I'm studying Peskin and Schroeder chapter 5. At the beginning of section 5.1, the book tries to compute S matrix of $e^+e^-\rightarrow \mu^+\mu^-$. Using the Feynman from section 4.8, we can draw a Feynman diagram (bottom of page 131) and write down its amplitudes

However I think this amplitude comes from the following term in the perturbation expansion(maybe up to a sign):

$$\bigg\langle 0\bigg|\,a_{k,r} b_{k',r'} (-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu(-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu \,a_{p,s}^\dagger b_{p',s'}^\dagger\bigg|0\bigg\rangle$$

My question is: since we are dealing with two different species of particles here, do the annilation operators and creation operators $a_{k,r},b_{k',r'},a_{p,s},b_{p',s'}$ act on the same Hilbert space? If not, then the contraction procedure no longer works and in effect the expression does not make sense.

Also in the expansion of $\psi$, what do the creation operators $a^\dagger, b^\dagger$ create? Electrons or muons?

I'm studying Peskin and Schroeder chapter 5. At the beginning of section 5.1, the book tries to compute S matrix of $e^+e^-\rightarrow \mu^+\mu^-$. Using the Feynman from section 4.8, we can draw a Feynman diagram (bottom of page 131) and write down its amplitudes

However I think this amplitude comes from the following term in the perturbation expansion  (maybe up to a sign):

$$\bigg\langle 0\bigg|\,a_{k,r} b_{k',r'} (-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu(-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu \,a_{p,s}^\dagger b_{p',s'}^\dagger\bigg|0\bigg\rangle.$$

My question is: since we are dealing with two different species of particles here, do the annilation operators and creation operators $a_{k,r},b_{k',r'},a_{p,s},b_{p',s'}$ act on the same Hilbert space? If not, then the contraction procedure no longer works and in effect the expression does not make sense.

Also in the expansion of $\psi$, what do the creation operators $a^\dagger, b^\dagger$ create? Electrons or muons?

I'm studying Peskin and Schroeder chapter 5. At the beginning of section 5.1, the book tries to compute S matrix of $e^+e^-\rightarrow \mu^+\mu^-$. Using the Feynman from section 4.8, we can draw a Feynman diagram (bottom of page 131) and write down its amplitudes

However I think this amplitude comes from the following term in the perturbation expansion(maybe up to a sign):

$$\bigg\langle 0\bigg|\,a_{k,r} b_{k',r'} (-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu(-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu \,a_{p,s}^\dagger b_{p',s'}^\dagger\bigg|0\bigg\rangle$$

My question is: since we are dealing with two different species of particles here, do the annilation operators and creation operators $a_{k,r},b_{k',r'},a_{p,s},b_{p',s'}$ act on the same Hilbert space? If not, then the contraction procedure no longer works and in effect the expression does not make sense.

I'm studying Peskin and Schroeder chapter 5. At the beginning of section 5.1, the book tries to compute S matrix of $e^+e^-\rightarrow \mu^+\mu^-$. Using the Feynman from section 4.8, we can draw a Feynman diagram (bottom of page 131) and write down its amplitudes

However I think this amplitude comes from the following term in the perturbation expansion(maybe up to a sign):

$$\bigg\langle 0\bigg|\,a_{k,r} b_{k',r'} (-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu(-i)\int_{\mathbb{R}^4}e\overline{\psi}\gamma^\mu\psi A_\mu \,a_{p,s}^\dagger b_{p',s'}^\dagger\bigg|0\bigg\rangle$$

My question is: since we are dealing with two different species of particles here, do the annilation operators and creation operators $a_{k,r},b_{k',r'},a_{p,s},b_{p',s'}$ act on the same Hilbert space? If not, then the contraction procedure no longer works and in effect the expression does not make sense.

Also in the expansion of $\psi$, what do the creation operators $a^\dagger, b^\dagger$ create? Electrons or muons?