from wave functions to quantum ﬁeldspoly › ~paganini › ch2.pdf · chapter 2 from wave...

Chapter 2

From wave functions to quantumfields

Few references:F. Halzen and D. Martin, “Quarks & Leptons”, Chapter 3 and 5, Wiley.P. Fayet, “Champs relativistes”, Chapter 8, 10, 11, les editions de l’ecole polytechnique.

The purpose of this chapter is to clearly define the mathematical objects thatdescribe particles of various nature: bosons (spin 0 and spin 1) or spin 1/2fermions. This is an important step toward the calculation of Feynman dia-gram.

2.1 Schrodinger equation

2.1.1 Correspondence principle

In quantum mechanics, the correspondence principle states that a quantum of momentum ~p is awave with a wavevector ~k through the relation postulated by De Broglie ~p = ~~k and its energyE to the angular frequency !: E = ~! (initially postulated by Einstein for photons). Then,Schrodinger expressed the phase of a plane wave as a complex phase factor using:

(~x, t) = Ne�i(!t�~k.~x) = Ne� i~ (Et�~p.~x) = Ne� i

~pµxµ(2.1)

Realizing that such wave is an eigenstate of the operators:

H = i~ @@t

, ~P = �i~~r (2.2)

with the corresponding eigenvalue E and ~p, leads immediately to the identification of H as the

energy operator and ~P as the momentum operator, or in a covariant way (see definition 1.16and 1.15):

Pµ = i@µ = i@

@xµ=

H

� ~P

!(2.3)

37

38 From wave functions to quantum fields

Pµ being the operator 4-momentum. Moreover, the conservation of the energy (non relativistic)

E = p2

2m + V , V being a potential, implies the Schrodinger equation:

H =1

2m~P 2 + V ) i~ @

@t (~x, t) = � ~2

2m~r2 (~x, t) + V (~x, t) (2.4)

The plane wave is solution of the equation for free particles (namely with V =0).

2.1.2 4-current

In what follows we are going to use probability density ⇢ and probability current (i.e. densityflux) ~j, since we are generally interested in moving particles. Probabilities are conserved quanti-ties in the sense that @

@t

Runiverse ⇢d

3~x = 0 which leads to the local version through the continuityequation:

@⇢

@t+ ~r.~j = 0 (2.5)

Indeed, this equation simply states that the rate of decrease of the number of particles in avolume V: � @

@t

RV ⇢ d3~x must be equal to the total flux of particles escaping from that volumeH

S~j.~n dS, ~n, being a unit vector normal to the element dS of the surface S enclosing the volume

V. Because of Gauss theorem, the closed path integral is equal toRV~r.~j d3~x so that:

� @

@t

ZV⇢ d3~x =

ZV~r.~j d3~x

Since this equation must be true for any volume V, we find the continuity equation 2.5. In orderto clearly identify ⇢ and ~j, we first conjugate the Schrodinger equation and multiply by :

�i~ (~x, t)@

@t ⇤(~x, t) = � ~2

2m (~x, t)~r2 ⇤(~x, t) + V ⇤(~x, t) (~x, t)

while multiplying by ⇤(~x, t) the Schrodinger equation we obtain:

i~ ⇤(~x, t)@

@t (~x, t) = � ~2

2m ⇤(~x, t)~r2 (~x, t) + V ⇤(~x, t) (~x, t)

Subtracting the 2 previous results leads to:

i~( ⇤ @

@t +

@

@t ⇤) = � ~2

2m( ⇤~r2 � ~r2 ⇤) ) i~ @

@t( ⇤) +

~2

2m~r.( ⇤~r � ~r ⇤) = 0

A simple comparison with the continuity equation 2.5 allows the identification:

⇢ = | |2 , ~j = i~

2m( ~r ⇤ � ⇤~r ) (2.6)

Therefore, for a plane wave (equ. 2.1), the number of particles per unit volume is ⇢ = |N |2 and~j = |N |2 ~p

m = |N |2~v.

P.Paganini Ecole Polytechnique Physique des particules

Spin 0 particles 39

2.2 Spin 0 particles

2.2.1 Klein-Gordon equation

We can apply the correspondence principle equations 2.2 to the relativistic equation for energy:E2 = |~p|2 + m2 (going back to natural units). Thus:

� @2

@t2� = �r2�+ m2� ) (⇤ + m2)� = 0 (2.7)

which is the formula of the Klein-Gordon equation for a free particle. What are the density andthe current with such equation? Following the same procedure as for the Schrodinger equation(namely taking the conjugate, multipling by � and subtracting the result to the Klein-Gordonequation times �⇤, we obtain:

@2

@t2�� r2�+ m2� = 0

⇤�! @2

@t2�⇤ � r2�⇤ + m2�⇤ = 0

⇥��! � @2

@t2�⇤ � �r2�⇤ + m2�⇤� = 0

@2

@t2�� r2�+ m2� = 0

⇥�⇤��! �⇤ @2

@t2�� ⇤r2�+ m2�⇤� = 0

) �⇤ @2

@t2�� @2

@t2�⇤ � �⇤r2�+ �r2�⇤ = 0

) @@t(�

⇤ @@t�� @

@t�⇤) + ~r.(�~r�⇤ � �⇤~r�) = 0

The last equation can be multiplied by any constant. In order to find a formula similar to theSchrodinger case, we multiply it by i, so that finally:

⇢ = i(�⇤ @

@t��

@

@t�⇤) , ~j = i(�~r�⇤ � �⇤~r�) (2.8)

Let us examine the consequences with plane wave 2.1. Applying the Klein-Gordon equation onplane wave:

�E2�+ |~p|2�+ m2� = 0 ) E = ±p

|~p|2 + m2

The 2 values for the energy are possible: +E and �E where E =p

|~p|2 + m2. Now, injectingthe plane wave into the density and current equation:

⇢ = 2E|N |2 , ~j = 2~p|N |2

or in a more compact way:

jµ =

✓⇢~j

◆= 2|N |2

✓E~p

◆= 2|N |2pµ (2.9)

Using the negative solution for the energy and injecting it into the density, we see that thedensity itself can be negative! It excludes the interpretation of the density as a probabilitydensity!

2.2.2 Re-interpretation of the 4-current

The 2 solutions for the energy means that both wave functions �1

= Ne�i(Et�~p.~x) and �2

=Ne�i(�Et�~p.~x) are solution of the Klein-Gordon equation leading to the 4-current:

jµ(�1

) = 2|N |2✓

E~p

◆, jµ(�

2

) = 2|N |2✓

�E~p

◆

P.Paganini Ecole Polytechnique Physique des particules avancee


The probability interpretation is impossible. However, Pauli and Weisskopf suggested (6 yearsafter the development of Dirac’s equation) to interpret the 4-current as a 4-current chargedensity. In order to do so, the direction of the current has to change as the charge changes. Itis possible if the direction of the momentum is changed. So let us define the 2 wave functions:

�(+)

~p = Ne�i(Et�~p.~x) = Ne�ip.x , �(�)

~p = Ne�i((�E)t�(�~p).~x) = Ne+ip.x

where p.x is a shorthand notation for pµxµ. The 2 functions �(+)

~p and �(�)

~p are still solution ofthe Klein-Gordon equation and now lead to the 4-current:

jµem = i(�⇤@µ�� @µ�⇤) ) jµ

em(�(±)

~p ) = ±2|N |2✓

E~p

◆(2.10)

Hence, the 2 wave functions �(±)

~p can now be interpreted as states of opposite electric charges.At the time of Pauli and Weisskopf interpretation, Dirac had already proposed his equation andCarl D. Anderson had discovered the anti-electron, namely the positron in 1932. It was then

logical to consider �(�)

~p as a state of an antiparticle with a positive energy. Antiparticle means

a charge conjugate state to the positive-energy state (�(�)

~p = �(+)

~p

⇤) with the same mass, with

opposite charge and reversed momentum direction. There is no place here for negative energiesor propagation backward in time. We can just say that an antiparticle can be represented as aparticle as soon as pµ is reversed. It is going to be even more clear with a quantized field.

