Negative Binomial Coherent States

Transcription

1 Negative Binomial Coherent States Superstatistics of Quanta T.S. Biró 1 1 Heavy Ion Research Group Hung.Acad.Sci. Research Centre for Physics, Budapest October 6, 2015 Talk given by T. S. Biró October 7-9, 2015, Leibnitz, Austria, EU. Biró Hadronization: NBS 1 / 40

2 Our Goal Is to construct particular Coherent States designed to describe NB n distribution and non-extensive Tsallisean statistical weights Biró Hadronization: NBS 2 / 40

3 Outline 1 Statistical Weight from n-distribution 2 3 Biró Hadronization: NBS 3 / 40

4 Outline 1 Statistical Weight from n-distribution 2 3 Biró Hadronization: NBS 4 / 40

5 of ideal reservoir fluctuations 1 T.S.Biró et.al. Entropy 16 (2014) 6497, arxiv: Ideal gas formula and thermodynamical limit: w LIM E (ω) = lim n E E/n = T Poisson average on n at fix E: Negative binomial (NB) average: ( 1 ω E ) n = e ω/t. (1) w POI E (ω) = e n ω/e. (2) ( we NBD (ω) = 1 + n k + 1 ) ω (k+1). (3) E Biró Hadronization: NBS 4 / 40

6 of ideal reservoir fluctuations 2 indico.kfki.hu/category/28 Expanding up to variance in n we obtain for general n fluctuations Exact for Poisson, Bernoulli, Negative Binomial T = E n (n 1), and q = n n 2 (4) w E (ω) = ( 1 + (q 1) ω ) 1 q 1 T (5) In general T and q are related to expectation values of derivatives of the EoS S(E) over reservoir fluctuations. Biró Hadronization: NBS 5 / 40

7 of general reservoir fluctuations Einstein: phase space Ω(E) = e S(E) S(E) and its derivatives give 1 T = S (E), and q = 1 + T 2 T 2 1 C (6) with C = de/dt total heat capacity and S (E) = 1/CT 2. Biró Hadronization: NBS 6 / 40

8 Trends in Tsallis fits to data G. Wilk s collection, Erice School on Complexity, 2015 Fits: T AA = (q 1) GeV; T pp = (q 1) GeV. AA: assume constant variance, T 2 /T 2 = σ 2 and ideal C = E/T. Then we obtain T = E ( σ 2 (q 1) ) pp: assume fix parameter NBD, (q 1) = 1/(k + 1), n = f (k + 1), T = E/ n, and obtain T = E (q 1) f Biró Hadronization: NBS 7 / 40

9 Trends in Tsallis fits to data G. Biro s collection, master thesis, 2015 Biró Hadronization: NBS 8 / 40

10 State labels Operator eigenstate Outline 1 Statistical Weight from n-distribution 2 State labels Operator eigenstate 3 Biró Hadronization: NBS 9 / 40

11 State labels Operator eigenstate Definition Consider a coherent state defined by z = pn (t) e inθ n (7) n=0 with z = te iθ. ( nonlinear coherent state ) It overlaps with the n-quantum state: n z 2 = p n (t) 0. (8) Biró Hadronization: NBS 9 / 40

12 State labels Operator eigenstate Normalization It is normalized, z z = n,m m p m p n e i(n m)θ n = n p n (t) = 1. (9) The expectation value of a function of the number operator is z ϕ(n) z = n ϕ(n) p n (t). (10) This ensures that p n (t) is a probability distribution in n! Biró Hadronization: NBS 10 / 40

13 State labels Operator eigenstate Completeness We construct a complete set: d 2 z π z z = = 0 0 dt dt n 2π 0 dθ 2π pn p m e i(n m)θ m n n,m p n (t) n n = n n n = 1. (11) It is satisfied only if dt p n (t) = 1. This makes p n (t) to a probability distribution function of t! 0 Biró Hadronization: NBS 11 / 40

