Lagrangian Formalism for the New Dirac Equation
OUSMANE MANGA Adamou
Department of Physics, Faculty of Sciences
Abdou Moumouni University of Niamey, P.O. Box 10662, Niger [email protected]
SAMSONENKO Nicolai Vladimirovich
Department of Theoretical Physics, Russian Friendship University 3, Ordjonokidze,117923 Moscow, Russia
MOUSSA Aboubacar
Department of Mathematics and Computer Sciences, Faculty of Sciences Abdou Moumouni University of Niamey P.O. Box 10662, Niger
Abstract
The expressions for the energy-impulse tensor and the spin operators are obtained for the particle described by the new Dirac equation in the framework of the Lagrange formalism.
Mathematics Subject Classification: 83C57, 47B15, 47B25
1. Introduction
The new relativistic wave equation proposed by Dirac in 1971 (see [3]) is not symmetric in term of positive and negative values of energy. This equation describes a spinless particle with positive energy, internal structure and non-zero rest mass. The equation has the following form:
r
0 r
m q 0
x α x β ψ
⎧ ∂ + ∂ + ⎫ =
⎨∂ ∂ ⎬
⎩ ⎭ (1)
where αr ( r=1 , 2 , 3 ) are real 4 × 4 matrices and β is an antisymmetric matrix given by
0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
β
⎛ ⎞
⎜ ⎟
⎜ ⎟
=
⎜− ⎟
⎜ − ⎟
⎝ ⎠
(2)
Note that β2 = −1. The quantity q is a colon vector. The symbol q ~ will denote a line vector ( q ,q ,q ,q )1 2 3 4 where q ,q1 3 = p1 and q ,q2 4 = p2 are the dynamic variables of two harmonic oscillators describing the internal structure of the particle. The quantities q ( aa =1 ,2 ,3 ,4 ) satisfy following commuting law
[
q ,qa b]
=q qa b−q qb a =iβab (3)The wave function ψ is one-component and depends on x , x0 r and two commuting quantities qa (for example q1 and q2). The matrices αr( r=1 ,2 ,3 )
and β satisfy the Clifford-Dirac algebra relations:
r s s r rs
r r
2
0
α α α α δ
α β βα
+ =
Introducing the notations
x
μ μ
∂ ∂ ≡
∂ and α0 =I the unity matrix, the equation (1) takes the form:
(
αμ∂ +μ mβ ψ)
q =0 (μ=0,1,2,3 ) (5)The multiplying equation (5) by matrix β at the right we get:
(
α βμ ∂ −μ m q)
ψ =0 (μ =0,1,2,3 ) (6)2. Lagrangian
The new Dirac equation, like the old one (see [4]) can be considered as an equation of some field. This field is described by functions qaψ and ψqb. With a corresponding Lagrange density function we can obtain the field equation (1) using variational method.
From the Lagrange density function LD (see [1]) given by:
(
~ ~)
~D
1 1
L q q q q q m q
4 2
μ μ
μ μ
ψ α β ψ ψ α β ψ ψ ψ
= − ∂ − ∂ + (7)
we can obtain the Lagrange equation in term of the field function ψq~ as following:
(
D ~)
D~L L
0 xμ μψq ψq
⎛ ∂ ⎞ ∂
∂ ⎜ ⎟ − =
⎜ ⎟
∂ ⎝∂ ∂ ⎠ ∂ , (8)
where D
~
L 1 1
q mq
q 4 2
μ μ
α β ψ ψ
ψ
∂
= − ∂ +
∂ ,
(
D)
~
L 1
q
x q 4
μ μ μ
μ
α β ψ
ψ
⎛ ∂ ⎞
∂ ⎜ ⎟ = ∂
⎜ ⎟
∂ ⎝∂ ∂ ⎠ .
So we obtain the equation (6).
3. Basic physical quantities
coordinates vanishing the variation of the action correspond s dynamic invariants, i.e. time-conserved combinations of field functions and their derivatives (see [2]).
