arXiv:1211.3586v2 [hep-th] 13 Dec 2012
N-Galilean conformal algebras and quantum theory with higher order time derivatives K. Andrzejewski∗, J. Gonera, P. Kosi´ nski Department of Theoretical Physics and Computer Science, University of L´ od´z, Pomorska 149/153, 90-236 L´ od´z, Poland
Abstract It is shown that centrally extended N-Galilean conformal algebra, with N-odd, is the maximal symmetry algebra of the Schr¨ odinger equaN +1 tion corresponding to the free Lagrangian involving 2 -th order time derivatives.
It is well known that Schr¨odinger group is the maximal symmetry group of free classical motion, while its central extension is the maximal symmetry group of the Schr¨odinger equation of free particle [1]. Recently, Gomis and Kamimura [2] have showed that the free higher-derivative theory d2n ~q = 0. dt2n
(1)
defined by the Lagrangian m L= 2
dn ~q dtn
2
,
(2)
where ~q is the coordinate in d-dimensional Euclidean space and m is a ”mass” parameter of dimension kg · s2(n−1) has a symmetry described by N-Galilean ∗
e-mail:
[email protected]
1
conformal algebra (N-GCA) with N = 2n − 1 (for more information about N-GCA and its relations with higher order time derivatives, see [5]-[10] and references therein). Moreover, they showed that its quantum counterpart, that is the Schr¨odinger equation i∂t ψ = Hψ, where
ψ = ψ(t, ~q1 , . . . , ~q
N+1 2
)
(3a)
N−1
H=
2 X
1 2 ~p N+1 2m 2
p~j ~qj+1 +
j=1
(3b)
is the Ostrogradski Hamiltonian [3] of (2), exhibits centrally extended NGCA symmetry. In the previous paper [4] the authors showed that N-GCA is the maximal symmetry algebra of the Lagrangian (2). Here, we generalize Niederer’s work [1] and show that the centrally extended N-GCA is the maximal symmetry algebra of the Schr¨odinger equation (3). In order to do this let us recall that the Lie algebra of the maximal Lie group which does not change equation (3) under the change of ψ ψ(t, ~q1 , . . . , ~q
N+1 2
) → (Tg ψ)(t, ~q1 , . . . , ~q
= fg (g −1 (t, ~q1 , . . . , ~q
N+1 2
N+1 2
)=
))ψ(g −1(t, ~q1 , . . . , ~q
N+1 2
))
(4)
consist of the operators X N+1
− iX(t, ~q1 , . . . , ~q
N+1 2
)=
2 X
~aj ∂~j + a∂t + c
(5)
j=1
satisfying the following equation [P, X] = iλP,
P = i∂t − H N+1 2
for a certain function λ = λ(t, ~q1 , . . . , ~q notation ~qj = (qαj ), ~aj = (ajα ),
(6)
). Let us introduce the following
∂~j = (∂jα ),
etc.
(7)
where α = 1, . . . , d and (if the contrary is not stated explicitly) repeated indices i, j, k etc. (α, β, γ etc.) denote summation from 1 to N 2−1 (from 1 to d, respectively). 2
In the case of the Hamiltonian (3b) condition (6) implies the following set of equations for coefficients of the operator X λ = ∂t a + qαk+1 ∂kα a N+1 2
δβα = ∂ αN+1 aβ 2
(8a) N+1 2
+ ∂ βN+1 aα
(8b)
2
1 α (∂ N+1 ∂ αN+1 )c 2 2 2m N+1 N+1 1 1 (∂ βN+1 ∂ βN+1 )aα 2 + ∂ αN+1 c + iqβk+1 ∂kβ aα 2 + 2m m 2 2 2
0 = i∂t c + iqαk+1 ∂kα c + N+1
0 = i∂t aα 2 0 = ∂ αN+1 a
(8c) (8d) (8e)
2
0 = ∂ αN+1 ajβ ,
j = 1, . . . ,
2
N −1 2
λqβj+1 = ∂t ajβ + qαk+1 ∂kα ajβ − aj+1 β ,
(8f)
j = 1, . . . ,
N −1 2
(8g)
Our main task is to show that the general solution of the above set of equations gives centrally extended N-GCA. In order to simplify our considerations we assume that N > 3. The case N = 3 is simpler and can be obtained in the same way. First, let us note that (8a) and (8e) imply ∂ βN+1 ∂ αN+1 λ = 0
(9)
2
2
On the other hand differentiating (8g) (for j =
N −1 ) 2
with respect to ∂ αN+1 2