14 State labels Operator eigenstate CS as eigenstate to some operator Eigenstate with eigenvalue z to F z = ag(ˆn) z = z z. (12) Here a is an annihilating (a is a creating) operator, and ˆn = a a is the number operator. Its action on the CS: F z = g(n) np n e inθ n 1, (13) n=1 can be re-indexed to F z = g(n + 1) (n + 1)p n+1 e i(n+1)θ n, (14) n=0 Biró Hadronization: NBS 12 / 40

15 State labels Operator eigenstate Recursion law Compare this with z z = t e iθ pn e inθ n, (15) n=0 to conclude that p n (t) = t ng(n) 2 p n 1(t). (16) Biró Hadronization: NBS 13 / 40

16 State labels Operator eigenstate n-probability Symmetric generalized binomial distributions, J.Math.Phys. 54 (2013) The recursion is solved by p n (t) = p 0 (t) tn n! n g(j) 2. (17) Here p 0 (t) can be obtained from the normalization condition. j=1 Also the completeness constraint, dt p n (t) = 1, has to be checked. 0 Biró Hadronization: NBS 14 / 40

17 Outline 1 Statistical Weight from n-distribution 2 3 Biró Hadronization: NBS 15 / 40

18 Traditional CS The most known CS is defined by g(n) = 1. This results in a Poisson in n and Euler-Gamma in t: p n (t) = tn n! e t, (18) and z is an eigenstate to the F = a annihilator. Biró Hadronization: NBS 15 / 40

19 NB coherent states The negative binomial distribution (NBD), ( ) n + k p n (t) = (t/k) n (1 + t/k) n k 1, (19) n is well normalized in n and, as an Euler-Beta distribution, also in t. From the recursion one obtains g(n) 2 = t n p n 1 p n = k + t k + n, (20) so this state satisfies a k + z 2 k + a a z = z z. (21) Biró Hadronization: NBS 16 / 40

20 7 sources of NBD 1 Phase space cell statistics (Biró) 2 Squeeze parameter (Varró, Jackiw) 3 Wave packet statistics (Pratt, Csörgő, Zimányi) 4 Temperature superstatistics (Beck, Wilk) 5 KNO + pqcd (Dokshitzer, Dremin, Hegyi, Carruthers) 6 Tsallis/Rényi entropy canonical state (Rényi, Tsallis,...) 7 Glittering Glasma (Gelis, Lappi, McLerran) Biró Hadronization: NBS 17 / 40

21 1. Phase Space Cell Statistics Probability to find n particles in k cells, if altogether we have thrown N particles into K cells: P n = ( k+n n Pólya distribution ) ( K k+n n ) N n ( K +N+1 N ). Necessary limit: K, N while f = N/K kept finite. Prediction: k + 1 = n / f N part, if f is universal. Here k is the number of observed phase space cells: from which the detected n particles seem to come. Biró Hadronization: NBS 18 / 40

22 2. Negative Binomial State (NBS) Neg.Binom. state as a nonlinear coherent state: z, k = pn (k) e inθ n (22) n=0 with p n (k) = ( k + n n ) f n (1 + f ) n k 1. (23) NBS annihilated a z, k = f (k + 1) e iθ z, k + 1. (24) Biró Hadronization: NBS 19 / 40

23 3. Wave packet statistics small size limit: k = 0 NBS HBT with onefold filled bosonic states gives a correlation factor of 2 at zero relative momentum. With M-fold occupation of the same state it reduces to C 2 (0) = M = k + 1. (25) The logarithmic cumulants for an NBS, defined by G(z) = p n z n and ln G(z) = C n (z n 1), are C n = k + 1 n n=1 ( f 1 + f ) n. (26) This is a (k + 1)-fold overload of the simple Bose case, given if k = 0. Biró Hadronization: NBS 20 / 40

24 4. Superstatistics Beck, Wilk Thermodynamical β-fluctuation and n-fluctuations are related by Poisson transform: 0 γ(β)e βω dβ = ( 1 ω ) n Pn (E). (27) E n=0 In this way β 2 / β 2 = 1/(k + 1). Biró Hadronization: NBS 21 / 40

