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If we consider following $S-$matrix element:$$\left\langle\mathbf{p}_1 \mathbf{k}_2|T\{\phi(x_1) \phi(x_2)\}| 0\right\rangle_0 $$ where $\phi$ denote Klein-Gordon field, and apply the convention in Peskin and Schroeder's book: $$\langle\mathbf{p}| \phi(x)=e^{+i p \cdot x} \tag{4.94}$$

Now I find following basic fact in book:

(1) if $\mathbf{p}_1$ and $\mathbf{k}_2$ are two $\textbf{distinguishable}$ out particle states: $$\left\langle\mathbf{p}_1 \mathbf{k}_2|T\{\phi(x_1) \phi(x_2)\}| 0\right\rangle_0 = 2\cdot e^{+i p_1 \cdot x_1} e^{+i k_2 \cdot x_2} $$

(2) if $\mathbf{p}_1$ and $\mathbf{k}_2$ are two $\textbf{indistinguishable}$ out particle states: $$\left\langle\mathbf{p}_1 \mathbf{k}_2|T\{\phi(x_1) \phi(x_2)\}| 0\right\rangle_0 = e^{+i p_1 \cdot x_1} e^{+i k_2 \cdot x_2} $$

And the factor $2$ work as symmetry factor in Feynman diagram calculation.

But I am troubled for a long time understanding case (1) and (2) from $\textbf{Mathematica viewpoint}$, why from math part their have a factor $2$ difference here?

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I want answer my post myself.

Actually, my analysis in post about the two cases is wrong!!!

In both case (1) and (2), $$ \left\langle\mathbf{p}_1 \mathbf{k}_2\left|T\left\{\phi\left(x_1\right) \phi\left(x_2\right)\right\}\right| 0\right\rangle_0=2 \cdot e^{+i p_1 \cdot x_1} e^{+i k_2 \cdot x_2}$$

I shouldn't confused about the symmetry factor of Feynman diagram and property of indistinguishable particle.

For a more general situation: $$\frac{1}{(n+2m)!}\left\langle\mathbf{p}_1 \mathbf{p}_2 \cdots \mathbf{p}_{n}\left|T\left\{\phi\left(x_1\right) \phi\left(x_2\right) \cdots \phi(x_{n+2m})\right\}\right| 0\right\rangle_0 $$

In this case, the overall factor is (regardless if the particle state distinguishable or not)

$$S=\frac{1}{(n+2m)!}A_{n+2m}^n (2m-1)!! $$

where $$(2m-1)!!=(2m-1)(2m-3)\cdots 1$$ comes form contraction of remaining $\phi$.

What we physically care is that, when we calculate cross section or decay rate, we need divide some number for identical out states.

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