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PASSING THE BOUNDARY BETWEEN THE PARITY BREAKING MEDIUM AND VACUUM BY VECTOR PARTICLES

Sergey Kolevatov
Saint Petersburg state university

June, 28 2013

in collaboration with A.A. Andrianov
S. Kolevatov (SPbSU) 28.06.13 1 / 19

Intro duction. The statement of the problem. Two vacua. Bogolub ov transformation. Classical solutions. Spatial CS vector (axion stars). Time-like CS vector (collision of heavy ions). Conclusion.

S. Kolevatov (SPbSU)

28.06.13

2 / 19

Intro duction

hrk inergy

S. Kolevatov (SPbSU)

28.06.13

3 / 19

Intro duction

hrk inergy

Fermi-LAT, PAMELA, AMS-2: anomalous excess of e +e -.

S. Kolevatov (SPbSU)

28.06.13

3 / 19

Intro duction

hrk inergy

Fermi-LAT, PAMELA, AMS-2: anomalous excess of e +e -. arXiv:1304.1483 [hep-ph]

S. Kolevatov (SPbSU)

28.06.13

3 / 19

Intro duction

hrk inergy

Fermi-LAT, PAMELA, AMS-2: anomalous excess of e +e -. arXiv:1304.1483 [hep-ph]

revy ion physis

NA60, PHENIX: abnormal yield of lepton pairs (e , Е)
S. Kolevatov (SPbSU) 28.06.13 3 / 19

The statement of the problem

L=- +
1 2

F (x )F (x ) - 1 F 4 m2 A (x )A (x ) + AЕ (x
1 4

Е

(x )
Е

)

FЕ (x ) ac (x ) B (x ) + 1 B 2 (x 2

),

S. Kolevatov (SPbSU)

28.06.13

4 / 19

The statement of the problem

L=- +
1 2

F (x )F (x ) - 1 F 4 m2 A (x )A (x ) + AЕ (x
1 4

Е

(x )
Е

)

FЕ (x ) ac (x ) B (x ) + 1 B 2 (x 2

),

We consider a slowly varying classical pseudoscalar background of the kind,

ac (x A (x
)+

)=

x

(- З x ) )

m2 A (x ) = A (x A (x ) + m2 A (x ) = 0

for З x < 0; for З x > 0.

S. Kolevatov (SPbSU)

28.06.13

4 / 19

Construction of the chiral p olarization vectors

S

= D+

k k

2 +
Е

k

2

- З k (

k

+

k

);

D ( З k )2 -
Е (0) Б

2

k

2

Transversal polarizations are,
2D 2 Scalar and longitudinal polarizations,

Е Б

S

Б

i

Е

k

D

-1 2

;

Е (k ) = Б

Е (k ) S

k

Е

k

2

,

Е (k ) L

D

k

1 2 -2

k

2

Е -

k

Е

З

k

S. Kolevatov (SPbSU)

28.06.13

5 / 19

Construction of the chiral p olarization vectors

S

= D+

k k

2 +
Е

k

2

- З k (

k

+

k

);

D ( З k )2 -
Е (0) Б

2

k

2

Transversal polarizations are,
2D 2 Scalar and longitudinal polarizations,

Е Б

S

Б

i

Е

k

D

-1 2

;

Е (k ) = Б

Е (k ) S

k

Е

k

2

,

Е (k ) L

D

k

1 2 -2

k

2

Е -

k

Е

З

k

ptil g vetorF Е = (H, -x , H, H)

S. Kolevatov (SPbSU)

28.06.13

5 / 19

Construction of the chiral p olarization vectors

S

= D+

k k

2 +
Е

k

2

- З k (

k

+

k

);

D ( З k )2 -
Е (0) Б

2

k

2

Transversal polarizations are,
2D 2 Scalar and longitudinal polarizations,

Е Б

S

Б

i

Е

k

D

-1 2

;

Е (k ) = Б

Е (k ) S

k

Е

k

2

,

Е (k ) L

D

k

1 2 -2

k

2

Е -

k

Е

З

k

ptil g vetorF Е = (H, -x , H, H) X dispersion lws

k1 k1Б
=

L=

k10

=

2 - -

m

2

-

2 k 2 k
28.06.13 5 / 19

2 -

m

2

k

2

x

2 -

S. Kolevatov (SPbSU)

^ k

= ( ,

^ k2 , k3 ), x

= (x0 ,

^^ x2 , x3 ) : k З x

= -

x0 + k2 x2 + k3 x3 .

