Conceptual difficulties with the q-averages in nonextensive statistical mechanics

Sumiyoshi Abe1,2,3 1

Department of Physical Engineering, Mie University, Mie 514-8507, Japan*)

2

Institut Supérieur des Matériaux et Mécaniques Avancés, 44 F. A. Bartholdi, 72000 Le Mans, France 3

Abstract

Inspire Institute Inc., McLean, Virginia 22101, USA

The q-average formalism of nonextensive statistical mechanics proposed in

the literature [C. Tsallis, R. S. Mendes, and A. R. Plastino, Physica A 261, 534 (1998)] is critically examined considering several pedagogical examples. It is shown that this formalism has a number of conceptual difficulties unlike the normal averages.

PACS number(s): 05.20.-y, 05.90.+m Keywords: normal average, q-average, nonextensive statistical mechanics

________________________________ *) Permanent address

1

1.

Introduction In the last decade, a lot of works have been done on so-called nonextensive statistical

mechanics [1,2], which has been expected to be a generalization of Boltzmann-Gibbs statistical mechanics and to describe certain complex systems. Currently, a comprehensive list of references is available at http://www.tsallis.cat.cbpf.br/biblio.htm. This theory is primarily based on a kind of the maximum entropy method, where the form of entropy and definition of average are deformed respectively as follows [1]:

 1 W S q [ p] = ( p n ) q − 1 , ∑  1 − q  n =1 

(1)

W

Q

q

= ∑ Qn Pn( q ) ,

(2)

n =1

where Pn( q ) is referred to as the escort distribution defined in terms of the original distribution {p n}n =1, 2, ..., W as

Pn( q ) =

∑

( pn ) q W

( pm ) q m =1

,

(3)

q a positive constant, W the number of accessible states and {Qn} n =1, 2, ..., W a certain quantity of interest. Eq. (2) is called the q-average. Together with the normalization constraint,

∑

pn =

W n =1

p n = 1, the maximum- S q -distribution is obtained to be

(

)

1 e q −( β / c q )(Qn − Q˜ q ) , Z q (β )

(4)

2

where

e q ( x ) = [1 + (1 − q ) x ] +

1 / (1− q )

(

with

the

notation

[a] + = max{0, a} ,

)

W W Z q ( β ) = ∑ n =1 e q −( β / c q )(Qn − Q˜ q ) and c q = ∑ n =1 ( p n ) q . Eq. (4) is termed the q-

exponential distribution, and Q˜ q in Eq. (4) is calculated in terms of p n in Eq. (4) itself in a self-referential manner. A basic requirement is that this scheme tends to the ordinary maximum entropy method [3] in the limit q → 1. There is a naive question why q in Eqs. (2) and (3) is the same as that in Eq. (1). In other words, why not

∑ = ∑

W

< Q > f (q)

n =1 W

Qn ( p n ) f ( q )

(5)

( p m ) f (q) m =1

with any f (q ) satisfying f (q ) → 1 (q → 1) ? In this article, we discuss conceptual difficulties with the q-average in Eq. (2). Combined with recent works in Ref. [4], we shall conclude that the q-average cannot be employed and the normal average, Q = ∑ n =1 Qn p n , should be used. W

2.

Lottery This can be seen as the simplest example, which shows how it is unreasonable to

employ the q-average for the q-exponential distribution. Suppose you are going to sell 869 lots. And among 869, there are 100 first prizes with prize money 1000 Yen each, 144 second prizes with 500 Yen, 225 third prizes with 100 Yen and finally 400 fourth prizes with 0 Yen. The total amount of money you have to

3

prepare is 0 × 400 + 100 × 225 + 500 × 144 + 1000 × 100 = 197500 Yen. Therefore, if you sell each piece of lot more than 197500 / 869 = 227.272727... Yen, then your pocket will be OK. The corresponding probability distribution is given by

p 4 = N / 9 (first prize), p 3 = 4 N / 25 (second prize), p 2 = N / 4 (third prize), p1 = 4 N / 9 (fourth prize). where N = 900 / 869 is the normalization constant. You see that using this probability distribution you obtain the following normal average number of the prize money, M :

M = 0 × p1 + 100 × p 2 + 500 × p 3 + 1000 × p 4 ,

which is precisely your previously obtained average value, 227.272727... Yen. Now, notice that the above probability distribution has the form

pn =

N (1 + n / 2) 2

( n = 1, 2, 3, 4 ),

which can be rewritten in the form of the q-exponential distribution

p n = N e q ( − n) with q = 3 / 2 . If you use the q-average for the q-exponential distribution as in Ref. [1], then you have

4

M

q =3/ 2

=

1

∑

4 n =1

( pn ) 3/ 2

  400  0 ×  869 

3/ 2

225  + 100 ×   869 

144  +500 ×   869 

which is M

3/ 2

3/ 2

3/ 2

100  + 1000 ×   869 

3/ 2

 , 

= 156.1015387... Yen.