2.2.3 Few words about the quantized field

In this section, we just want to give some basic ideas without demonstration. We refer to theQuantum Field Theory courses for the details. See for example [4].

So far, we have implicitly considered that the Klein-Gordon equation describes a single rela-tivistic particle of spin 0. Indeed, if it were describing particles with non-zero spin, the solutionsof the Klein-Gordon equation should be able to distinguish particles with di↵erent spin projec-tion. However, there is no place in the solution for such extra-degree of freedom. Thus, theKlein-Gordon equation seems consistent with the description of a single relativistic particle ofspin 0. However, as soon as we are in the relativistic regime, this interpretation cannot hold.Indeed, imagine a single particle of mass m in a box of size L. Because of Heisenberg uncer-tainty, �p � ~/L. Since the particle is relativistic, E ⇡ pc and hence �E � ~c/L. But thisuncertainty can potentially exceed the threshold of production from the vacuum of a pair ofparticle anti-particule �E = 2mc2. We can then conclude, that as soon as a particle is localizedwithin a distance called the Compton wavelength of order � = ~/mc (forgetting the factor 2),the probability to detect a pair of particle anti-particle becomes large. Thus, the notion of asingle particle within a volume becomes a non-sense. The formalism must be able to describestates of any number of particles. This is precisely what does the quantum field theory bychanging the wave functions with operators.

Let us use N = 1 as the normalization of the wave functions

�(±)

~p (x) = e�i(±p.x) (2.11)

The wave functions �(±)

~p (x) constitute a basis, and hence any wave function can be expressedas:

�(x) =

Zd3~p

(2⇡)32Ep

hap�

(+)

~p (x) + b⇤p�

(�)

~p (x)i

, Ep =p

|~p|2 + m2 (2.12)


Few words about the quantized field 41

where the volume of integration is large enough (infinity) to consider the quantization of mo-mentum to be continuous. ap and b⇤

p are the complex amplitudes of the eigenmodes of � andcorrespond to the Fourier transform coe�cients of the wave functions basis.

The quantisation of the wave function (classical field) transforms � in an operator acting on theHilbert space of state vectors. � still obeys the same dynamic equation (Klein Gordon) but ap

and b⇤p are now operators. To better emphasize this, b⇤

p is now written b†p. The quantum field

with its hermitian adjoint are then:

�(x) =R d3~p

(2⇡)

3

2Ep

hap�

(+)

~p (x) + b†p�

(�)

~p (x)i

�†(x) =R d3~p

(2⇡)

3

2Ep

ha†

p�(+)⇤~p (x) + bp�

(�)⇤~p (x)

i (2.13)

One can show that ap and a†p obey the same commutation rules as the one of the harmonic

oscillator and thus, are respectively annihilation and creation operator. a†p creates a quantum

of excitation associated to the plane wave �(+)

~p which plays the role of a propagation mode

with vector ~p (or ~k). This quantum is interpreted as a particle of mass m propagating with

momentum ~p. bp and b†p form another set of annihilation and creation operators associated

to the plane wave �(�)

~p and similarly, b†p creates a particle with the same mass m as a†

p. Asannihilation and creation operators they satisfy the commutation relations:

hap0 , a†

p

i=hbp0 , b†

p

i= (2⇡)32Ep �(3)(~p 0 � ~p)⇥

ap0 , ap

⇤=⇥bp0 , bp

⇤=ha†

p0 , a†p

i=hb†p0 , b

†p

i= 0h

a†p0 , b

†p

i=hap0 , b†

p

i=ha†

p0 , bp

i= 0

(2.14)

The vacuum state |0i is the state with 0 particle properly normaized h0|0i = 1. With ournormalization1, we have:

ap |0i = 0

a†p |0i = |1a, ~pi

ap |1a, ~pi = |0i

where |1a, ~pi correspond to a state with 1 particle of type a having a momentum ~p. Similarrelations are obtained applying |0i on b operators. Since the operators a and b commute,bp |1a, ~pi = 0. one can then show that the hamiltonian, momentum and charge operators are:

H =R d3~p

(2⇡)

3

2EpEp (a†

pap + b†pbp)

P =R d3~p

(2⇡)

3

2Ep~p (a†

pap + b†pbp)

Q =R d3~p

(2⇡)

3

2Ep(a†

pap � b†pbp)

1In some textbooks, the operators used are of the kind a†~p i.e for the 3-momentum. The connection to the

notation used here is a†p =

p2Epa

†~p .



Let us see the value of the energy, momentum and charge of the state |1a, ~pi.

H |1a, ~pi =R d3~p 0

(2⇡)

3

2E0p

E0p (a†

p0ap0 + b†p0bp0) |1a, ~pi

=R d3~p 0

(2⇡)

3

2E0p

E0p a†

p0ap0a†p |0i

=R d3~p 0

(2⇡)

3

2E0p

E0p a†

p0([ap0a†p] + a†

pap0) |0i=R d3~p 0

(2⇡)

3

2E0p

E0p a†

p0(2⇡)32Ep �(3)(~p 0 � ~p) |0i= Ep a†

p |0i= Ep |1a, ~pi

The same procedure can be followed for P and Q with the states |1a, ~pi and |1b, ~pi and we find:

H |1a, ~pi = Ep |1a, ~pi , H |1b, ~pi = Ep |1b, ~piP |1a, ~pi = ~p |1a, ~pi , P |1b, ~pi = ~p |1b, ~piQ |1a, ~pi = 1 |1a, ~pi , Q |1b, ~pi = �1 |1b, ~pi

Hence, the 2 states |1a, ~pi and |1b, ~pi have the same energy, same momentum (and thus samemass since m =

pE2 � |~p|2) but opposite charge: |1b, ~pi is just the antiparticle of |1a, ~pi (and

vise-versa). Moreover, since we can apply the creation operator several times, we can havestates like |na, ~pi with na > 1. Several particles in the same quantum state of propagation isonly possible for bosons (Pauli exclusion principle). Here is another way of realizing the bosonnature: consider a Fock space with only 2 particles. The wave function is related to:

~p1

,~p2

( ~x1

, ~x2

) = h ~x1

, ~x2

|1, ~p1

; 1, ~p2

i= (h ~x

1

| ⌦ h ~x2

|)(|1, ~p1

i ⌦ |1, ~p2

i)= (h ~x

1

| ⌦ h ~x2

|)a†p1

a†p2

|0i= (h ~x

1

| ⌦ h ~x2

|)a†p2

a†p1

|0i= h ~x

1

, ~x2

|1, ~p2

; 1, ~p1

i= ~p

1

,~p2

( ~x2

, ~x1

)

As expected, the wave function of two identical bosons is symmetric under the interchange ofthe two particles.

We notice that a spatial rotation or a change of reference frame doesn’t a↵ect the internalstructure of the field since it is a scalar field. Hence, the theory describes particles with spin 0that can have an electric charge. A remark about neutral particles: according to the expressionof Q, the charge of a state of a single particle can be zero if a = b. Hence �(x) = �†(x): thescalar field is hermitian. The particle is then its own antiparticle: a well known example is the⇡0.

