Download From wave functions to quantum fields

Chapter 2
From wave functions to quantum
fields
Few references:
F. Halzen and D. Martin, “Quarks & Leptons”, Chapter 3 and 5, Wiley.
P. Fayet, “Champs relativistes”, Chapter 8, 10, 11, les éditions de l’école polytechnique.
The purpose of this chapter is to clearly define the mathematical objects that
describe particles of various nature: bosons (spin 0 and spin 1) or spin 1/2
fermions. This is an important step toward the calculation of Feynman diagram.
2.1
2.1.1
Schrödinger equation
Correspondence principle
In quantum mechanics, the correspondence principle states that a quantum of momentum p~ is a
wave with a wavevector ~k through the relation postulated by De Broglie p~ = ~~k and its energy
E to the angular frequency !: E = ~! (initially postulated by Einstein for photons). Then,
Schrödinger expressed the phase of a plane wave as a complex phase factor using:
(~x, t) = N e
i(!t ~k.~
x)
= Ne
i
(Et
~
p
~.~
x)
= Ne
i
p xµ
~ µ
(2.1)
Realizing that such wave is an eigenstate of the operators:
Ĥ = i~
@
,
@t
ˆ
P~ =
~
i~r
(2.2)
with the corresponding eigenvalue E and p~, leads immediately to the identification of Ĥ as the
ˆ
energy operator and P~ as the momentum operator, or in a covariant way (see definition 1.16
and 1.15):
!
Ĥ
@
Pˆµ = i@µ = i µ =
(2.3)
ˆ
@x
P~
37
38
From wave functions to quantum fields
Pˆµ being the operator 4-momentum. Moreover, the conservation of the energy (non relativistic)
p2
E = 2m
+ V , V being a potential, implies the Schrödinger equation:
1 ~ˆ 2
P + V̂
2m
Ĥ =
)
i~
~2 ~ 2
r (~x, t) + V (~x, t)
2m
@
(~x, t) =
@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. density
flux) ~j, since we are generally
interested in moving particles. Probabilities are conserved quantiR
@
3 x = 0 which leads to the local version through the continuity
ties in the sense that @t
universe ⇢d ~
equation:
@⇢ ~ ~
+ r.j = 0
(2.5)
@t
Indeed, this equation
simply states that the rate of decrease of the number of particles in a
R
@
3~
volume V: @t
⇢
d
x
must be equal to the total flux of particles escaping from that volume
V
H
~j.~n dS, ~n, being a unit vector normal to the element dS of the surface S enclosing the volume
S
R
~ ~j d3 ~x so that:
V. Because of Gauss theorem, the closed path integral is equal to V r.
@
@t
Z
3
⇢ d ~x =
V
Z
V
~ ~j d3 ~x
r.
Since this equation must be true for any volume V, we find the continuity equation 2.5. In order
to clearly identify ⇢ and ~j, we first conjugate the Schrödinger equation and multiply by :
i~ (~x, t)
⇤ (~
x, t)
while multiplying by
i~
⇤
@
@t
(~x, t)
⇤
~2
~2
(~x, t)r
2m
(~x, t) =
⇤
⇤
(~x, t) + V
(~x, t) (~x, t)
the Schrödinger equation we obtain:
~2
2m
@
(~x, t) =
@t
⇤
~ 2 (~x, t) + V
(~x, t)r
⇤
(~x, t) (~x, t)
Subtracting the 2 previous results leads to:
i~(
⇤
@
@t
+
@
@t
⇤
)=
~2
(
2m
⇤~ 2
~2
r
r
⇤
) ) i~
@
(
@t
⇤
)+
~2 ~
r.(
2m
⇤~
r
~
r
⇤
)=0
A simple comparison with the continuity equation 2.5 allows the identification:
⇢ = | |2
,
~
~j = i ~ ( r
2m
⇤
⇤~
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~ = |N |2~v .
m
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Ecole Polytechnique
Physique des particules
Spin 0 particles
2.2
2.2.1
39
Spin 0 particles
Klein-Gordon equation
We can apply the correspondence principle equations 2.2 to the relativistic equation for energy:
E 2 = |~
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 and
the current with such equation? Following the same procedure as for the Schrödinger equation
(namely taking the conjugate, multipling by and subtracting the result to the Klein-Gordon
equation times ⇤ , we obtain:
⇤
@2
@t2
r 2 + m2 = 0 !
@2 ⇤
@t2
@2
@t2
r2
⇤
+ m2
r2
+
)
m2
)
@
@t (
⇥
@2 ⇤
2
@t
⇥ ⇤
@2
⇤
= 0 ! @t2
@2 ⇤
⇤ @2
@t2
@t2
@ ⇤
⇤ @
)+
@t
@t
⇤
=0 !
r2
⇤ r2
⇤
+ m2
⇤
=0
m2 ⇤
+
=0
2
⇤
+ r
=0
⇤
⇤
~
~
~
r.( r
r )=0
⇤ r2
The last equation can be multiplied by any constant. In order to find a formula similar to the
Schrödinger 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 on
plane wave:
p
E 2 + |~
p|2 + m2 = 0 ) E = ± |~
p|2 + m2
p
The 2 values for the energy are possible: +E and E where E = |~
p|2 + m2 . Now, injecting
the 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 |2 pµ
(2.9)
Using the negative solution for the energy and injecting it into the density, we see that the
density itself can be negative! It excludes the interpretation of the density as a probability
density!
2.2.2
Re-interpretation of the 4-current
The 2 solutions for the energy means that both wave functions 1 = N e i(Et p~.~x) and
N e i( Et p~.~x) are solution of the Klein-Gordon equation leading to the 4-current:
✓
◆
✓
◆
E
E
µ
2
µ
2
j ( 1 ) = 2|N |
,
j ( 2 ) = 2|N |
p~
p~
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Physique des particules avancée
40
From wave functions to quantum fields
The probability interpretation is impossible. However, Pauli and Weisskopf suggested (6 years
after the development of Dirac’s equation) to interpret the 4-current as a 4-current charge
density. In order to do so, the direction of the current has to change as the charge changes. It
is 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)
where p.x is a shorthand notation for pµ xµ . The 2 functions
the Klein-Gordon equation and now lead to the 4-current:
µ
jem
= i(
⇤ µ
@
@
µ ⇤
))
(±)
µ
jem
( p~ )
(+)
p
~
= ±2|N |
and
2
✓
E
p~
= N e+ip.x
( )
p
~
are still solution of
◆
(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 and
Carl 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 energies
or propagation backward in time. We can just say that an antiparticle can be represented as a
particle 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 the
Quantum Field Theory courses for the details. See for example [4].