(without summation) and using (8f) and (8b) we obtain N−1 N+1 3 ∂ αN−1 aα 2 = λ + qα 2 ∂ αN+1 λ 2 2 2
(10)
Next, we differentiate the above equation with respect to ∂ γN+1 and use (9) 2
together with (8f) to obtain ∂ γN+1 λ = 0
(11)
2
Differentiating eq. (8b) with respect to ∂ γN+1 and combining equations ob2 tained by cyclic permutations of α, β, γ we arrive at N+1
∂ γN+1 ∂ αN+1 aβ 2 = 0 2
2
3
(12)
and consequently N+1
aβ 2 =
N+1 λ N+1 qβ 2 + Uβα qα 2 + dβ 2
(13)
N+1
where U = −U T and dβ do not depend on qα 2 . Next, we will show inductively that for each j = 1, . . . , N 2+1 we have N +2 β j ∂j aα = − j λδαβ + Uαβ , (14a) 2 ∂jβ a = 0,
and if j > 1 then ∂jβ akα = 0 for k = 1, . . . , j − 1
(14b)
Indeed, due to (8e),(8f) and (13) for j = N 2+1 the above assertion holds. Assume that it is true for fixed j. Then, by virtue of (8a) and the induction hypothesis, for index j we have ∂jα ∂jβ λ = 0
(15)
On the other hand, differentiating (8g) for j − 1 with respect to ∂jγ and using the induction hypothesis we arrive at γ ∂j−1 aαj−1 = (
N +4 − j)λδαγ + Uαγ + qαj ∂jγ λ 2
(16)
Putting γ = α in (16) and differentiating with respect to ∂jβ , one gets by (15) and (14b) ∂jβ λ = 0 (17) As a result eq. (16) takes the form of (14a) for j − 1 and, due to (8a), β ∂j−1 a = 0. Moreover, differentiating (8g) for k = 1, . . . , j − 2 with respect β to ∂jβ , by the induction hypothesis and (17) one obtains ∂j−1 akα = 0 for k = 1, . . . , j − 2 which completes the proof of (14). Additionally, by virtue of (14b) and (8a) we conclude that a and λ are functions of t only. Differentiating eq. (14a) with respect to ∂jβ (without summation) and combining with equations obtained by cyclic permutations of α, β, γ we get Uβα is also only function of t. Now, differentiating (8g) for j = 1, . . . , N 2−1 with respect to ∂jγ (without summation) and using (14a) we obtain a recurrence formula for ∂jγ aαj+1 γ ∂t ∂jγ ajα + ∂j−1 ajα = ∂jγ aj+1 α | {z } 0 for j=1
4
(18)
which, due to (14a), can be explicitly solved and the final result reads ∂jγ aαj+1 = (−j + N + 1)
j λ˙ γ δ + j U˙ αγ , 2 α
j = 1, . . . ,
N −1 2
(19)
γ Similarly, differentiating (8g) with respect to ∂j−1 we obtain a recurrence γ N −1 j+1 formula for ∂j−1 aα , j = 2, . . . , 2 which solution is of the form .. γ ∂j−1 aj+1 α
jλ (j − 1)j .. γ = (−2j + 3j(N + 2) − 3N − 4) δαγ + U α, 12 2 2
(20)
Let us now study the behaviour of c. Differentiating (8d) with respect to ∂ δN+1 ∂ γN+1 and using (19) together with (14a) we find that the third order 2
2 N+1
derivative of c with respect to qα 2 is zero, so N+1
N+1
N+1
2 c = cαβ qβ 2 + cα2 qα 2 + c3 1 qα
(21)
N+1
βα α 3 2 . Substituting, (21) into (8c) where cαβ 1 = c1 , c2 , c do not depend on qα N+1 2 and comparing terms with q ’s one gets the following set of equations (γ
0 = ∂ N−1 c1αβ)
(22a)
2
(β
α)
αβ 0 = qγk +1 ∂kγ′ cαβ 1 + ∂ N−1 c2 + ∂t c1
(22b)
0 = qγk +1 ∂kγ′ cα2 + ∂ αN−1 c3 + ∂t cα2
(22c)
′
2
′
2
0 = qγk +1 ∂kγ′ c3 + i∂t c3 + ′
1 αα c m 1
(22d)
where index with subscript ” ’ ” runs from 1, . . . , N 2−3 and (α, . . .) is symmetrization over the enclosed indices. On the other hand, differentiating (8d) with respect to ∂ γN+1 and next substituting (13) for j = N 2+1 and (19) for j=
N −1 2
2
one gets cαβ 1 = −
mi ˙ γα (N + 1)2 λδ 16 U˙ αβ = 0
(23a) (23b)
Consequently c1 ’s are functions of t only, using this fact, (19),(8d) and (20) for j = N 2−1 we obtain ∂ γN−1 c2α = − 2
im .. γ 2 λδ (N − 1)(2N + 3) 48 α 5
(24)
Substituting (24) and (23) into (22b), one finds that ..
