Labelle P. Supersymmetry DeMYSTiFied

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Supersymmetry Demystified

4.6 The QED Lagrangian in Terms of

Weyl Spinors

Later in the book we will be constructing theories directly in terms of Weyl spinors,

and it will be very useful to be able to make the connection with the familiar la-

grangians of quantum electrodynamics (QED), quantum chromodynamics (QCD),

and so on, which are usually all expressed in terms of four-component Dirac spinors.

So let’s see what the QED lagrangian looks like when expressed in terms of Weyl

spinors.

Let’s ﬁrst consider the Dirac lagrangian for an electron (so the subscript p will

be replaced by e), which we can obtain directly from Eq. (2.28):

Dirac

= 



i∂

− m





= χ

†

¯σ

i∂

+ η

†

i∂

− m



†

+ χ

†



Again, it is useful to write the Dirac spinor in terms of left-chiral spinors only,

using Eq. (4.6). This is so because we will later write all the interactions of the

standard model and of the minimal supersymmetric standard model in terms of

left-chiral spinors. We already took care of the mass term in Eq. (4.11), so let’s

concentrate on the kinetic energy:



∂



= i 

†

∂



= i



(−iσ

)χ

†





0 1

1 0



0¯σ

∂





iσ

†T



= i χ

∂

†T

+i χ

†

¯σ

∂

(4.12)

which may be written more elegantly as (see Exercise 4.3)



∂



= iχ

†

¯σ

∂

+i χ

†

¯σ

∂

(4.13)

So the Dirac lagrangian is ﬁnally, using Eq. (4.11),

Dirac

= iχ

†

¯σ

∂

+i χ

†

¯σ

∂

− m(χ

· χ

+ ¯χ

· ¯χ

) (4.14)

In Chapter 13 we will build the supersymmetric generalization of this lagrangian.

CHAPTER 4 The Physics of Spinors

EXERCISE 4.3

Derive Eq. (4.13) from Eq. (4.12). Hint: You will need Eqs. (A.36), and (A.5) and

an integration by parts.

Let’s now include the coupling of the fermion with the electromagnetic ﬁeld.

This is obtained from the Dirac lagrangian by replacing the partial derivative ∂

the gauge covariant derivative D

deﬁned by

≡ ∂

+iq A

where A

is the gauge (photon) ﬁeld, and q is the electric charge of the particle.

For the electron, q =−e (we use the convention that e is positive), so

= ∂

−ieA

In addition, we must add the kinetic energy of the photon, i.e.,

−

μν

The two new terms coupling the Weyl spinors to the gauge ﬁeld can be obtained

from the two terms of Eq. (4.12) by replacing ∂

by −ieA

Terms containing A

= e χ

†T

+ e χ

†

¯σ

=−e χ

†

¯σ

+ e χ

†

¯σ

(4.15)

where in the second step we have used Eqs. (A.36) and (A.5) in the way described

in Exercise 4.3.

Note that it would have been incorrect to replace ∂

by −ieA

in Eq. (4.14) (see

Exercise 4.4).

EXERCISE 4.4

Explain why we would have obtained the wrong coupling of the gauge ﬁeld with the

spinors had we replaced ∂

by −ieA

in Eq. (4.14) instead of doing it in Eq. (4.12).

Adding Eq. (4.15) to the Dirac lagrangian Eq. (4.14), we ﬁnally obtain the QED

lagrangian expressed in terms of left-chiral Weyl spinors:

QED

= i χ

†

¯σ

∂

+i χ

†

¯σ

∂

− eA

†

¯σ

+ eA

†

¯σ

−m(χ

· χ

+ ¯χ

· ¯χ

) −

μν

(4.16)

Supersymmetry Demystified

There is something quite interesting going on here. In the derivation of the kinetic

term of the positron, we have used Eq. (A.36) and an integration by parts. Each

operation introduces a minus sign, so the ﬁnal kinetic energy of the positron has

the same sign as the kinetic energy of the electron (which did not require any

manipulation). On the other hand, in the case of the coupling of the positron to

the electromagnetic ﬁeld, we used Eq. (A.36) but, of course, did not perform any

integration by parts (there are no derivatives!), so we ended up with the particle

and antiparticle spinors having couplings to the electromagnetic ﬁeld of opposite

signs. This, of course, was to be expected because these two states have opposite

charges! What is interesting is to see how we started from the Dirac lagrangian

written in terms of a Dirac spinor, in which there is only one coupling to the

electromagnetic ﬁeld, and ended up with an expression containing two couplings

to A

with opposite signs once we expressed the theory in terms of particle and

antiparticle Weyl spinors.

