A Creation and annihilation operator

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The discussion centers on the action of creation and annihilation operators on complex exponential functions in the context of quantized fields. It clarifies that while these operators commute with the exponential factors, they do not act on them in the traditional sense. The creation operator \(\hat{a}^{\dagger}_{\textbf{p}}\) and annihilation operator \(\hat{a}_{\textbf{p}}\) are linear operators defined in Fock space, whereas the exponential functions are treated as numerical values. Thus, the operators do not directly influence the exponential terms \(e^{i\textbf{p} \cdot \textbf{x}}\) and \(e^{-i\textbf{p} \cdot \textbf{x}}\). This distinction is crucial for understanding their roles in quantum field theory.
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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##.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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