So far, we have implicitly considered that the Klein-Gordon equation describes a single relativistic particle of spin 0. Indeed, if it were describing particles with non-zero spin, the solutions
of the Klein-Gordon equation should be able to distinguish particles with di↵erent spin projection. However, there is no place in the solution for such extra-degree of freedom. Thus, the
Klein-Gordon equation seems consistent with the description of a single relativistic particle of
spin 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 uncertainty, p
~/L. Since the particle is relativistic, E ⇡ pc and hence E
~c/L. But this
uncertainty can potentially exceed the threshold of production from the vacuum of a pair of
particle anti-particule E = 2mc2 . We can then conclude, that as soon as a particle is localized
within 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 a
single particle within a volume becomes a non-sense. The formalism must be able to describe
states of any number of particles. This is precisely what does the quantum field theory by
changing 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 expressed
as:
Z
h
i
p
d3 p~
(+)
⇤ ( )
(x) =
a
(x)
+
b
(x)
,
E
=
|~
p|2 + m2
(2.12)
p
p
p
p
~
p
~
(2⇡)3 2Ep
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Physique des particules
Few words about the quantized field
41
where the volume of integration is large enough (infinity) to consider the quantization of momentum to be continuous. ap and b⇤p are the complex amplitudes of the eigenmodes of and
correspond to the Fourier transform coefficients of the wave functions basis.
The quantisation of the wave function (classical field) transforms in an operator acting on the
Hilbert 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)
=
† (x)
=
R
R
d3 p
~
(2⇡)3 2Ep
d3 p
~
(2⇡)3 2Ep
h
(+)
†
p
~ (x) + bp
h
(+)⇤
a†p p~ (x) + bp
ap
i
( )
p
~ (x) i
( )⇤
(x)
p
~
(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 . As
annihilation and creation operators they satisfy the commutation relations:
h
i h
i
ap0 , a†p = bp0 , b†p = (2⇡)3 2Ep (3) (~
p 0 p~)
h
i
h
i
⇥
⇤ ⇥
⇤
ap0 , ap = bp0 , bp = a†p0 , a†p = b†p0 , b†p = 0
h
i h
i h
i
a†p0 , b†p = ap0 , b†p = a†p0 , bp = 0
(2.14)
The vacuum state |0i is the state with 0 particle properly normaized h0|0i = 1. With our
normalization1 , we have:
ap |0i
=0
†
ap |0i
= |1a , p~i
ap |1a , p~i = |0i
where |1a , p~i correspond to a state with 1 particle of type a having a momentum p~. Similar
relations are obtained applying |0i on b operators. Since the operators a and b commute,
bp |1a , p~i = 0. one can then show that the hamiltonian, momentum and charge operators are:
H =
P
=
Q
=
R
R
R
d3 p
~
Ep (a†p ap + b†p bp )
(2⇡)3 2Ep
3
d p
~
p~ (a†p ap + b†p bp )
(2⇡)3 2Ep
d3 p
~
(a† a
b†p bp )
(2⇡)3 2Ep p p
In some textbooks, the operators used are of the kind ap†~ i.e for the 3-momentum. The connection to the
p
notation used here is a†p = 2Ep ap†~ .
1
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Ecole Polytechnique
Physique des particules avancée
42
From wave functions to quantum fields
Let us see the value of the energy, momentum and charge of the state |1a , p~i.
R d3 p~ 0
†
†
0
H |1a , p~i = (2⇡)
~i
3 2E 0 Ep (ap0 ap0 + bp0 bp0 ) |1a , p
p
R d3 p~ 0
†
†
= (2⇡)3 2E 0 Ep0 ap0 ap0 ap |0i
R d3 p~ 0 p 0 †
†
†
= (2⇡)
3 2E 0 Ep ap0 ([ap0 ap ] + ap ap0 ) |0i
p
R d3 p~ 0
0 †
3
(3) (~
= (2⇡)
p 0 p~) |0i
3 2E 0 Ep ap0 (2⇡) 2Ep
p
= Ep a†p |0i
= Ep |1a , p~i
The same procedure can be followed for P and Q with the states |1a , p~i and |1b , p~i and we find:
H |1a , p~i = Ep |1a , p~i , H |1b , p~i = Ep |1b , p~i
P |1a , p~i = p~ |1a , p~i
, P |1b , p~i = p~ |1b , p~i
Q |1a , p~i = 1 |1a , p~i
, Q |1b , p~i = 1 |1b , p~i
Hence, the 2 states
p |1a , p~i and |1b , p~i have the same energy, same momentum (and thus same
mass since m = E 2 |~
p|2 ) but opposite charge: |1b , p~i is just the antiparticle of |1a , p~i (and
vise-versa). Moreover, since we can apply the creation operator several times, we can have
states like |na , p~i with na > 1. Several particles in the same quantum state of propagation is
only possible for bosons (Pauli exclusion principle). Here is another way of realizing the boson
nature: consider a Fock space with only 2 particles. The wave function is related to:
~1 , x~2 )
p
~1 ,~
p2 (x
= hx~1 , x~2 |1, p~1 ; 1, p~2 i
= (hx~1 | ⌦ hx~2 |)(|1, p~1 i ⌦ |1, p~2 i)
= (hx~1 | ⌦ hx~2 |)a†p1 a†p2 |0i
= (hx~1 | ⌦ hx~2 |)a†p2 a†p1 |0i
= hx~1 , x~2 |1, p~2 ; 1, p~1 i
= p~1 ,~p2 (x~2 , x~1 )
As expected, the wave function of two identical bosons is symmetric under the interchange of
the two particles.
We notice that a spatial rotation or a change of reference frame doesn’t a↵ect the internal
structure of the field since it is a scalar field. Hence, the theory describes particles with spin 0
that can have an electric charge. A remark about neutral particles: according to the expression
of Q, the charge of a state of a single particle can be zero if a = b. Hence (x) = † (x): the
scalar 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 antiparticle. The global e↵ect of is to decrease the total charge of the system by 1 unit. On the
contrary, 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. Such
quadratic 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) creation of a pair particle-antiparticle, 4) annihilation of a pair particle-antiparticle.
e
Finally, one may wonder what is the connection between the usual wave functions ± (x) =
and the quantum field ˆ(x) of formula 2.13? Since usual plane wave functions deal
i(±p.x)
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Physique des particules
Spin 1/2 particles
43
with a single particle, one has to project the field into a single state particle. More precisely, it
is easy to check that:
h0| ˆ(x)|1a , p~i = e ip.x
h1b , p~ | ˆ(x)|0i = e+ip.x
For instance,
h0| ˆ(x)|1a , p~i =
=
=
=
h
R
(+)
( )
d3 p
~0
h0|ap0 |1a , p~i ~0 (x) + h0|b†p0 |1a , p~i ~0 (x)
(2⇡)3 2Ep0 h
p
p
i
R d3 p~ 0
(+)
( )
†
† †
0 ap |0i
h0|a
(x)
+
h0|b
a
|0i
(x)
0
p
3
0
p
p
(2⇡) 2E
p~0
p~i0
R d3 p~ 0 p h
(+)
3
(3)
0
~
h0|0i (2⇡) 2Ep0
(p p~) ~0 (x)
(2⇡)3 2Ep0
p
(+)
ip.x
p
~ (x) = e
i
and similarily for h1b , p~ | ˆ(x)|0i.
