Chapter 17 The bilinear covariant fields of the Dirac - Physics Quest

Chapter 17

The bilinear covariant fields of the Dirac electron

−from my book:

Understanding Relativistic Quantum Field Theory

Hans de Vries

November 10, 2008

2 Chapter

Contents

17 The bilinear covariant fields of the Dirac electron 117.1 Bilinear fields of the 2d Dirac’s equation . . . . . . . . . . 217.2 Bilinear fields of the 4d Dirac’s equation . . . . . . . . . . 517.3 Lorentz transform of spin four vectors . . . . . . . . . . . . 917.4 Lorentz transform of the vector/axial fields . . . . . . . . . 1117.5 Lorentz transform of the scalar fields . . . . . . . . . . . . 1817.6 Lorentz transform of the tensor field . . . . . . . . . . . . . 1917.7 Overview of the bilinear field transforms . . . . . . . . . . 24

Chapter 17

The bilinear covariant fields ofthe Dirac electron

2 Chapter 17. The bilinear covariant fields of the Dirac electron

17.1 Bilinear fields of the 2d Dirac’s equation

The spinor representation of the Dirac equations embodies the electro dy-namic and kinematic properties of the electron like the charge/currentdensity and the components of the spin density. All these values can beextracted using 4x4 gamma matrices in various combinations. The re-sults are the so called bilinear covariant fields, The Lorentz scalar, pseudoscalar, vector and axial vector of the theory. We will these here first forthe simpler case of the 2d Dirac equation.

All results given here are directly applicable for the full 4d dirac equation.The only limitations stem from the use of a single spatial dimension, forinstance: The spin is always in the direction of the positive r-axis. TheDirac equation uses four 4x4 matrices corresponding with the four dimen-sions. Here we use two 2x2 Υ-matrices corresponding with t and r. TheDirac theory uses a 5th derived 4x4 matrix γ5 for practical reasons. Wewill use an Υ5 matrix here defined in an equivalent way: Υ5 = ΥtΥr. Thefollowing are

Υt =(

0 11 0

), Υr =

(0 1−1 0

), Υ5 =

(−1 0

0 1

)(17.1)

γt =(

0 II 0

), γi =

(0 σi

−σi 0

), γ5 =

(−I 0

0 I

)(17.2)

We see that the correspondence is complete, keeping in mind that σt = Iand σ5 = I. All we need to do to from the 2d Dirac equation to the full4d Dirac equation is to replace the ”1”s with the appropriate Pauli spinmatrices.

We now define the adjoint ψ of the wave-function ψ with the help of theadjoint ’spinors’ u and v (denoted by the bars) in exactly the same wayas in the 4d Dirac theory, where ψψ plays a similar role as Ψ∗Ψ in thenon-relativistic theory. For particle and anti-particle plane-wave solutionswe have respectively.

ψ = u e−Et/~+ipr, ψ = u e−Et/~+ipr (17.3)

ψ = v e+Et/~+ipr, ψ = v e+Et/~+ipr (17.4)

17.1 Bilinear fields of the 2d Dirac’s equation 3

Where u and v are defined by.

u = Υt u† =(

0 11 0

) (u∗Lu∗R

)=

(u∗Ru∗L

)(17.5)

v = Υt v† =(

0 11 0

) (v∗Lv∗R

)=

(v∗Rv∗L

)(17.6)

Generic Lorentz transformations

The advantage of first using a simpler equation is that it is easy to deriveall the bilinear expressions. We first derive the behavior of the scalars andvector components under a boost.(

uLuR

)boost⇒

(exp (−ϑ/2) uLexp (+ϑ/2) uR

)(17.7)

From this Lorentz transformed (boosted) spinor we can construct the com-binations containing terms like ψ∗LψL. Since the exponents drops out theproduct here (=1) we can work with terms like u∗LuL instead.

u∗RuL + u∗LuR = 2m = 2mu∗RuL − u∗LuR = 0 = 0u∗RuR + u∗LuL = 2m coshϑ = 2m γu∗RuR − u∗LuL = 2m sinhϑ = 2m βγ

(17.8)

We can express these as bilinear combinations with the help of the adjointspinors and the Υ-matrices.

——————– generic transformations ——————–

u u = (2m) 1 u Υ5 u = (2m) 0u Υt u = (2m) γ u ΥtΥ5 u = (2m) βγu Υr u = (2m) βγ u ΥrΥ5 u = (2m) γ

(17.9)

We see that uu is a Lorentz scalar independent of the reference framewhich can be associated with the mass of the particle. The bottom two


rows can be combined into 2-vectors, JV the vector current and JA theaxial current.

