How do you write the Wightman function $\langle\phi(t_1)\phi(t_2)\rangle$ for a massive scalar field in position space?

Question

For a free real scalar field $\phi(t,\mathbf{x})$, we define the Wightman function as: $$ W(t_1,t_2) \equiv \langle 0 | \phi(t_1,\mathbf{x}_1) \phi(t_2,\mathbf{x}_2) | 0 \rangle $$ I'm suppressing the position labels for clarity. And the Feynman propagator is defined as: $$ F(t_1,t_2) \equiv \langle 0 | \mathcal{T}\big( \phi(t_1,\mathbf{x}_1) \phi(t_2,\mathbf{x}_2) \big) | 0 \rangle $$ where $\mathcal{T}$ is the time-ordering "operator". This can be expressed in terms of the Wightman function as $$ F(t_1,t_2) = \Theta(t_1-t_2) W(t_1,t_2) + \Theta(t_2-t_1) W(t_2,t_1) $$

These functions have the properties $$ W(t_1,t_2) = W(t_2,t_1)^{\ast} \\ F(t_1,t_2) = F(t_2,t_1) $$

I believe that it follows from this (correct me if I'm wrong): $$ W(t_1,t_2) = \mathrm{Re}\big[ F(t_1,t_2) \big] + i \mathrm{sign}(t_1-t_2) \mathrm{Im}\big[ F(t_1,t_2) \big] \\ F(t_1,t_2) = \mathrm{Re}\big[ W(t_1,t_2) \big] + i \mathrm{sign}(t_1-t_2) \mathrm{Im}\big[ W(t_1,t_2) \big] \\ $$

It is well known that the position-space representation of $F$ for a massive $m\neq 0$ field is: $$ F(t_1,t_2)=\lim_{\epsilon \to 0^{+}} \frac{m}{4\pi^2 \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 + i \epsilon } } K_{1}\left( m \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 + i \epsilon } \right) $$ where $K_1$ is the modified Bessel function of the second kind (of order $1$).

QUESTION: Is the following true? $$ W(t_1,t_2)=\lim_{\epsilon \to 0^{+}} \frac{m}{4\pi^2 \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 } } K_{1}\left( m \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 } \right) $$

I would be inclined to believe this, since for a massless $m=0$ field, the corresponding functions are: $$ F(t_1,t_2) = \lim_{\epsilon \to 0^{+}} \frac{1}{4\pi^2} \frac{1}{-(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 + i \epsilon} \\ W(t_1,t_2) = \lim_{\epsilon \to 0^{+}} \frac{1}{4\pi^2} \frac{1}{-(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2} $$

Answer 1

The answer is positive and the result can be proved separately form the expression you quoted for the Feynman propagator, just expanding smeared quantum fields into definite-momentum modes and computing weak limits ($\epsilon$-prescriptions). However a formal proof of how the formula of Feynman propagator gives rise to that of the two-point function can be obtained as follows (there are mathematical details to fix) From $$ F(t_1,t_2) = \Theta(t_1-t_2) W(t_1,t_2) + \Theta(t_2-t_1) W(t_2,t_1) \tag{1} $$ you have, where the bar denotes the complex conjugation, $$ \overline{F(t_1,t_2)} = \Theta(t_1-t_2) \overline{W(t_1,t_2)} + \Theta(t_2-t_1) \overline{W(t_2,t_1)}\:. $$ However, assuming the quantum fields Hermitian, you also have $$\overline{W(t_1,t_2)}= W(t_2,t_1)$$ so that $$ \overline{F(t_1,t_2)} = \Theta(t_1-t_2) {W(t_2,t_1)} + \Theta(t_2-t_1) {W(t_1,t_2)}\:.\tag{2} $$ Multiplying (1) for $\Theta(t_1-t_2)$ and (2) for $\Theta(t_2-t_1)$ we obtain $$ \Theta(t_1-t_2)F(t_1,t_2) = \Theta(t_1-t_2) W(t_1,t_2) \tag{1'} $$ and $$ \Theta(t_2-t_1)\overline{F(t_1,t_2)} = \Theta(t_2-t_1) {W(t_1,t_2)}\:.\tag{2'} $$ Summing side-by-side $$\Theta(t_1-t_2)F(t_1,t_2) + \Theta(t_2-t_1)\overline{F(t_1,t_2)} = (\Theta(t_1-t_2) + \Theta(t_2-t_1))W(t_1,t_2)\:. $$ In other words $$W(t_1,t_2) = \Theta(t_1-t_2)F(t_1,t_2) + \Theta(t_2-t_1)\overline{F(t_1,t_2)}\tag{3}\:.$$ It is not difficult that this identity produces $$ W(t_1,t_2)=\lim_{\epsilon \to 0^{+}} \frac{m}{4\pi^2 \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 } } K_{1}\left( m \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 } \right)\tag{4} $$ from $$ F(t_1,t_2)=\lim_{\epsilon \to 0^{+}} \frac{m}{4\pi^2 \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 + i \epsilon } } K_{1}\left( m \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 + i \epsilon } \right)\:. $$ In particular since both the square root and $K_1$ have the property that $\overline{f(z)}= f(\overline{z})$ in the considered branch of their domains (which are Riemann surfaces) $$ \overline{F(t_1,t_2)}=\lim_{\epsilon \to 0^{+}} \frac{m}{4\pi^2 \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 - i \epsilon } } K_{1}\left( m \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 - i \epsilon } \right)\:. $$ If $t_2-t_1>0$, we can replace $i \epsilon $ for $2i \epsilon(t_2-t_1)$ producing $$ \Theta(t_2-t_1)\overline{F(t_1,t_2)}=\lim_{\epsilon \to 0^{+}} \frac{m\Theta(t_2-t_1)}{4\pi^2 \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 - 2i \epsilon (t_2-t_1)} } K_{1}\left( m \sqrt{ -(t_1-t_2)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2 - 2i \epsilon (t_2-t_1)} \right)\:. $$ Namely, $$ \Theta(t_2-t_1)\overline{F(t_1,t_2)}=\lim_{\epsilon \to 0^{+}} \frac{m\Theta(t_2-t_1)}{4\pi^2 \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2} } K_{1}\left( m \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2} \right)\:. $$ The same argument analogously produces $$ \Theta(t_1-t_2)F(t_1,t_2)=\lim_{\epsilon \to 0^{+}} \frac{m\Theta(t_1-t_2)}{4\pi^2 \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2} } K_{1}\left( m \sqrt{ -(t_1-t_2-i\epsilon)^2 + |\mathbf{x}_1-\mathbf{x}_2|^2} \right)\:. $$ Using (3), we find just your identity (4).