λ=0
(25)
Thus, by virtue of (20), we have γ ∂j−1 aj+1 = 0, α
j = 2, . . . ,
N −1 2
(26)
A simple consequence of (26) is ∂1γ ajα = 0,
j = 3, . . . ,
N +1 2
(27)
Indeed, differentiating (8g) with respect to ∂1γ we have (∂t + qβk+1∂kβ )(∂1γ ajα ) = ∂1γ aj+1 α . This together with eq. (26) for j = 2 imply (27). Now, we show that for each k = 2, . . . , N 2−1 we have γ ∂j−k ajα = 0,
j = k + 1, . . . ,
N +1 2
(28)
Indeed, for k = 2 eq. (28) holds due to (26). Assume now (28) is true for k − 1; we will show that it holds for k. Differentiating (8g) with respect to γ ∂j−k+1 , by the induction hypothesis, one obtains γ γ ∂j−k ajα = ∂j−k+1 aj+1 α ,
j = k + 1, . . . ,
N +1 2
(29)
but for j = k + 1, due to (27), we have ∂1γ ak+1 = 0 which proves (28). α N+1
Let us note that (28) implies ∂kγ aα 2 = 0 for k = 1, . . . , N 2−3 ; therefore N+1
aα 2 =
N+1 λ N+1 (N + 3)(N − 1) N−1 qα 2 + λ˙ qα 2 + Uαβ qβ 2 + fα (t) 2 8
(30)
Substituting this in (8d) we find that c2α is a function of t only, more precisely we have c2α = −imf˙α (31) Furthermore, the partial derivatives of c3 are expressed in terms of c2 : ∂kα c3 = (−1)
N−2k+1 2
) 2 ( N−2k+1 2 cα ,
∂t
6
k = 1, . . . ,
N −1 2
(32)
this can be proved inductively differentiating eq. (22d) with respect to ∂ αN−1 −l 2
for l = 0, . . . , N 2−3 and using eqs. (22c), (31). Differentiating (22d) with respect to ∂1γ and using eq. (32) one gets ( N+1 ) 2 2 cα
∂t
=0
(33)
Thus, by (31), we have N−1
N−1
c2α =
2 X
Akα tk ,
k=0
2 i X tk+1 +D fα = Akα m k=0 k+1
(34)
where Akα and D are some constants. Summarizing, due to (22d) and (23a), (32) we have N−2k′ −1 N−2k′ +1 (N + 1)2 ) 2 k ′ +1 ( 2 cα + C c3 = dλ − (−1) qα ∂0 2 16
(35)
where C is a constant. Now, it remains to find the dependence ajα of t. By virtue of (8g) and equations (14),(19) and (26) we have aj+1 = ∂t ajα − fαj , α
j = 1, . . . ,
N −1 2
(36)
˙
where fαj = λ(j − N2 )qαj+1 + λ2 (j − N − 2)(j − 1)qαj − Uαβ qβj+1 and by definition ..