Now that we have a better understanding of the operators, looking at the field formula 2.13,we see that � destroys a positive charge or creates a negative charge corresponding to the an-tiparticle. The global e↵ect of � is to decrease the total charge of the system by 1 unit. On thecontrary, the e↵ect of �† is to increase the total charge of the system by 1 unit. A term involving�†(x)�(x) (appearing in the Lagrangian) doesn’t change the global charge of the system. Suchquadratic term can describe at the spacetime x any of the following charge-conserving processes:1) creation and annihilation of a particle, 2) creation and annihilation of an antiparticle, 3) cre-ation of a pair particle-antiparticle, 4) annihilation of a pair particle-antiparticle.

Finally, one may wonder what is the connection between the usual wave functions �±(x) =e�i(±p.x) and the quantum field �(x) of formula 2.13? Since usual plane wave functions deal


Spin 1/2 particles 43

with a single particle, one has to project the field into a single state particle. More precisely, itis easy to check that:

h0|�(x)|1a, ~pi = e�ip.x

h1b, ~p |�(x)|0i = e+ip.x

For instance,

h0|�(x)|1a, ~pi =R d3~p 0

(2⇡)

3

2E0p

hh0|ap0 |1a, ~pi�(+)

~p0 (x) + h0|b†p0 |1a, ~pi�(�)

~p0 (x)i

=R d3~p 0

(2⇡)

3

2E0p

hh0|ap0a†

p|0i�(+)

~p0 (x) + h0|b†p0a

†p|0i�(�)

~p0 (x)i

=R d3~p 0

(2⇡)

3

2E0p

hh0|0i (2⇡)32Ep0�(3)(~p0 � ~p)�(+)

~p0 (x)i

= �(+)

~p (x) = e�ip.x

and similarily for h1b, ~p |�(x)|0i.

2.3 Spin 1/2 particles

2.3.1 Dirac equation

In 1928, P.A.M Dirac proposed his famous equation in order to avoid the solution with a negativeenergy and a negative density probability of the Klein-Gordon equation. It was 6 years before thecorrect interpretation of the Klein-Gordon equation. Even if it was not his initial goal, Diracfound that his equation was able to describe particles and antiparticles with half spin unit.Starting from the Klein-Gordon equation, he realized that the solution with negative energywas due to the second derivative @2/@t2 leading to a probability density (old interpretation)involving a single derivative @/@t, and thus allowing a negative probability. Hence he looked foran equation having a @/@t dependency as in the Schrodinger equation. The equation should becovariant under Lorentz transformations, and hence, the dependency must be also linear withr. Moreover,we want the wave function to still satisfy the Klein-Gordon equation:

(@µ@µ + m2) = 0

So basically, the idea is to factorize the previous equation:

(�@ � im)(��@� + im) = 0

Where �µ and �� are a priori 2 sets of 4 numbers. If satisfies

(�@ � im) = 0 (2.15)

or

(��@� + im) = 0 (2.16)

then the Klein-Gordon equation will be satisfied as well. Equations 2.15 or 2.16 fulfill thelinearity condition with the derivatives. Let us develop and identify the di↵erent terms:

(@µ@µ + m2) = (�@ � im)(��@� + im) = (��@@� + im(�@ � ��@�) + m2)



In order to cancel the linear term with m, we see that � = �. The identification with theD’Alembertian term (@µ@µ) then imposes:

@µ@µ = ��@@�

namely:

@2

0

� @2

1

� @2

2

� @2

3

= (�0)2@2

0

+ (�1)2@2

1

+ (�2)2@2

2

+ (�3)2@2

3

+(�0�1 + �1�0)@0

@1

+ (�0�2 + �2�0)@0

@2

+ (�0�3 + �3�0)@0

@3

+(�1�2 + �1�2)@1

@2

+ (�1�3 + �1�3)@1

@3

+(�2�3 + �3�2)@2

@3

If the �’s were complex numbers, the first line of the equality would impose �0 = ±1 and�k=1,2,3 = ±i. However, It would be impossible to cancel the last 3 lines. Dirac then proposedto interpret the �’s as matrices satisfying:

(�0)2 = l1 , (�k=1,2,3)2 = � l1�µ�⌫ + �⌫�µ = 0 for µ 6= ⌫

which can be compacted with:

{�µ, �⌫} = 2gµ⌫ l1 (2.17)

where {} is the anticommutator {a, b} = ab + ba. The relation 2.17 is known as the Cli↵ordalgebra. What is the minimal rank of the �’s? Knowing that Tr(AB) = Tr(BA) and since�0�0 = 1 we have

Tr(�k) = Tr(�k�0�0) = Tr(�0�k�0) = �Tr(�0�0�k) = �Tr(�k)

where the negative sign is due to the � matrices properties. Thus Tr(�k) = 0. Similarly,

Tr(�0) = �Tr(�0�k�k) = �Tr(�k�0�k) = Tr(�k�k�0) = �Tr(�0)

where we applied successively, �k�k = � l1, Tr(AB) = Tr(BA), �µ�⌫ + �⌫�µ = 0 and �k�k = � l1.So, Tr(�µ) = 0. But the trace of a matrix is equal to the sum of its eigenvalues. Since �0�0 = l1,the eigenvalues of �0 are ±1. Similarly, the eigenvalues of �k are ±i. The trace being 0, theremust be as many positive eigenvalues as negative ones. Hence the rank n is necessarily an evennumber. Rank n = 2 is not possible, because there are only 3 independent traceless matricesin the vector space generated by the 2 ⇥ 2 matrices. But we need 4 independent matrices!(because of the dimension of the Minkowski space). Hence, the �µ’s with minimal rank are 4⇥4hermitian matrices. Several choices are possible and the most common one, called the Dirac’srepresentation is:

�0 =

✓l1 0000 � l1

◆, �i =

✓00 �i

��i 00

◆(2.18)

where the symbols l1 and 00 denote respectively the 2⇥ 2 unit matrix and zero matrix, and �i arethe Pauli spin matrices seen in the first chapter:

�1 =

✓0 11 0

◆, �2 =

✓0 �ii 0

◆, �3 =

✓1 00 �1

◆


4-current and the adjoint equation 45

Actually, any matrix U�U�1 where U is a 4 ⇥ 4 unitary matrix is a valid choice. We can easilycheck that the definition of the �’s matrices implies for the hermitian conjugates2:

�0† = �0 , �1† = ��1 , �2† = ��2 , �3† = ��3 (2.19)

or in more concise:�µ† = �0�µ�0 (2.20)

We can now come back to the 2 “square root” equations 2.15 and 2.16 of the Klein-Gordonequation. By convention, the Dirac equation is 2.16 (with � = �) multiply by i (remember thati@µ = Pµ):

(i/@ � m) = 0 with /@ = �µ@µ (2.21)

Since the �’s are 4 ⇥ 4 matrices, the wave function is actually a wave function with 4 compo-nents:

=

0BB@

1

2

3

4

1CCA

It is called a Dirac-spinor or 4-spinor. At this point, we can already begin to see that the extramultiplicity is likely to have something to do with an angular momentum degree of freedom assuggested by the presence of the Pauli matrices in the �’s definition.

