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Minimalisation of uncertainty relations in noncommutative quantum mechanics

Minimalisation of uncertainty relations in noncommutative quantum mechanics
Minimalisation of uncertainty relations in noncommutative quantum mechanics

a r X i v :h e p -t h /0208213v 1 29 A u g 2002

Minimalization of uncertainty relations in noncommutative quantum mechanics

Katarzyna Bolonek Piotr Kosi′n ski ?

Department of Theoretical Physics II

University of L ′o d′z Pomorska 149/153,90-236 L ′o d′z

Poland

February 7,2008

Abstract

The explicit construction of states saturating uncertainty relations following from basic commutation rules of NCQM is given both in Fock space and coordinate representation.

1Introduction

There are strong indications coming from the study of brane con?gurations in string theory or matrix model of M-theory that noncommutative spaces are of some importance for very high energy physics[1].As a result,there appeared a large number of papers devoted to the study of?eld theories on such spaces[2].In order to reveal the important aspects of quantum theory on noncommutative spaces one should tend to simplify the systems under consideration as much as possible.By considering the low-energy limit of one-particle sector of?eld theory on noncommutative space one arrives at what is called noncommutative quantum mechanics.Again various aspects of it have been studied recently[3]-[23].In particular,in[23]we considered single-particle quantum mechanics on noncommutative plane de?ned by the following commutation rules

[?x i,?x j]=iθ?ij I(1a)

[?x i,?p j]=i δij I i,j=1,2(1b)

[?p i,?p j]=0;(1c) here we can assumeθ>0without loosing generality.

By standard arguments,eqs.(1)result in the following uncertainty rela-tions

?x1?x2≥θ

(2b)

2

?x2?p2≥

inequalities (2).

Finally,some

basic facts concerning the standard coherent states are collected in Appendix.

2Representations of the basic algebra

It is not di?cult to ?nd irreducible representation of the algebra (1).In fact,this algebra is equivalent to standard Heisenberg-Weyl algebra:

?x i ≡?x i +

θ

√2

?ij )?p j )

a ?i ≡

12

(?x i +(?iδij +θ√√

2

i

(7)

It is often convenient to work with the modi?ed creation/anihilation opera-tors carrying de?nite angular momentum.To this end we de?ne

a ±≡

1

2(a 1?ia 2)a ?±≡12

(a ?

1±ia ?2);(8)

The new basis is

|n +,n ? =

1n +!

1n ?!

(a ?+)n +(a ??)n ?

|0 (9)

In terms of new variables the angular momentum operator reads

?L =?i ?ij a ?i

a j = (a ?+a +?a ??a ?)(10)

The angular momentum of the state (9)equals (n +?n ?).

3Saturating uncertainty relations

Let us ?rst ?nd all vectors saturating the uncertainty relation (2a).The

relevant commutation rule (1a)resembles the one concerning ?x 1and ?p 1,with ?p 1replaced by ?x 2and replaced by θ.Therefore,it is not surprising that we can use the same strategy as described in Appendix once the appropriate creation/anihilation operators are found.To this end we de?ne

b ≡ 2θ 1+θ

2 a ?+

b ?≡

2θ 1+θ2

a +

c ≡ 2θ 1+θ2 a ??

c ?≡ 2θ 1+θ

2 a ?

(11)One easily veri?es that b ,c ,b ?,c ?form the set of independent creation/ani-hilation operators.

The key point is that b –operators are related to ?x –operators in the stan-dard way

b ≡1

2θ(?x 1+i ?x 2)b ?

12θ

(?x 1?i ?x 2)(12)

Therefore,we can repeat the procedure outlined in Appendix to ?nd the states saturating (2a).They read

|z,γ φ=e ?

1

4

ln γ((b ?)2?b 2)e

zb ?

|φ (13)

where |φ is arbitrary state such that

b |φ =0

(14)

The ”vacuum”state is by far not unique;it may contain an arbitrary number of c –excitations.