JV = u Υµ u = 2m( γ, βγ ) (17.10)

JA = u ΥµΥ5 u = 2m( βγ, γ ) (17.11)

JV transforms as an energy/momentum vector while JA transforms as aspin vector. Currents associated with the left chiral component ψL andthe right chiral component ψR can be expressed like.

JL = u∗LuL =12u Υµ

(1−Υ5

)u (17.12)

JR = u∗RuR =12u Υµ

(1 + Υ5

)u (17.13)

The generic transformations can thus be mapped on the contravariantvectors for momentum (E, p) and spin s since these transform in the sameway. We will see that this holds for four-vectors as well in the case of theDirac equation.

(uLuR

)boost⇒

( √(E − pr)√(E + pr)

)(17.14)

—————– energy-momentum and spin —————–

u u = 2 m u Υ5 u = (4m/~) 0u Υt u = 2 E u ΥtΥ5 u = (4m/~) st

u Υr u = 2 pr u ΥrΥ5 u = (4m/~) sr(17.15)

Alternatively, we can map the charge/current density J or magnetic mo-mentum spin density µ on the bilinear expressions.

—– charge-current density, magnetic moment density —–

u u = (2m/q) q u Υ5 u = 2m/(geµB) 0u Υt u = (2m/q) jt u ΥtΥ5 u = 2m/(geµB) µt

u Υr u = (2m/q) jr u ΥrΥ5 u = 2m/(geµB) µr

(17.16)


17.2 Bilinear fields of the 4d Dirac’s equation

We can now go from the 2d bilinear covariant fields to the 4d bilinearcovariant fields. The spin is introduced via the Pauli spin matrices σi

when we replace the 2x2 Υ-matrices with the 4x4 γ-matrices:

Υt =(

0 11 0

), Υr =

(0 1−1 0

), Υ5 =

(−1 0

0 1

)(17.17)

γt =(

0 II 0

), γi =

(0 σi

−σi 0

), γ5 =

(−I 0

0 I

)(17.18)

The bilinear covariants of the 4d Dirac equation are listed below.

Bilinear expression transforms like a:

ψψ 1× scalarψγµψ 4× vectorψσµνψ 6× antisymmetric tensorψγµγ5ψ 4× axial vectorψγ5ψ 1× pseudoscalar

(17.19)

Where the relation between ψ and ψ is given by

ψ = γt ψ† =(

0 II 0

) (ψ†Lψ†R

)=

(ψ†Rψ†L

)(17.20)

With † denotes the complex conjugate. Next to the four we already knewfrom the 2d Dirac equation, the two Lorentz scalars and the two Lorentzvectors, we see a fifth form that represents a quantity which transforms asa tensor. This quantity is, not surprisingly, associated with the spin andoccurs in spin interactions.

σµν =i

2[γµ, γν ] =

i

2(γµγν − γνγµ) (17.21)


We see from (17.19) that there are in total 16 different 4x4 matrices whichcan be sandwiched in the bilinear expression, including I the unity ma-trix. With these 16 matrices one can construct any 4x4 matrix as a linearcombination.

In this section we’ll have a general discussion about the physical meaningof the Dirac bilinears. The relevant insight-full derivations will then followin the subsequent sections.

The Lorentz scalar and pseudo scalar

The Lorentz scalar and pseudo scalar are the same in any reference frameunder general Lorentz transform (boosts and rotations). The scalar ψψ =2m represents the mass of the particle. The relation with the pseudo scalarbecomes clearer when we show the explicit dependence on the left and rightchiral components ψL and ψR.

Bilinear expression LR decomposition transforms like a:

ψψ ψ†LψR + ψ†RψL scalarψγ5ψ ψ†LψR − ψ†RψL pseudo scalar

(17.22)

A pseudo scalar changes sign under parity inversion, the sign reversal ofspacial axis. The quantity ψγ5ψ is always zero in the 2d theory. We willshow in sections further on that the same hold for the 4d theory undervery general conditions.

The condition that ψγ5ψ = 0 has important consequences. Dirac showed inhis first paper on his equation in 1928 that the orbital angular momentumand the spin angular momentum are only separately conserved if ψγ5ψ iszero.