2.3
2.3.1
Spin 1/2 particles
Dirac equation
In 1928, P.A.M Dirac proposed his famous equation in order to avoid the solution with a negative
energy and a negative density probability of the Klein-Gordon equation. It was 6 years before the
correct interpretation of the Klein-Gordon equation. Even if it was not his initial goal, Dirac
found 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 energy
was 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 for
an equation having a @/@t dependency as in the Schrödinger equation. The equation should be
covariant under Lorentz transformations, and hence, the dependency must be also linear with
r. 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:
(
Where
µ
and

@
im)(
@ + im) = 0
are a priori 2 sets of 4 numbers. If

(
@
satisfies
im) = 0
(2.15)
@ + im) = 0
(2.16)
or
(
then the Klein-Gordon equation will be satisfied as well. Equations 2.15 or 2.16 fulfill the
linearity condition with the derivatives. Let us develop and identify the di↵erent terms:
(@ µ @µ + m2 )
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
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@ ) + m2 )
Physique des particules avancée
44
From wave functions to quantum fields

In order to cancel the linear term with m, we see that
D’Alembertian term (@ µ @µ ) then imposes:
@ µ @µ =

=
.
The identification with the
@ @
namely:
@02
@12
@22
@32 = ( 0 )2 @02 + ( 1 )2 @12 + ( 2 )2 @22 + (
+( 0 1 + 1 0 )@0 @1 + ( 0 2 +
+( 1 2 + 1 2 )@1 @2 + ( 1 3 +
+( 2 3 + 3 2 )@2 @3
3 )2 @ 2
3
2 0 )@ @
0 2
1 3 )@ @
1 3
+(
0 3
+
3 0 )@
0 @3
’s were complex numbers, the first line of the equality would impose 0 = ±1 and
= ±i. However, It would be impossible to cancel the last 3 lines. Dirac then proposed
to interpret the ’s as matrices satisfying:
If the
k=1,2,3
0 )2
(
µ ⌫
=1l
+ ⌫
µ
, ( k=1,2,3 )2 = 1l
=0
for µ 6= ⌫
which can be compacted with:
{
µ
,
⌫
} = 2g µ⌫ 1l
(2.17)
where {} is the anticommutator {a, b} = ab + ba. The relation 2.17 is known as the Cli↵ord
algebra. What is the minimal rank of the ’s? Knowing that Tr(AB) = Tr(BA) and since
0 0 = 1 we have
k
Tr(
) = Tr(
k 0 0
) = Tr(
where the negative sign is due to the
Tr(
0
)=
Tr(
0 k k
0 k 0
)=
Tr(
0 0 k
)=
matrices properties. Thus Tr(
)=
Tr(
k 0 k
) = Tr(
k k 0
)=
Tr(
k)
k
)
= 0. Similarly,
Tr(
0
)
where we applied successively, k k = 1,
l Tr(AB) = Tr(BA), µ ⌫ + ⌫ µ = 0 and k k = 1.
l
µ
So, Tr( ) = 0. But the trace of a matrix is equal to the sum of its eigenvalues. Since 0 0 =1,
l
the eigenvalues of 0 are ±1. Similarly, the eigenvalues of k are ±i. The trace being 0, there
must be as many positive eigenvalues as negative ones. Hence the rank n is necessarily an even
number. Rank n = 2 is not possible, because there are only 3 independent traceless matrices
in 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 ⇥ 4
hermitian matrices. Several choices are possible and the most common one, called the Dirac’s
representation is:
✓
◆
✓
◆
i
1l 00
00
0
i
=
,
=
(2.18)
i
00 1l
00
where the symbols 1l and 0 denote respectively the 2 ⇥ 2 unit matrix and zero matrix, and
the Pauli spin matrices seen in the first chapter:
✓
◆
✓
◆
✓
◆
0 1
0
i
1 0
1
2
3
=
,
=
,
=
1 0
i 0
0
1
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Physique des particules
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 easily
check that the definition of the ’s matrices implies for the hermitian conjugates2 :
0†
=
0
1†
,
1
,
µ†
=
=
2†
2
=
3†
,
=
3
(2.19)
or in more concise:
0 µ 0
(2.20)
We can now come back to the 2 “square root” equations 2.15 and 2.16 of the Klein-Gordon
equation. By convention, the Dirac equation is 2.16 (with = ) multiply by i (remember that
i@µ = P̂µ ):
(i@/ m) = 0
with @/ = µ @µ
(2.21)
Since the ’s are 4 ⇥ 4 matrices, the
nents:
wave function is actually a wave function with 4 compo0
1
1
B
=B
@
2
3
4
C
C
A
It is called a Dirac-spinor or 4-spinor. At this point, we can already begin to see that the extra
multiplicity is likely to have something to do with an angular momentum degree of freedom as
suggested 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 Schrödinger equations.
This time however, the complex conjugate must be replaced by the hermitian conjugate because
of the presence of matrices. The hermitian conjugate of 2.21 is
where the hermitian adjoint
†
( i@µ
† µ†
†
⇤
1,
m
†
)=0
(2.22)
is:
=(
⇤
2,
⇤
3,
⇤
4)
(2.23)
Now, using equality 2.19, equation 2.22 becomes:
( i@0
and multiplying by
0
† 0 0
† k 0
( i@0
+i@k
† 0
+ i@k
† k
†
m
)=0
from the right side:
m
† 0
)=
(i@0
† 0 0
+i@k
† 0 k
+m
† 0
)=
(i@µ
† 0 µ
+m
† 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 ¯
2
µ
+m¯= 0
A† = (A⇤ )t where A⇤ means complex conjugate and
P.Paganini
t
(2.25)
the transpose matrix.