2.3.2 4-current and the adjoint equation

In order to find the 4-current, let us proceed as for the Klein-Gordon or Schrodinger equations.This time however, the complex conjugate must be replaced by the hermitian conjugate becauseof the presence of matrices. The hermitian conjugate of 2.21 is

(�i@µ †�µ† � m †) = 0 (2.22)

where the hermitian adjoint † is:

† = ( ⇤1

, ⇤2

, ⇤3

, ⇤4

) (2.23)

Now, using equality 2.19, equation 2.22 becomes:

(�i@0

†�0 + i@k †�k � m †) = 0

and multiplying by �0 from the right side:

(�i@0

†�0�0+i@k †�k�0�m †�0) = �(i@

0

†�0�0+i@k †�0�k+m †�0) = �(i@µ

†�0�µ+m †�0) = 0

where the anticommutation relations have been used. Now let us define the row Dirac adjoint:

= †�0 = ( ⇤1

, ⇤2

, � ⇤3

, � ⇤4

) (2.24)

so that satisfies the adjoint equation:

i@µ �µ + m = 0 (2.25)

2A† = (A⇤)t where A⇤ means complex conjugate and t the transpose matrix.



We can now derive the continuity equation by multiplying the previous equation from the rightby , equation 2.21 from the left by and add the result:

i(@µ )�µ + i �µ@µ = i@µ( �µ ) = 0

so that the conserved 4-current satisfying the previous continuity equation is:

jµ = �µ (2.26)

leading to the density:

⇢ = �0 = †�0�0 = † ) ⇢ =4X

i=1

| i|2 (2.27)

This time the density is always positive and can be interpreted as a probability density.

As for boson, we can define a charge current by simply put by hands the charge of the particleq = ±1:

jµq = qjµ = q �µ (2.28)

Contrary to the boson case, changing the sign of p won’t change jµ. Here, we have to explicitlyput the correct charge.

2.3.3 Free-particles solutions

The way the Dirac’s equation was built shows that a solution of the equation is also a solutionof the Klein-Gordon equation, or more precisely, each component of the 4-spinor is a solutionof the Klein-Gordon equation. Then, it is natural to look for solutions in which the space-timebehaviour is the one of the plane-wave:

(x) = ue�ip.x (2.29)

where, this time, u is a 4-spinor that doesn’t depend on x and p.x the 4-momentum product(we omit the index µ for simplicity). A general solution of the Dirac’s equation can alwaysbe expressed as a linear combination of plane-wave solutions. Injecting 2.29 into the Dirac’sequation 2.21, we get the so-called momentum space Dirac equation:

(i�µ ⇥ �ipµ � m)u = 0 ) (/p � m)u = 0 (2.30)

Let us write this 4-spinor as a 2-component spinor:

u =

✓ua

ub

◆(2.31)

ua and ub are bi-spinor.


Free-particles solutions 47

2.3.3.1 Solutions for particle at rest

It is very instructive to first examine the solutions for a particle at rest where only p0 = E isnon zero. Equation 2.30 simplifies to:

E�0u � mu = 0 ) E

✓ua

�ub

◆= m

✓ua

ub

◆)

⇢Eua = mua

Eub = �mub

The equation satisfied by ua corresponds to 2 independent orthogonal solutions

✓10

◆and

✓01

◆

both with E = m while the 2 independent orthogonal solutions for ub have E = �m. Comingback to 2.29, the 4 solutions are then:

1

=

0BB@

1000

1CCA e�imt ,

2

=

0BB@

0100

1CCA e�imt ,

3

=

0BB@

0010

1CCA e+imt ,

4

=

0BB@

0001

1CCA e+imt (2.32)

1

and 2

having E = m > 0 and 3

and 4

having E = �m < 0. Finally, even in Dirac’sequation we have negative energy solutions! However, this time, they don’t lead to a negativeprobability density. These negative energy solutions will be interpreted in the next section assolutions for antiparticles.

2.3.3.2 General solutions

Let us come back to the general case with ~p 6= 0. Injecting 2.31 into 2.30, we have:

(�µpµ � m)

✓ua

ub

◆= (�0E � ~�.~p � m)

✓ua

ub

◆

=

✓✓E.l1 0000 �E.l1

◆�✓

00 ~�.~p�~�.~p 00

◆�✓

m.l1 0000 m.l1

◆◆✓ua

ub

◆

=

✓(E � m).l1 �~�.~p

~�.~p �(E + m).l1

◆✓ua

ub

◆

(2.33)so ua and ub must satisfy the 2 coupled equations:

ua =~�.~p

E � mub (2.34)

ub =~�.~p

E + mua (2.35)

These equations impose a constraint on E and ~p. Indeed, let us inject 2.35 into 2.34:

ua =(~�.~p)2

E2 � m2

ua

but:

~�.~p =

✓0 11 0

◆px +

✓0 �ii 0

◆py +

✓1 00 �1

◆pz =

✓pz px � ipy

px + ipy �pz

◆(2.36)



And thus:(~�.~p)2 = |~p|2 l1

so that, the non trivial solution (ua 6= 0) requires:

|~p|2E2 � m2

= 1 ) E = ±p

|~p|2 + m2

ua is a spinor with 2 components: thus, we can choose for ua any 2 orthogonal spinors �1

and �2

,ub will have then to satisfy 2.35. Similarly, we can decide to chose first ub from any 2 orthogonalspinors �

1

and �2

, and then ua will have to satisfy 2.34. In both cases, the relativistic energyconstraint must be fulfilled. Hence, 4 independent solutions of the Dirac’s equation are givenby these 4-spinors:

u1

= N1

✓�

1

~�.~pE+m �

1

◆, u

2

= N2

✓�

2

~�.~pE+m �

2

◆, u

3

= N3

✓ ~�.~pE�m �

1

�1

◆, u

4

= N4

✓ ~�.~pE�m �

2

�2

◆

(2.37)where Ni are normalization factors of the spinors. We have to identify which 4-spinors correspondto positive or negative energy. In the limit where ~p ! 0, we should recover the solutions for aparticle at rest, so that we can conclude that u

1

and u2

are the solutions with E = +p

|~p|2 + m2

while u3

and u4

are the solutions with E = �p

|~p|2 + m2.

2.3.3.3 Interpretation of negative energies

Historically, Dirac considered that the vacuum was full (!) of negative energy states satisfyingthe Pauli exclusion principle. This theory is usually referred as the “Dirac sea”. A hole in thenegative energy states was then interpreted as an antiparticle with positive energy and oppositecharge. This interpretation was successful in predicting the positron (discovered by Anderson in1932) but had many drawbacks. For instance, the vacuum being full of fermions with negativeenergy states didn’t give any explanations why bosons couldn’t populate the vacuum as well(which would be dramatic since they are not concerned by the Pauli exclusion principle). Infact, Stuckelberg in 1941 and Feynman in 1948 proposed the correct interpretation still valid withquantum field theory and briefly mentioned in the previous case of the Klein-Gordon equation.Consider the 2 solutions u

3

and u4

with negative energy in 2.37. They can be rewritten:

u3

= N3

� ~�.~p

|E|+m �1

�1

!, u

4

= N4

� ~�.~p

|E|+m �2

�2

!

and both are associated to a propagation term e�i(�|E|t�~p.~x). But this term can be simplywritten e�i(|E|(�t)�~p.~x), so that it looks as a particle travelling backward in time form (t

2

, x1

) to(t

1

, x2

) where t2

> t1

as shown on schema b of figure 2.1. The picture is clearly not equivalentto an electron travelling “normally” from (t

1

, x2

) to (t2

, x2

) as depicted in schema a. Moreover,the propagation term is no more Lorentz invariant. In order to restore the invariance, we alsohave to change direction of the momentum (to have a Lorentz scalar p.x) so that we obtain theschema c except that the final point in the spacetime is exchanged with the initial point.Feynman’s approach consists in interpreting schema c as the antiparticle of schema a withpositive energy and travelling “normally” from (t

1

, x2

) to (t2

, x2

). We can now define the twonew spinors v

1

and v2

for the antiparticle, interpreting E as a positive quantity and invertingthe momentum:

u3,4(�E, �~p)e�i((�E)t�(�~p).~x) = v

1,2(E, ~p)e+i(Et�~p.~x)