We can write Eq. (4.16) even more elegantly by introducing two gauge covariant

derivatives acting on the left-chiral particle and antiparticle spinors. Actually, it is

customary to use a common symbol D

for both derivatives, with the understanding

that when it is acting on the electron Weyl spinor χ

, it is deﬁned as

≡ (∂

−ieA

)χ

(4.17)

whereas when it is acting on the positron Weyl spinor χ

, it carries the opposite

charge:

≡ (∂

+ieA

)χ

(4.18)

With this convention, we may write Eq. (4.16) as

QED

= iχ

†

¯σ

+i χ

†

¯σ

− m(χ

· χ

+ ¯χ

· ¯χ

) −

μν

(4.19)

In Chapter 13 we will build the supersymmetric generalization of this lagrangian.

4.7 Adding a Mass: Majorana Spinors

Now consider taking the two right-chiral states η

and η

to be one and the same,

as illustrated in Figure 4.6.

Identifying η

and η

only makes sense if the particle is its own antiparticle.

Therefore, this quantum ﬁeld cannot carry quantum numbers that can distinguish a

CHAPTER 4 The Physics of Spinors

¯p

CPT

Figure 4.6 The two states of a massive Majorana fermion (η

is deﬁned to be equal

to η

particle from an antiparticle (e.g., it must have no electric charge or lepton number).

In this case, there are only two degrees of freedom, which we can choose to be χ

and η

with

= η

= i σ

†T

where we have used Eq. (4.3).

It’s easy to check that with this choice, the antiparticle left-chiral state χ

indeed the same as the particle left-chiral state χ

. Using Eq. (4.5) with the particle

and antiparticle labels switched, we ﬁnd

≡−iσ

†T

=−iσ



iσ

†T



†T

=−iσ

(iσ

)

∗

=−iσ

(iσ

)χ

= χ

Instead of expressing everything in terms of χ

and η

, we may, of course, choose

to express everything in terms of the left-chiral ﬁeld χ

and its hermitian conjugate

†

. Note that in terms of degrees of freedom, there is no difference with a massless

Weyl spinor. Here, we have all four quantities χ

,η

, and χ

, but they all can

Supersymmetry Demystified

be expressed in terms of χ

and χ

†

using

= χ

= η

= iσ

†T

We now can assemble χ

and η

into a four-component spinor that is called a

Majorana spinor 









iσ

†T



(4.20)

By construction, a Majorana spinor is unchanged by charge conjugation:



= 

which simply expresses the fact that χ

= χ

and η

= η

. In fact, we could have

used this as our starting point (and many authors do) in deﬁning a Majorana spinor;

i.e., we could have started from a Dirac spinor and imposed the antiparticle states

to be the same as the particle states.

Let us write down the lagrangian for a free Majorana particle. We insert

Eq. (4.20) into the Dirac lagrangian, but we also include an overall factor of 1/2

for reasons that will become clear later:





i∂

− m





(4.21)

The theory of a Majorana particle is usually left in this form, but it is instructive to

expand it in terms of the two-component spinors χ

and χ

†

¯σ

i∂

(−iσ

)σ

(iσ

) i∂

†T

−



(−iσ

)χ

+ χ

†

iσ

†T



†

¯σ

i∂

¯σ

μT

i∂

†T

−

(χ

· χ

+ ¯χ

· ¯χ

) (4.22)

where we have again used Eq. (A.5). This expression can be simpliﬁed further using

an integration by parts on the second term to write it as

¯σ

μT

i∂

†T

=−



i∂



¯σ

μT

†T

(4.23)

CHAPTER 4 The Physics of Spinors

This is actually exactly equal to the ﬁrst term of Eq. (4.22), as can be shown using

Eq. (A.36):

−



i∂



¯σ

μT

†T

†

¯σ

i∂

The Majorana lagrangian (4.22) therefore can be simpliﬁed to

= χ

†

¯σ

i∂

−

(χ

· χ

+ ¯χ

· ¯χ

) (4.24)

Note the difference between the Majorana lagrangian and the Dirac lagrangian of

Eq. (4.14), ignoring the difference of normalization of the mass terms for now.

