In contrast to the matrix notation, the product of two identical gamma matrices is written as the unit scalar .... DµDÂ
Covariant gauge transformation of Dirac equation to get the spin adjusted Klein-Gordon equation. Originally appeared at: http://sites.google.com/site/peeterjoot/math2011/diracCovariant.pdf Peeter Joot —
[email protected] Sept 1, 2011
diracCovariant.tex
Contents 1
Motivation.
1
2
Geometric Algebra notation.
2
3
Gauge transforming the Dirac equation.
2
4
Some checks 4.1 Gauge covariant derivative form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Verifying the space time split into energy and spatial momentum operators. . . . . .
3 4 4
5
A last word on covariant gauge transformations.
6
1. Motivation. In §36.4 of [1] is a covariant treatment of the gauge transformed Dirac equation, with the end goal of finding the Klein-Gordon relation with the spin terms required for electromagnetic phenomena. Two typos to start things off, and (36.63-64) should be respectively pµ → pµ − eAµ
(1)
Dµ = ∂µ + ieAµ .
(2)
Other than this there are no typos till the end where a factor of two is lost, and we should have e D µ Dµ − σµν Fµν + m2 ψ = 0 (3) 2 It’s slightly tempting to re-derive this, with the inclusion of the h¯ and c factors. However, a bit of play using the Geometric Algebra (GA) operators discussed in [2] proves more productive. This text introduces a purely GA formalism for the Dirac equation which I’ll not use here. Instead I’ll use the more conventional Feynman slash notation so that a four vector with coordinates aµ is written with its basis as µ µ a = a γµ = aµ γ
1
(4)
2. Geometric Algebra notation. We require the GA dot, and wedge operators 1 (5) ( ab + ba) = aµ bµ 2 1 1 µ ν (6) = (ab −b a) = a b γ[µ γν] . a ∧ b 2 2 In contrast to the matrix notation, the product of two identical gamma matrices is written as the unit scalar value (or grade zero product), instead of a using an explicit four by four identity matrix representation. We similarly label the product of different basis elements, for example γ0 γ1 as a grade two element, or bivector. Thus the dot product is the grade zero term of the multivector product ab, and the wedge product is the grade zero term of the same. More generally, for a product of multivectors A and B the grade selection operator is defined as = a · b
h ABin .
(7)
This is an abstract notation encoding the instructions to select just the n grade elements of the multivector product if they exist. In this notation, the product of vectors splits into scalar and bivector terms which may also be expressed as the dot and wedge products . = hab i0 + hab i2 = a · b + a ∧ b ab
(8)
3. Gauge transforming the Dirac equation. Our electron equation is p − mc) ψ = 0 (
(9)
After a gauge transformation e
p → p − A, c
(10)
this is e (11) p − A − mc ψ = 0. c We left multiply by the conjugate quantity p − ce A + mc, and our task is now to reduce operator equation e e 0= p − A + mc p − A − mc ψ D c e ce E = p − A + mc p − A − mc ψ D c e c e 0E (12) + p − A + mc p − A − mc ψ c c D E1 e e + p − A + mc p − A − mc ψ . c c 2 We have two contributions for the scalar parts, one is the product of mc scalars, and the other is the dot product of the vectors
2
E e e e e 2 p − A + mc p − A − mc ψ = p − A · p − (13) c c c c A ψ − (mc) ψ. 0 The grade one (four vector) components sum to zero since those are only the scalar times four vector portions and have opposing signs D
E e e e e p − A + mc p − A − mc ψ = p − A mc ψ + mc p − (− ) ( ) c c c c A ψ = 0. 1
D
(14)
Only the vector products can contribute to the grade two portion of the multivector product, so retain only the wedges between those E e e e e A + mc p − A − mc ψ = p − A ∧ p − p − c c cA ψ c 2 e2 e = p ∧ pψ − p ∧ Aψ + A ∧ pψ + 2 A ∧ Aψ c c e p ∧ A)ψ + ( pψ) ∧ A + A ∧ pψ = − ( c e p ∧ A)ψ. = − ( c
D
Here braces have been used to denote the range of operation of our differential operator p. µ With ∂ = γ ∂µ , our momentum operator is p = i¯h∂,
(15)
allowing us to write p ∧ A = i¯h∂ ∧ A For the electromagnetic field bivector lets write F = ∂ ∧ A =
1 Fµν γµ ∧ γν . 2
(16)
So the grade two terms of 12 are D
e
p − A + mc c
e
E
ie¯h
p − A − mc ψ = − Fψ. c c 2
Assembling all results we have e ie¯h e 2 p− A − F − (mc) ψ = 0. p− A · c c c
(17)
(18)
4. Some checks There are two tasks that remain. One is to verify that this reproduces the result expressed in terms of the gauge covariant derivative Dµ . The other is to verify that this also reproduces the earlier result with an explicit split into energy and momentum operators. 3
4.1. Gauge covariant derivative form With p = i¯hγµ ∂µ our dot product takes the form e e e µ e µ A i¯ h ∂ − Aµ p − A · p − A = i¯ h ∂ − µ c c c c ie µ ie 2 µ = −h¯ ∂ + A ∂µ + Aµ c¯h h¯ c
We thus write ie Aµ , c¯h differing from the text only by the inclusion of h¯ and c factors. Since we have Dµ = ∂ µ +
σµν =
1 µ γ ∧ γν , i
(19)
(20)
we have or
−h¯ 2 D µ Dµ +
e¯h µν σ Fµν − (mc)2 ψ = 0, 2c
e µν m2 c2 D Dµ − ψ = 0. σ Fµν + 2 2c¯h h¯ With c = h¯ = 1, this reproduces 3, the (corrected) result (36.78) from the text.