Physique des particules M1 HEP X Pascal Paganini LLR-IN2P3-CNRS 21

3) Spin 1/2 particles�

x1 � x2 �

t1 �

t2 �

x1 � x2 �

t1 �

t2 �

x1 � x2 �

t1 �

t2 �a) � b) � c) �

Figure 2.1: Feynman approach of the antiparticles: left, a particle travelling from x1

to x2

. Centre:time is reversed. Right: antiparticle as a particle with �t and �~p.

with:

v1

= N 01

✓ ~�.~pE+m �

1

�1

◆, v

2

= N 02

✓ ~�.~pE+m �

2

�2

◆(2.38)

Injecting ve+ip.x into the Dirac equation 2.21, we see that v satisfies the momentum space Diracequation:

(/p + m)v = 0 (2.39)

2.3.3.4 Solutions with normalization

So far, we left over the normalization of the spinors. As for the boson case, we are going tonormalize to 2E particles per unit volume 3. Using the density 2.27, we require:Z

V=1

† = u†u = 2E

where u can be u1

, u2

, v1

or v2

. Doing the job for vi=1,2 as an example:

u†u = |N 0i |2((

~�.~pE+m �i)†,�†

i )

✓ ~�.~pE+m �i

�i

◆

= |N 0i |2(�

†i (

~�.~pE+m)† ~�.~p

E+m�i + �†i�i)

= |N 0i |2(�

†i (

~�.~pE+m)2�i + 1)

= |N 0i |2(

|~p|2(E+m)

2

+ 1)

= |N 0i |2 2E

E+m

where moving from the second line to the third, we used (~�.~p)† = ~�.~p and �†i�i = 1. Thus, we

conclude N 0i =

pE + m. And similarly for Ni. Hence the 4 solutions of the Dirac equation are:

E = +p

|~p|2 + m2

1

=p

E + m

✓�

1

~�.~pE+m �

1

◆e�ip.x,

2

=p

E + m

✓�

2

~�.~pE+m �

2

◆e�ip.x

¯

1

=p

E + m

✓ ~�.~pE+m �

1

�1

◆e+ip.x,

¯

2

=p

E + m

✓ ~�.~pE+m �

2

�2

◆e+ip.x

(2.40)

3When we chose N = 1 in 2.11, it implies for the 4-current 2.10, jµ = ±2

✓E~p

◆and hence 2E particles per

unit volume



where 1

and 2

correspond to a particle and ¯

1

and ¯

2

to an antiparticle.

Example of explicit formula: let us consider for example, �1

= �2

as a spin-up state and�

2

= �1

as a spin-down state4 (and thus eigenstates of �3

):

�1

= �2

=

✓10

◆, �

2

= �1

=

✓01

◆

The 4 solutions become:

1

=p

E + m

0BB@

10pz

E+mpx+ipyE+m

1CCA e�ip.x,

2

=p

E + m

0BB@

01

px�ipyE+m

� pzE+m

1CCA e�ip.x

¯

1

=p

E + m

0BB@

px�ipyE+m

� pzE+m

01

1CCA e+ip.x,

¯

2

=p

E + m

0BB@

pzE+m

px+ipyE+m

10

1CCA e+ip.x

(2.41)

where we used equality 2.36.

2.3.3.5 Interpretation in terms of spin and helicity

For a particle or an antiparticle, we always have two solutions degenerated in energy. Theremust be an operator that commutes with the energy operator HD of the Dirac’s equation whoseeigenvalues would distinguish the solutions. First, let us determine HD. Expanding the Dirac’sequation 2.21, we have:

i�0

@

@t + (i~�.~r � m) = 0

Multiplying from the left by �0, and using ~P = �i~r we get:

i@

@t = HD

with the Dirac’s Hamiltonian:

HD = �0~�. ~P + �0m =

✓00 ~�~� 00

◆. ~P + �0m (2.42)

Spin states: Now, we have to find an operator that commutes with HD. Let us try to see ifthe spin operators are satisfactory. A simple generalization of the operators 1.41 based on Paulimatrices for the 4-spinors case is clearly:

~S =1

2~⌃ =

1

2

✓~� 0000 ~�

◆(2.43)

4with this choice solution for particle and antiparticle with the same index, will have a similar interpretationin term of spin.



However, HD and ~S don’t commute because of the presence of ~P . Indeed, for instance with[HD, Sz], we would have a term5:

h~�. ~P,�z

i=h�xPx + �yPy + �zPz,�z

i= [�x,�z]Px + [�y,�z]Py 6= 0

In case of zero momentum, HD reduced to �0m that commutes with ~S and the explicit solutions2.41 (with px = py = pz = 0) are eigenstates of Sz:

~p = 0 ) Sz 1

= +1

2

1

, Sz 2

= �1

2

2

, ˆSz ¯

1

= +1

2

¯

1

, ˆSz ¯

2

= �1

2

¯

2

Note that the spin operator for the antiparticles is defined as:

ˆS = �S

It is justified because with antiparticles, the operator returning the physical momentum (positive

energy, appropriate direction for ~p) is changed from ~P ! � ~P (because antiparticles have a

propagation mode e+ip.x), so that ~L = ~r ^ ~P ! �~L. Thus the conservation of the total angular

momentum ~J = ~L + ~S, requires ~S ! � ~S. Finally, we see that 1

and ¯

1

would have spin upwhile

2

and ¯

2

would have spin down when px = py = pz = 0. This result also holds if the(anti)particle travels in the z-direction (px = py = 0, pz = ±|~p|).

Helicity states: For any ~p, the spin alone cannot distinguish the 2 degenerated solutions(since it’s not a conserved number because the spin operator in general does not commute with

the Hamiltonian). One can show that the total angular momentum ~J = ~L + ~S with ~L = ~r ^ ~Pdoes commute. It is fortunate because otherwise, the total angular momentum would not beconserved! Another possibility is the so-called helicity defined as the projection of the spin onthe direction of the momentum. Hence, the operator is simply defined for fermion spinor as:

h = ~S.~P

|~p| =1

2~⌃.

~P

|~p| =1

2

0@ ~�.

ˆ~P|~p| 00

00 ~�.ˆ~P

|~p|

1A (2.44)

Note that for antifermions, the same operator can be used since ~S ! � ~S and ~P ! � ~P . Thehelicity operator clearly commutes with HD, so there exists a basis of 4-spinors which is botheigenstates of HD and of the helicity. Since the spin on any axis (and thus on the direction of themomentum) can only be ±1/2, the helicity quantum number6 � can only be ±1/2. Dependingon �, the state is called:

� = �1

2

) Left handed helicity� = +1

2

) Right handed helicity

5I use indi↵erently �x for �1

, �y for �2

and �z for �3

.6Note that some authors, define the helicity as twice that number in order to have an integer for both fermions

and bosons.



After some math (see exercise 2.3), one can determine the general solution for helicity states:

+

1

2

=p

E + m

0BBB@

cos ✓2

ei� sin ✓2

|~p|E+m cos ✓

2

|~p|E+mei� sin ✓

2

1CCCA e�ip.x, � 1

2

=p

E + m

0BBB@

� sin ✓2

ei� cos ✓2

|~p|E+m sin ✓

2

� |~p|E+mei� cos ✓

2

1CCCA e�ip.x

+

¯

1

2

=p

E + m

0BBB@

|~p|E+m sin ✓

2

� |~p|E+mei� cos ✓

2

� sin ✓2

ei� cos ✓2

1CCCA e+ip.x, �¯

1

2

=p

E + m

0BBB@

|~p|E+m cos ✓

2


2

cos ✓2

ei� sin ✓2

1CCCA e+ip.x

(2.45)where ✓ and � are the polar angles.

