Quantum Field Theory I

2.2. CANONICAL QUANTIZATION 69
For infinite number of DOF, i.e. for the same set of commutation relations,
but with i,j = 1,...,∞ the situation is dramatically different:
• existence of |0〉 annihilated by all a i operators is not guaranteed
• the Fock representation, nevertheless, does exist
• there are infinitely many representations non-equivalent to the Fock one
• Hamiltonians and other operators are usually ill-defined in the Fock space
Let us discuss these four point in some detail.
As to the existence of |0〉, for one oscillator the proof is notoriously known
fromQMcourses. Itisbasedonwell-knownpropertiesoftheoperatorN = a + a:
1
¯. N |n〉 = n|n〉 ⇒ Na|n〉 = (n−1)a|n〉 (a+ aa = [a + ,a]a+aa + a = −a+aN)
2
¯.0 ≤ ‖a|n〉‖2 = 〈n|a + a|n〉 = n〈n|n〉 implying n ≥ 0, which contradicts 1¯
unless the set of eigenvalues n contains 0. The corresponding eigenstate is |0〉.
For a finite number of independent oscillators the existence of the common
|0〉 (∀i a i |0〉 = 0) is proven along the same lines. One considers the set of
commuting operators N i = a + i a i and their sum N = ∑ i a+ i a i. The proof is
basically the same as for one oscillator.
For infinite number of oscillators, however, neither of these two approaches
(nor anything else) really works. The step by step argument proves the statement
for any finite subset of N i , but fails to prove it for the whole infinite set.
The proof based on the operator N refuses to work once the convergence of the
infinite series is discussed with a proper care.
Instead of studying subtleties of the breakdown of the proofs when passing
from finite to infinite number of oscillators, we will demonstrate the existence of
the so-called strange representations of a + i ,a i (representations for which there
is no vacuum state |0〉) by an explicit construction (Haag 1955). Let a + i ,a i
be the creation and annihilation operators in the Fock space with the vacuum
state |0〉. Introduce their linear combinations b i = a i coshα + a + i sinhα and
b + i
= a i sinhα +a + i
coshα. Commutation relations for the b-operators are the
same as for the a-operators (check it). Now let us assume the existence of a
state vector |0 α 〉 satisfying ∀i b i |0 α 〉 = 0. For such a state one would have
0 = 〈i|b j |0 α 〉 = 〈i,j|0 α 〉coshα+〈0|0 α 〉δ ij sinhα
which implies 〈i,i|0 α 〉 = const (no i dependence). Now for i being an element
of an infinite index set this constant must vanish, because otherwise
the norm of the |0 α 〉 state comes out infinite (〈0 α |0 α 〉 ≥ ∑ ∞
∑ i=1 |〈i,i|0 α〉| 2 =
∞
i=1 const2 ). And since const = −〈0|0 α 〉tanhα the zero value of this constant
implies 〈0|0 α 〉 = 0. Moreover, vanishing 〈0|0 α 〉 implies 〈i,j|0 α 〉 = 0.
Itisstraightforwardtoshowthatalso〈i|0 α 〉 = 0(0 = 〈0|b i |0 α 〉 = 〈i|0 α 〉coshα)
and finally 〈i,j,...|0 α 〉 = 0 by induction
〈i,j,...| b k |0 α 〉 = 〈i,j,...|
} {{ } } {{ }
n n+1
0 α 〉coshα+〈i,j,...| 0 α 〉sinhα
} {{ }
n−1
But 〈i,j,...| form a basis of the Fock space, so we can conclude that within
the Fock space there is no vacuum state, i.e. a non-zero normalized vector |0 α 〉
satisfying ∀i b i |0 α 〉 = 0.