Є roEtukelerg solution

APS (x
Е

)=

^ d k ( 2 -
3

k - m
2

2

3

)

u

^ k,

r (x ) = [ ( 2 ) 2

k10

]

-1/2

e

r =1 ^ ) exp{ r (k

Е Е a k , r u k , r (x ) + a^ , r u k ,r (x ) ^ ^ ^ k

^^ i k10 x1 + i k З x

}

( r = 1, 2, 3 )

ghernEimons solution

A (x CS v
^ k

)=

^ dk

A

=Б,

L

2 (k1A ( ,
1 -2

k

))

c

^ ^ ^ ^ k ,A v k A (x ) + c k ,A v k A (x )

3 A (x ) = (2 ) 2k1A

(k ) exp{ A

^^ ik1A x1 + i k З x =0

} (A = L, Б)

AЕ (x PS
S. Kolevatov (SPbSU)

) - AЕ (x ) | Зx CS

=0

28.06.13

6 / 19

Bogolub ov Transformations
3

v

^ k ,A

(x ) = ^

relations between the creation-destruction operators are,
a k, r = ^
A
=Б,L

s

=1

^^ ^^ sA (k ) u k ,s (x ) - sA (k ) u k , (x ) ^ ^s ^^ ^ rA (k ) c k ,A - (k ) c k ,A rA ^
^ ^ Ar (k ) a k , r + Ar (k ) a , r ^ ^ k

c

3
^ k ,A =

r

=1

There are two dierent Fock vacua,
a k , r |0 = 0 ^

c

^ k ,A | = 0

S. Kolevatov (SPbSU)

28.06.13

7 / 19

Bogolub ov Transformations
3

v

^ k ,A

(x ) = ^

relations between the creation-destruction operators are,
a k, r = ^
A
=Б,L

s

=1

^^ ^^ sA (k ) u k ,s (x ) - sA (k ) u k , (x ) ^ ^s ^^ ^ rA (k ) c k ,A - (k ) c k ,A rA ^
^ ^ Ar (k ) a k , r + Ar (k ) a , r ^ ^ k

c

3
^ k ,A =

r

=1

There are two dierent Fock vacua,
a k , r |0 = 0 ^

c

^ k ,A | = 0

^^ ^ 0 | a p,s c k ,A | 0 = (k - p ) As (k ) ^ ^ The latter quantity can be interpreted as the relative probability amplitude that particle is transmitted from the left face to the right face.
S. Kolevatov (SPbSU) 28.06.13 7 / 19

Vacuum as a coherent state

|0 k = ^

p ,m , l

=0

fpml

(c

^ k

,+

)p (

c k ,- )m (c ^ p !m !l !

^ k ,L

)l | k ^

To nd fpml we use the equality a k , r |0 k = 0. ^ ^

S. Kolevatov (SPbSU)

28.06.13

8 / 19

Vacuum as a coherent state

|0 k = ^

p ,m , l

=0

fpml

(c

^ k

,+

)p (

c k ,- )m (c ^ p !m !l !

^ k ,L

)l | k ^

To nd fpml we use the equality a k , r |0 k = 0. ^ ^
|0 k = exp ^
^ ^ ^ r + (k ) 2 (k ) 2 (k ) 2 (c ^ ) + r - (c ^ ) + rL (c ^ ) | k ^ ^ ^ ^ 2r + (k ) k ,+ 2r - (k ) k ,- 2rL (k ) k ,L ^ ^ ^ -A1 (k ) 2 -A2 (k ) 2 -A3 (k ) 2 (a ^ ) + (a ^ ) + (a ^ ) |0 k ^ ^ ^ ^ 2A1 (k ) k , 1 2A2 (k ) k , 2 2A3 (k ) k , 3

| k = exp ^

S. Kolevatov (SPbSU)

28.06.13

8 / 19

Vacuum as a coherent state

In the correct normalization, 0|0 = 1, | = 1. ^ Going to the continuum limit for k ,
^ rA (k ) 2 2 ^ ^ |0 = exp (c ^ ) (k1A (k )) d k | ^ k ,A A=Б,L 2rA (k ) (k ) ^ -Ar 2 ^ (a )2 d k |0 | = exp ( 2 - m2 - k ) ^ ^ 2Ar (k ) k , r r =1,2,3

S. Kolevatov (SPbSU)

28.06.13

9 / 19

Classical solutions

Е = (H, - , H, H)

A

+

m2 A

+ 1

(-x1 )

A

=0

S. Kolevatov (SPbSU)

28.06.13

10 / 19

Classical solutions

Е = (H, - , H, H)

A

+

m2 A

+ 1

(-x1 )

