s-scattering Length by Bold DiagMC

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1 s-scattering Length by Bold DiagMC For a 3D particle of the mass m in the potential U(r, the s-scattering length a can be expressed as the zeroth Fourier component of the pseudo-potential Ũ (we set h = 1: a = m Ũ(q = 0. (1 2π In Fourier representation, the pseudo-potential is related to the potential U by the integral equation Ũ(q = U(q U(q q 1 G (0 (q 1 Ũ(q 1 d 3 q 1 /(2π 3, (2 where G (0 (q = 2m/q 2 (3 is the (zero-frequency Green s function (propagator of a free particle. The theory also covers the case of two particles (with masses m 1 and m 2 in a central potential, with the mass m being the reduced mass: m = m 1 m 2 /(m 1 + m 2. Our goal is to find a by solving Eq. (2 simulating Ũ(q with Bold Diagrammatic Monte Carlo (BDMC. We start with rescaling the variables. Introducing we get [also using (3] f(q = (m/2πũ(q, u(q = (m/2πu(q, (4 u(q q1 f(q 1 f(q = u(q 2π 2 q1 2 d 3 q 1, (5 a = f(0. (6 In what follows, we confine ourselves with the case of a spherically symmetric potential U(r U(r, when u(q u(q, f(q f(q, and Eq. (5 simplifies to f(q = u(q 1 1 dχ u q π 2 + q1 2 2qq 1χ f(q 1 dq 1. (7 1 0 Setting q = 0 in Eq. (7, we get the formula a = u(0 (2/π u(q f(q dq, (8 which will prove very useful for estimating a by Monte Carlo procedure. 1

2 BDMC procedure. Our configuration space consists of three sectors (diagram types 0, 1, and 2. Type-0 diagram is just a number D 0 ; it has no physical meaning and is used for normalization purposes. Type-1 (type-2 diagrams correspond to the first (second term in the r.h.s. of (7. Note that, in the context of diagrammatic Monte Carlo, the word diagram stands for the integrand, not integral. Hence, just changing internal/external variables means changing the diagram. The updating scenario involves the following moves. Update for the type-0 diagram: Move 0 1. Propose to move (from the type-0 diagram to type-1 diagram, seeding q by the function Ω 0 1 (q. Accept with probability P (acc 0 1 (q. Note that, in the type-0 sector, there is only one update. Hence, as opposed to all other updates, the addressing probability is identically equal to 1. Updates for type-1 diagrams: Move 1 0. Address this update with probability p (addr 1 0. If the update is addressed, accept with probability P (acc 1 0 (q. Move 1 2. Address this update with probability p (addr 1 2. If the update is addressed, propose to move to type-2 diagram with q inherited from the type-1 diagram and q 1 and χ seeded by the function Ω 1 2 (q, q 1, χ. Accept with probability P (acc 1 2 (q, q 1, χ. Variables-1. Address this update with probability 1 p (addr 1 0 p (addr 1 2. If the update is addressed, propose a new value, q, (of the variable q seeded by the function Ω 1 (q, q. Accept with probability P (acc 1 (q, q. Updates for type-2 diagrams: Move 2 1. Address this update with probability p (addr 2 1. If the update is addressed, propose to move to type-1 diagram with q inherited from the type-2 diagram. Accept with probability P (acc 2 1 (q, q 1, χ. 2

3 Variables-2. Address this update with probability 1 p (addr 2 1. If the update is addressed, propose new values, (q, q 1, of the variables (q, q 1, seeding them with the function Ω 2 (q, q 1, χ, q, q 1 (acc. Accept with probability P 2 (q, q 1, χ, q, q 1. The acceptance ratios are readily constructed with the detailed-balance equation. Below we present generic expressions along with particular examples. Moves 0 1 and 1 0. R 0 1 = p(addr 1 0 u(q D 0 Ω 0 1 (q, R 1 0 = D 0 Ω 0 1 (q p (addr 1 0 u(q. (9 For example, seed q uniformly within the interval (0, q 0. Then Ω 0 1 (q = θ(q 0 q/q 0 and R 0 1 = p(addr 1 0 q 0 u(q, R 1 0 = θ(q 0 q D 0 D 0 p (addr 1 0 q 0 u(q. (10 Moves 1 2 and 2 1. R 1 2 = p(addr q u 2 + q1 2 2qq 1χ f(q p (addr π u(q Ω 1 2 (q, q 1, χ 1 2, R 2 1 = R (11 For example, seed q 1 and χ uniformly within the intervals (0, q 0 and ( 1, 1, respectively. Then Ω 1 2 (q, q 1, χ = θ(q 0 q 1 /(2q 0 and 2q 0 u q 2 + q R 1 2 = p(addr 1 2 2qq 1χ f(q p (addr, (12 π u(q 1 2 R 2 1 = θ(q 0 q 1 p(addr 1 2 p (addr 2 1 π u(q 2q 0 u q 2 + q1 2 2qq. (13 1χ f(q 1 Variables-1. This is a standard single-variable Metropolis-Hastings walk: R = u(q Ω 1 (q, q u(q Ω 1 (q, q. (14 3

