18.01.2015 Views

Quantum Field Theory I

Quantum Field Theory I

Quantum Field Theory I

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

42 CHAPTER 1. INTRODUCTIONS<br />

the scalar representation of the Poincaré group<br />

Investigations of the Poincaré group representations are of vital importance<br />

to any serious attempt to discuss relativistic quantum theory. It is, however,<br />

not our task right now. For quite some time we will need only the simplest<br />

representation, the so-called scalar one. All complications introduced by higher<br />

representations are postponed to the summer term.<br />

Let us consider a Hilbert space in which some representation of the Poincaré<br />

group is defined. Perhaps the most convenient basis of such a space is the<br />

one defined by the eigenvectors of the translation generators P µ , commuting<br />

with each other. The eigenvectors are usually denoted as |p,σ〉, where p =<br />

(p 0 ,p 1 ,p 2 ,p 3 ) stands for eigenvalues of P. In this notation one has<br />

P µ |p,σ〉 = p µ |p,σ〉<br />

and σ stands for any other quantum numbers. The representation of space-time<br />

translations are obtained by exponentiation of generators. If U (Λ,a) is the<br />

element of the representation, corresponding to the Poincaré transformation<br />

x → Λx+a, then U (1,a) = e −iaP . And since |p,σ〉 is an eigenstate of P µ , one<br />

obtains<br />

U(1,a)|p,σ〉 = e −ipa |p,σ〉<br />

And how does the representation of the Lorentz subgroup look like Since the<br />

p µ is a fourvector, one may be tempted to try U (Λ,0)|p,σ〉 = |Λp,σ〉. This<br />

really works, but only in the simplest, the so-called scalar representation, in<br />

which no σ is involved. It is straightforward to check that in such a case the<br />

relation<br />

U (Λ,a)|p〉 = e −i(Λp)a |Λp〉<br />

defines indeed a representation of the Poincaré group.<br />

Let us remark that once the spin is involved, it enters the parameter σ and<br />

the transformation is a bit more complicated. It goes like this: U (Λ,a)|p,σ〉 =<br />

∑<br />

σ ′ e −i(Λp)a C σσ ′ |Λp,σ ′ 〉 and the coefficients C σσ′ do define the particular representation<br />

of the Lorentz group. These complications, however, do not concern<br />

us now. For our present purposes the simplest scalar representation will be<br />

sufficient.<br />

Now it looks like if we had reached our goal — we have a Hilbert space<br />

with a representation of the Poincaré group acting on it. A short inspection<br />

reveals, however, that this is just the rather trivial case of the free particle. To<br />

see this, it is sufficient to realize that the operator P 2 = P µ P µ commutes with<br />

all the generators of the Poincaré group, i.e. it is a Casimir operator of this<br />

group. If we denote the eigenvalue of this operator by the symbol m 2 then the<br />

irreducible representations of the Poincaré group can be classified by the value<br />

of the m 2 . The relation between the energy and the 3-momentum of the state<br />

|p〉 is E 2 − ⃗p 2 = m 2 , i.e. for each value of m 2 we really do have the Hilbert<br />

space of states of a free relativistic particle. (The reader is encouraged to clarify<br />

him/herselfhow should the Hilbert space ofthe states of free relativistic particle<br />

look like. He/she should come to conclusion, that it has to be equal to what we<br />

have encountered just now.)<br />

Nevertheless, this rather trivial representation is a very important one — it<br />

becomes the starting point of what we call the particle-focused approach to the<br />

quantum field theory. We shall comment on this briefly in the next paragraph.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!