25 5. pqcd Dokshitzer, Dremin, Hegyi, Carruthers KNO scaling + GLAP give nearly NBD with constant k, related to Λ QCD and expressed by n-variance. NBD is in fact slightly violated. Biró Hadronization: NBS 22 / 40

26 6. Canonical Accepting Tsallis or Rényi entropy as a formula, the usual canonical constraint on the average energy leads to w(ω) = ( 1 + (q 1) ω ) 1 q 1 T (28) Using further assumptions about reservoir fluctuations, further entropy formulas can be constructed, as expectation values of formal logarithms, behaving additively (ARC). Biró Hadronization: NBS 23 / 40

27 7. Glittering Glasma: k-fold ropes Gelis, Lappi, McLerran k = κ(n 2 c 1)Q 2 s R 2 /2 is about the number of tubes, makes NBS with this parameter. Biró Hadronization: NBS 24 / 40

28 NB coherent states The negative binomial distribution (NBD), ( ) n + k p n (t) = (t/k) n (1 + t/k) n k 1, (29) n is well normalized in n and, as an Euler-Beta distribution, also in t. From the recursion one obtains g(n) 2 = t n p n 1 p n = k + t k + n, (30) so this state satisfies a k + z 2 k + a a z = z z. (31) Biró Hadronization: NBS 25 / 40

29 su(1,1) structure in NBD Rapidity-like notation: t/k = sinh 2 ζ; ( ) k + n p n (t) = sinh 2n ζ cosh 2n 2k 2 ζ. n (k ) + n ( z = cosh k 1 (ζ) tanh(ζ)e iθ) n n. (32) n n=0 Using velocity v = tanh(ζ) the overlap between two NBD Coh.States: [ z 1 z 2 2 = 1 + γ1 2 γ2 2 v 1 e iθ 1 v 2 e iθ 2 2] k 1. (33) 2+1 dim relative velocity vector separates Biró Hadronization: NBS 26 / 40

30 2+1 dim vector representation K = γ 1 γ 2 ( v 1 v 2 ) = 1 k (γ 2 z 1 γ 1 z 2 ). (34) Overlap written this way z 1 z 2 2 = [ k γ 2z 1 γ 1 z 2 2 ] k 1 e z 1 z 2 2. (35) with γ i = 1 + z i 2 /k. Particle properties of the vector K : K = 1 m 1 m 2 ( E 2 P1 E 1 P2 ) (36) Its parallel component does not Lorentz transform: v K = v K. Biró Hadronization: NBS 27 / 40

31 Negative Binomial States in our own notation with z k = kf e iθ and z k, k = p n (k) = pn (k)e inθ n (37) n=0 ( k + n n ) f n (1 + f ) n k 1 (38) provides an average photon number n = f (k + 1). excited geometric state: n over k intermediate number state: f /(1 + f ) eigenstate with complex eigenvalue Biró Hadronization: NBS 28 / 40

32 The Effect of Ladder Operator Effect of annihilating a quantum: a z k, k = pn e inθ n n 1 = n=1 (n + 1)pn+1 e iθ e inθ n. (39) Consider now ( ) ( ) (n+1) k + n + 1 f n+1 (1+f ) (n+1) k 1 k n = f (k + 1) n + 1 k + 1 f n (1+f ) n (k+1) 1. (40) Here we recognize z k+1 = f (k + 1)e iθ as a factor and arrive at NBS annihilated n=0 a z k, k = z k+1 z k+1, k + 1. (41) Biró Hadronization: NBS 29 / 40

33 NBS as eigenstate of what? The action of another operator ˆN + k + 1 z, k = (k + 1)(1 + f ) z, k + 1, (42) based on the relation ( ) k + n (k n) k This helps to recognize: NBS eigenvalue equation ( k n = (k + 1) k + 1 ). (43) ( ˆn f (ˆn + k + 1) 1 + f e iθ + k + 1 a) z, k = 0. (44) Biró Hadronization: NBS 30 / 40