The inversion of parity separately from time inversion depends on thereverence frame if the quantity does not propagate on the light-cone. Theonly value which remains the same under sign change is the value 0.


The Lorentz vector and axial vector

The Lorentz vector and axial vector transform like charge/current densityand axial current density respectively. The decomposition in ψL and ψRshows that both can be expressed with two independent chiral terms whichdescribe the chiral currents JR and JL.


jV = ψγµψ ψ†RσµψR + ψ†Lσ

µψL vectorjA = ψγµγ5ψ ψ†Rσ

µψR − ψ†LσµψL axial vector

(17.23)

So we can simply write jV = jR + jL and jA = jR− jL. We have made useof a notation above involving σµ and σµ which are defined by.

σµ =(σt + σx + σy + σz

), σµ =

(σt − σx − σy − σz

)(17.24)

Where σµ is typically related to ψL while σµ pairs with ψR. This notationalso allows us to write the 4d Dirac equation in a more compact form. Forexample.

i

(0 σµ

σµ 0

)∂µ(ψLψR

)= m

(ψLψR

)(17.25)

The vector quantity jV transforms like a contra-variant four vector. Thetime component transforms as γ (like energy) and the spatial componentstransform like βγ (like momentum). The time component transform likecharge-density. The total charge is a Lorentz scalar, it is the same in allreference frames. The charge density however transforms like γ since thetotal volume becomes Lorentz contracted with γ. The spatial componenttransforms like electric current-density. The total current depends only onthe speed β in any given reference frame. The current-density transformslike βγ where the extra factor γ comes again from the Lorentz contractionof the total volume.


The axial vector quantity jV transforms like an (inertial) spin-four-vector.The time component is zero in the rest-frame of the spin and transformslike −βγ. The spatial inertial spin-components transform like γ along thedirection of the boost while the spin-components orthogonal to the boostdo not transform.

The time component of jA transforms like axial (electric) current density.The total axial current is Lorentz invariant

jV = jR + jL, jA = jR − jL (17.26)

The Lorentz (anti-symmetric) tensor

− ψσµνψ =

0 −Px −Py −PzPx 0 −Mz My

Py Mz 0 −Mx

Pz −My Mx 0

(17.27)

Bilinear expression LR decomposition transforms like the:

Mk = ψσijψ ψ†LσkψR + ψ†Rσ

kψL Magnetization fieldP j = ψσ0jψ iψ†Lσ

jψR − iψ†RσjψL Polarization field

(17.28)

∂νFµν = µoj

ν (17.29)

∂ν(ψσµνψ

)=

2mejν (17.30)

17.3 Lorentz transform of spin four vectors 9

17.3 Lorentz transform of spin four vectors

We will do a quick review here of the Lorentz transform of the classicalaxial current density vector to remind us what we should expect from theboost operations on spinors. Dirac’s theory should give us the same results.The relativistic spin four-vector is equal to the classical three componentvector in the rest-frame with the time-component set to zero

Figure 17.1: Lorentz transform of the spin four vector

Image 17.1 shows a circular current in its rest-frame (left), boosted parallelto the spin-axis (middle) and boosted orthogonal to the spin-axis (right).Recall that the magnetic moment is given by.

µ = IA (17.31)

The total circular current is Lorentz invariant just as the total charge. Thecurrent times the area determines the (total) magnetic moment. Nothingchanges if the boost is parallel to the rest spin. However, the area Adecreases with γ in case of a boost orthogonal to the spin (left image). So,the magnetic moment µ is decreases in the orthogonal case while it staysunchanged for the parallel boost.


~µ ′ =1γ~µ⊥ + ~µ‖ (17.32)

The axial current density transforms with an extra factor γ due to Lorentzcontraction. The axial current density is constant in the orthogonal casewhile it increases with γ for the parallel boost. The result is that for ultra-relativistic speeds the spin is always aligned with the velocity (except forthe case where the spin is exactly orthogonal to the velocity)

~jA′ = ~jA⊥ + γ ~jA‖ (17.33)

The inertial spin ~s transforms in the same way as the axial current density,because energy/momentum transforms with an extra factor γ comparedto the total charge/current. The time components of the four-vectors ~µ,~JA and ~s are per definition zero in the rest-frame. In other frames theyare defined by the general law for the invariant of vector transformation.(Which implies that the constants are −|~µ|2, −| ~JA|2 and −|~s|2).