The Dirac lagrangian contains the kinetic terms of two distinct states, χ

and χ

and it connects them through the mass term. By contrast, the Majorana lagrangian

contains a single kinetic term and uses only χ

and its hermitian conjugate to make

up the mass term.

We see why we had included a factor of 1/2 in front of the kinetic term in

Eq. (4.21): This was in order to obtain the canonical normalization after combining

the two kinetic terms of Eq. (4.22) into a single term. As for the factor of 1/2 in front

of the mass term, it is chosen so that the parameter m appearing in the lagrangian

is indeed the mass of the particle, as we will demonstrate in Section 4.9.

4.8 Dirac Spinors and Parity

You might wonder why we consider Dirac spinors at all, given that Weyl spinors

(or, equivalently, Majorana spinors) are more fundamental and simpler. Why bother

with Dirac ﬁelds at all? The answer has to do with parity. Parity transforms χ

and η

into each other. Obviously, a theory containing a single Weyl spinor (or,

equivalently, a single Majorana spinor) cannot be parity-invariant. In order to build

a parity-invariant theory, such as QED or QCD, one needs both chiral states of the

particle, and it is obviously convenient to group these states into a four-component

spinor, which turns out to be nothing other than a Dirac spinor. Of course, the

CPT-conjugate states are also needed to complete the theory.

Consider the Dirac mass term written in the form (4.9) (before expressing η

terms of χ

†

−m



†

+ χ

†



Supersymmetry Demystified

This is clearly parity-invariant because it is invariant under the exchange of η and

χ. On the other hand, a Majorana mass term obviously violates parity (there is no

ﬁeld to switch with the χ

!).

4.9 Definition of the Masses of Scalars

and Spinor Fields

If you are used to working with Dirac spinors, the presence of the factor of 1/2

in the mass terms of the Majorana and Weyl spinors may be somewhat unsettling.

Before discussing this issue, let us ﬁrst talk about scalar ﬁelds. The lagrangian of

a free real scalar ﬁeld A is

∂

A ∂

A −

(4.25)

How do we know that the parameter m is indeed the mass of the particle? By

working out the equation of motion of the ﬁeld and comparing it with the Klein-

Gordon equation. For a scalar ﬁeld φ, the Euler-Lagrange equation is

∂



∂L

∂(∂

φ)



−

∂L

∂φ

= 0 (4.26)

Applying this to Eq. (4.25), we simply get

∂

A + m

A = 0

which is the Klein-Gordon equation for a particle of mass m. The operator ∂

∂

often represented by the symbol

 and is called the d’Alembertian operator.

Our result shows that the factor of 1/2 was necessary in Eq. (4.25) for m to

be the mass of the particle. Now consider instead a complex scalar ﬁeld φ whose

lagrangian is given by

= ∂

φ∂

†

− m

φφ

†

Note that there are no factors of 1/2 now! We want to check that the m appearing

there is indeed the mass of the particle. This time, in applying the Euler-Lagrange

CHAPTER 4 The Physics of Spinors

equation, we treat φ and φ

†

as independent ﬁelds. Applying Eq. (4.26), we get

∂

†

+ m

†

= 0

which is again the Klein-Gordon for the ﬁeld φ

†

with mass m. On the other hand,

the Euler-Lagrange equation for φ

†

yields again the Klein-Gordon equation, but

this time for the ﬁeld φ, still with mass m. It is clear why the factor of 1/2 is required

in the real scalar ﬁeld lagrangian and not in the case of the complex scalar ﬁeld: In

the ﬁrst case, the real scalar ﬁeld appears twice in each term of the lagrangian.