A remark: beware that the helicity is not Lorentz invariant in general. Actually, fora massive particle, one can always find a reference frame where the particle appears to reverseits relative direction of motion.

2.3.4 Operations on spinors

2.3.4.1 Charge conjugation

In classical electrodynamic, the motion of a charged particle with a charge q in an electromagneticfield Aµ = (V, ~A) is obtained by making the substitution:

~p ! ~p � q ~A , E ! E � qV

The quantum version is then simply from the correspondence principle:

pµ ! pµ � qAµ ) i@µ ! i@µ � qAµ ) @µ ! @µ + iqAµ (2.46)

so that the Dirac’s equation becomes:

[�µ(i@µ � qAµ) � m] = 0 (2.47)

Now the charge conjugate state: 0 = C (2.48)

is supposed to satisfy:[�µ(i@µ + qAµ) � m] 0 = 0 (2.49)

In order to find C, let us start by conjugating 2.47:

[�µ⇤(�i@µ � qAµ) � m] ⇤ = 0⇥�0(�i@

0

� qA0

) + �1(�i@1

� qA1

) � �2(�i@2

� qA2

) + �3(�i@3

� qA3

) � m⇤ ⇤ = 0

In order to have the correct sign in front of all �’s we can multiply by �2 from the left and usethe anticommutation relation:⇥

��0(�i@0

� qA0

) � �1(�i@1

� qA1

) � �2(�i@2

� qA2

) � �3(�i@3

� qA3

) � m⇤�2 ⇤ = 0

[�µ(i@µ + qAµ) � m] �2 ⇤ = 0


Operations on spinors 53

Thus, �2 ⇤ seems a good candidate for 0. If represents the wave function of an electron, weexpect 0 to be the one of a positron. Let us check for instance with the electron described by

1

in 2.41:

�2 ⇤ =p

E + m

0BB@

0 0 0 �i0 0 i 00 i 0 0�i 0 0 0

1CCA26640BB@

10pz

E+mpx+ipyE+m

1CCA e�ip.x

3775

⇤

= �ip

E + m

0BB@

px�ipyE+m

� pzE+m

01

1CCA e+ip.x

It would be the positron solution described by ¯

1

except that there is this extra �i. Hence, byusing:

C = i�2 ⇤ (2.50)

we get the correct description.

2.3.4.2 Parity

The parity transformation changes the space coordinates into their opposite:

x = (t, ~x) ! x0 = (t, �~x) (2.51)

How the spinors satisfying the Dirac’s equation 2.21 must be transformed to still satisfy:

(x) ! 0(x0) = P (x) with (i�µ@0µ � m) 0(x0) = 0

Let us express the Dirac’s equation with the new coordinates:

(i�µ@µ � m) = 0

(i�0

@@t0

@t0

@t + i�1

@@x0

@x0

@x + i�2

@@y0

@y0

@y + i�3

@@z0

@z0

@z � m) = 0

(i�0

@@t0 � i�1

@@x0 � i�2

@@y0 � i�3

@@z0 � m) = 0

We need to revert the sign in front of �i=1,2,3. Thus, let us multiply by �0 from the left and usethe anticommutation algebra:

(i�0

@@t0 + i�1

@@x0 + i�2

@@y0 + i�3

@@z0 � m)�0 = 0

(i�µ@0µ � m) 0(x0) = 0

where we identify:

0 = �0 ) P = �0 (2.52)

Now, according to spinors 2.40 solution of the Dirac’s equation, we see that:

u(�~p) = P u(~p) , v(�~p) = �P v(~p)

meaning that the intrinsic parity is +1 for fermion and �1 for antifermion:

⌘Pf = +1 , ⌘P

¯f = �1 (2.53)



2.3.4.3 Chirality

consider the so-called chirality matrix:

�5 = i�0�1�2�3 (2.54)

which in the Dirac representation reads:

�5 =

✓00 l1l1 00

◆(2.55)

It’s easy to show that �5 satisfies:

(�5)2 = 1 , �5

†= �5 (2.56)

Since (�5)2 = 1, the eigenvalues of �5 are ±1. The eigenstates of �5 are called chirality statesand are denoted with the Left L and Right R label for a reason that is going to be clarifiedin few lines. Eigen-spinors with positive (negative) chirality have right (left) chirality and theopposite for eigen-antispinors:

�5uR = +uR , �5uL = �uL , �5vR = �vR , �5vL = +vL (2.57)

The reason why chiral antispinors have opposite eigenvalues with respect to chiral spinor isbecause we want that chirality states match the helicity states in the massless approximation(when m = 0, for both particles and anti-particles, a left-handed chiral state is equal to a left-handed helicity state as shown few lines below). Note that due to the anticommutation algebraof the �’s matrices, we have: �

�5, �µ

= 0 (2.58)

And hence, the commutator with the Dirac’s Hamiltonian 2.42 reduces to:⇥HD, �5

⇤= m

⇥�0, �5

⇤which is di↵erent than zero, except when m = 0. Therefore, for massive particles, the chiralityeigenvalues are not conserved and the physical states having a defined energy (so eigenstates ofthe hamiltonian) will be a mixture of chirality states.

Now, Consider the helicity states 2.45 in the ultrarelativistic limit E � m or equivalently whenm = 0:

u+

1

2

⇡p

E

0BB@

cos ✓2

ei� sin ✓2

cos ✓2

ei� sin ✓2

1CCA , u� 1

2

⇡p

E

0BB@

� sin ✓2

ei� cos ✓2

sin ✓2

�ei� cos ✓2

1CCA

v+

1

2

⇡p

E

0BB@

sin ✓2

�ei� cos ✓2

� sin ✓2

ei� cos ✓2

1CCA , v� 1

2

⇡p

E

0BB@

cos ✓2

ei� sin ✓2

cos ✓2

ei� sin ✓2

1CCA

(2.59)

It is easy to check that these helicity states are now eigenstates of �5 using the representation2.55. Hence, in the limit E � m, the helicity states coincide with the chirality states:

u+

1

2

= uR , u� 1

2

= uL , v+

1

2

= vR , v� 1

2

= vL (2.60)


Operations on spinors 55

and we now understand why the chiral states are called left or right handed. Looking at 2.59, wesee that, in the limit E � m, left handed fermions are equivalent to right handed antifermions(the di↵erence of sign has no physical e↵ects). We will see later that the weak interaction viacharged current (mediated by W bosons) only acts on the chirality left-handed states of particlesand chirality right-handed states of antiparticles. Thus, it is useful to define the projectors onthese states:

PL =1

2(1 � �5) , PR =

1

2(1 + �5) (2.61)

It’s trivial to check that PL and PR have the projectors properties: P 2

L = PL, P 2

R = PR,PLPR = 0, PL + PR = l1. And using 2.59 and 2.60, we see that:

PR uR = uR PR uL = 0 PL uR = 0 PL uL = uL

PR vR = 0 PR vL = vL PL vR = vR PL vL = 0

meaning for example that for spinors, PR projects out right handed states while for antispinors,it projects out left handed states. Therefore, for any (anti)spinors, we can always write:

u = uL + uR uL = PL u uR = PR uv = vL + vR vL = PR v vR = PL v

(2.62)

Now consider the parity transformation of the projector on left-handed state7

PPLP † = �0

1

2(1 � �5)�0† =

1

2(1 + �5)�0�0 = PR

meaning that the chirality changes under a parity transformation as we would expect from itsdenomination (chirality comes from the greek, meaning “hand”: a left hand in a mirror becomesa right hand).