(21)
µ
(22)
4.2. Verifying the space time split into energy and spatial momentum operators. Temporarily working with c = h¯ = 1 we have D µ Dµ = D0 D0 + D k Dk
= (∂t + ieφ)2 − (∂k + ieAk )2 With E = i∂t and (p)k = −i∂k = i∂k , this is D µ Dµ = (−iE + ieφ)2 − (−ip + ieA)2
= −( E − eφ)2 + (p − eA)2 . Restoring c and h¯ , and multiplying throughout by h¯ 2 , 21 becomes 1 e 2 e¯h µν 2 2 2 − 2 ( E − eφ) + p − A − σ Fµν + m c ψ = 0. c c 2c
(23)
Now let’s expand the σµν Fµν product in terms of the matrices used in the text previously. That is 4
σµν Fµν = σ00 F00 + σ0k F0k + σ j0 Fj0 + σ jk Fjk
= 2σ0k F0k + σ jk Fjk
Since F0k = ∂0 Ak − ∂k A0
= − ∂0 A k − ∂ k A0 = (E)k , and 1 0 j ( γ γ − γ j γ0 ) 2i = −iγ0 γ j
σ0k =
= iγ j γ0 0 σj 1 0 =i −σj 0 0 −1 0 −σj =i −σj 0 = −iα j So 2σ0k F0k = −2iα · E. For the magnetic terms we have Fjk = ∂ j Ak − ∂k A j
= −∂ j Ak + ∂k A j = emkj Bm . For σ jk , considering the j 6= k case (since it is zero otherwise), we have
5
(24)
1 j k (γ γ − γk γ j ) 2i 1 = γ j γk i 0 σj 0 σk = −i −σj 0 −σk 0 σj σk 0 =i 0 σk σj σm 0 2 = i e jkm . 0 σm
σ jk =
So σ jk Fjk = −e jkn σ0 n emkj Bm
= 2δmn σ0 n Bm = 2σ 0 · B. Thus the space time split of the field is σµν Fµν = −2iα · E + 2σ 0 · B
1 e 2 e¯h 2 0 2 2 − 2 ( E − eφ) + p − A − −iα · E + σ · B + m c ψ = 0. c c c
(25) (26)
This recovers (36.22) from the text (once one adds back in the h¯ and c and fixes the sign error in the electric field term). 5. A last word on covariant gauge transformations. Usually discussions of gauge invariance involve a Lagrangian. Such a beastie for the Dirac equation has not yet been discussed yet in the text. Suppose we just apply the phase transformation to the Dirac equation of the same form that we use with the Klein-Gordon Lagrangian e
ψ → ψei h¯ c θ .
(27)
Observe that if the dimensions of ∂µ θ are those of the vector potential Aµ , then eθ/¯hc is dimensionless. Now lets transform the Dirac equation, utilizing the freedom to adjust the phase arbitrarily e
( p − mc)ψ → ( p − mc)ψei h¯ c θ
e
= (i¯hγµ ∂µ − mc)ψei h¯ c θ ie i h¯ec θ µ =e i¯hγ ∂µ θ + ∂µ − mc ψ h¯ c
6
With the usual assignment ∂µ θ = Aµ ,
(28)
this becomes e e ( p − mc)ψ → ei h¯ c θ p − A − mc ψ. c The bigger question of why 28 is the usual transformation remains. I don’t know the answer for that one. It is common in Lagrangian transformation discussions for the Klien-Gordon equation, and I’m guessing it’s something better justified in places I’ve not yet read. So in the limit where θ → 0, the Dirac equation is transformed as e ( p − mc)ψ → p − A − mc ψ, (29) c and we can say that our momentum operator transforms e
p → p − A. c
(30)
The gauge covariant derivative is then just what we get when we factor out an iγµ from the gauge transformed momentum operator e e µ p − A = i¯hγ ∂µ + i Aµ c c¯h e = i¯h ∂ + i A h¯ c So with e Aµ , c¯h
(31)
e A, c¯h
(32)
Dµ = ∂ µ + i or D = ∂ + i we define e
p − A = i¯h D = i¯hγµ Dµ c
(33)
References [1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009. 1 [2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003. 1
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