µ2t /c

2 − µ2x − µ2

y − µ2z = constant (17.34)

Magnetic moment density versus charge density

Note that an attempt to relate the circular current, which causes the mag-netic moment in case of the (point) spin density, directly with the chargedensity would run into problems. The charge would have to move at speeds>> c when the radius decreases to below the Compton radius to zero incase of a (theoretical) point spin.

One way out is to realize that the spin current can be both a positivecharge density rotating counter-clockwise as well as a negative charge den-sity rotating clockwise. So, one could for instance consider the magneticmoment to be augmented by a magnetic vacuum polarization in this way.

This problem doesn’t occur in case of the inertial spin since the inertialmomentum increases to ∞ when approaching c. There is no minimumradius in case of the inertial spin.

17.4 Lorentz transform of the vector/axial fields 11

17.4 Lorentz transform of the vector/axial fields

We will first look into two elementary cases: A particle with spin z-upgetting a boost in the x-direction and in the z-direction, orthogonal andparallel to spin. Following this we will derive a general formula.

Boost in the x-direction orthogonal to the spin

Figure 17.2: Transform of the chiral components under an x-boost

The spinor representing a spin up in the z-direction is (1, 0). From equa-tion (11.94) we see that it transforms into ( cosh(ϑ/2), sinh(ϑ/2 ) under aboost ϑ in the x-direction. We write out the components of JR = u∗Rσ

µuR

u∗RσtuR =

(coshϑ/2sinhϑ/2

)∗( 1 00 1

)(coshϑ/2sinhϑ/2

)= γ (17.35)

u∗RσxuR =

(coshϑ/2sinhϑ/2

)∗( 0 11 0

)(coshϑ/2sinhϑ/2

)= βγ (17.36)

u∗RσyuR =

(coshϑ/2sinhϑ/2

)∗( 0 −ii 0

)(coshϑ/2sinhϑ/2

)= 0 (17.37)

u∗RσzuR =

(coshϑ/2sinhϑ/2

)∗( 1 00 −1

)(coshϑ/2sinhϑ/2

)= 1 (17.38)


These are the four components of the current jR = u∗RσµuR which is

associated with the right chiral component ψR. We have omitted the factorm, and used γ = coshϑ and βγ = sinhϑ. We are going to repeat this nowfor jL = u∗Lσ

µuL = u∗L( σt, −σx, −σy, −σz )uL, so that we can constructthe vector current jV = jR + jL and the axial current jA = jR − jL

The results are:

jR = m(

γ βγ 0 1)

jL = m(

γ βγ 0 −1)

jV = 2m(

γ βγ 0 0)

jA = 2m(

0 0 0 1) (17.39)

The four currents are drawn in figure 17.2. We see that the vector currentjV can be written as 2(E, px, 0, 0) as it should if it has to present theenergy/momentum current. The axial current has a component 1 in thez-direction. Indeed, a boost orthogonal to the spin should not change thespin. jA is the same as it is in the rest-frame.

Next we check the results for this elementary case with the expressionsfor time-like, spin-like and light-like transformations as we require for thevarious currents.

— Lorentz transforms of the vector, axial and chiral currents —

Momentum: j2V t − j2V x − j2V y − j2V z = (2m)2

Spin (axial): j2At − j2Ax − j2Ay − j2Az = −(2m)2

Left chiral: j2Lt − j2Lx − j2Ly − j2Lz = 0

Right chiral: j2Rt − j2Rx − j2Ry − j2Rz = 0

(17.40)

By inserting the results of (17.39) into table (17.40) we can check that ourelementary example indeed behaves as we require. The fact that jR andjL transform light-like implies that they propagate with the speed of lightjust as in the case of the 2d Dirac equation.


Boost in the z-direction parallel to the spin

The z-up spinor (1, 0) becomes ( exp(ϑ/2), 0 ) as a result from a boost inthe z-direction parallel to the spin, see equation (11.99) Writing out thefour components of the current JR = u∗Rσ

µuR gives us:

u∗RσtuR =

(exp ϑ

20

)∗( 1 00 1

)(exp ϑ

20

)= expϑ (17.41)

u∗RσxuR =

(exp ϑ

20

)∗( 0 11 0

)(exp ϑ

20

)= 0 (17.42)

u∗RσyuR =

(exp ϑ

20

)∗( 0 −ii 0

)(exp ϑ

20

)= 0 (17.43)

u∗RσzuR =

(exp ϑ

20

)∗( 1 00 −1

)(exp ϑ

20

)= expϑ (17.44)

Repeating this for jL = u∗LσµuL allows us to calculate the vector and axial

currents.

jR = m(

expϑ 0 0 + expϑ)

jL = m(

expϑ 0 0 − expϑ)

jV = 2m(

γ 0 0 βγ)

jA = 2m(

βγ 0 0 γ) (17.45)

The vector current jV can be written as 2(E, 0, 0, pz) as it should be. Theaxial current jA has a time component which is zero in the rest-frame. Thez-component of the axial current (spin) has grown from 1 to γ as expectedfor a boost parallel to the spin.

