• Aucun résultat trouvé

Quantisation and the Spin-Statistics Theorem

This is the so called Weyl equation. Now we look on the subspace Eig(p, p0) ofH(0,1/2,+).

The Weyl equation is then given by

(1−σ3+= 0, (4.58)

which is solved by ξ2 = 0. The space of solutions is one-dimensional and is due to (B.89) not changed by null rotations (which means for α = 0). So indeed the null rotations are represented trivially in our representation. It shows also that this spinor has helicity λ = +1/2. It remains as an exercise for the reader to show that the analogue calculations for the dotted spinors give helicity λ=−1/2.

4.3 Quantisation and the Spin-Statistics Theorem

Now we want to quantise the fields above. We have to fulfil the postulates given in the beginning of this chapter. The one-particle wave functions are in general not the solution of the problem to quantise relativistic particles. The reasons are explained already in chapter 3:

A mathematical reason is that we cannot restrict the fields to the Hilbert spaces of positive energy since the interactions “scatter” the wave functions from the positive energy space to the negative one which means that the Hilbert spaces are in general given as the orthogonal sum of the positive and the negative energy space leading to a closed Hilbert space under time evolution. The physical reason is nowadays obvious since the production and annihilation of particle antiparticle pairs are a well known fact. Thus the relativistic quantum mechanics is necessarily a many-particle theory.

To quantise the theory we take the case of scalar particles as a guideline. There we have found the following “recipes” to solve the problems of the negative energy states and the zero point energy:

• The negative energy states are identified with the anti-causal propagation of particles which can be reinterpreted as the causal propagation of an antiparticle with positive energy. This is known as the Feynman-Stueckelberg formalism.

• To renormalise the vacuum to energy 0 we have to normal-order the local operators representing densities of additive observables. This densities are expressed with help of the fields by quantising the corresponding Noether currents of symmetries of the classical action.

Since we have done this quantisation procedure for scalar fields in great detail in chapter 3 we come immediately to the important case of Dirac fields.

4.3.1 Quantisation of the spin-1/2 Dirac Field

As shown in section 4.1. in the massive case these fields are completely determined by the Dirac equation which is is the field equation of motion

i/∂µψ=mψ. (4.59)

Since there are no constraints necessary to obtain the spin-1/2 field the theory is completely determined by the action, which is given by the Lagrangian

L = ¯ψ(i/∂−m)ψ. (4.60)

From this we read off the canonical energy-momentum (3.46) tensor

Θµν = ¯ψiγµνψ−δνµL. (4.61) We like now to quantise this theory. For this purpose we need the total energy of the field configuration to be sure to fulfil postulate 2, namely that the energy is bounded from below.

The Hamiltonian is given with help of (4.51):

H = Z

d3~xΘ00 = Z

d3~xψγ¯ 0(i~γ∇+m)ψ. (4.62) For fields fulfilling the free Dirac equation (4.59) we can write this in the shorter form

H = Z

d3~xψi∂¯ tψ. (4.63)

Now we use the fact that there are known only bosons and fermions in nature which means that the multi-particle Fock space states are totally symmetric or antisymmetric respectively under exchange of single-particle states contained in this multi-particle states. In chapter 1 we have given a nice path integral argument for this.

Since in the case of free fields the Lagrangian is bilinear in the field operators the local observables are represented by bilinear forms of field operators too. Thus the microcausality condition (4.1) can be fulfilled for bosonic as well as fermionic quantisation. This can be seen from the simple formula

[AB,X]=A[B,C]+ [A,B]C=A[B,C]+−[A,B]+C. (4.64) From nature we know that spin-1/2 particles. Thus we try to quantise the Dirac fields as a fermionic field, i.e., we introduce particle annihilation operators a(~p, σ) and antiparticle annihilation operators b(~p, σ) fulfilling the anti-commutator relations

ha(~p1, σ1),a(~p2, σ2)i

+(3)(~p1−~p2σ1σ2, h

b(~p1, σ1),b(~p2, σ2)i

+(3)(~p1−~p2σ1σ2, (4.65) with all other anti-commutators vanishing.

Now we express the field operators in terms of the energy-momentum eigenstates (4.23):

ψ(x) =

Z d3~p p2ω(~p)(2π)3

h

a(~p, σ)u+(p, σ) exp(−ipx) +b(~p, σ)u(−p, σ) exp(ipx)i

p0=ω(~p), (4.66) where we have incorporated the Feynman-Stueckelberg formalism.

