Wick contraction between two scalar fields

Question

I have a short question about Wick contraction.

It is given that $$\phi\left(x\right) = \phi^{+}\left(x\right) + \phi^{-}\left(x\right)\tag{1}$$ where: $$\phi^{+}\left(x\right) = \int \frac{d^3p}{\left(2\pi\right)^3} \frac{1}{\sqrt{2\omega_{\textbf{p} }}}\hat{a}^{\dagger}_{\textbf{p}}e^{-ip\cdot x}\tag{2}$$ $$\phi^{-}\left(x\right) = \int \frac{d^3p}{\left(2\pi\right)^3} \frac{1}{\sqrt{2\omega_{\textbf{p} }}}\hat{a}_{\textbf{p}}e^{ip\cdot x}. \tag{3}$$

I have to prove the following: $$T \left\{\phi_{1}\phi_{2} \right\} - N \left\{\phi_{1} \phi_{2} \right\} = \Delta_{F}\left(x_{1} - x_{2} \right)\tag{4}$$ where $\phi_{i} \equiv \phi\left(x_{i}\right)$. Note that $T$ is the time and $N$ is the normal ordering operator.

Writing out the time ordering operator and normal ordering operator (in terms of $\phi_{1}^{+}$, $\phi_{1}^{-}$, $\phi_{2}^{+}$ and $\phi_{2}^{-}$) and subtracting them from each other yields: $$T \left\{\phi_{1}\phi_{2} \right\} - N \left\{\phi_{1} \phi_{2} \right\} = \theta\left(t_{1} - t_{2}\right)\left[\phi_{1}^{-}, \phi_{2}^{+} \right] + \theta\left(t_{2} - t_{1}\right)\left[\phi_{2}^{-}, \phi_{1}^{+} \right]\tag{5}$$ where $\theta$ is the Heaviside function. This expression is correct.

We also know that: $$\Delta_{F}\left(x_{1} - x_{2} \right) = \theta\left(t_{1} - t_{2}\right)D\left(x_{1} - x_{2}\right) + \theta\left(t_{2} - t_{1}\right)D\left(x_{2} - x_{1}\right).\tag{6}$$

Comparing the two equations, I come to the conclusion that: $$D\left(x_{1} - x_{2}\right) = \left[\phi_{1}^{-}, \phi_{2}^{+} \right]\tag{7}$$ $$D\left(x_{2} - x_{1}\right) = \left[\phi_{2}^{-}, \phi_{1}^{+} \right].\tag{8}$$

I don't understand this. The left-hand side is an integral (a complex number) while the right-hand side are two operators?

The reason why I don't understand is as follows:

We defined that: $$ D\left(x_{1} - x_{2}\right) \equiv \langle 0|\phi_{1}\phi_{2} |0\rangle.\tag{9}$$ Writing $\phi_{1}$ and $\phi_{2}$ in terms of $\phi^{+}$ and $\phi^{-}$ gives: $$D\left(x_{1} - x_{2}\right) = \langle 0|\phi_{1}^{-}\phi_{2}^{+} |0\rangle.\tag{10}$$ The other terms are zero because it contains $\widehat{a}$ on the right-hand side or $\widehat{a}^{\dagger}$ on the left-hand side. Acting the vacuum state, $|0\rangle$ or $\langle 0|$ will be zero. To be specific: $\widehat{a}|0\rangle = 0$, $\langle 0| \widehat{a}^{\dagger} = 0$.

We can also write: $$D\left(x_{1} - x_{2}\right) = \langle 0|\left[\phi_{1}^{-}, \phi_{2}^{+}\right] |0\rangle\tag{11}$$ where $\langle 0|\phi_{2}^{+}\phi_{1}^{-} |0\rangle = 0$.

It seems weird to say that: $$D\left(x_{1} - x_{2}\right) = \langle 0|\left[\phi_{1}^{-}, \phi_{2}^{+}\right] |0\rangle = \left[\phi_{1}^{-}, \phi_{2}^{+} \right].\tag{12}$$

The question is, why is it true that $$D\left(x_{1} - x_{2}\right) = \left[\phi_{1}^{-}, \phi_{2}^{+} \right]~?\tag{13}$$

Answer 1

Firstly, $[\phi^+(x_1),\phi^-(x_2)]$ is not two operators, note that

$$[\phi^+(x_1),\phi^-(x_2)]\approx [a^\dagger (\vec p),a(\vec q)]=(2\pi)^3\delta^{(3)}(\vec p-\vec q),\tag{1}$$

where I've omitted the rest of the integral, the right hand side of this is a number (of sorts, but certainly not an operator). Regarding your final point, note that for any number $c\in \Bbb C$ the following is true

$$\langle 0|c|0\rangle=c\>\cdot\>\langle0|0\rangle=c\>\cdot\> 1 \tag{2}$$

because the vacuum is normalised to unity. It therefore follows that since the commutator of the fields is just a number, the expectation value on the vacuum is just equal to the commutator itself.

Answer 2

The main points are:

1. There is an implicitly written identity operator $\hat{\bf 1}$ on the right-hand side of OP's eq. (4).

2. The operators in OP's eqs. (7), (8), (12) & (13) should be sandwiched between the vacuum bra and the vacuum ket. See also this related Phys.SE post.