Now, a question is if you sell the lots with reference to M

3/ 2

= 156.1015387... . Yen.

If you do so, then your pocket can seriously be damaged. So far, you considered only the average. There is another important quantity, which is the variance or its square root, i.e., the standard deviation. Using the normal average, you obtain for the standard deviation, σ ≡ whereas using the q-average, σ q = 3 / 2 ≡

M2

M2 − M

2

= 327.7773887... Yen,

− M

2

= 282.0710854... Yen.

3/ 2

3/ 2

Clearly, the latter is smaller than the former. This is natural, since it is nothing but a general feature of Eq. (5): the larger the value of f (q ) (> 1) is, the smaller the generalized standard deviation, σ f ( q ) , is. The q-average makes fluctuation small around the unreliable average value. The above trivial example suggests that the use of the q-average for the q-exponential distribution is definitely in doubt.

3.

Experiments Imagine an experimentalist, who does not need nonextensive statistical mechanics.

He does measurements of a certain physical quantity of a statistical-mechanical system and obtains a probability distribution of a positive random variable, x, that turns out to

5

be well fitted by the Zipf-Mandelbrot distribution

f ( x) =

A (1 + λ x ) α

(6)

over a finite range, say [a, b]. Here, α and λ are observed positive constants, and A is a normalization factor. He may naively calculate the average value of a physical quantity, Q ( x ) , using the normal definition

b

Q = ∫ d x Q ( x) f ( x).

(7)

a

There is nothing wrong here. On the other hand, the distribution in Eq. (6) can be rewritten as the q-exponential distribution with the following identifications:

α=

1 , q −1

λ = (q − 1) β .

(8)

In fact, with these identifications, Eq. (6) is expressed as follows:

f ( x ) = Ae q ( − β x ) .

(9)

Then, according to Ref. [1], one should use, instead of Eq. (7), the following expression for the q-exponential distribution:

6

Q

∫ =

b

d x Q ( x ) [ f ( x )] q

a

q

∫

b

a

d x' [ f ( x' )] q

.

(10)

The experimentalist will never understand this: he will not see why he cannot use Eq. (7) but has to use the expression

Q

∫ =

b

d x Q ( x ) [ f ( x )]1+1/ α

a

q

∫

b

a

d x' [ f ( x' )]1+1/ α

.

(11)

Furthermore, suppose he does another experiment and obtains this time the following distribution of a positive random variable, x:

−x /x

F ( x) =

Be 0 , (1 + κ x )γ

(12)

where x 0 , κ , and γ are observed positive constants, and B is a normalization factor. Eq. (12) asymptotically tends to the q-exponential distribution for large values of x but contains the exponential factor. Should he still use the q-average with q = 1 + 1 / γ following Ref. [1]? It seems that no legitimate answers to this question can be made for the q-average.

4.

Engineering Suppose that an engineer, who does not know/need nonextensive statistical

mechanics, is analyzing diffusion in a porous medium. He is trying to model it using the

7

following nonlinear Fokker-Planck equation for his physical distribution p ( x, t ) of diffusing particles:

∂ p ( x, t ) ∂ D ∂ 2 pν ( x, t ) = − [k x p ( x, t )] + , ∂t ∂x 2 ∂x2

(13)

where k is a positive constant and D is a generalized diffusion coefficient. He may find an exact analytic solution of this equation, which has the form

p ( x, t ) =

[

1 2 1 − (ν − 1) B (t ) ( x − ξ (t )) Z (t )

]

1 / (ν −1) +

,

(14)

where Z (t ) , B (t ) , and ξ (t ) are somewhat involved. He is interested especially in the case when ν < 1, because Eq. (14) is an asymptotically power-law distribution in this case. Then, he may try to evaluate the shifted variance as follows:

( x − ξ ) 2 (t ) =

∞

∫

d x ( x − ξ ) 2 p ( x, t ) .

(15)

−∞

Clearly, this variance is finite if and only if ν > 1 / 3 . Soon after this discovery, he may find that the solution in Eq. (14) has already been obtained in Ref. [5]. In this reference, the following identification is made:

ν = 2 − q.

(16)

And the solution can be written as

8

p ( x, t ) =

(

)

1 2 e q − B ( t ) ( x − ξ ( t )) . Z (t )

(17)

Then, he may look at the literature and find that, in the field of nonextensive statistical mechanics, the distribution in Eq. (17) is called the q-Gaussian distribution and maximizes the Tsallis entropy (in the unit scale)

∞  1  q S q [ p] =  ∫ d x [ p ( x, t )] − 1 1 − q −∞ 

(18)

under the constraints on the normalization condition and the shifted q-variance

(x − ξ)

∫ (t ) =

∞

−∞

2 q

d x ( x − ξ ) 2 [ p ( x, t )] q

∫

∞

−∞

d x' [ p ( x' , t )] q

,

(19)

provided that time-dependence is assumed not to be too violent. Of course, he understands that the range of ν for finite ( x − ξ ) 2

q

becomes widened

from 1 / 3 < ν < 1 to −1 < ν < 1 , if Eq. (19) is used. However, he wonders why he cannot use Eq. (15) but must calculate

∫

∞

−∞

d x ( x − ξ ) 2 [ p ( x, t )] 2 −ν

∫

∞

−∞

d x' [ p ( x' , t )] 2 −ν

from his physical distribution p ( x, t ) , because basically he is perfectly fine only with the nonlinear Fokker-Planck equation in Eq. (13) and does not need any entropic

9

approach.