Since from (4.40) we have

¯

u±(±p, σ)γ0u±(±p, σ) =u±(±p, σ)u±(±p, σ) = 2ω(~p) (4.67)

4.3 · Quantisation and the Spin-Statistics Theorem

we find by inserting this ansatz the equal-time anti-commutator relation hψ(t, ~x),ψ(t, ~y)i

+(3)(~x−~y)ˆ1, (4.68) where ˆ1 is the 4×4 unity matrix in the Dirac-spinor space.

The Hamiltonian can also be calculated in a straightforward way. With application of normal-ordering the quantised form of (4.63) reads

H=X

σ

Z

d3~pω(~p)[Na(~p, σ) +Nb(~p, σ)], (4.69) where the normal-ordering implies signs from interchanging two fermionic field operators due to the anti-commutators used for fermionic quantisation. This change of sign is crucial for the energy to be positive definite. Thus we have a very important result: The spin-1/2 field is necessarily to be quantised in terms of fermions to get a positive definite energy for the free particles. We shall show below that the observables built with help of the invariant bilinear forms of the field operators, especially the energy and momentum densities, fulfil the requirement of micro causality.

The number operators are given as usual defined by

Na(~p, σ) =a(~p, σ)a(~p, σ), Nb(~p, σ) =b(~p, σ)b(~p, σ). (4.70) The momentum operator is given by

pk= Z

d3~xΘ0k=X

σ

Z

d3~p pk[Na(~p, σ) +Nb(~p, σ)]. (4.71) At last we calculate the electrical charge which is given by the Noether charge of the symmetry under U(1) transformations

δψ= iqψδα, δψ¯=−iqψδα, δx¯ = 0 withq, δα∈R. (4.72) The Noether current is given by (3.41):

jµ=qψγ¯ µψ. (4.73)

Its quantised version is given with help of normal-ordering

jµ=q: ¯ψγµψ: (4.74)

leading to the operator representing the electric charge Q=q

Z

d3~x: ¯ψγ0ψ:=qX

σ

Z

d3~p[Na(~p, σ)−Nb(~p, σ)] (4.75) which shows that the a-particles have electric charge +q and theb-particles −q. Thus from the anti-commutator relations we find that particles and antiparticles have the same quantum numbers except for the electric charge which has opposite sign.

Now we want to show that the microcausality condition is fulfilled. To this end we calculate the anti-commutator of the Dirac field operators for arbitrary space-time points:

ψ(x1),ψ(x¯ 2)

+ = X

σ12

Z d3p~1 p2ω(p~1)(2π)3

Z d3p~2 p2ω(p~2)(2π)3

a(p1, σ1)u+(p1, σ1) exp(−ip1x1) +b(p1, σ1)u(−p1, σ1) exp(+ip1x1), a(p2, σ2)¯u+(p2, σ2) exp(ip2x2) +b(p2, σ2)¯u(−p2, σ2) exp(−ip2x2)

+. (4.76) Using the anti-commutator relations for the creation and annihilation operators (4.65) we find

X

σ

Z d3~p p2ω(~p)(2π)3

u+(p, σ)¯u+(p, σ) exp[ip(x1−x2)]+

+u(−p, σ)¯u(−p, σ) exp[−ip(x1−x2)] .

(4.77)

Herein we have to calculate the matrix 2ˆρ+(p,0) =X

σ

u+(p, σ)¯u+(p, σ) (4.78)

which is two times the polarisation matrix for particles. We know that this matrix can only depend on the on-shell momentum p. Further we know thatu is a solution of the free Dirac equation in momentum representation (4.22) with the upper sign on the right hand side.

By direct calculation we verify

γ0γµγ0µ† (4.79)

and together with (4.22) we find

¯

u+(p, σ)(/p−m) = 0. (4.80)

The end of the story is that the polarisation matrix fulfils the equation

(/p−m)ˆρ+(p,0) = ˆρ+(p,0)(/p−m) (4.81) which has the unique solution

ˆ

ρ+(p,0) =N(/p+m) (4.82)

where the normalisation constantN is given with help of our normalisation convention (4.39) 2 tr ˆρ+(p,0) =X

σ

¯

u+(p, σ)u+(p, σ) = 4m⇒N = 1

2. (4.83)

So our final result is

ˆ

ρ+(p,0) = 1

2(/p+m). (4.84)

In the same line of arguing one obtains for the polarisation matrix for unpolarised antiparticles ˆ

ρ(p,0) = 1 2

X

σ

u(p, σ)¯u(p, σ) = 1

2(/p−m). (4.85)