Let us now turn our attention to spinor ﬁelds. Roughly speaking, a Majorana

spinor is to a Dirac ﬁeld what a real scalar ﬁeld is to a complex scalar ﬁeld.

Indeed, a Majorana spinor obeys what is essentially a reality condition for spinors,



= 

. By analogy with scalar ﬁelds, we would expect the free Majorana

lagrangian to contain a factor of 1/2 relative to the Dirac lagrangian. This is, of

course, a purely hand-waving argument, but we can demonstrate that it is indeed

correct by working out the Euler-Lagrange equations, which we write again as

applied to the spinor components χ

∂



∂L

∂(∂

)



−

∂L

∂χ

= 0 (4.27)

We now apply this to the lagrangian Eq. (4.24) [we will drop the label p on

the spinors because there are no antiparticle spinors in Eq. (4.24) and therefore no

possibility of confusion]. The only thing we must be careful about is that derivatives

with respect to spinor components are Grassmann quantities, so they anticommute

with spinors. For example,

∂

∂χ

(χ ·χ) =

∂

∂χ

[χ

(−iσ

)

]

=−



∂χ



(iσ

)

+ χ

(iσ

)

∂χ

where we picked up a minus sign in the second term by passing the derivative

through the spinor component χ

. Now we use

∂χ

= δ

Supersymmetry Demystified

where the δ is just the usual Kronecker delta; i.e., δ

= δ

= 1, whereas δ

= δ

0. We then have

∂

∂χ

χ ·χ = δ

(−iσ

)

+ χ

(iσ

)

= (−iσ

)

+ χ

(iσ

)

= (−iσ

)

+ χ

(−iσ

)

=−2(iσ

)

(4.28)

Of course,

∂

∂χ

¯χ · ¯χ = 0

Using Eq. (4.28), we ﬁnd

∂L

∂χ

= im(σ

)

(4.29)

This takes care of the second term of Eq. (4.27). Now consider the ﬁrst term:

∂



∂

∂(∂

)

(χ

†

¯σ

i∂

χ)



=−i



∂

†



( ¯σ

)

where the minus sign came from passing the derivative with respect to ∂

χ to the

right of χ

†

, and we did not show explicit spinor indices to simplify the notation.

This may be written as

−i



∂

†



( ¯σ

)

=−i ( ¯σ

μT

)

∂

†T

(4.30)

Substituting Eqs. (4.29) and (4.30) into Eq. (4.27), the Euler-Lagrange equation

for χ is ﬁnally

−i ¯σ

μT

∂

†T

−imσ

χ = 0

Using the fact that σ

= 1 and multiplying by i, we may write this as

¯σ

μT

∂

†T

+ m σ

χ = 0

CHAPTER 4 The Physics of Spinors

If we multiply this from the left by yet another σ

and use Eq. (A.8), we ﬁnally get

the Euler-Lagrange equation for χ:

∂

†T

+ mχ = 0 (4.31)

On the other hand, a similar calculation shows that the Euler-Lagrange equation

for χ

†

¯σ

∂

χ −m χ

†T

= 0 (4.32)

If we now isolate χ

†T

from Eq. (4.32) and substitute it into Eq. (4.31), we ﬁnd

¯σ

∂

χ +m

χ = 0 (4.33)

where we again used σ

= 1. With the help of Eq. (A.9), we may write the ﬁrst

term as

¯σ

∂

χ = 2∂

∂

χ −σ

¯σ

∂

But, after renaming the indices μ ↔ ν in the last term, we get

¯σ

∂

χ = 2∂

∂

χ −σ

¯σ

∂

which implies

¯σ

∂

χ = ∂

∂

χ (4.34)

Using this in Eq. (4.33), we ﬁnally get that χ obeys

∂

χ +m

χ = 0

the Klein-Gordon equation for a particle of mass m. One can easily show that χ

†

obeys the same equation. This completes the proof that with the factor of one-half

included in Eq. (4.24), the parameter m is indeed the mass of the fermion.

4.10 Adding a Mass: Weyl Spinor

It is sometimes stated in the literature that it is not possible to write a mass term

for a theory containing only a left-chiral particle (together with its right-chiral

antiparticle), that it is necessary to introduce a right-chiral particle state. This is