I am currently working on exercise 3.7(c) of Peskin and Schroeder's An Introduction to Quantum Field Theory. The problem statement is:

Show that any Hermition, Lorentz scalar local operator built from $\psi(x)$ and $\phi(x)$ and their conjugates have $CPT = + 1$

The solution manual states:

" Any Lorentz-scalar hermitian local operator $O(x)$ constructed from $\psi(x)$ or $\phi(x)$ can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and contains either $\psi(x)$ or $\phi(x)$ only"

How do I prove this statement? To me, this statement seems false. Why cannot I have something that is a mixture of $\psi(x)$ or $\phi(x)$ (i.e. contains both $\psi(x)$ or $\phi(x)$)?

Next the solution states:

" and for operators constructed from $\phi(x)$, we note that all such operators can be decomposed further into a product (including Lorentz inner product) of operators of the form $(\partial_{\mu_1}· · · \partial_{\mu_m}\phi^\dagger)(\partial_{\mu_1}· · · \partial_{\mu_n}\phi) + c.c$"

As before, why is this true? Why can it not contain products of $\phi$'s and $\phi^\dagger$'s? Is Lorentz invariance broken or something, and if so why?

I am currently working on exercise 3.7c of Peskin and Schroeder's An Introduction to Quantum Field Theory. The problem statement is:

Show that any Hermition, Lorentz scalar local operator built from $\psi(x)$ and $\phi(x)$ and their conjugates have $CPT = + 1$

The solution manual states:

" Any Lorentz-scalar hermitian local operator $O(x)$ constructed from $\psi(x)$ or $\phi(x)$ can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and contains either $\psi(x)$ or $\phi(x)$ only"

How do I prove this statement? To me, this statement seems false. Why cannot I have something that is a mixture of $\psi(x)$ or $\phi(x)$ (i.e. contains both $\psi(x)$ or $\phi(x)$)?

Next the solution states:

" and for operators constructed from $\phi(x)$, we note that all such operators can be decomposed further into a product (including Lorentz inner product) of operators of the form $(\partial_{\mu_1}· · · \partial_{\mu_m}\phi^\dagger)(\partial_{\mu_1}· · · \partial_{\mu_n}\phi) + c.c$"

As before, why is this true? Why can it not contain products of $\phi$'s and $\phi^\dagger$'s? Is Lorentz invariance broken or something, and if so why?

I am currently working on exercise 3.7(c) of Peskin and Schroeder's An Introduction to Quantum Field Theory. The problem statement is:

Show that any Hermition, Lorentz scalar local operator built from $\psi(x)$ and $\phi(x)$ and their conjugates have $CPT = +1$

The solution manual states:

" Any Lorentz-scalar hermitian local operator $O(x)$ constructed from $\psi(x)$ or $\phi(x)$ can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and contains either $\psi(x)$ or $\phi(x)$ only"

How do I prove this statement? To me, this statement seems false. Why cannot I have something that is a mixture of $\psi(x)$ or $\phi(x)$ (i.e. contains both $\psi(x)$ or $\phi(x)$)?

Next the solution states:

" and for operators constructed from $\phi(x)$, we note that all such operators can be decomposed further into a product (including Lorentz inner product) of operators of the form $(\partial_{\mu_1}· · · \partial_{\mu_m}\phi^\dagger)(\partial_{\mu_1}· · · \partial_{\mu_n}\phi) + c.c$"

As before, why is this true? Why can it not contain products of $\phi$'s and $\phi^\dagger$'s? Is Lorentz invariance broken or something, and if so why?

I am currently working on exercise 3.7(c) of Peskin and Schroeder's An Introduction to Quantum Field Theory. The problem statement is:

Show that any Hermition, Lorentz scalar local operator built from $\psi(x)$ and $\phi(x)$ and their conjugates have $CPT = + 1$

The solution manual states:

" Any Lorentz-scalar hermitian local operator $O(x)$ constructed from $\psi(x)$ or $\phi(x)$ can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and contains either $\psi(x)$ or $\phi(x)$ only"

How do I prove this statement? To me, this statement seems false. Why cannot I have something that is a mixture of $\psi(x)$ or $\phi(x)$ (i.e. contains both $\psi(x)$ or $\phi(x)$)?

Next the solution states:

" and for operators constructed from $\phi(x)$, we note that all such operators can be decomposed further into a product (including Lorentz inner product) of operators of the form $(\partial_{\mu_1}· · · \partial_{\mu_m}\phi^\dagger)(\partial_{\mu_1}· · · \partial_{\mu_n}\phi) + c.c$"

As before, why is this true? Why can it not contain products of $\phi$'s and $\phi^\dagger$'s? Is Lorentz invariance broken or something, and if so why?