Consider the following infinitesimal transformation of the field functions and coordinates:
a a a a
, , , ,
a a a a
x ,
xμ μ xμ xμ
μ μ μ μ
δ
ϕ ϕ ϕ δϕ
ϕ ϕ ′ ϕ δϕ
′
⎯⎯→ = +
′
⎯⎯→ = +
⎯⎯→ = +
(9)
where a a a a
q , a 1,2 ,3 ,4 ;
q , a 5,6 ,7 ,8 .
ϕ ψ
ϕ ψ
= =
= =
The variations δxμ and δϕa are expressed in terms of the linearly independent infinitesimal transformation parameters δωn using the formulas:
( n ) n 1 n s
a a( n ) n 1 n s
xμ Xμ
δ δω
δϕ Ψ δω
≤ ≤
≤ ≤
= =
∑
∑
(10)Note that δϕa,μ is not a derivate of δϕa, i.e. operators ∂ ∂/ x and δ do not commute. The fact is that δϕa is a variation of the field function as by changing its shape and by the argument.
Denote the variations of the form of the field functions as:
(
)
, ,
a a a a a a( n ) a ( n ) n n
x X
τ τ τ
τ
δϕ =ϕ ϕ′ − =δϕ ϕ δ− =
∑
Ψ −ϕ δω (11)The operations δ and ∂ ∂/ x do commute. We now define the variation of the action:
D D
I L dx L dx
δ =
∫
′ ′−∫
,where
,
D D a a D D
L′ =L (ϕ ϕ′, μ′)=L +δL . The total variation δLD is equal to:
,
D D D
D a , a D
a a
L L dL
L L x
dx
μ
μ μ
μ
δ δϕ δϕ δ δ
ϕ ϕ
∂ ∂
= + = +
∂ ∂ .
Here δLD is the variation of LD due to the variations of the forms ϕa and ϕa,μ: ,
D D
D a , a
a a
L L
L μ μ
δ δϕ δϕ
ϕ ϕ
∂ ∂
= +
In result we obtain: D
D D D D D
x
dL d
I ( L L x )( 1 )dx L dx ( L ( L x ))dx
dx x dx
μ μ μ μ μ μ δ δ = +δ + δ +∂ − = δ + δ ∂
∫
∫
∫
(13) Using the Lagrange equations (8) we transform δLD into the following form:D D D
D , a , a , a
a a a
L L L
L
xμ μ μ xμ xμ μ
δ δϕ δϕ δϕ ϕ ϕ ϕ ⎛ ∂ ⎞ ∂ ⎛ ∂ ⎞ ∂ ∂ ∂ = ⎜ ⎟ + = ⎜ ⎟ ∂ ⎝∂ ⎠ ∂ ∂ ∂ ⎝∂ ⎠
Substituting this expression for δLD in (13) we obtain:
(
)
(
)
D a D , a , Da( n ) a ( n ) n D ( n ) n ,
n n
a
, D
a( n ) a ( n ) D ( n ) n ,
n a
L
I L x dx
x
L d
X L X dx
dx
L d
X L X dx
dx μ μ μ τ τ μ μ μ τ τ μ μ μ δ δϕ δ ϕ Ψ ϕ δω δω ϕ Ψ ϕ δω ϕ ⎛ ∂ ⎞ ∂ = ⎜ + ⎟ ∂ ⎝∂ ⎠ ⎡∂ ⎤ = ⎢ − + ⎥ ∂ ⎣ ⎦ ⎡∂ ⎤ = ⎢ − + ⎥ ∂ ⎣ ⎦
∫
∑
∑
∫
∑∫
Since δI =0, then by the linear independence of transformation parameters δωn and the arbitrariness of the integration domain, we have:
(
,)
D
a( n ) a ( n ) D ( n ) ,
a L d
X L X 0
dx τ τ μ μ μ Ψ ϕ ϕ ⎡∂ − + ⎤= ⎢∂ ⎥
⎣ ⎦ (14)
We introduce the notations
(
,)
D
( n ) , a( n ) a ( n ) D ( n ) a
L
X L X
μ τ τ μ μ Θ Ψ ϕ ϕ ∂ = − + ∂ (15)
Then (14) takes the form: ( n ) d
0 dx
μ μ
Θ = (16)
So
( n ) d dx 0 dx μ μ Θ =
∫
(17)1 2
( n ) ( n )
d μ μ d μ μ 0
σ σ
σ Θ − σ Θ =
∫
∫
(18)Here dσμ is the projection of the element of surface σ in a 3-plane perpendicular to the axis xμ. This equation shows that the surface integrals
n ( n )
C ( ) dxμ μ
σ
σ =
∫
Θdo not depend on the surface σ. a) Energy-impulse tensor
Consider an infinitesimal space-time translation xμ′ = xμ+δxμ. Choosing δxμ as the transformation parameters, we have:
xμ Xνμ ν Xνμ xν
ν ν
δ =
∑
δω =∑
δ , i.e. Xνμ =δμν.Since the field functions are not converted, then Ψaν =0.