This elementary case (boost parallel to the spin) is identical to the 2dDirac equation case, as we can see by comparing (17.45) with (17.9)

Inserting (17.45) into the table 17.40 confirms that jR and jL transformlight-like. jV transforms as an energy/momentum vector (as well as acharge/current density vector), and jA transforms as a spin.


The derivation of the general Lorentz transformation

For the general boost formula we’ll take an arbitrary spinor ξ in its rest-frame and give it a boost in an arbitrary direction. For a plane-wavesolution the left and right 2-spinors are equal in the rest frame. After aboost they become generally different.

ψ =(ξξ

)e−iEot/~ boost

⇒

(ξ′Lξ′R

)e−iEt/~+i~p·~x/~ (17.46)

We recall equation (11.91) here for the general boost operator.

ψL

ψR

boost⇒

exp

(− ϑ/2 σi

)ψL

exp(

+ ϑ/2 σi)ψR

(17.47)

Where i denotes the x, y or z-direction. For the right and left chiralcomponents we can write the 2-spinor boost transform as. (R = +, L = −)

ξ′R/Lboost

=exp

(±ϑ

2σi)ξ = I cosh (

ϑ

2)ξ ± σi sinh (

ϑ

2)ξ (17.48)

The currents corresponding with the right and left chiral components are.

jµR = ξ′∗R σµ ξ′R = ξ′∗R ( σt, +σx, +σy, +σz ) ξ′R (17.49)

jµL = ξ′∗L σµ ξ′L = ξ′∗L ( σt, −σx, −σy, −σz ) ξ′L (17.50)

Working this out for jµR gives us.

jµR =(ξ∗σµξ

)cosh2

(ϑ

2

)+(ξ∗(σiσµσi)ξ

)sinh2

(ϑ

2

)+

+(ξ∗(σiσµ + σµσi)ξ

)sinh

(ϑ

2

)cosh

(ϑ

2

)(17.51)

Note that we have removed the transformed ξ′R here. The only spinors inthe formula above are the rest-frame ξ here. We have used the importantidentity. (

σβ ξ)∗·(σλ ξ

)= ξ∗

(σβ σλ

)ξ (17.52)


Since products of pauli sigma matrices are also sigma matrices we willobtain an expression containing only terms like ξ∗σiξ which represent theprojection of the spin pointer on the i-axis. This will make it easy tointerpret the final results.

Next we want to remove the ϑ/2 arguments and replace them with hy-perbolic functions which carry ϑ itself as argument since we can readilyinterpret these with the relativistic transformation factors γ and β.

jµR =12

(ξ∗(σµ + σiσµσi)ξ

)cosh (ϑ) +

12

(ξ∗(σµ − σiσµσi)ξ

)+

+12

(ξ∗(σiσµ + σµσi)ξ

)sinh (ϑ) (17.53)

It’s advantageous to separate the time component from the spatial com-ponents here because σiσ0 = σ0σi while σiσj = −σjσi. The terms in theabove expression will either cancel or otherwise simplify depending on µbeing either 0 or 1, 2, 3.

We replace the hyperbolic functions cosh and sinh with γ and βγ. TheLorentz scalars ξ∗ξ are replaced by m and the projections like ξ∗σiξ arereplaced by explicit dot-products as m(s · xi), where s is the unit vector inthe direction of the spin. This all now leads us to.

j0R = m(γ + (s · xi) βγ

)jjR = m

(δijβγ + δij s · xj + δij s · xi γ

)j0L = m

(γ − (s · xi) βγ

)jjL = m

(δijβγ − δij s · xj − δij s · xi γ

)j0V = 2m γ jjV = 2m δijβγ

j0A = 2m (s · xi) βγ jjA = 2m ( δij s · xj + δij s · xi γ )

(17.54)

Where δij is the opposite of δij , it is 1 when i 6= j. With the help of theabove equations (17.54) we can check the behavior of these currents underan arbitrary boost.