Ecole Polytechnique
Physique des particules avancée
)=0
46
From wave functions to quantum fields
We can now derive the continuity equation by multiplying the previous equation from the right
by , equation 2.21 from the left by ¯ and add the result:
i(@µ ¯)
µ
+i¯
µ
= i@µ ( ¯
@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
=
=
†
)⇢=
4
X
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 particle
q = ±1:
jqµ = qj µ = q ¯ µ
(2.28)
Contrary to the boson case, changing the sign of p won’t change j µ . Here, we have to explicitly
put 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 solution
of the Klein-Gordon equation, or more precisely, each component of the 4-spinor is a solution
of the Klein-Gordon equation. Then, it is natural to look for solutions in which the space-time
behaviour 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 always
be expressed as a linear combination of plane-wave solutions. Injecting 2.29 into the Dirac’s
equation 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.
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Ecole Polytechnique
Physique des particules
Free-particles solutions
2.3.3.1
47
Solutions for particle at rest
It is very instructive to first examine the solutions for a particle at rest where only p0 = E is
non zero. Equation 2.30 simplifies to:
✓
◆
✓
◆
⇢
ua
ua
Eua = mua
0
E u mu = 0 ) E
=m
)
ub
ub
Eub = mub
✓ ◆
✓ ◆
1
0
The equation satisfied by ua corresponds to 2 independent orthogonal solutions
and
0
1
both with E = m while the 2 independent orthogonal solutions for ub have E = m. Coming
back to 2.29, the 4 solutions are then:
0 1
0 1
0 1
0 1
1
0
0
0
B 0 C imt
B 1 C imt
B 0 C +imt
B 0 C +imt
B Ce
B Ce
B Ce
B Ce
,
,
,
(2.32)
1 =@
2 =@
3 =@
4 =@
0 A
0 A
1 A
0 A
0
0
0
1
and 2 having E = m > 0 and 3 and 4 having E = m < 0. Finally, even in Dirac’s
equation we have negative energy solutions! However, this time, they don’t lead to a negative
probability density. These negative energy solutions will be interpreted in the next section as
solutions for antiparticles.
1
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:
✓
◆
✓
◆
ua
ua
µ
0
( pµ m)
= ( E ~ .~
p m)
ub
ub
✓✓
◆ ✓
◆ ✓
◆◆ ✓
◆
E.l
1
00
00
~ .~
p
m.l
1 00
ua
=
00
E.l
1
~ .~
p◆ ✓00 ◆
00 m.l
1
ub
✓
(E m).l
1
~ .~
p
ua
=
~ .~
p
(E + m).l
1
ub
(2.33)
so ua and ub must satisfy the 2 coupled equations:
ua =
~ .~
p
ub
E m
(2.34)
ub =
~ .~
p
ua
E+m
(2.35)
These equations impose a constraint on E and p~. Indeed, let us inject 2.35 into 2.34:
ua =
(~ .~
p) 2
ua
E 2 m2
but:
~ .~
p=
P.Paganini
✓
0 1
1 0
◆
px +
✓
0
i
i
0
◆
py +
✓
1
0
0
1
◆
pz =
Ecole Polytechnique
✓
pz
px ipy
px + ipy
pz
◆
(2.36)
Physique des particules avancée
48
From wave functions to quantum fields
And thus:
(~ .~
p)2 = |~
p|2 1l
so that, the non trivial solution (ua 6= 0) requires:
p
|~
p| 2
= 1 ) E = ± |~
p|2 + m2
2
m
E2
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 orthogonal
spinors 1 and 2 , and then ua will have to satisfy 2.34. In both cases, the relativistic energy
constraint must be fulfilled. Hence, 4 independent solutions of the Dirac’s equation are given
by these 4-spinors:
✓
◆
✓
◆
✓ ~ .~p
◆
✓ ~ .~p
◆
1
2
1
2
E m
E m
u1 = N1
, u2 = N2
, u3 = N3
, u4 = N4
~ .~
p
~ .~
p
E+m
1
E+m
2
1
2
(2.37)
where Ni are normalization factors of the spinors. We have to identify which 4-spinors correspond
to positive or negative energy. In the limit where p~ ! 0, we should recover the solutions
for a
p
2
particle at rest, so that we can conclude that up
p| + m2
1 and u2 are the solutions with E = + |~
2
2
while u3 and u4 are the solutions with E =
|~
p| + m .
2.3.3.3
Interpretation of negative energies
Historically, Dirac considered that the vacuum was full (!) of negative energy states satisfying
the Pauli exclusion principle. This theory is usually referred as the “Dirac sea”. A hole in the
negative energy states was then interpreted as an antiparticle with positive energy and opposite
charge. This interpretation was successful in predicting the positron (discovered by Anderson in
1932) but had many drawbacks. For instance, the vacuum being full of fermions with negative
energy 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). In
fact, Stückelberg in 1941 and Feynman in 1948 proposed the correct interpretation still valid with
quantum field theory and briefly mentioned in the previous case of the Klein-Gordon equation.
Consider the 2 solutions u3 and u4 with negative energy in 2.37. They can be rewritten:
!
!
u3 = N3
~ .~
p
|E|+m
1
, u 4 = N4
1
~ .~
p
|E|+m
2
2
and both are associated to a propagation term e i( |E|t p~.~x) . But this term can be simply
written e i(|E|( t) p~.~x) , so that it looks as a particle travelling backward in time form (t2 , x1 ) to
(t1 , x2 ) where t2 > t1 as shown on schema b of figure 2.1. The picture is clearly not equivalent
to an electron travelling “normally” from (t1 , 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 also
have to change direction of the momentum (to have a Lorentz scalar p.x) so that we obtain the
schema 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 with
positive energy and travelling “normally” from (t1 , x2 ) to (t2 , x2 ). We can now define the two
new spinors v1 and v2 for the antiparticle, interpreting E as a positive quantity and inverting
the momentum:
u3,4 ( E, p~)e i(( E)t ( p~).~x) = v1,2 (E, p~)e+i(Et p~.~x)
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Ecole Polytechnique
Physique des particules
Free-particles solutions
49
a)
t2
b)
t2
t1
t1
x1
c)
t2
t1
x2
x1
x2
x1
x2
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 =
N10
✓
~ .~
p
E+m
1
1
◆
, v2 =
N20
✓
~ .~
p
E+m
2
◆
2
(2.38)
Physique des particules
M1 HEP X
Pascal Paganini LLR-IN2P3-CNRS
Injecting ve+ip.x into the Dirac equation 2.21, we see that v satisfies the momentum space Dirac
equation:
(p
(2.39)
/ + m)v = 0
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 to
normalize to 2E particles per unit volume 3 . Using the density 2.27, we require:
Z
†
= u† u = 2E
V=1
where u can be u1 , u2 , v1 or v2 . Doing the job for vi=1,2 as an example:
✓ ~ .~p
◆
†
~ .~
p
i
†
0
2
†
u u = |Ni | (( E+m i ) , i ) E+m
† ~ .~
p † ~ .~
p
i ( E+m ) E+m i +
† ~ .~
p 2
i ( E+m ) i + 1)
|~
p |2
0
2
|Ni | ( (E+m)
2 + 1)
2E
0
2
|Ni | E+m
= |Ni0 |2 (
= |Ni0 |2 (
=
=
i
†
i i)
where moving from
the second line to the third, we used (~ .~
p)† = ~ .~
p and †i i = 1. Thus, we
p
conclude Ni0 = E + m. And similarly for Ni . Hence the 4 solutions of the Dirac equation are:
p
E = + |~
p|2 + m2
✓
✓
◆
◆
p
p
1
2
ip.x
E+m
e
,
E+m
e ip.x
1 =
2 =
~ .~
p
~ .~
p
1
2
E+m
E+m
(2.40)
✓ ~ .~p
◆
✓ ~ .~p
◆
p
p
E + m E+m 1
e+ip.x ,
E + m E+m 2
e+ip.x
1̄ =
2̄ =
1
3
2
When we chose N = 1 in 2.11, it implies for the 4-current 2.10, j µ = ±2
unit volume
P.Paganini
Ecole Polytechnique
✓ ◆
E
and hence 2E particles per
p
~
Physique des particules avancée
21
50
where
From wave functions to quantum fields
1
and
2
correspond to a particle and
1̄
and
2̄
to an antiparticle.