With in mind, and in this particular case, from (15) we get a second order tensor:
(
)
(
)
(
)
D D
( n ) a b D
a b
~ ~
D
L L
T q q L
q q
1
q q q q L
4
μ ν ν
μν μ μ νμ
ν ν
μ μ νμ
Θ ψ ψ δ
ψ ψ
ψ α β ψ ψ α β ψ δ
∂ ∂
⎯⎯→ = − ∂ − ∂ +
∂ ∂ ∂ ∂
= ∂ − ∂ +
As for qψ and ~
q
ψ satisfying the field equations we have LD≡0, then
(
~ ~)
1
T q q q q
4
ν ν
μν = ψ α β ψμ ∂ − ∂ψ α β ψμ (19)
Hence
~ ~
00
1 q q
T q q
4 t t
i
2 t t
ψ ψ
ψ β β ψ
ψ ψ
ψ ψ
⎛ ∂ ∂ ⎞
= ⎜ − ⎟
∂ ∂
⎝ ⎠
∂ ∂
⎛ ⎞
= ⎜ − ⎟
∂ ∂
⎝ ⎠
(20)
In the same way we obtain the expressions for the remaining components of the tensor:
n0
n n i
T , n 1 , 2 , 3
2 x x
ψ ψ
ψ ψ
⎛ ∂ ∂ ⎞
= ⎜ − ⎟ =
∂ ∂
⎝ ⎠ (21)
The conserved quantity in this case is: 3
0
Pμ =
∫
T d xμ{
}
2 2 2 2
1 2 1 2 1 1 2 2 1 2 0 3
1
( x ,q ,q ) k exp q q ip ( q q ) 2ip q q /( p p ) exp ip x
2
μ
μ μ
ψ = ⎧⎨− ⎣⎡ + + − − ⎤⎦ + ⎬⎫× −
⎩ ⎭
(22) and the normalisation condition
3
d x 1
ψψ =
∫
we obtain the expression for the energy
3 3
0 00 0 0
P =
∫
T d x=∫
pψψd x= pand 3-momentum
3 3
r r0 r r
P =
∫
T d x=∫
pψψd x= p . b) Angular momentum tensor and spin tensor Consider the infinitesimal 4-rotationx a x ,
x x
a , ν
μ μ μ μν
μ μ μ ν
μν ′
⎯⎯→ = + ′ ⎯⎯→
∂ ∂ = ∂ + ∂ (23)
where aμν = −aνμ. Thus, in this case deal with a 6-parameter transformations group. Then
(
)
x X a x a x a
x a x a
x x a
μ ρσ μν σν
μ ρσ ν ν σμ
ρ σ
σν σν
ν σμ ν σμ σ ν σ ν
σν ν σμ σ νμ σ ν
δ δ
δ δ
δ δ
<
< >
<
= = = =
= + =
= −
∑
∑
∑
∑
Consequently, we have:
Xρσμ =xσ ρμδ −xρ σμδ (24)
We now find an expression for Ψaρσ. Since ϕa′ =ϕ δϕa+ a, then the requirement of relativistically invariant field equations in coordinate transformation (23), the field functions are converted as following [3]:
(
)
qψ′ = −1 βN qψ , where N 1a 4
ρσ ρ σ
α βα
= ,
hence
dc eb
a ab b ad ce b
1
( N ) a
4
ρσ
ρ σ
δϕ = − β ϕ = −
∑
β α β α ϕWe also have
a a a , ( a 1 , 2 , 3 , 4 )
ρσ ρσ
Comparing these expressions for δϕa, we get: dc eb
a ad ce b
1 4
ρσ ρ σ
Ψ = − β α β α ϕ (25)
Given the transformation law for functions ϕ%b, we get a similar expression for bρσ
Ψ :
mn k
b m n kb
1 4
ρσ ρ σ
Ψ = ϕ α β α βl
l
% (26)
In this case, the tensor Θ( n )μ transforms into the tensor Mρσμ :