The representation given by b ,b ?,c ,c ?is unitary equivalent to that de?ned by a ±,a ?±.In fact,one can check that

b =W a ?W ?,b ?=W a ??W

?

c =W a +W ?

,c ?

=

W a ?+W

?

(15)

where

W =e

1

θ

)(a +a ??a ?+a ?

?)

(16)

This can be seen by using the results of [23].However,we prefer to give a

straightforward proof.De?ne for any t ∈R

W (t )=e t (a +a ??a ?

+a ?

?)

(17)

and

b (t )≡W (t )a ?W ?(t )

c ?(t )≡W (t )a ?+W ?

(t )

(18)

Then b (0)=a ?,c ?(0)=a ?+while simple computation gives

˙b

(t )˙c ?(t ) =

0110 b (t )c ?(t ) (19)

Therefore

b (t )

c ?(t )

= exp

t t

a ?

a ?+ ==

cosh t sinh t

sinh t

cosh t a ?a ?+

(20)

For t =1

θ

we arrive at (15).Eqs.(15),together with the results presented in Appendix allow us to conclude that the states saturating (2a)are linear combinations (with respect to n +but with z ,γ?xed)of the states

|z,γ,n+ =e?14lnγ(a2??(a??)2)e za??|n+,0 (21) Let us note that W commutes with?L.This implies that the states z=0,γ=1are eigenstates of?L.This conclusion is rather obvious:real and imaginary parts of z are related to expectation values of?x1,?x2(which should be zero from rotational invariance)while expectation values of?x21,resp.?x22 are proportional toγ,resp.1

2

(22) We follow the same strategy.First,de?ne new creation/anihilation operators

d=a1+

4

(a2?a?2)

e=a2+

4 (a1?a?1)

(23)

which obey

d=1

2

(?x1+i?p1)(24)

Unitary equivalence of old and new operators,

d=T a1T?

e=T a2T?

(25) is obtained by choosing T in the form(cf.[23])

T=e iθ

2|z|2T e?1

2|z|2T?e?1

4Coordinate representation

For the variables?x i,?p i we use standard representation

?x i=x i

?p i=?i

?

2?ij

?

?x i

(30) The stateψsaturating(2a)obeys

(?x1?α)ψ=?iγ(?x2?β)ψ(31) which,due to eqs.(30),takes the form

θ

?x1+i

?

√γx2

e?1γ+γx22 ?z x1γ?i√

√γβ;f is an arbitrary function such thatψis normalizable. In particular,the eigenstate of?L corresponding to the eigenvalue m reads

ψ(x1,x2)=

2m+1

√m!θm+1θ(34)

One can check explicitly that ?x21 = ?x22 =θ

2 ?x21+?x22 =1

4 2

(?p21+?p22)?

θ

the right-hand side is the combination of harmonic oscillator and angular momentum.Standard reasoning gives for the spectra

?x 21

+?x 22

+θ2

?L

:θ(2n ?+1)?L

: (n +?n ?)(36)

The states saturating (2a)

correspond

to

n

?

=0;but n +?n ?=m ,i.e.

m =n +≥0.

Let us look for the states saturating (2a).The relevant equation

?x 1ψ1=?iγ1?p 1ψ1

(37)

reads

iθ?x 2

+γ1

?

θ

x 2)e

?

3x 21

2θ2

+

ix 1x 2

θ

x 1)e

?

3x 22

2θ2

?

ix 1x 2

θx 1

θ

x 1

=

5

θ2

x 1?i

x 2

θ

?

θ

x 1so the right-hand

side must also;this is,however,imposible as one can immedietely check.

One can also ask whether (39)((40))can be an eigenstate of ?L

provided an appropriate choice of f 1(f 2)has been made.Again we check that this is impossible inserting (39)into the eigenequation

?Lψ

= mψ(42)

Let us ?nally insert eq.(33)

into eq.(37).The resulting equation for the function f reads

f ′ x 1γ+i √f x 1

γ

+i √γθ x 1?iγx 2+z √2+γ1 γ γ√√

γx 2)implies

γ1

θ

x 1

γ

+i √γ

γx 2

(45)

Inserting this back to (33)we conclude that ψis nonnormalizable.This

shows that also (2a)and (2b)cannot be simultaneously saturated.

