Comment on "Uncertainty relations in terms of the Tsallis entropy

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by  Iwo Bialynicki-Birula

Comment on ”Uncertainty relations in terms of the Tsallis entropy”
Iwo Bialynicki-Birula∗ and L ukasz Rudnicki†
Center for Theoretical Physics, Polish Academy of Sciences
Al. Lotniko´w 32/46, 02-668 Warsaw, Poland
We prove that the inequality used recently by Wilk and W lodarczyk [Phys. Rev. A 79, 062108
(2009)] to find a better lower bound in the uncertainty relations for the R´enyi entropies is invalid.
Thus, the problem of improving the bound given in our paper [Phys. Rev. A 74, 052101 (2006)]
remains unsolved.
It has been shown in our earlier paper [1] that the of the length 2a of the line segment used in Eq. (25) of
0 uncertainty relation restricting the values of the R´enyi Ref. [3] and is given by the following formula:
1 entropies for position H(x) and momentum H(p) has the
0 β α 2a2
following form:
2 ln =1 ln2 2lna. (6)
− e − −

an Hα(p)+Hβ(x) ≥−12 1lnαα + 1lnββ −ln δπx~δp ,(1) The authors have chosen a = 1, but any other choice is
J (cid:18) − − (cid:19) (cid:18) (cid:19) equally good. We would like to stress that in any case

6 where δxandδp determine the binsizes for positionand the bound shouldalwaystend to 0,whenthe sizes ofthe
2 momentum. The parameters α and β are assumed to be bins δx and δp tend to . We shall illustrate this fact
positive and they are constrained by the relation: with a simple example o∞f a Gaussian wave function:

]
h 1 1
p α + β =2. (2) ψ(x)= 1 1/e4ip0(x−x0/2)/~exp (x−x0)2 , (7)
- πσ2 − 2σ2
t (cid:18) (cid:19) (cid:20) (cid:21)
n The R´enyi entropy is defined in a standard way,
a and its Fourier transform
u 1
H = ln pα . (3)
q α 1 α k σ2 1/4 σ2(p p )2
[ Inthe limit, whenα 1−andβ(cid:16)X1,th(cid:17)is uncertaintyre- ψ˜(p)=(cid:18)π~2(cid:19) e−ix0(p−p0/2)/~exp(cid:20)− 2−~2 0 (cid:21).
1 lationreducestotheu→ncertainty→relationfortheShannon (8)
v
entropies [2]
6
Let us choose now two infinitely large bins ( ,0) and
6 −∞
2δxδp (0, ) for both position and momentum. The left hand
6 H(p)+H(x) ln . (4) ∞
4 ≥− eh side of the uncertainty relation (5) for this choice of the
(cid:18) (cid:19)
. wave function is equal to
1
The bounds in Eqs. (1) and (4) are certainly not satu-
0 rated. Evenworse,theybecomenegativeforlargevalues H(p)+H(x) =2ln2 (1 Erf(δ))ln(1 Erf(δ))
0 of δxδp/~, whereas the left hand side is always positive. − − −
1 (1+Erf(δ))ln(1+Erf(δ)), (9)
: Therefore,an improvementof these bounds is highly de- −
v
i sired. where δ = x0p0/~ and we have chosen, for simplicity,
X The bound proposed by Wilk and Wl odarczyk in the
σ = x /δ. For large values of δ, the expression on the
0
uncertainty relation (4) has the form p
r righthandsideof(9)tendsto0. Thelimitδ means
a →∞
that we localize the Gaussians (7) and (8) so far to the
2 δxδp
H(p)+H(x) ln . (5) right that its presence in the left bin is negligible.
≥− eh+δxδp
(cid:18) (cid:19) Theexistenceofacounterexamplemeansthatthereis
This lower bound has the attractive property of being an error in the proof. We shall now explain where the
always positive but it cannot be correct because in the authors made a mistake. In the originalderivationgiven
limit δxδp it does not approach 0. As a matter in [1] a crucial role is played by the inequality (Eq. (21)
of fact, if on→e f∞ollows the procedure proposed in [3], the of Ref. [1]) which was rewritten in [3] in the form
lower bound in (5) is determined only up to an additive
1/α 1/β
constant. This constant depends on an arbitrary choice
pα η(α,β) xβ , (10)
− k! ≥− l!
k l
X X
where
∗[email protected]
†[email protected] β 1/2α 2βδxδp 1−1/α
η(α,β)= , (11)
α h
(cid:18) (cid:19) (cid:18) (cid:19)
2
and the numbers p and x are the probabilities to find after the change of variables the wave functions are no
k l
themomentumandthe positioninthe k-thandl-thbin, longer related by the Fourier transformation, there is no
respectively. As was shown in [1], this inequality fol- reasonto expect that the inequality (12) holds. To drive
lows from the Babenko-Beckner inequality for the p and the point home, we give a straightforwardcounterexam-
q norms of a function and its Fourier transform. It is ple proving that the inequality (12) is indeed wrong.
important to stress that the inequality (10) holds only To this end, let us consider an arbitrary normalized
because the probabilities p and x are defined in terms even function ψ(x) localized on a small segment of the
k l
ofwavefunctions thatareconnected by Fourier transfor- real axis,
mation.
s δt
The decisive role in the derivation presented in [3] is x a= x x , (16)
played by the following inequality (Eq. (31) of Ref. [3]) | |≤ 1 δtx
where the parameters s > 0 a−nd 1 > δt > 0 have the
which is patterned after our inequality (10): x x
same meaning as in Ref. [3]. For example, we could use
1/α 1/β the following function ψ(x)
p′α η(α,β) x′β , (12)
− k k ! ≥− l l ! ψ(x)= θ(a−|x|). (17)
X X √2a
where
Its Fourier transform is
1/2α
β
η(α,β)= (2βδt δt )1−1/α. (13) ~ sin(ap/~)
α x p ψ˜(p)= . (18)
(cid:18) (cid:19) aπ p
r
However, the authors seem to have missed the crucial
This functionoccupies twobins ofsizeδt inthe t vari-
point in our derivation. They failed to notice that the x x
able. The Fourier transform (18) of this function is also
probabilitiesp′ andx′ thatentertheirinequalityarenot
k l an even function.
related by the Fourier transformation. Of course, one
Next, we choose in the inequality (12) the value of
may always change arbitrarily the integration variables
δt = 1. This means that we divided the whole range
in the correct inequality (10). However, one should also p
( 1,1)oft alsointotwosegments. Inthenotationof[3],
change the integration measure and this has not been − p
the value ofk is equalto 1. Under these assumptions
done correctly in Ref. [3]. max
the probabilities p′ and x′ are all equal to 1/2 and the
To make this Comment complete, we invoke the full k l
inequality (12) reads
definitions of p′ and x′. As the first step, the authors
k l
compactifytheinfiniterangesofxandpbythefollowing 1/2α
change of variables: 21−α 1/α β (2βδt )1−1/α 21−β 1/β,(19)
x
− ≥− α
x p (cid:18) (cid:19)
t = , t = . (14) (cid:0) (cid:1) (cid:0) (cid:1)
x x +s p p +s or equivalently
x p
| | | |
Thenewvariablesvarybetween-1and1andtheparam- δtx 1 (2α 1)22αα−−12 1/4. (20)
eters s and s are free to choose. Next, they define p′ ≥ 8α − ≥
x p k
and x′l with respect to the bins corresponding to these Choosing a sufficiently small arbitrary parameter δtx we
new variables, i.e., shall clearly violate this inequality. Thus, the inequal-
ity (12) does not hold and all conclusions based on this
(k+1)δtp s
p′ = dt ρ˜[p(t )] p , (15a) inequality in Ref. [3] are not valid.
k p p (1 t )2
Zkδtp −| p|
(l+1)δtx s
x′ = dt ρ[x(t )] x , (15b) ACKNOWLEDGMENTS
l x x (1 t )2
Zlδtx −| x|
whereρ˜(p)andρ(x)aretheoriginalprobabilitydensities This research was partly supported by the grant from
and δt and δt are the bin sizes in the new variables. the Polish Ministry of Science and Higher Education.
p x
Extra factors arise from the change of variables. Since
[1] I. Bialynicki-Birula, Phys. Rev.A 74, 052101 (2006).
[2] I. Bialynicki-Birula, Phys. Lett. 103A, 253 (1984).
[3] G. Wilk and Z. W lodarczyk, Phys. Rev. A 79, 062108
(2009).