(
)
(
)
D
( n ) a a
a D
b b D
b
L
M q X
q
L
q X L X
q μ μ τ τ ρσ μ ρσ ρσ τ τ μ ρσ ρσ ρσ μ Θ Ψ ψ ψ Ψ ψ ψ ∂ ⎡ ⎤ ⎯⎯→ = ⎣ − ∂ ⎦+ ∂ ∂ ∂ ⎡ ⎤ + ⎣ − ∂ ⎦+ ∂ ∂ (27) where
(
)
(
)
a a a
q X q x x x x q
τ τ τ ρ σ ρσ σ ρτ ρ στ σ ρ ψ ψ δ δ ψ ∂ = ∂ − = ∂ − ∂ . So
(
)(
)
(
)(
)
(
)
(
)
(
)
D D a b a b D DD a b
a b
~ ~ 2
L L
M x x q x x q
q q
L L
L x x
q q
1 1
x T x T q q q q
16 16
x T x T S
μ σ ρ σ ρ ρσ μ ρ σ μ ρ σ σ ρμ ρ σμ μ ρσ μ ρσ σ ρμ ρ σμ ρ σ μ μ ρ σ μ σ ρμ ρ σμ ρσ ψ ψ ψ ψ δ δ Ψ Ψ ψ ψ ψ α βα βα β ψ ψ α β α βα ψ ∂ ∂ = ∂ − ∂ + ∂ − ∂ + ∂ ∂ ∂ ∂ ∂ ∂ + − + = ∂ ∂ ∂ ∂ = − + + = = − + (28)
Here x Tσ ρμ−x Tρ σμ is the orbital angular momentum of the particle. The tensorSρσμ corresponds to the spin angular momentum of the particle.
Consider the spatial part of the spin angular momentum: 0 1 ~
S q q
8
ρσ = − ψ α βα ψρ σ
We can write the antisymmetric form in term of indices ρ and σ:
(
)
0 1 ~
S q q
16
ρσ = − ψ α βαρ σ −α βασ ρ ψ
(
)
~ 3
1
S q q d x
16
ρσ = −
∫
ψ α βαρ σ −α βασ ρ ψFrom this we can define the spin operators as follows:
(
)
~ ~
1
ˆS q q
16
1 i
q q g
8 4
ρσ ρ σ σ ρ
ρ σ ρσ
α βα α βα
α βα
= − − =
= − +
(29)
4. Conclusion
Thus the Lagrange formalism allows us to obtain expressions for all physical quantities, and these expressions are identical to those formulas obtained by Dirac without using the variational method. The resulting formulas give the opportunity to generalize the considered case of classical field for a more interesting, from a physical point of view, case of quantized field, i.e. carry out the procedure of second quantization. However, it should be emphasized that in this approach remains an unsolved problem of including the interaction of the field with known physical fields (see, for example [4]).
References
[1] N.N. Bogolubov, D.V. Shirkov, Introduction to the theory of quantized fields (in Russian), Hauka, Moscow, 1984.
[2] J.E. Castillo H. and A.H. Salas, A Covariant Relativistic Formalism for the New Dirac Equation, Adv. Studies Theor. Phys., Vol. 5, no. 8, (2011), 399-404.
[4] P.A.M. Dirac, The Quantum Theory of the Electron, Proc. R. Soc. A.117 (1928), 610-624.