Example of explicit formula: let us consider for example, 1 =
4
2 = 1 as a spin-down state (and thus eigenstates of 3 ):
✓ ◆
✓ ◆
1
0
, 2= 1=
1 = 2 =
0
1
The 4 solutions become:
0
1
1̄
=
=
p
p
B
E + mB
@
0
B
E + mB
@
1
0
pz
E+m
px +ipy
E+m
px ipy
E+m
pz
E+m
0
1
where we used equality 2.36.
2.3.3.5
1
C
C e
A
ip.x ,
2
1
C +ip.x
C e
,
A
2̄
=
p
=
0
B
E + mB
@
p
0
B
E + mB
@
0
1
px ipy
E+m
pz
E+m
pz
E+m
px +ipy
E+m
1
0
2
as a spin-up state and
1
C
C e
A
1
ip.x
(2.41)
C +ip.x
C e
A
Interpretation in terms of spin and helicity
For a particle or an antiparticle, we always have two solutions degenerated in energy. There
must be an operator that commutes with the energy operator ĤD of the Dirac’s equation whose
eigenvalues would distinguish the solutions. First, let us determine ĤD . Expanding the Dirac’s
equation 2.21, we have:
@
~ m) = 0
i 0
+ (i~ .r
@t
ˆ
~ we get:
Multiplying from the left by 0 , and using P~ = ir
i
@
@t
ˆ
~ .P~ +
0
= ĤD
with the Dirac’s Hamiltonian:
ĤD =
0
m =
✓
00 ~
~ 00
◆
ˆ
.P~ +
0
m
(2.42)
Spin states: Now, we have to find an operator that commutes with ĤD . Let us try to see if
the spin operators are satisfactory. A simple generalization of the operators 1.41 based on Pauli
matrices for the 4-spinors case is clearly:
✓
◆
1 ~ 00
ˆ 1 ~ˆ
~
S= ⌃=
(2.43)
2
2 00 ~
4
with this choice solution for particle and antiparticle with the same index, will have a similar interpretation
in term of spin.
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Physique des particules
Free-particles solutions
51
~ˆ don’t commute because of the presence of P~ˆ . Indeed, for instance with
However, ĤD and S
[HD , Sz ], we would have a term5 :
h
~ˆ
~ .P,
z
i
=
h
ˆ +
x Px
ˆ +
ˆ
y Py
z Pz ,
z
i
=[
x,
ˆ +[
z ] Px
y,
ˆ 6= 0
z ] Py
~ˆ and the explicit solutions
In case of zero momentum, ĤD reduced to 0 m that commutes with S
2.41 (with px = py = pz = 0) are eigenstates of Ŝz :
p~ = 0 )
Ŝz
1
=+
1
2
1
,
Ŝz
2
=
1
2
2
,
S̄ˆz
1̄
=+
1
2
1̄
,
S̄ˆz
2̄
=
1
2
2̄
Note that the spin operator for the antiparticles is defined as:
S̄ˆ =
Ŝ
It is justified because with antiparticles, the operator returning the physical momentum (positive
ˆ
ˆ
energy, appropriate direction for p~) is changed from P~ ! P~ (because antiparticles have a
~ˆ = ~rˆ ^ P~ˆ ! L.
~ˆ Thus the conservation of the total angular
propagation mode e+ip.x ), so that L
ˆ ~ˆ ~ˆ
~ˆ ! S.
~ˆ Finally, we see that 1 and 1̄ would have spin up
momentum J~ = L
+ S, requires S
while 2 and 2̄ 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
ˆ ~ˆ ~ˆ
~ˆ = ~rˆ ^ P~ˆ
the Hamiltonian). One can show that the total angular momentum J~ = L
+ S with L
does commute. It is fortunate because otherwise, the total angular momentum would not be
conserved! Another possibility is the so-called helicity defined as the projection of the spin on
the direction of the momentum. Hence, the operator is simply defined for fermion spinor as:
0
1
~ˆ
ˆ
ˆ
P
~
~
~
.
0
0
~ˆ P = 1 ⌃.
~ˆ P = 1 @ |~p|
A
ĥ = S.
~ˆ
P
|~
p|
2 |~
p|
2
00
~ . |~
p|
(2.44)
~ˆ ! S
~ˆ and P~ˆ ! P~ˆ . The
Note that for antifermions, the same operator can be used since S
helicity operator clearly commutes with ĤD , so there exists a basis of 4-spinors which is both
eigenstates of ĤD and of the helicity. Since the spin on any axis (and thus on the direction of the
momentum) can only be ±1/2, the helicity quantum number6 can only be ±1/2. Depending
on , the state is called:
= 12 ) Left handed helicity
= + 12 ) Right handed helicity
5
I use indi↵erently x for 1 , y for 2 and z for 3 .
Note that some authors, define the helicity as twice that number in order to have an integer for both fermions
and bosons.