4 For example, for uniformly seeding q within an interval (λ 1 q, λq, we have R = u(q q u(q q. (15 Variables-2. Similar to the previous one. u q 2 + q 1 2 2q q 1 χ f(q 1 Ω 2 (q, q 1, χ, q, q 1 R = q u 2 + q1 2 2qq. (16 1χ f(q 1 Ω 2(q, q 1, χ, q, q 1 For example, for uniformly seeding q and q 1 within the intervals (λ 1 q, λq and (λ 1 q 1, λq 1, respectively, we have R = u q 2 + q 1 2 2q q 1 χ f(q 1 q q 1 q u 2 + q1 2 2qq. (17 1χ f(q 1 q q 1 Square-Well Potential For definiteness, consider the square-well potential U(r = { U, if r < r, 0, if r r. (18 Here we have with U(q = 4πU q 3 [sin(r q r q cos(r q], (19 u(q = 3u 0 sin(r q r q cos(r q (r q 3, (20 u 0 u(0 = 2 3 mu r 3. (21 It is convenient to measure a, u, u 0, and f in units of r. Correspondingly, q is measured in units of 1/r, and the problem (7 becomes dimensionless. 4

5 The dimensionless function u(q of the dimensionless variable q is defined by the dimensionless parameter u 0 : Useful asymptotic relations for u(q are u(q = 3u 0 sin q q cos q q 3. (22 u(q u 0 (1 q 2 /2 at q 0, (23 u(q 3u 0 cos q q 2 at q. (24 Estimators The simplest method of restoring the function f(q from the BDMC statistics is histogramming. Split the interval (0, of the variable q into small intervals of equal size δq. Each interval s is responsible for corresponding bin of the histogram. We thus will be using the symbol bin s for both the bin s and the interval it corresponds to. The size δq is supposed to be small enough so that the integral of f over the bin can be approximated as bin s f(q dq f(q 0 q (q 0 bin s. (25 The content, H s, of the s-th bin of the histogram (bin s is defined as follows. Each type-1/type-2 diagram with the momentum q bin s adds +1 ( 1 to H s, if the diagram is positive (negative. Obviously, non-physical type-0 diagrams contribute nothing to the histogram. We are in a position to derive estimators. In what follows, the precise meaning of the expression A b, with A a Monte Carlo observable and b certain quantity, is A approaches b in the statistical limit. Let S tot be the total number of Monte Carlo steps, i.e., the total number of elements in the Markov chain up to a given moment of simulation. Then H s 1 Z f(q dq S Z 1 f(q 0 q (q 0 bin s, (26 tot bin s where Z = D 0 + u(q dq + 1 π dχ dq dq 1 u q 2 + q1 2 2qq 1χ 5 f(q 1. (27

6 Now let S 0 be the total number of type-0 configurations. Observe that S 0 S tot D 0 Z. (28 In principle, this allows us to estimate Z as Z = D 0 S tot /S 0. A more elegant way to proceed, though, is to simply exclude Z between (28 and (26, arriving at an explicit estimator for f(q 0 (note that S tot drops out as well: D 0 S 0 H s bin s f(q dq f(q 0 q (q 0 bin s. (29 Finally, with (8 taken into account, we get the following integral estimator for a: u(0 2D 0 u(q s H s a (q s bin s. (30 πs 0 s Note that the estimator (30 is much better than a naive estimator a = f(0 D 0 H 0 S 0 q (31 (with H 0 the bin corresponding to q = 0, because, due to the smallness of q, the statistical noise in H 0 is much larger than that in the sum over all the bins in (30. 6