Fall Term Exam, N^{0}. 6 (solution by D. Freed)

Problem:

Let O(x) be a local operator which annihilates the vacuum:

(1)  O(x)|Ω⟩ = 0.

Show at least at a physical level of rigor that O(x) = 0.

Now extend this slightly. Suppose that for some distinct points x and y⟨Ω|O^{∗}(y)O(x)|Ω⟩ = 0. (O^{∗} is the hermitian adjoint of O.) Prove that O = 0.

Solution:

We recall that in a quantum field theory the correlation function (better distribution) F (x_{1}, …, x_{n}) = ⟨O_{1}(x_{1})O_{2}(x_{2}) … O_{n}(x_{n})⟩ is initially defined for x_{1}, …, x_{n} distinct points in Minkowski space V. It is the boundary value of an analytic function on a connected domain D ⊂ (V_{ℂ})^{×n} where D includes:

(1) The subspace of E^{×n} consisting of n distinct points where E is Euclidean space; and

(2) an open set in V^{×n} consisting of n spacelike separated points.

(The precise description of some admissible open sets in (2) is given in the lemma of Jost.) The restriction of F to either of the two sets described determines F.

Now suppose O is an operator which annihilates the vacuum: O(x)|Ω⟩ = 0. We would like to show that O = 0, which means that any correlation function involving O vanishes identically. For this we restrict the correlation function to a set of the form (2) above, and then

⟨O_{1}(x_{1}) ⋯ O(x) ⋯ O_{n}(x_{n})⟩ = ⟨O_{1}(x_{1}) ⋯ O_{n}(x_{n})O(x)⟩
= ⟨Ω|O_{1}(x_{1}) ⋯ O_{n}(x_{n})O(x)|Ω⟩
= 0.

For the second assertion, we suppose ⟨\overline{O(y)}O(x)⟩ = 0 for distinct points x, y in Euclidean space; we want to show that any correlation function with O vanishes (in Euclidean space). Let π denote the perpendicular bisecting hyperplane to the segments joining x and y, and denote by \overline{p} the reflection of p in π. Thus y = \overline{x}. Now let

O′ = O_{1}(p_{1}) ⋯ O_{n}(p_{n}),
\overline{O′} = \overline{O_{1}(\overline{p}_{1})} ⋯ \overline{O_{n}(\overline{p}_{n})},

denote some product of operators. We must show ⟨O(x)O′⟩ = 0. Consider

⟨(\overline{O(\overline{x})} + λO′) (O(x) + λ\overline{O′})⟩.

By reflection positivity this is nonnegative. Set

λ = \frac{−Re⟨O(x)O′⟩}{⟨O′\overline{O′}⟩}

to conclude

(Re⟨O(x)O′⟩^{2} ≤ ⟨\overline{O(\overline{x})}O(x)⟩ ⟨O′\overline{O′}⟩ = 0.

Thus Re⟨O(x)O′⟩ = 0. The arguments applied to \sqrt{−1}O′ show the imaginary part vanishes as well.