6
P.Paganini
Ecole Polytechnique
Physique des particules avancée
52
From wave functions to quantum fields
After some math (see exercise 2.3), one can determine the general solution for helicity states:
0
1
0
1
cos 2✓
sin 2✓
B
C
B
C
ei sin 2✓
ei cos 2✓
p
p
B
C
B
C
ip.x ,
E
+
m
e
=
E
+
m
1 =
1
B
C
B
C e ip.x
|~
p
|
|~
p
|
✓
+2
2
sin
@ E+m cos 2✓ A
@
A
E+m
2
|~
p|
|~
p|
✓
✓
i
i
E+m e sin 2
E+m e cos 2
=
+ 1̄2
p
0
B
B
E + mB
@
1
|~
p|
✓
E+m sin 2
|~
p|
✓
i
E+m e cos 2
sin 2✓
i
e cos 2✓
C
C +ip.x
,
C e
A
1̄
2
p
=
0
B
B
E + mB
@
|~
p|
✓
E+m cos 2
|~
p|
✓
i
E+m e sin 2
cos 2✓
i
e sin 2✓
1
C
C +ip.x
C e
A
(2.45)
where ✓ and are the polar angles.
A remark: beware that the helicity is not Lorentz invariant in general. Actually, for
a massive particle, one can always find a reference frame where the particle appears to reverse
its 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 electromagnetic
~ is obtained by making the substitution:
field Aµ = (V, A)
p~ ! p~
~
qA
E!E
,
qV
The quantum version is then simply from the correspondence principle:
pµ ! p µ
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
= Ĉ
(2.48)
is supposed to satisfy:
[
µ
(i@µ + qAµ )
m]
0
=0
(2.49)
In order to find Ĉ, let us start by conjugating 2.47:
⇥
0(
i@0
qA0 ) +
1(
i@1
qA1 )
2(
i@2
[
qA2 ) +
µ⇤ (
i@µ
3 ( i@
3
qAµ )
qA3 )
m]⇤
m
⇤
⇤
=0
=0
In order to have the correct sign in front of all ’s we can multiply by 2 from the left and use
the anticommutation relation:
⇥ 0
⇤
1 ( i@
2 ( i@
3 ( i@
( i@0 qA0 )
qA1 )
qA2 )
qA3 ) m 2 ⇤ = 0
1
2
3
[ µ (i@µ + qAµ ) m] 2 ⇤ = 0
P.Paganini
Ecole Polytechnique
Physique des particules
Operations on spinors
53
Thus, 2 ⇤ seems a good candidate for 0 . If represents the wave function of an electron, we
expect 0 to be the one of a positron. Let us check for instance with the electron described by
1 in 2.41:
2
⇤
0
0
p
B 0
= E + mB
@ 0
i
0
0
i
0
1 20
i
6B
0 C
C 6B
0 A 4@
0
0
i
0
0
1
1
0
C
Ce
A
pz
E+m
px +ipy
E+m
It would be the positron solution described by
using:
Ĉ
1̄
=i
3⇤
0
p
B
i E + mB
@
7
5 =
ip.x 7
px ipy
E+m
pz
E+m
except that there is this extra
0
1
1
C +ip.x
Ce
A
i. Hence, by
⇤
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:
0
(x) !
(x0 ) = P̂ (x)
with (i
µ 0
@µ
m) 0 (x0 ) = 0
Let us express the Dirac’s equation with the new coordinates:
(i
@t0
0 @
@t0 @t
+i
(i
i=1,2,3 .
We need to revert the sign in front of
the anticommutation algebra:
(i
0 @
@t0
(i
µ@
µ
1 @
2 @
3 @ @z 0
+
i
+
i
0
0
0
@x @x
@y @y
@z @z
0 @
1 @
2 @
3 @
i
i
i
0
0
0
@t
@x
@y
@z 0
+i
@y 0
@x0
1 @
@x0
m)
m)
m)
Thus, let us multiply by
+i
2 @
@y 0
0
0
=0
=0
=0
0
from the left and use
+ i 3 @z@ 0 m) 0
=0
(i µ @µ0 m) 0 (x0 ) = 0
where we identify:
=
) P̂ =
0
(2.52)
Now, according to spinors 2.40 solution of the Dirac’s equation, we see that:
u( p~) = P̂ u(~
p)
,
v( p~) =
meaning that the intrinsic parity is +1 for fermion and
⌘fP = +1
P.Paganini
,
⌘fP¯ =
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P̂ v(~
p)
1 for antifermion:
1
(2.53)
Physique des particules avancée
54
2.3.4.3
From wave functions to quantum fields
Chirality
consider the so-called chirality matrix:
5
=i
0 1 2 3
(2.54)
00 1l
1l 00
◆
(2.55)
,
5†
which in the Dirac representation reads:
5
It’s easy to show that
5
=
✓
satisfies:
(
5 2
) =1
=
5
(2.56)
Since ( 5 )2 = 1, the eigenvalues of 5 are ±1. The eigenstates of 5 are called chirality states
and are denoted with the Left L and Right R label for a reason that is going to be clarified
in few lines. Eigen-spinors with positive (negative) chirality have right (left) chirality and the
opposite for eigen-antispinors:
5
uR = +uR ,
5
uL =
uL ,
5
vR =
5
vR ,
vL = +vL
(2.57)
The reason why chiral antispinors have opposite eigenvalues with respect to chiral spinor is
because 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 lefthanded helicity state as shown few lines below). Note that due to the anticommutation algebra
of 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 chirality
eigenvalues are not conserved and the physical states having a defined energy (so eigenstates of
the hamiltonian) will be a mixture of chirality states.
Now, Consider the helicity states 2.45 in the ultrarelativistic limit E
m or equivalently when
m = 0:
0
0
1
1
cos 2✓
sin 2✓
p B ei sin ✓ C
p B ei cos ✓ C
2 C,
2 C
u
⇡
EB
u+ 1 ⇡ E B
1
✓
@ cos ✓ A
@
A
2
2
sin
2
2
✓
✓
i
i
e sin 2
e cos 2
(2.59)
0
1
0
1
sin 2✓
cos 2✓
p B ei cos ✓ C
p B ei sin ✓ C
2 C,
2 C
v+ 1 ⇡ E B
v
EB
1 ⇡
✓
@
A
@ cos ✓ A
2
2
sin
2
2
ei cos 2✓
ei sin 2✓
It is easy to check that these helicity states are now eigenstates of 5 using the representation
2.55. Hence, in the limit E
m, the helicity states coincide with the chirality states:
u+ 1 = uR , u
2
P.Paganini
1
2
= uL , v + 1 = v R , v
2
Ecole Polytechnique
1
2
= vL
(2.60)
Physique des particules
Operations on spinors
55
and we now understand why the chiral states are called left or right handed. Looking at 2.59, we
see 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 via
charged current (mediated by W bosons) only acts on the chirality left-handed states of particles
and chirality right-handed states of antiparticles. Thus, it is useful to define the projectors on
these states:
1
1
5
PL = (1
) ,
PR = (1 + 5 )
(2.61)
2
2
It’s trivial to check that PL and PR have the projectors properties: PL2 = PL , PR2 = PR ,
PL PR = 0, PL + PR =1.
l 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 = u L + u R u L = PL u u R = PR u
v = vL + vR vL = PR v vR = PL v
(2.62)
Now consider the parity transformation of the projector on left-handed state7
P̂ PL P̂ † =
01
2
(1
5
)
0†
1
= (1 +
2
5
)
0 0
= PR
meaning that the chirality changes under a parity transformation as we would expect from its
denomination (chirality comes from the greek, meaning “hand”: a left hand in a mirror becomes
a right hand).
