# Why are these distributions positive?

I am trying to understand some calculations in a paper by Sidney Coleman. He is showing that certain distributions are positive. The paper can be found here. What I am talking about is happening at the bottom of page 262. (Which is 4 pages into the document.)

We are looking at distributions in 2d space (scalar products evaluate as $a^\nu g_{\nu \mu} b^\mu=a^0b^0 -a^1b^1$): $$F(k)=\int d^2x\ e^{i k x} \langle0|\phi(x)\phi(0)|0\rangle \tag{11a}$$ $$F_{\mu}(k)=\int d^2x\ e^{i k x} \langle0|j_\mu(x)\phi(0)|0\rangle \tag{11b}$$ $$F_{\mu\nu}(k)=\int d^2x\ e^{i k x} \langle0|j_\mu(x)j_\nu(0)|0\rangle \tag{11c}$$

He says that $\int d^2x\ h(x) (aj_0(x)+b\phi(x))|0\rangle$ must have positive norm for any test function $h$ and any $a, b \in \mathbb C$. Ok, so far so good.

He then says that this implies $F_{00}$ and $F$ are positive distributions, which follows from a calculation.

It seems extremely straitforward, and if you look at $F_{00}$ and $F$ it seems clear that this must follow from considering $b=0$ for $F_{00}$ and $a=0$ for $F$.

However I cannot do the calculation and am feeling like an idiot. How does it work?

(Side point, it is also a bit unclear to me how these distributions act on test functions, do I take the Fourier transform of a function then multiply with $F(k)$ and integrate, or are we just taking the distribution associated to the (hopefully $L^1_\text{loc}$) function $F$?)

## 1 Answer

In the end it turns out that the key was a physical consideration. These correlation functions $\langle0|\phi(x)\phi(y)|0\rangle$ are Lorentz-invariant, which implies they only depend on the distance between $x$ and $y$. With that in mind the calculation is simple. From now on write $\langle0|\phi(x)\phi(0)|0\rangle=f(x)$.

In the case of $F$, let $g(k)$ be a positive test function, so write $g(k)=|\tilde h(k)|^2$ for some test function $\tilde h$. Then if $\tilde h(k)=\mathcal F(h)(k)$ is the Fourier transform of some Schwartz function $h$, we have $g(k)=\tilde h(k)\overline{\tilde h(k)}=\mathcal F(h *\overline{h \circ \iota})(k)$, where $\iota(x)=-x$. Putting that to use gives us

\begin{align} F(g)&=\int d^2k\int d^2x\ e^{ikx}\mathcal F\left(h *\overline{h \circ \iota}\right)(k) f(x)=\int d^2k \ \mathcal F\left(h *\overline{h \circ \iota}\right) \mathcal F(f)(k)\\ &=\int d^2k \ \mathcal F\left(h *\overline{h \circ \iota}*f\right)(k)=\left(h *\overline{h \circ \iota}*f\right)(0) \end{align}

Evaluating the convolutions:

$$F(g)=\left(h *\overline{h \circ \iota}*f\right)(0)=\int d^2x\ d^2x'\ h(x)\ (\overline{h\circ \iota})(0-x + x')\ f(x')$$

Doing a substitution $y=x-x'$ and putting in $f(x-y)=\langle0|\phi(x)\phi(y)|0\rangle$ then gives:

$$F(g)=\left(\int d^2x \ h(x) \langle0|\phi(x)\right)\left(\int d^2y \ \overline{h(y)} \phi(y)|0\rangle\right)$$

Which must be positive since it is the scalar product of $\left(\int d^2y \ \overline{h(y)} \phi(y)|0\rangle\right)$ with itself.

Thus $F(g)$ is positive whenever $g$ is positive and $F$ is a positive distribution.

For $F_{00}$ the proof is identical, except $\phi(x)$ is replaced by $j_0(x)$.