Spin states, helicity states and chirality states are often confused. We invite the reader toconsult the reference [5] for detailed explanations of the di↵erences.

2.3.4.4 Useful formulas and completeness relations

Momentum space equations: We already established the momentum space equations (2.30and 2.39):

(/p � m)u = 0 , (/p + m)v = 0 (2.63)

We wish to see which equations satisfy u and v. For instance, taking the hermitian conjugateof 2.30:

u†(�µ†pµ � m) = 0u†(�0�µ�0pµ � m) = 0

where we used 2.20. Multiplying from the right by �0 and using (�0)2 = 1:

u†(�0�µpµ � m�0) = 0

7The transformation of a wave function under P is P while the transformation of an operator A is PAP †.Hence, if a state |ai is transformed in |a0i = U |ai under a unitary transformation U (i.e. U† = U�1 becauseha|bi = ha0|b0i ) U†U = 1), the variable ha|A|bi must be equal to ha0|A0|b0i and thus: ha|A|bi = ha|U†A0U |bi andso for any state |ai or |bi. We thus have: UAU† = A0. Requiring ha0|A0|b0i = ha|A|bi, is equivalent for a classicalscalar field (temperature for instance), as saying �0(x0) = �(x): the temperature at a rotated point seen by therotated field is the same as the temperature at the initial point.



Applying the same procedure for v we finally get:

u(/p � m) = 0 , v(/p + m) = 0 (2.64)

Normalisations: Using the normalized spinors appearing 2.40, and following a similar ap-proach as for the spinors normalization, we have:

uiuj = u†i�

0uj = (E + m)

✓�†

i ,(~�.~p)†

E + m�†

i

◆✓l1 0000 �l1

◆✓�j

~�.~pE+m �j

◆

uiuj = (E + m)⇣�†

i ,~�.~p

E+m�†i

⌘✓ �j

� ~�.~pE+m �j

◆

= (E + m)⇣�ij � |~p|2

(E+m)

2

�ij⌘

And finally:uiuj = 2m �ij , vivj = �2m �ij (2.65)

with a similar calculus for v.

Completeness relations: Following same kind of calculus, it’s easy to show:

Xi=1,2

uiui = /p + m ,Xi=1,2

vivi = /p � m (2.66)

These matrices are extensively used with Feynman diagram.


As for the scalar case, we give here basic ideas without rigorous demonstration. Consult [4]for the details. We insist again, that we have to give up with the notion of a relativistic singleparticle described by the Dirac equation (see explanations in Klein-Gordon section).

The wave functions:ur(p)e�ipx , vr(p)e+ipx

with r = 1, 2 constitute a basis of solutions of the Dirac’s equation. Hence, a general functioncan be written:

(x) =

Zd3~p

(2⇡)32Ep

Xr=1,2

⇥c~p,rur(p)e�ipx + d⇤

~p,rvr(p)e+ipx⇤

where c~p,r and d⇤~p,r are complex Fourier coe�cients. Quantization of the field transforms these

numbers in operators:

(x) =

Zd3~p

(2⇡)32Ep

Xr=1,2

hc~p,rur(p)e�ipx + d†

~p,rvr(p)e+ipxi

(2.67)

(x) =

Zd3~p

(2⇡)32Ep

Xr=1,2

hc†~p,rur(p)e+ipx + d~p,rvr(p)e�ipx

i(2.68)



We obviously wish to interpret c†~p,r as an operator creating a fermion with a given momentum

~p and polarization r. Now consider the wave function of a state with 2 such particles:

~p1

,~p2

( ~x1

, ~x2

) = h ~x1

, ~x2

|1, ~p1

, r1

; 1, ~p2

, r2

i= (h ~x

1

| ⌦ h ~x2

|)c†p1

,r1

c†p2

,r2

|0i

Clearly if the c† operators commute, we will end-up with a symmetric function as for the bo-son case. The only way to get an asymmetric function is to postulate that fermion operatorsanticommute. In that case:

~p1

,~p2

( ~x1

, ~x2

) = (h ~x1

| ⌦ h ~x2

|)c†p1

,r1

c†p2

,r2

|0i= �(h ~x

1

| ⌦ h ~x2

|)c†p2

,r2

c†p1

,r1

|0i= � ~p

1

,~p2

( ~x2

, ~x1

)

With the normalization we use, the complete set of anticommutation rules then reads:

{cp0,r0 , c†p,r} = {dp0,r0 , d†

p,r} = (2⇡)32Ep �(3)(~p 0 � ~p)�r0r

{cp0,r0 , cp,r} = {dp0,r0 , dp,r} = {c†p0,r0 , c

†p,r} = {d†

p0,r0 , d†p,r} = 0

{c†p0,r0 , d

†p,r} = {cp0,r0 , d†

p,r} = {c†p0,r0 , dp,r} = 0

(2.69)

The interpretation of the operators is the following: c†~p,r creates a fermion with momentum ~p

and polarization r, while c~p,r destroys it. And d†~p,r creates a antifermion with momentum ~p and

polarization r, while d~p,r destroys it.We can easily check that we can’t create 2 particles or antiparticles in the same states: since{c†

~p,r, c†~p,r} = 2c†

~p,rc†~p,r = 0. Thus c†

~p,rc†~p,r |0i = |0i.

The hamiltonian, momentum and charge operators expressed in terms of creation annihilationoperators are then:

H =R d3~p

(2⇡)

3

2Ep

Pr=1,2 Ep (c†

p,rcp,r + d†p,rdp,r)

P =R d3~p

(2⇡)

3

2Ep

Pr=1,2 ~p (c†

p,rcp,r + d†p,rdp,r)

Q =R d3~p

(2⇡)

3

2Ep

Pr=1,2 q (c†

p,rcp,r � d†p,rdp,r)

As for the scalar field, the energy, momentum and charge of a state with 1 fermion:

|1~p, ri = c†p,r |0i (2.70)

is Ep, ~p and q while for the antifermion

|1~p, ri = d†p,r |0i (2.71)

we have Ep, ~p and �q.

Finally, as for scalar fields, one may wonder what is the connection between the usual wavefunctions ur(p)e�ip.x, vr(p)e+ip.x and the quantum field (x) of formula 2.67? The answer isvery similar to the scalar case and can be easily checked using the commutation rules of theoperators:

h0| (x)|1~p, ri = ur(p)e�ip.x

h1~p, r | (x)|0i = vr(p)e+ip.x



A last word concerning the handedness of the Dirac field. By definition, a field L is left-handed (for chirality) and R right-handed if:

PL L = L , PR L = 0 , PR R = R , PL R = 0

Let us apply the left-handed chiral projector PL on the field (2.67):

L = PL =R d3~p

(2⇡)

3

2Ep

Pr=1,2

hc~p,rPLur(p)e�ipx + d†

~p,rPLvr(p)e+ipxi

=R d3~p

(2⇡)

3

2Ep

Pr=1,2

hc~p,ruL,r(p)e�ipx + d†

~p,rvR,r(p)e+ipxi

We see that L annihilates a particle associated with uL, a (chiral) left-handed particle butcreates an anti-particle associated with vR, a (chiral) right handed antiparticle8. Naturally, L creates a left-handed particle and annihilates a right-handed antiparticle. Similarly, R

annihilates a right-handed particle and creates a left-handed antiparticle while R creates aright-handed particle and annihilates a left-handed helicity antiparticle. For massless particles,the chirality of the particle or antiparticle created/annihilated by the field is equal to its helicity.