fαj = 0 for j < 1. Since λ = 0, the explicit solution of (36) is of the form (j−1) 1 aα
ajα = ∂t
− ∂t fαj−2 − fαj−1,
j = 2, . . . ,
N +1 2
(37) ( N−1 ) 1 2 aα
Comparing eq. (37) for j = N 2+1 with eq. (30) we obtain that fα = ∂t and consequently, by virtue of (14) and (25) N−3
a1α
=
2 X
k=0
Bαk tk
N + λqα1 + Uαβ qβ1 + 2
Z
Z
. . . fα dt | {z } N−1 2
7
(38)
where Gkα are some constants. Coming back to (37) we arrive at N−1 −j 2
ajα =
X l=0
N−1
N−1
−j N+3 2 2 X i X k!tk−j+ 2 j+l−1 (j + l − 1)! l k t + + Bα A C l tl l! m k=0 α (k − j + N 2+3 )! l=0
λ˙ λ + (j − 1)(N − j + 2)qαj−1 + (N − 2j + 2)qαj + Uαβ qβj 2 2 −j
N+1 −j 2
t + D N +1 2
(39) Shifting B’s by C’s or D, after some indices manipulations we find that ajα , for j = 1, . . . , n take the form N−1
ajα
=
2 X
l=j−1
N−1
Bαl
N+3 2 i X l! k!tk−j+ 2 l−j+1 k t + A (l − j + 1) m k=0 α (k − j + N 2+3 )!
(40)
λ λ˙ + (j − 1)(N − j + 2)qαj−1 + (N − 2j + 2)qαj + Uαβ qβj 2 2 Due to eq.(25) λ = 2Et + F thus (8a) yields a = Et2 + F t + G. Summarizing, we see that all a’s and c depend on some constants C, E, F, G, A’s, B’s and U’s. Thus the maximal symmetry algebra of (3) is finitedimensional and its basis is obtained by selecting the coefficient related to these constants. After troublesome indices manipulations, we find all generators: iG H = −i∂t (41a) iF
N+1 2 X 1 N D = −it∂t − i ( − j + 1)qαj ∂jα − i (N + 1)2 d 2 16 j=1
(41b)
iE N+1 2 X 1 K = −it2 ∂t − i (N + 1)2 dt − i (j − 1)(N − j + 2)qαj−1∂j 8 i=1
2
N+1 (N + 1) N+1 qα 2 qα 2 −it(N − 2j + 2)qαj ∂jα − m 8
8
(41c)
iBαl , for l = 0, . . . , N 2−1 Plα
= −il!(
l X k=0
−
(j− N+1 )! α 2 Aj− N+1 , mj! 2
N−1
Pjα
= −ij!
2 X
l=0
for j =
tl−k α ∂ ) (l − k)! k+1
N +1 ,...,N 2
j X N+1 tj−l α tj−k ∂l+1 − mj! , qαN −k+1(−1) 2 +k (j − l)! (j − k)! N+1
i β U 2 α
C m
(41d)
k=
(41e)
2
Jβα = −i(qβj ∂jα − qαj ∂jβ )
(41f)
Z=m
(41g)
The obtained generators agree with the ones from [2] and do satisfy N-GCA commutation rules with central charge Z = m. Thus centrally extended NGCA is in fact the maximal symmetry algebra of the Sch¨odinger equation corresponding to the free theory with higher order time derivatives. Acknowledgments The discussions with Cezary Gonera and Pawel Ma´slanka are gratefully acknowledged. This work is supported in part by MNiSzW grant No. N202331139.
References [1] U.Niederer, Helv. Phys. Acta. 45 (1973), 802; [2] J. Gomis, K. Kamimura, Phys. Rev. D85 (2012), 045023; [3] M. Ostrogradski, Mem. Acad. St. Petersburg 4 (1850), 385; [4] K. Andrzejewski, J. Gonera, preprint, arXiv:1209.5884 (2012) [5] J. Negro, M.A. del Olmo, A. Rodriguez-Marco, J. Math. Phys. 38 (1997), 3786, 3810; [6] J. Lukierski, P.C. Stichel, W.J. Zakrzewski, Phys. Lett. A357 (2006), 3810; 9
[7] A.V. Galajinsky, I. Masterov, Phys. Lett. B702 (2011), 265; [8] C. Duval, P.A. Horvathy, J. Phys. A44 (2011), 335203; [9] S. Fedoruk, E. Ivanov, J. Lukierski, Phys. Rev. D83 (2011), 085013; [10] K. Andrzejewski, J. Gonera, P. Ma´ slanka, Phys. Rev. D86 (2012), 065009; [11] A.V. Galajinsky, I. Masterov, to appear in Nucl. Phys. B866 (2013), 212;
10