With respect to the vector current jV we find for the energy j0V = 2E, andfor the momentum jjV = δij 2pj , the momentum is in the direction of theboost due to the δij .

The axial current jV contains only spin-projections like (s · xi). For thespatial of the four-vector we recall equation (17.33) for the transform ofthe magnetic moment: ~µ′ = ~µ⊥ + γ ~µ‖. The boost only affects the spincomponent parallel to the boost which increases by a factor γ. We see thisback in jjA where the spin component parallel to the boost is s · xiγ andthe components orthogonal to the boost are s · xj .

The expressions (17.54) contain terms like δij and δij because we startedwith the Pauli matrices of the principle axis σi expressing boosts in thedirection of the principle axis. We can generalize this to arbitrary boostin vector algebra.

Lorentz transforms of the Vector and Axial currents

j0R = m(

1 + s · ~β)γ ~jR = m

(~βγ + s⊥ + s‖γ

)j0L = m

(1− s · ~β

)γ ~jL = m

(~βγ − s⊥ − s‖γ

)j0V = 2m γ ~jV = 2m ~βγ

j0A = 2m (s · ~β) γ ~jA = 2m ( s⊥ + s‖γ )

(17.55)

Where s‖ = (β · s) β and s⊥ = s − s‖ are the parallel and orthogonalcomponents of the spin unit vector s in the rest frame with respect to ~β.The general expressions of (17.55) confirm that the quantum mechanicalDirac vector current and axial current transform in exactly the same wayas their classical counter parts.

We will check the expressions against the table 17.40. The general expres-sions for the left-chiral jL and right-chiral jR current is light-like whichmeans that they propagate with c. The condition for jR to transformlightlike, like a massless component is given by:


j2Rt − j2Rx − j2Ry − j2Rz = 0 (17.56)

If we apply this to our results then we get for the time component and thespatial component repectively.

(jRt)2 = m2

(γ + |s‖| βγ

)2

(17.57)

(jjR)2 = m2

(~β + s‖

)2

γ2 + m2

(s⊥

)2

(17.58)

Where the spatial component has two orthogonal components parallel andtransversal to the direction of motion. Collecting all the terms in oneexpression gives us.

(jRt)2 − (jjR)2 =

m2

((1−β2)γ2 + 2( s‖ · ~β − s‖ · ~β )γ − s2‖(1−β

2)γ2 − s2⊥

)(17.59)

With (1− β2)γ2 = 1 per definition this becomes.

(jRt)2 − (jjR)2 = m2

(1 − s2‖ − s2⊥

)= 0 (17.60)

Which is 0 since |s| = 1. We have hereby proved that jR transformslight-like. The equivalent prove for jL just needs some sign changes. Thiscompletes our prove for the light like behavior of the chiral components.

In the specific case in which one zeroes out either the left chiral or the rightchiral part of the four spinor, then the vector current and axial currentbecome equal jA = jV or reversed jA = −jV . The spin is in parallelor anti-parallel to the momentum, in all reference frames. This is onlypossible for light-like transforming particles. In this specific case this iseasy to prove. We have shown here that this property actually holds ingeneral.


Reflection of the chiral current by spin

(to be finished)

Figure 17.3: Reflection of the chiral current by the spin

17.5 Lorentz transform of the scalar fields


ψ ψ ψ∗L ψR + ψ∗R ψL scalariψγ5ψ iψ∗L ψR − iψ∗R ψL pseudo scalar

(17.61)

ξ′Rboost

=exp

(+ϑ

2σi)ξ, ξ′L

boost=

exp(−ϑ

2σi)ξ (17.62)

Working this out for J jRL = ψ∗R ψL and J jLR = ψ∗L ψR gives us.

J jRL = ξ∗ξ , J jLR = ξ∗ξ (17.63)

ψ ψ = 2m, iψγ5ψ = 0 (17.64)

17.6 Lorentz transform of the tensor field 19

17.6 Lorentz transform of the tensor field

The antisymmetric tensor ψσµνψ is the most complex of the Dirac bilinearquantities. It can be expressed in the form of γ commutators.

σµν =i

2[ γµ, γν ] , with [γµ, γν ] = (γµγν − γνγµ) (17.65)

The various 4x4 matrices are equal to the general Lorentz transformationgenerator Jµν for boosts Ki and rotations Li

Jµν =i

2σµν =

0 −Kx −Ky −Kz

Kx 0 −Lz LyKy Lz 0 −LxKz −Ly Lx 0

(17.66)

With all combinations written out in Pauli σ matrices we get.

