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Feynman Propagator

Physics Asked by Little Physicist on May 14, 2021

I have learned two ways of representing Feynman Propagator, which are as follows:

The first method is to represent it in the form of commutation relations:
$$[a^{-}(x), a^{+}(y)] = iG(x-y)$$

Where $a^{-}$ and $a^{+}$ represent annihilation and creation operators, respectively. $iG(x-y)$ represents the motion of virtual particle from $y$ to $x$.

The second representation is in the form of Expectation Value of Time Ordering Operator (T), and it is as follows:

$$langle0|T{a^{-}(x) a^{+}(y)}|0rangle = iG(x-y)$$

I do understand Physics, but I cannot apprehend the mathematical nature of the Feynman Propagator. According to the first expression, it is defined by the commutation of two operators, which makes the result (Feynman Propagator) an operator. According to the second expression, it is defined as the time ordering operator’s expectation value, which means it is a scalar. Can someone help me out?

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