5.

Quantum theory Consider an isolated quantum system. Its state is represented by ψ , which can be

expanded in terms of the complete set of eigenstates

ψ = ∑ n a n u n , where the expansion coefficients condition:

∑

n

{u } n

{a } n

n

n

of some observable as

satisfy the normalization

a n* a n = 1 . A state of the system is described by the expansion coefficients,

largely. For example, consider two states, ψ

and ψ ' . They have the following

expansions: ψ = ∑ n a n u n and ψ ' = ∑ n a ' n u n . The difference between ψ and

ψ ' is reflected in the difference between the expansion coefficients. The quantum-mechanical average of some observable, Q, in the state ψ is given by

Q = ∑ m, n a n* a m u n Q u m . This bilinear structure with respect to the expansion coefficients precisely represents the probabilistic interpretation of quantum theory. One may also consider a pure-state density matrix, ρ 0 = ψ ψ = ∑ m, n a n* a m u m u n .

(

)

Using this, one has Q = Tr Q ρ 0 . Now, a crucial question is the following. Is the q-average, Q

q

= Tr (Q ρ q ) / Tr ( ρ q ) ,

consistent with quantum theory? The point is that in the probabilistic interpretation of quantum theory the bilinear form a n* a m is essential as mentioned above. This bilinear structure has to be respected in any quantum-mechanical calculation. Let us construct a mixed state using a general

10

trace-preserving positive quantum-mechanical operation

ρ 0 = ψ ψ → ρ = Φ( ρ 0 ) = ∑ Vk ρ 0 Vk+ = ∑ a n* a m ∑ Vk u m u n Vk+ , k

m, n

provided that the trace-preserving condition,

∑

k

(20)

k

Vk+ Vk = I , is imposed. Then, the

(

)

quantum-mechanical average is still of the bilinear form: Q = Tr Q ρ 0 → Tr (Q ρ )

= ∑ a n* a m u n Φ (Q) u m . m, n

On the other hand, Q

q

is profoundly inconsistent with this quantum-mechanical

principle.

6.

Comment on the escort-distribution representation The above four stories cause a serious problem for the researchers of nonextensive

statistical mechanics, since they want to establish a statistical-mechanical bridge between the q-exponential distributions and the maximum Tsallis entropy principle. If the q-average is still to be employed, then the one and only solution to the problem may be to identify the physically measured q-exponential distribution with the escort distribution. However, the situation is very unsatisfactory. If an observed physical distribution is the escort distribution, Pn( q ) = ( p n ) q / ∑ m =1 ( p m ) q , then the value of the Tsallis entropy W

should be calculated using it. The result for this is

11

1  W (q) Sq =  ∑ Pn 1 − q  n =1

(

)

1/ q

  

−q

 − 1 . 

(21)

Unfortunately, this quantity is not an entropy any more, since it fails to be concave with respect to Pn( q ) [6].

7.

Conclusion: Farewell to the q-average formalism These considerations lead us to the conclusion that the q-average formalism suffers

from fundamental conceptual difficulties and is not acceptable. One need employ the normal definition of average and identify an observed physical distribution with the original distribution, {p n} n =1, 2, ..., W , not its associated escort distribution. This conclusion is actually supported strongly by another recent physical discussion based on the generalized H-theorem [7].

Acknowledgment This work was supported in part by a Grant-in-Aid for Scientific Research from the Japan Society for the Promotion of Science.

References [1] C. Tsallis, R. S. Mendes, and A. R. Plastino, Physica A 261, 534 (1998). [2] Nonextensive Statistical Mechanics and Its Applications, edited by S. Abe and Y. Okamoto, Lecture Notes in Physics Vol. 560 (Springer-Verlag, Heidelberg, 2001)

12

[3] E. T. Jaynes: Papers on Probability, Statistics and Statistical Physics, edited by R. D. Rosenkrantz (Kluwer, Dortrecht, 1989). [4] S. Abe, Europhys. Lett. 84, 60006 (2008); J. Stat. Mech. P07027 (2009). [5] C. Tsallis and D. J. Bukman, Phys. Rev. E 54, R2197 (1996). [6] S. Abe, Phys. Lett. A 275, 250 (2000). [7] S. Abe, Phys. Rev. E 79, 041116 (2009).

13