Graduate Creation and annihilation operator

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SUMMARY

The discussion centers on the role of creation operator \(\hat{a}^{\dagger}_{\textbf{p}}\) and annihilation operator \(\hat{a}_{\textbf{p}}\) in the context of the quantized field in the Schrödinger picture. It is established that these operators do not act on the complex exponentials \(e^{i\textbf{p} \cdot \textbf{x}}\) and \(e^{-i\textbf{p} \cdot \textbf{x}}\) directly, as they commute with these factors. Instead, the operators are linear operators defined within the Fock space, while the exponential functions are treated as scalar quantities.

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Sebas4
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TL;DR
Does the annihilation/creation operator on the complex exponent?
Hey, I have a short question.
The quantized field in Schrödinger picture is given by:
\hat{\phi} \left(\textbf{x}\right) =\int \frac{d^{3}p}{\left(2\pi\right)^3} \frac{1}{\sqrt{\omega_{2\textbf{p}}}}\left(\hat{a}_{\textbf{p}}e^{i\textbf{p} \cdot \textbf{x}} + \hat{a}^{\dagger}_{\textbf{p}}e^{-i\textbf{p} \cdot \textbf{x}}\right)

My question is, does the the annihilation \hat{a}_{\textbf{p}} and creation \hat{a}^{\dagger}_{\textbf{p}} operator act on e^{i\textbf{p} \cdot \textbf{x}} and e^{-i\textbf{p} \cdot \textbf{x}} respectively? In other words: does the annihilation/creation operator on the complex exponent?
 
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Sebas4 said:
does the the annihilation a^p and creation a^p† operator act on eip⋅x and e−ip⋅x respectively?

No.
 
Well, it does in the sense that ##a## and ##a^\dagger## commute with these factors.
 
No, they don't. The creation and annihilation operators are linear operators defined in the Fock space. The expeonential functions are numbers; ##\vec{x}, \vec{p} \in \mathbb{R}^3##.
 

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