Spin states, helicity states and chirality states are often confused. We invite the reader to
consult 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 (2.30
and 2.39):
(p
(p
(2.63)
/ m)u = 0 ,
/ + m)v = 0
We wish to see which equations satisfy ū and v̄. For instance, taking the hermitian conjugate
of 2.30:
u† ( µ† pµ m)
=0
†
u ( 0 µ 0 pµ m) = 0
where we used 2.20. Multiplying from the right by
u† (
0 µ
pµ
m
0
0
and using (
0 )2
= 1:
)=0
The transformation of a wave function under P̂ is P̂ while the transformation of an operator A is P̂ AP̂ † .
Hence, if a state |ai is transformed in |a0 i = U |ai under a unitary transformation U (i.e. U † = U 1 because
ha|bi = ha0 |b0 i ) U † U = 1), the variable ha|A|bi must be equal to ha0 |A0 |b0 i and thus: ha|A|bi = ha|U † A0 U |bi and
so for any state |ai or |bi. We thus have: U AU † = A0 . Requiring ha0 |A0 |b0 i = ha|A|bi, is equivalent for a classical
scalar field (temperature for instance), as saying 0 (x0 ) = (x): the temperature at a rotated point seen by the
rotated field is the same as the temperature at the initial point.
7
P.Paganini
Ecole Polytechnique
Physique des particules avancée
56
From wave functions to quantum fields
Applying the same procedure for v we finally get:
ū(p
/
m) = 0
,
v̄(p
/ + m) = 0
(2.64)
Normalisations: Using the normalized spinors appearing 2.40, and following a similar approach as for the spinors normalization, we have:
✓
◆✓
◆✓
◆
p) † †
1l 00
j
† 0
† (~ .~
ūi uj = ui uj = (E + m) i ,
~ .~
p
00 1l
E+m i
E+m j
ūi uj
= (E + m)
= (E + m)
And finally:
ūi uj = 2m
⇣
⇣
ij
† ~ .~
†
p
i , E+m i
⌘✓
|~
p |2
(E+m)2 ij
ij
,
v¯i vj =
⌘
j
~ .~
p
E+m
2m
j
◆
(2.65)
ij
with a similar calculus for v.
Completeness relations:
Following same kind of calculus, it’s easy to show:
X
ui ūi = p
/+m
i=1,2
,
X
vi v̄i = p
/
m
(2.66)
i=1,2
These matrices are extensively used with Feynman diagram.
2.3.5
Few words about the quantized field
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 single
particle 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 function
can be written:
Z
X ⇥
⇤
d3 p~
(x) =
cp~,r ur (p)e ipx + dp⇤~,r vr (p)e+ipx
3
(2⇡) 2Ep
r=1,2
where cp~,r and dp⇤~,r are complex Fourier coefficients. Quantization of the field transforms these
numbers in operators:
Z
i
X h
d3 p~
†
ipx
+ipx
(x) =
c
u
(p)e
+
d
v
(p)e
(2.67)
p
~,r r
p
~,r r
(2⇡)3 2Ep
r=1,2
¯(x) =
P.Paganini
Z
X h †
d3 p~
cp~,r ūr (p)e+ipx + dp~,r v̄r (p)e
(2⇡)3 2Ep
r=1,2
Ecole Polytechnique
ipx
i
(2.68)
Physique des particules
Few words about the quantized field
57
We obviously wish to interpret cp†~,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:
= hx~1 , x~2 |1, p~1 , r1 ; 1, p~2 , r2 i
= (hx~1 | ⌦ hx~2 |)c†p1 ,r1 c†p2 ,r2 |0i
~1 , x~2 )
p
~1 ,~
p2 (x
Clearly if the c† operators commute, we will end-up with a symmetric function as for the boson case. The only way to get an asymmetric function is to postulate that fermion operators
anticommute. In that case:
= (hx~1 | ⌦ hx~2 |)c†p1 ,r1 c†p2 ,r2 |0i
= (hx~1 | ⌦ hx~2 |)c†p2 ,r2 c†p1 ,r1 |0i
=
~2 , x~1 )
p
~1 ,~
p2 (x
~1 , x~2 )
p
~1 ,~
p2 (x
With the normalization we use, the complete set of anticommutation rules then reads:
{cp0 ,r0 , c†p,r } = {dp0 ,r0 , d†p,r } = (2⇡)3 2Ep (3) (~
p 0 p~) r0 r
{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: cp†~,r creates a fermion with momentum p~
and polarization r, while cp~,r destroys it. And dp†~,r creates a antifermion with momentum p~ and
polarization r, while dp~,r destroys it.
We can easily check that we can’t create 2 particles or antiparticles in the same states: since
{cp†~,r , cp†~,r } = 2cp†~,r cp†~,r = 0. Thus cp†~,r cp†~,r |0i = |0i.
The hamiltonian, momentum and charge operators expressed in terms of creation annihilation
operators are then:
H =
P
=
Q
=
R
R
R
d3 p
~
(2⇡)3 2Ep
d3 p
~
(2⇡)3 2Ep
3
d p
~
(2⇡)3 2Ep
P
P
P
r=1,2
Ep (c†p,r cp,r + d†p,r dp,r )
r=1,2
p~ (c†p,r cp,r + d†p,r dp,r )
r=1,2
q (c†p,r cp,r
d†p,r dp,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)
|1̄~
p, ri = d†p,r |0i
(2.71)
is Ep , p~ and q while for the antifermion
we have Ep , p~ and
q.