A

=0

A1 may be found in the whole space, ^ dk ~ A1 = (u1 ( , k2 , k3 )e ik10 x 3
(2 )

1

~ + u1 ( ,

k2 , k3 )e

-ik10 x1

^^ )e i k x

S. Kolevatov (SPbSU)

28.06.13

10 / 19

Classical solutions

Е = (H, - , H, H)

A

+

m2 A

+ 1

(-x1 )

A

=0

A1 may be found in the whole space, ^ dk ~ A1 = (u1 ( , k2 , k3 )e ik10 x 3
Solution for A ( = 0, 2, 3)
(2 )

1

~ + u1 ( ,

k2 , k3 )e

-ik10 x1

^^ )e i k x

A
~ A = ~ u

=

(2 )3

^ dk

~ A e

^^ ikx -ik10 x1

( ,

k2 , k3 )e

ik10 x1 + u ( , ~

k2 , k3 )e

,

x1

>0 ,

~ A v A ( ,

k2 , k3 )e

ik1A x1 + v ~ A ( ,

k2 , k3 )e

-ik1A x1

x1

<0
10 / 19

S. Kolevatov (SPbSU)

28.06.13

After the integration of the eld equations over x1 from - to ,
- -
2 k10
~ ~ (- u0 -u0 ik10
2

+ + +

A A

v0A -v0A ) = i (k ~ ~ 2 ik1A

A

v3A -v3A - ~ ~ ik1A

k3

A A A

v2A -v2A ) ~ ~ ik1A v3A -v3A ) ~ ~ ik1A

2 k10

~ ~ ( - u 2 -u ik10

v2A -v2A ) = -i (k ~ ~ 3 ik1A v3A -v3A ) = i ( ~ ~ ik1A A

A

v0A -v0A + ~ ~ ik1A

-

k

2 10

~ ~ (- u3 -u ik10

3

A

v2A -v2A + ~ ~ ik1A

k2

v0A -v0A ) ~ ~ ik1A

S. Kolevatov (SPbSU)

28.06.13

11 / 19

After the integration of the eld equations over x1 from - to ,
- -
2 k10
~ ~ (- u0 -u0 ik10
2

+ + +

A A

v0A -v0A ) = i (k ~ ~ 2 ik1A

A

v3A -v3A - ~ ~ ik1A

k3

A A A

v2A -v2A ) ~ ~ ik1A v3A -v3A ) ~ ~ ik1A

2 k10

~ ~ ( - u 2 -u ik10

v2A -v2A ) = -i (k ~ ~ 3 ik1A v3A -v3A ) = i ( ~ ~ ik1A A

A

v0A -v0A + ~ ~ ik1A

-

k

2 10

~ ~ (- u3 -u ik10

3

A

v2A -v2A + ~ ~ ik1A

k2

v0A -v0A ) ~ ~ ik1A

Moreover, all contributions from dierent polarizations to A are continuous,
~( ~( uA) + uA) ~ = v A

~ + v A

(A = Б, L).

wthing onditions
~( uA) ~( uA) ~ = (v A (

~ = (-v A (

1 2

1 2

k1A + k10 k10 k1A - k10 k10

~ ) - v A ( ~ ) + v A (

k1 k1

A- A+

k10 k10

k10 k10

)) ))
28.06.13 11 / 19

S. Kolevatov (SPbSU)

Escaping from the parity-breaking medium

Using the relations obtained before, it is possible to nd, which part is reected
~ v = ~ v L = 0 k1Б - k10 ~ v k1Б + k10 Б ~( ~ uL) = v L 2k1Б ~ v k10 + k1Б Б

Б

and which pass through the boundary,
~( uБ) =

S. Kolevatov (SPbSU)

28.06.13

12 / 19

Escaping from the parity-breaking medium

Using the relations obtained before, it is possible to nd, which part is reected
~ v = ~ v L = 0 k1Б - k10 ~ v k1Б + k10 Б ~( ~ uL) = v L 2k1Б ~ v k10 + k1Б Б
(M 2 -m2 )2 2 (M 2 -m2 )2 2

Б

and which pass through the boundary,
~( uБ) =

The dispersion laws in terms of invariant mass are M 2 = kЕ k Е :

S. Kolevatov (SPbSU)

k1L k1Б

= =

- -

m2 M2
28.06.13 12 / 19

e)etion oe0ient

kref

| = |

( (

M 2 -m
2 2

2 )2 2 )2

- -

M M

2 2

- +

(M 2 -m2 )2 2 (M 2 -m2 )2 2

- -

m m

2 2

| |

M 2 -m

Figure : The coecient of reection from the boundary for photons escaping from the broken-parity medium. The kinematically forbidden domain of invariant mass values is hatched.
S. Kolevatov (SPbSU) 28.06.13 13 / 19

Figure : The coecient of reection from the boundary for vector mesons escaping from the broken-parity medium. The kinematically forbidden domain of invariant mass values is hatched. We take = 300MeV

S. Kolevatov (SPbSU)

28.06.13

14 / 19

Entrance

k

e

1 1

e = 1(

2 k10 + k1+

k10 - k1+

+

k k

10 10

- +

k k

1- 1-

);

k

e

1 2

e = 1(

2 k10 + k

k

10

-

k

1+ 1+

-

k10 - k1 k10 + k1

- -

).

