Quantum Field Theory I

18 CHAPTER 1. INTRODUCTIONS
Remark: Let us recall that for two linear spaces U (basis e i , dimension m) and
V (basis f j , dimension n), the direct sum and product are linear spaces U ⊕V
(basis generated by both e i and f j , dimension m+n) and U⊗V (basis generated
by ordered pairs (e i , f j ), dimension m.n).
Remark: The fact that the Hilbert space of a system of two non-identical
particles is the direct product of the 1-particle Hilbert spaces may come as not
completely obvious. If so, it is perhaps agood idea to start from the question what
exactly the 2-particle system is (provided that we know already what the 1-particle
system is). The answer within the quantum physics is not that straightforward
as in the classical physics, simply because we cannot count the quantum particles
directly, e.g. by pointing the index finger and saying one, two. Still, the answer
is not too complicated even in the quantum physics. It is natural to think of a
quantum system as being 2-particle iff
a) it contains states with sharp quantum numbers (i.e. eigenvalues of a complete
system of mutually commuting operators) of both 1-particle systems, and this
holds for all combinations of values of these quantum numbers
b) such states constitute a complete set of states
This, if considered carefully, is just the definition of the direct product.
Remark: A triviality which, if not explicitly recognized, can mix up one’s mind:
H 1 ∩H 2 = ∅, i.e. the 2-particle Hilbert space contains no vectors corresponding
to states with just one particle, and vice versa.
Remark: The fact that multiparticle states of identical particles are represented
by either completely symmetric or completely antisymmetric vectors should be
familiar from the basic QM course. The former case is called bosonic, the latter
fermionic. In all formulae we will try, in accord with the common habit, to
treat these two possibilities simultaneously, using symbols like ± and ∓, where
the upper and lower signs apply to bosons and fermions respectively.
Remark: As to the basis vectors, our notation is not the only possible one. Another
widely used convention (the so-called occupation number representation)
denotes the basis vectors as |n 1 ,n 2 ,...〉, where n i is the number of particles in
the i-th 1-particle state. So e.g. |2,2,2,4,4〉 ⇔ |0,3,0,2,0,0,0,...〉, where the
LHS is in our original notation while the RHS is in the occupation number representation.
The main drawback of the original notation is that it is not unique,
e.g. |1,2,3〉 and ±|1,3,2〉 denotes the same vector. One should be therefore
careful when summing over all basis states. The main drawback of the occupation
number representation is typographical: one cannot write any basis vector
without the use of ellipsis, and even this may sometimes become unbearable (try
e.g. to write |49,87,642〉 in the occupation number representation).
Remark: The basis vectors |i,j,...,k〉 or |n 1 ,n 2 ,...〉 are not all normalized to
unity (they are, but only if all i,j,...,k are mutually different, i.e. if none of
n i exceeds 1). If some of the i,j,...,k are equal, i.e. if at least one n i > 1, then
the norm of the fermionic state is automatically zero (this is the Pauli exclusion
principle), while the norm of the bosonic state is √ n 1 ! n 2 ! .... Prove this.
Remark: A triviality which, if not explicitly recognized, can mix up one’s mind:
the vacuum |0〉 is a unit vector which has nothing to do with the zero vector 0.