We veri?ed explicitly that,for a given state ψ,at most one of the in-equalities (2a)–(2c)can be saturated;this con?rms the general theorems of [23].

Although there are no states saturating both (2b)and (2c),both lower bounds can be simultaneously approached as close as one wishes.To see this we select the state

ψ=

π

e ?δ(x 2

1+x 2

2)

(46)

Then ?Lψ=0, ?p 1 =0, ?x 1 =0,and

?p 21 ψ=δ 2

,

?x 21 =1

4

;(47)

consequently

(?x 1)2ψ(?p 1)2

ψ

= 2

4

(48)

By symmetry

(?x 2)2ψ(?p 2)2ψ

= 2

4

.(49)

(46)is normalizable for any δ>0.The bounds are saturated for δ→0;

however,the state (46)becomes nonnormalizable in the limit δ→0.

Appendix:Uncertainty principles and coher-ent states

First let us remind the general setting for uncertainty principles[24](for recent alternative approach see[25]).Given two observables?A,?B subject to commutation rule:

[?A,?B]=i?C,(50)

one can derive the following inequality(generalized Heisenberg principle)

(?A)ψ·(?B)ψ≥1

ψ|(?A? ?A ψI)2|ψ ,etc.;(52) (51)is saturated i?the following condition holds

(?A? ?A ψI)|ψ =?iγ(?B? B ψI)|ψ ,γ∈R(53) Acting with?A? ?A ψI on both sides of(53),using(50)and again(53)one arrives at

(?A? ?A ψI)2|ψ =?γ2(?B? ?B ψI)2|ψ +γ?C|ψ (54) or,on multiplying by|ψ from the left

(?A)2ψ+γ2(?B)2ψ=γ C ψ.(55)

(55),together with the saturated form of(51)gives(providedγ=0)

(?A)2ψ=γ

2γ C ψ(56)

which explains the meaning ofγ.

Let us apply this scheme to the standard Heisenberg relation

[?x,?p]=i (57) The relevant inequality reads

?x·?p≥

(58)is saturated i?

(?x?α)|ψ =?iγ(?p?β)|ψ ,α= ?x ψ,β= ?p ψ(59) Let us de?ne creation/anihilation operators(we work withω=1,m=1 units)

a≡1

2

(?x+i?p)

a?≡12 (?x?i?p)

[a,a?]=1.

(60) Hilbert space of states is spanned by the vectors

|n =1n!(a?)n|0 (61)

To?nd the general solution to(59)?rst note thatγ>0.In fact,γ=0 because?x?αI cannot have normalized eigenvectors(operators commuting to C–number have no normalized eigenvectors in their common invariant domain);forγ=0(56)givesγ>0.We start withγ=1.Eq.(59)can be rewritten as

a|ψ =z|ψ ,z=α+iβ

2

(62)

The eigenstates of the anihilation operators are called coherent states(cs). Vacuum state is the coherent state corresponding to z=0.In order to?nd other cs one de?nes,for any z∈C,the unitary operators

U(z)≡e za??ˉz a=e?1

2|z|2e za?|0 =e?1√

where

z =

12 αγ+iβ√√√

γ?

p a ?γ=

1

2

?x γ

?i √4

ln γ(a 2?(a ?)2)

,(68)

the following relations are obeyed

V (γ)aV ?(γ)=a γV (γ)a ?V ?(γ)=a ?γ

(69)

The solution to eq.

(59)can be now written as

|z,γ =V (γ)U (z )|0 ;

(70)the complex parameter z is related to the mean values of x and p while γ

describes their dispersions:

(?x )2=

γ

(71)

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