S. Kolevatov (SPbSU)

28.06.13

15 / 19

Time-like CS vector

Е = ( , H, H, H)

A

+

m2 A

+ (-x1 )

0

A

= 0.

S. Kolevatov (SPbSU)

28.06.13

16 / 19

Time-like CS vector

Е = ( , H, H, H)

A

+

m2 A

+ (-x1 )

0

A

= 0.

kref
hispersion lws

=|

k1A - k10 k1A + k10

|

k1L k1- k1+
= =

=

k10
2 2

=

2 -
2 2

m

2

-

2 k

2 - 2 -

m m

- -

k k

+ +

2

2 2

+ -

2 - 2 -

m m

2 2

+ +

2

4 4

2

2

S. Kolevatov (SPbSU)

28.06.13

16 / 19

In terms of invariant mass,

kref

| = |

(M 2 -m2 )2 2 (M 2 -m2 )2 2

- -

2 k 2 k

- +

(M 2 -m2 )2 2 (M 2 -m2 )2 2

+( +(

M M

2 2

- -

m m

2 2

)- )-

2 k 2 k

| |

Figure : The reection coecient for vector mesons escaping. The kinematically forbidden domain of invariant mass values is hatched. Two views at the three-dimensional graph are represented. For the vector meson we take = 300MeV .
S. Kolevatov (SPbSU) 28.06.13 17 / 19

rolem with the guge invrine

d 3 x Е A A
olution of this prolem

Е

Е = ( (-x1 ), - t (x1 ), 0, 0)

S. Kolevatov (SPbSU)

28.06.13

18 / 19

rolem with the guge invrine

d 3 x Е A A
olution of this prolem

Е

Е = ( (-x1 ), - t (x1 ), 0, 0)

wthing onditions

transversal polarization,
(Б) (Б) i ~ (u0 - v0 ) = 2k10 2 CБ (k2 , k3 ) ~ A - continuous ~ 11 ( Б) (Б) CБ (k2 ,k3 ) ~ ~ u0 + u0 = k3 (k2 -k3 C2A ) + CuБ (k2 , ( v3Б + v3Б = (CkБ-kk23,k32)A ) ~ ~ C 2
S. Kolevatov (SPbSU)

longitudinal,
(L) (L) u0 = v0 ~ ~ A - continuous ~ 11 (L) ( ) u0 + u0L) = CuL (k2 , k3 ) ~ ~ 2 v3L + v3L = CL (k ,k3 ) ~ ~
28.06.13 18 / 19

k3

Conclusion The main results for the spatial CS vector:

* The expression that relates two dierent physical vacua was found. It was shown that each of this vacua can be presented as a coherent state in terms of another one; * The relations that can be used to calculate the passage through or reection of incoming and outgoing particles of any polarization are obtained; * In particular, it was shown that transverse polarizations undergo strong reection up to total internal one at certain frequencies; * When a medium with broken parity is irradiated with photons, an additional rotation of circular polarizations can occur at the reection from the interface. * Reection coecients; * The gauge invariance can be restored.
28.06.13 19 / 19

For time-like CS vector:

S. Kolevatov (SPbSU)

arXiv:nucl-ex/0605007v1 10 May 2006

S. Kolevatov (SPbSU)

28.06.13

19 / 19

to 9

v ~2- v ~2+ v0- ~ v ~0+

= =

~ v2L =- =- ~ v0L

2 2 -k ~3- 2 -k 2 2 2 k2 k3 +i 2 -k ~3+ 2 2 -k2

k2 k3 -i

v v

= k2 v3L k3 ~ 2 k3 -ik2 2 -k ~ v3- 2 -k 2 2 2 k3 +ik2 2 -k ~ v3+ 2 2 -k2 ~ = - k3 v3L

(2)

S. Kolevatov (SPbSU)

28.06.13

19 / 19

to 16

S. Kolevatov (SPbSU)

28.06.13

19 / 19