34 Our NBS algebra In the previous equation the following operator occurs: K = ˆn + k + 1 a, K+ = K = a ˆn + k + 1. The commutator, [K, K + ] = (ˆn + 1)(ˆn + k + 1) ˆn(ˆn + k) = 2 ˆN + k + 1 = 2K 0 (45) defines K 0 = ˆn + (k + 1)/2. With α = (1 + f )/f e iθ we get OUR eigenvalue equation (α K K 0 ) z, k = k z, k. (46) Biró Hadronization: NBS 31 / 40

35 SU(1,1) algebra for NBS The commutators form an SU(1,1) algebra: Commutators [K 0, K + ] = K + [K 0, K ] = K [K, K + ] = 2K 0 (47) The Casimir operator is given as: Q = K 2 0 K 0 K + K. Biró Hadronization: NBS 32 / 40

36 NBS created from 0 : 1. preliminaries Operator identity e A+B = e λ/2 e A e B if [A, B] = λ (const.) (48) We choose A = α zg(ˆn) a and B = β z a 1 g(ˆn) A. Commutator: λ = [A, B] = z 2 αβ g(ˆn)a a 1 g(ˆn) + z 2 αβ a 1 g(ˆn) g(ˆn)a = z 2 αβ. (49) Biró Hadronization: NBS 33 / 40

37 NBS created from 0 : 2. evolution operator U = e Φ/2+A+B = e (Φ λ)/2 e A e B. (50) Here e B 0 = 0 due to B 0 = 0. U 0 = e (Φ αβ z 2 )/2 α n z n n=0 n! ( g(ˆn)a ) n 0. (51) Biró Hadronization: NBS 34 / 40

38 NBS created from 0 : 3. n-photon state We have ( g(ˆn)a ) n 0 = g(n)... g(1) n! n. (52) Regarding the form U 0 = un e inθ n, (53) n=0 with z = t e iθ, and using g(ˆn) = ξ ˆn + k we obtain ) ( ) n k + n u n = e (α Φ αβ t 2 ξ 2 t. (54) n Biró Hadronization: NBS 35 / 40

39 NBS created from 0 : 4. normalization For U 0 2 = 1 one needs (k+1) u n = e (1 Φ αβt α 2 ξ t) 2 = 1. (55) n=0 From this we express and gain α 2 ξ 2 t = 1 e 1 k+1 (Φ αβt) = 1 w, (56) negative binomial distribution ( ) n + k u n = w k+1 (1 w) n. (57) n Biró Hadronization: NBS 36 / 40

40 NBS created from 0 : 5. superstatistics For the superstatistics normalization, we utilize 0 ( k + n dt u n(t) = n ) ( ) dt w k+1 (1 w) n k + n = k n dw w k 1 (1 w) n = 1. (58) This is achieved if dt = k dw w 2, t = k or expressing w(t) if ( ) 1 w 1, (59) w = t/k = k t + k. (60) This identifies Φ as being Biró Hadronization: NBS 37 / 40

41 NBS created from 0 : 6. conclusion In this way α, β and ξ remain undetermined. A purposeful choice is α = β = 1/ t = 1/ z with ξ = 1/ t + k. The log of the evolution operator becomes in this case neither hermitic nor anti-hermitic ln U = k ln (1 + t/k) ˆn + k eiθ t + k a e iθ t + k a ˆn + k. (62) U may be connected to a Hamiltonian... Biró Hadronization: NBS 38 / 40

42 NBS created from 0 : 7. physics behind The anti-hermitic part gives a guess for the Hamiltonian i Hdτ = 1 2 (ln U (ln U) ) ( = e iθ f (ˆn) + 1 ) ( a e iθ a f (ˆn) + 1 ). f (ˆn) f (ˆn) (63) The hermitic part means non-conserved particle number... Biró Hadronization: NBS 39 / 40

43 A class of coherent states designed for a given n pdf is also a superstatistics in t = z 2 ; The NB coherent state is an eigenstate for the regularized phase operator. An su(1,1) structure is inherent in the NB coherent state. Hamiltonians combined from K, K 0 and K + operators are likely to produce NB distributed bosons. Biró Hadronization: NBS 40 / 40