σµν=

[0 00 0

] [σx 00 −σx

] [σy 00 −σy

] [σz 00 −σz

][−σx 0

0 σx

] [0 00 0

] [iσz 00 iσz

] [−iσy 0

0 −iσy

][−σy 0

0 σy

] [−iσz 0

0 −iσz

] [0 00 0

] [iσx 00 iσx

][−σz 0

0 σz

] [iσy 00 iσy

] [−iσx 0

0 −iσx

] [0 00 0

]

(17.67)

We will show that the bilinear ψσµνψ can be associated with the magneti-zation/ polarization tensor which describes the electromagnetic fields dueto the continuously distributed spin of the wave function.


− ψσµνψ =

0 −Px −Py −PzPx 0 −Mz My

Py Mz 0 −Mx

Pz −My Mx 0

(17.68)

We see the relation between the magnetization and polarization vector ifwe decompose the vectors in products involving the left and right chiralcomponents ψL and ψR. We will show that the magnetization and polar-ization vectors transform like the magnetic and electric field respectively.

Bilinear expression LR decomposition transforms like the:

Mk = ψσijψ ψ∗LσkψR + ψ∗Rσ

kψL Magnetic fieldP j = iψσ0jψ iψ∗Lσ

jψR − iψ∗RσjψL Electric field

(17.69)If ξ is the two component Pauli spinor representing the spin in the restframe then we can define the transform of ξ for the left and right chiralcomponent as follows.

ξ′Rboost

=exp

(+ϑ

2σi)ξ = I cosh (

ϑ

2)ξ + σi sinh (

ϑ

2)ξ (17.70)

ξ′Lboost

=exp

(−ϑ

2σi)ξ = I cosh (

ϑ

2)ξ − σi sinh (

ϑ

2)ξ (17.71)

Working this out for J jRL = ψ∗RσjψL and J jLR = ψ∗Lσ

jψR gives us.

J jRL =(ξ∗σjξ

)cosh2

(ϑ

2

)−(ξ∗(σiσjσi)ξ

)sinh2

(ϑ

2

)+

+(ξ∗(σiσj − σjσi)ξ

)sinh

(ϑ

2

)cosh

(ϑ

2

)(17.72)

J jLR =(ξ∗σjξ

)cosh2

(ϑ

2

)−(ξ∗(σiσjσi)ξ

)sinh2

(ϑ

2

)+

−(ξ∗(σiσj − σjσi)ξ

)sinh

(ϑ

2

)cosh

(ϑ

2

)(17.73)


Where we have eliminated the transformed ξ′R and ξ′L Pauli spinors here.The only Pauli spinor in the formulas above is the rest-frame ξ. We madeuse of the identity. (

σβ ξ)∗·(σλ ξ

)= ξ∗

(σβ σλ

)ξ (17.74)

Next we want to remove the ϑ/2 arguments and replace them with hy-perbolic functions which carry ϑ itself as argument since we can readilyinterpret these with the relativistic transformation factors γ and β.

cosh2(x) =12

(cosh(2x) + 1), sinh2(x) =12

(cosh(2x)− 1)

sinh(x) cosh(x) =12

sinh(2x) (17.75)

With the above standard identities we get.

J jRL =12

(ξ∗(σj − σiσjσi)ξ

)cosh (ϑ) +

12

(ξ∗(σj + σiσjσi)ξ

)+

+12

(ξ∗(σiσj − σjσi)ξ

)sinh (ϑ) (17.76)

J jLR =12

(ξ∗(σj − σiσjσi)ξ

)cosh (ϑ) +

12

(ξ∗(σj + σiσjσi)ξ

)+

− 12

(ξ∗(σiσj − σjσi)ξ

)sinh (ϑ) (17.77)

We replace the hyperbolic functions cosh and sinh with γ and βγ. TheLorentz scalars ξ∗ξ are replaced by m and the projections like ξ∗σiξ arereplaced by explicit dot-products as m(s · xi), where s is the unit vector inthe direction of the spin. This all now leads us to.

Products of pauli sigma matrices are also sigma matrices and we will obtainexpressions containing only terms like ξ∗σiξ which represent the projectionof the spin pointer on the i-axis in order to make it easier to interpret thefinal results.