2.4 Spin 1 particle: the photon

2.4.1 Solutions of Maxwell equations

In the first chapter, we have already seen that the Maxwell equations can be derived from theelectromagnetic tensor Fµ⌫ (see section 1.2.3.6) where:

Fµ⌫ = @µA⌫ � @⌫Aµ

depends on the 4-potential Aµ = (V/c, ~A) and satisfies:

@µFµ⌫ = µ0

j⌫

j⌫ being the 4-current. Hence, Aµ, satisfies the wave equation:

@µ(@µA⌫ � @⌫Aµ) = µ0

j⌫ ) ⇤A⌫ � @⌫(@µAµ) = µ0

j⌫ (2.72)

However, Aµ is not uniquely defined. Consider an arbitrary scalar function �. The new potential:

Aµ ! A0µ = Aµ + @µ� (2.73)

still leads to the same electromagnetic tensor. Transformation 2.73 is called a gauge transfor-mation. And because the electromagnetic tensor is invariant under this gauge transformation,the Maxwell equations will be as well. Now, � can be chosen so that:

@µA0µ = 0 (2.74)

This gauge choice is called Lorenz gauge (@µAµ + ⇤� = 0). Now the wave equation simplifiesto:

⇤Aµ = µ0

jµ (2.75)

8What is done here is not a proof. One should establish how the annihilation/creator operators behave underchirality operation. In addition, some authors called vL = PLv. It has the advantage of using consistent notationsin the expression of L field but has the drawback of being called “left” while the created particles is actuallyright-handed!



where the new electromagnetic field is written again without the ’ symbol. The solution for afree photon (jµ = 0) has a plane-wave form:

Aµ = N✏µe�ip.x

✏µ being a polarization 4-vector and N a normalization factor. A possible basis of 4-vectorspolarization requires 4 such vectors ✏�=0,1,2,3 with:

✏�=0

=

0BB@

1000

1CCA ✏�=1

=

0BB@

0100

1CCA ✏�=2

=

0BB@

0010

1CCA ✏�=3

=

0BB@

0001

1CCA

However, inserting the previous plane-wave into the wave equation 2.75 (with jµ = 0), we getp2 = 0, as expected since photons are massless. Therefore, the Lorenz gauge condition 2.74translates to:

✏.p = 0

Thus the polarization has apparently 3 degrees of freedom (4 components of 4-vector - 1 due tothe condition above). For instance, choosing the z-axis as the direction of propagation of thephoton, only a basis with ✏

1

, ✏2

and ✏0

+ ✏3

would be needed. Actually, it is not true. Nothingprevents us to redo the same kind of gauge transformation:

Aµ ! A0µ = Aµ + @µ�

but this time � must satisfies ⇤� = 0 since the field Aµ now satisfies the Lorenz gauge. Let uschoose:

� = iNae�ip.x

with a a constant that doesn’t depend on x. The new field definition then becomes:

A0µ = N(✏µ � apµ)e�ip.x

meaning that A0 has the new polarization vector ✏0µ = ✏µ �apµ. We can choose a so that ✏00 = 0,and forgetting again the ’, the field polarization vector satisfies now:

~✏.~p = 0

and therefore, there are now only 2 degrees of freedom. With a photon traveling along z,a basis with only ✏

1

, ✏2

would be enough. As expected from classical electromagnetism, thepolarization of a (free-)photon is orthogonal to the direction of motion. By convention, even ifthe photon travels in another direction, ✏

1

, ✏2

denote the two 4-vectors polarizations transverseto the direction of motion. The choice imposing ✏0 = 0 (which is clearly not covariant) is calledthe Coulomb gauge.


The photon polarization is orthogonal to the direction of motion. Any polarisation can then beexpressed as the linear combination of the 2 basis polarization-vectors ~✏(~p,� = 1) and ~✏(~p,� = 2)namely ~✏(~p, 1)↵~p,1 + ~✏(~p, 2)↵~p,2 . The general solution of the photon (classical) field is then:

Aµ(x) =

Zd3~p

(2⇡)32Ep

X�=1,2

✏µ(p,�)↵~p,�e�ip.x + ✏⇤µ(p,�)↵⇤~p,�e+ip.x



where Ep = |~p| and ✏0(p, 1) = ✏0(p, 2) = 0 because of the Coulomb gauge. The norm was chosenas before with the scalar and spin 1/2 cases. Notice that Aµ(x), the photon field is a real field,reflecting the fact that the photon is its own anti-particle.

The quantization of the field promotes the coe�cients ↵~p,� to operators but there are manysubtleties in the quantization procedure of the photon field mainly due to translation of gaugecondition in the quantum field language. The interesting reader is invited to consult the reference[6, p. 189-195] to know more. The field after quantization then reads:

Aµ(x) =

Zd3~p

(2⇡)32Ep

3X�=0

✏µ(p,�)↵~p,�e�ip.x + ✏⇤µ(p,�)↵†~p,�e+ip.x (2.76)

where the operator ↵†~p,� creates a photon of momentum ~p and polarization � and ↵~p,� is the

corresponding destructor. One of the subtleties we mentioned above leads to the (strange)commutation rules: h

↵~p0,�0 ,↵†~p,�

i= �g��0(2⇡)32Ep�(3)(~p0 � ~p)h

↵~p0,�0 ,↵~p,�

i=h↵†

~p0,�0 ,↵†~p,�

i= 0

(2.77)

It looks pretty similar to the scalar boson case (commutator) except that there is the extra�g��0 . Note also, that we sum-up on 4 polarization states: � = 0 is a “time-like” polarisation,� = 1, 2 are the 2 transverse polarizations while � = 3 is the longitudinal polarization. You seethat for the polarization � = 1, 2, 3, the corresponding operators do correspond to the ones ofthe harmonic oscillators as in the scalar case (�g�� = 1). However, it’s note the case for � = 0(�g

00

= �1)! In fact, after properly taking into account the gauge constraint, one can showthat for physical states (on which we can do measurements), only the transverse polarizationsmatter.

2.5 Exercises

Exercise 2.1 Using the spinors v1

and v2

given by 2.38, check that v1

e+ip.x and v2

e+ip.x satisfy(/p + m)v = 0.

Exercise 2.2 Covariance of the Dirac’s equation. We recall that establishing the covariance ofthe Dirac’s equation means that under a Lorentz transformation, (x) becomes 0(x0) = S (x)and satisfies the same mathematical form of the Dirac’s equation: (/@

0 � m) 0(x0) = 0. Thequestion is to establish the existence of the matrix S which should only depend on the Lorentztransformation parameters (i.e. � and �).

1. Show that S must satisfy S�µ = �⌫S�µ⌫ where �µ

⌫ is the Lorentz transformation matrix.

2. Restricting to a transformation along the x-axis, show that S =q

�+1

2

l1 +q

��1

2

�0�1

satisfies the previous relation.

Exercise 2.3 Determination of the helicity states. In the following, u� denotes a spinor for aparticle eigenstate of the helicity operator (equation 2.44) with an helicity � = ±1/2. We recallthe general formula of a u-spinor:

u� =p

E + m

✓��

~�.~pE+m ��

◆


Exercises 61

1. Show that ~�.~p �� = 2�|~p|��.

2. Using polar coordinates, show that �� can take the form for � = +1/2:

� 1

2

=

✓cos ✓

2

ei� sin ✓2

◆

3. Finally, show that the helicity state is given by:

+

1

2

=p

E + m

0BBB@

cos ✓2

ei� sin ✓2

|~p|E+m cos ✓

2


2

1CCCA e�ip.x

4. Using the same procedure, show that the 3 other helicity states are given by formula 2.45.


from wave functions to quantum ﬁeldspoly › ~paganini › ch2.pdf · chapter 2 from wave...

Documents