved in many ways! Fermion determinant approach(oldest), meron cluster, fermion bag approach 19
Solutions to the fermion sign problem Resummation over a class of world-line configurations The fermion determinant approach is clearly one well known solution! Z = dψ dψ e ψmψ = Sign C C W C = Det M We will use this result later in Fermion bag approach Can be positive!(?) 20
What about interacting fermions? 21
Massless Thirring Model S ψ, ψ = η ij ψ i ψ j ψ j ψ i + Uψ i ψ i ψ j ψ j ij The theory contains a U c 1 U f 1 symmetry U c (1) is a chiral symmetry ψ x e i( 1)x θ ψ x, ψ x ψ x e i( 1)x θ By a proper choice of the η ij we can get massless Dirac fermions. In the lattice QCD literature these are called staggered fermions. 22
Quantum Critical Point Classical XY critical behavior quantum critical point A similar quantum critical point is of interest in the physics of Graphene 23
Conventional Approach S ψ, ψ = η ij ψ i ψ j ψ j ψ i + Uψ i ψ i ψ j ψ j ij Using the Hubbard Stratanovich transformation e Uψ iψ i ψ j ψ j = dφ 2π e U η ij e iφ ψ i ψ j e iφ ψ j ψ i We can then write S ψ, ψ, φ = η ij 1 + Ue iφ ij ψ i ψ j η ij 1 + Ue iφ ij ψ j ψ i ij 24
Thus we have written the action as fermion bilinear S ψ, ψ, φ = ψ i M φ ij ψ j So we can now write Z = dφ dψ dψ e ψ im φ ij ψ j = dφ Det(M φ ) Positive determinant: sign problem solved! But 25
S ψ, ψ, φ = η ij 1 + Ue iφ ij ψ i ψ j η ij 1 + Ue iφ ij ψ j ψ i ij When U U c ~1, fluctuation due to the auxiliary field could produce zero modes 0 0 0 0 = zero weight (odd number of sites) 26
But at the cost of the following: For large U, the matrix M has a large number of small eigenvalues and zero modes. The problem has become completely non-local! (M is V x V matrix) Monte Carlo Algorithms become inefficient! Massless limit is usually difficult (like QCD)! 27
every theoretical physicist who is any good knows six or seven different theoretical representations for exactly the same physics. R. P. Feynman 28
Fermion bag approach I Instead of the Hubbard Stratanovich, consider writing Z = dψ dψ e S 0(ψ,ψ) 1 + Uψ i ψ i ψ j ψ j <ij> Bond free fermions Z = U b ij Sign C, b W( C, b ) [b] ij free fermions Positive defined Here free fermions hop on a lattice not touched by b=1 bonds. = Det Q([b] Z = U b ij Det(Q b ) S. C. PRD 82:025007,2010 [b] ij 29
Det Q([b] = Det(Q B i ) i e Uψ iψ i ψ j ψ j = 1 + Uψ i ψ i ψ j ψ j = + i j i j fermions are free inside certain regions Bag Model At large U the bags are small, so fermions are confined in small regions. 30
S τ L Average bag size Lattice size 30 30 matrix Efficient at large U The effort to compute the determinant is optimal 31
Features Determinantal Monte Carlo Ratio of the determinant = inverse of matrix element needs to be computed at each local step Size of the matrix is optimal at large U depends on the interaction, does not scale with volume Small U, size scales with the volume, and is not optimal 32
Fermion bag approach II Z = dψ dψ e ψdψ+uψ xψ x ψ x+α ψ x+α = dψ dψ 1 + Uψ x ψ x ψ x+α ψ x+α e ψdψ x,α V/2 = dψ dψ U N ψ x1 ψ x1 ψ x2 ψ x2 ψ x2n ψ x2n e ψdψ x N=1 What does the partition function of this theory mean? S. C. & A. Li 33
Examples of 1 bond dψ dψ (U ψ 1 ψ 1 ψ 2 ψ 2 ) e ψdψ = S 12 S 21 U Wick contraction S 12 Free fermion propagator from site 1 to site 2 1 2 Det 0 S 12 S 21 0 = S 12 S 21 Massless fermion 2a translational invariant S 12 = S 21 Anti-PBC 34
Examples of 2 bonds dψ dψ (U 2 ψ 1 ψ 1 ψ 2 ψ 2 ψ 3 ψ 3 ψ 4 ψ 4 ) e ψdψ = U 2 (S 12 S 21 S 43 S 34 S 12 S 23 S 34 S 41 S 14 S 43 S 32 S 21 + S 14 S 41 S 23 S 32 ) + 35
Det 0 S 12 S 14 S 32 S 34 S 21 S 23 S 41 S 43 0 = S 12 S 21 S 43 S 34 S 12 S 23 S 34 S 41 S 14 S 43 S 32 S 21 + S 14 S 41 S 23 S 32 The partition function can be rewritten as Det Z = 0 M M T 0 x V/2 N (DetM) 2 U N 36
U 9 Z = DetM 0 1 + U# + U 2 # + + U V 2# 9 9 matrix determinant Compared to conventional approach V V matrix determinant Efficient for small number of bonds or at small U 37
U c Close to U c N ~7000 for 40 3 lattice size N ~4000 for 30 3 lattice size 38
Benchmark of the algorithm Determinant calculation by SuperLU, 4000 4000 matrix costs 30 secs on single core Generate one independent configuration on a 30 3 volume costs 1.5 days on single core Measurement of chiral condensation and its susceptibility becomes straightforward Massless limit is not a problem Results are coming within few weeks 39
Conclusion Fermion bag approach offer an alternative and powerful approach to fermion lattice field theories Applicable in general where sign problem are solved in conventional approach It is also determinant Monte Carlo method where the size of the matrix is optimized by interactions It can be used to solve some new sign problems Potential applications in the studies of graphene, unitary fermion gas, nuclear effective field theory! The potential of the method remains largely unexplored. 40
Backup slides 41
4D QED Wilson fermions γ 5 D W γ 5 = D W κ < κ c = 0.5 eigenvalues of D W are either real or come in complex conjugate pairs κ > κ c Negative eigenvalues Sign problem 42
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Fermion bag approach Integrate U(1) gauge Integrate Grassmann variables using the identity 44
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Solution to the sign problem 46
Model without sign problem 47
Dynamical fermion mass generation In QCD quarks acquire a mass through dynamics! How? The MIT Bag model provides a very intuitive way to understand it Can this picture emerge starting from a QCD partition function? 48
multilevel resummations 0109:010,2001 Lüscher, Weisz JHEP A Clever representations world-line approach : complex scalar field theory with chemical potential Combine resummation and clever representations meron cluster S. Chandrasekharan, U. Wiese PRL. 83 (1999) 3116-3119 Fermion bag approach 49