Finally, as for scalar fields, one may wonder what is the connection between the usual wave
functions ur (p)e ip.x , vr (p)e+ip.x and the quantum field ˆ(x) of formula 2.67? The answer is
very similar to the scalar case and can be easily checked using the commutation rules of the
operators:
h0| ˆ(x)|1~
p, ri = ur (p)e ip.x
h1̄~
p, r | ˆ(x)|0i = vr (p)e+ip.x
P.Paganini
Ecole Polytechnique
Physique des particules avancée
58
From wave functions to quantum fields
A last word concerning the handedness of the Dirac field. By definition, a field
handed (for chirality) and R right-handed if:
PL
L
=
L
, PR
L
= 0 , PR
R
=
R
, PL
R
L
is left-
=0
Let us apply the left-handed chiral projector PL on the field (2.67):
h
i
R d3 p~ P
ipx + d† P v (p)e+ipx
=
P
=
c
P
u
(p)e
L
L
p
~,r L r
r=1,2
p
~,r L r
(2⇡)3 2Ep
h
i
R d3 p~ P
†
ipx
= (2⇡)3 2Ep r=1,2 cp~,r uL,r (p)e
+ dp~,r vR,r (p)e+ipx
We see that L annihilates a particle associated with uL , a (chiral) left-handed particle but
creates 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 a
right-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 the
electromagnetic tensor F µ⌫ (see section 1.2.3.6) where:
F µ⌫ = @ µ A⌫
@ ⌫ Aµ
~ and satisfies:
depends on the 4-potential Aµ = (V /c, A)
@µ F µ⌫ = µ0 j ⌫
j ⌫ being the 4-current. Hence, Aµ , satisfies the wave equation:
@µ (@ µ A⌫
@ ⌫ Aµ ) = µ0 j ⌫ ) ⇤A⌫
@ ⌫ (@µ Aµ ) = 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 transformation. And because the electromagnetic tensor is invariant under this gauge transformation,
the Maxwell equations will be as well. Now, can be chosen so that:
@µ A0µ = 0
(2.74)
This gauge choice is called Lorenz gauge (@µ Aµ + ⇤ = 0). Now the wave equation simplifies
to:
⇤Aµ = µ0 j µ
(2.75)
8
What is done here is not a proof. One should establish how the annihilation/creator operators behave under
chirality operation. In addition, some authors called vL = PL v. It has the advantage of using consistent notations
in the expression of L field but has the drawback of being called “left” while the created particles is actually
right-handed!
P.Paganini
Ecole Polytechnique
Physique des particules
Few words about the quantized field
59
where the new electromagnetic field is written again without the ’ symbol. The solution for a
free photon (j µ = 0) has a plane-wave form:
Aµ = N ✏µ e
ip.x
✏µ being a polarization 4-vector and N a normalization factor. A possible basis of 4-vectors
polarization requires 4 such vectors ✏ =0,1,2,3 with:
0 1
0 1
0 1
0 1
1
0
0
0
B0C
B1C
B0C
B0C
C
B C
B C
B C
✏ =0 = B
@0A ✏ =1 = @0A ✏ =2 = @1A ✏ =3 = @0A
0
0
0
1
However, inserting the previous plane-wave into the wave equation 2.75 (with j µ = 0), we get
p2 = 0, as expected since photons are massless. Therefore, the Lorenz gauge condition 2.74
translates to:
✏.p = 0
Thus the polarization has apparently 3 degrees of freedom (4 components of 4-vector - 1 due to
the condition above). For instance, choosing the z-axis as the direction of propagation of the
photon, only a basis with ✏1 , ✏2 and ✏0 + ✏3 would be needed. Actually, it is not true. Nothing
prevents us to redo the same kind of gauge transformation:
Aµ ! A0µ = Aµ + @ µ
but this time
choose:
must satisfies ⇤ = 0 since the field Aµ now satisfies the Lorenz gauge. Let us
= iN ae
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, the
polarization of a (free-)photon is orthogonal to the direction of motion. By convention, even if
the photon travels in another direction, ✏1 , ✏2 denote the two 4-vectors polarizations transverse
to the direction of motion. The choice imposing ✏0 = 0 (which is clearly not covariant) is called
the Coulomb gauge.
2.4.2
Few words about the quantized field
The photon polarization is orthogonal to the direction of motion. Any polarisation can then be
expressed 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:
Z
X
d3 p~
Aµ (x) =
✏µ (p, )↵p~, e ip.x + ✏⇤µ (p, )↵p⇤~, e+ip.x
3
(2⇡) 2Ep
=1,2
P.Paganini
Ecole Polytechnique
Physique des particules avancée
60
From wave functions to quantum fields
where Ep = |~
p| and ✏0 (p, 1) = ✏0 (p, 2) = 0 because of the Coulomb gauge. The norm was chosen
as 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 coefficients ↵p~, to operators but there are many
subtleties in the quantization procedure of the photon field mainly due to translation of gauge
condition 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:
µ
A (x) =
Z
3
X
d3 p~
✏µ (p, )↵p~, e
(2⇡)3 2Ep
ip.x
+ ✏⇤µ (p, )↵p†~, e+ip.x
(2.76)
=0
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
i
†
↵ ~0 0 , ↵p~, = g 0 (2⇡)3 2Ep (3) (~
p0 p~)
h p,
i h
i
(2.77)
↵p~0 , 0 , ↵p~, = ↵†~0 0 , ↵p†~, = 0
p,
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 see
that for the polarization = 1, 2, 3, the corresponding operators do correspond to the ones of
the harmonic oscillators as in the scalar case ( g = 1). However, it’s note the case for = 0
( g00 = 1)! In fact, after properly taking into account the gauge constraint, one can show
that for physical states (on which we can do measurements), only the transverse polarizations
matter.
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 of
the Dirac’s equation means that under a Lorentz transformation, (x) becomes 0 (x0 ) = S (x)
0
and satisfies the same mathematical form of the Dirac’s equation: (@/
m) 0 (x0 ) = 0. The
question is to establish the existence of the matrix S which should only depend on the Lorentz
transformation parameters (i.e. and ).
1. Show that S must satisfy S
µ
=
⌫S
µ
⌫
where
µ
⌫
is the Lorentz transformation matrix.
q
q
+1
1 0 1
2. Restricting to a transformation along the x-axis, show that S =
1
l
+
2
2
satisfies the previous relation.
Exercise 2.3 Determination of the helicity states. In the following, u denotes a spinor for a
particle eigenstate of the helicity operator (equation 2.44) with an helicity = ±1/2. We recall
the general formula of a u-spinor:
✓
◆
p
u = E+m
~ .~
p
E+m
P.Paganini
Ecole Polytechnique
Physique des particules
Exercises
1. Show that ~ .~
p
61
= 2 |~
p|
.
2. Using polar coordinates, show that
can take the form for
✓
◆
cos 2✓
1 =
2
ei sin 2✓
3. Finally, show that the helicity state is given by:
0
cos 2✓
B
p
ei sin 2✓
B
E
+
m
1 =
B
|~
p|
+2
@ E+m cos 2✓
|~
p|
✓
i
E+m e sin 2
1
C
C
C e
A
= +1/2:
ip.x
4. Using the same procedure, show that the 3 other helicity states are given by formula 2.45.
P.Paganini
Ecole Polytechnique
Physique des particules avancée