J jLR = m(−δij s · xk βγi + δij s · xj + δij s · xi γ

)J jRL = m

(δij s · xk βγi + δij s · xj + δij s · xi γ

)P j = −2m

(δij s · xk βγi

)M j = 2 m

(δij s · xj + δij s · xi γ

)(17.78)

Where δij is the opposite of δij , it is 1 when i 6= j. The expressions (17.78)contain terms like δij and δij because we started with the Pauli matricesof the principle axis σi expressing boosts in the direction of the principleaxis. We can generalize this to arbitrary boost in vector algebra.

Lorentz transform of the anti-symmetric tensor components

~JLR = m(s⊗ βγi + s‖ + s⊥γ

)~JRL = m

(−s⊗ βγi + s‖ + s⊥γ

)~P = 2m ( s⊗ βγi )

~M = 2m(s‖ + s⊥γ

)(17.79)

We have made use of the shorthand notations s‖, s⊥ and s⊗ for the com-ponents of the unit vector s of the spin in the particles rest frame withregard to the boost β. They are defined by.

s‖ = ( β · s ) β parallel component with regard to ~β

s⊥ = ( β × s )× β orthogonal component with regard to ~β

s⊗ = ( β × s ) 90o rotated orthogonal component(17.80)

Where s = s‖ + s⊥ is always true.


The standard Lorentz transform of the EM-field under a boost in the x-direction is.

E′x = Ex B′x = BxE′y = γ(Ey − βBz) B′y = γ(By + βEz)E′z = γ(Ez + βBy) B′z = γ(Bz − βEy)

(17.81)

Which we can rewrite for an arbitrary boost to a form which also uses ourshort hand notation for the various components.

Lorentz transform of the electromagnetic field

E′ = E‖ + E⊥ γ + B⊗ βγB′ = B‖ + B⊥ γ − E⊗ βγ

(17.82)

Where we have made use of the following shorthand notations.

E‖ = ( β · E ) β parallel component with regard to ~β

E⊥ = ( β × E )× β orthogonal component with regard to ~β

E⊗ = ( β × E ) 90o rotated orthogonal component

B‖ = ( β · B ) β parallel component with regard to ~β

B⊥ = ( β × B )× β orthogonal component with regard to ~β

B⊗ = ( β × B ) 90o rotated orthogonal component(17.83)

We can now compare the transformation of the magnetization and polar-ization vectors from the rest frame to a boosted frame, one-to-one withthe general Lorentz transform of the electromagnetic field field.

We see that only the magnetization vector ~M is non-zero in the rest-frame where ~M = 2ms. The polarization vector ~P is zero in the restframe. Under a boost ~M is transformed like the magnetic field B and thepolarization ~P is a result from the boosted magnetization like the electricfield E arises from the boosted magnetic field B.


17.7 Overview of the bilinear field transforms

Lorentz transforms of the bilinear Dirac fields from the (local) rest framewhere s is the unit (rest) spin vector under a boost ~β with the assumptionthat the chiral components are equal: ψR=ψL, in the (local) rest-frame.

Bilinear scalar- and pseudo scalar field transform

ψψ = 2m |φ|2 ψγ5ψ = 0 (17.84)

Bilinear vector- and axial vector field transform

ψγµψ = ψψ

γ

~βx γ

~βy γ

~βz γ

ψγµγ5ψ = ψψ

γ(s · ~β)

(s⊥ + s‖γ)x

(s⊥ + s‖γ)y

(s⊥ + s‖γ)z

(17.85)

Bilinear tensor field transform

ψσµνψ =

ψψ

2

0 − ( s⊗ βγ )x − ( s⊗ βγ )y − ( s⊗ βγ )z

( s⊗ βγ )x 0 i(s‖ + s⊥γ)z −i(s‖ + s⊥γ)y

( s⊗ βγ )y −i(s‖ + s⊥γ)z 0 i(s‖ + s⊥γ)x

( s⊗ βγ )z i(s‖ + s⊥γ)y −i(s‖ + s⊥γ)x 0

(17.86)

σµν =i

2(γµγν − γνγµ)

s‖ = ( β · s ) β parallel component with regard to ~β

s⊥ = ( β × s )× β orthogonal component with regard to ~β

s⊗ = ( β × s ) 90o rotated orthogonal component

Chapter 17 The bilinear covariant fields of the Dirac - Physics Quest

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Transcript of Chapter 17 The bilinear covariant fields of the Dirac - Physics Quest