Given commutation relation for fields, derive commutation relations for 3-momentum valued creation and annihiliation vectors [closed]

Question

In @Jorge Lavín's solution to this question, Annihilation and creation operator - $\phi$ and $\pi$ for Klein-Gordon Field, can someone explain to me why in the first line of his answer, we need to distinguish between the momentum of the Fourier transform,

$$ \phi(p) = \int d^3x e^{-i p \cdot x} \phi(x) $$

and the momentum in the expression for $\phi(x)$,

$$ \phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 E_p}} [a_p e^{i p\cdot x} + a_p {}^\dagger e^{-i p \cdot x}] $$

when we plug the second equation into the first.

It is a bit unclear to me.

Answer 1

I see it in this way: in the expression for $\phi(x)$ we are integrating over all possible momenta in the phase space. So in that expression there are all the possible momenta, integrating it's just a way of summing

$$\phi(x) \propto a_{p_1}e^{ip_1 x}+a_{p_1}^\dagger e^{-ip_1 x}+ a_{p_2}e^{ip_2 x}+a_{p_2}^\dagger e^{-ip_2 x}+\cdots$$

In the Fourier transform of $\phi(x)$ the momentum is a single variable, you choose one and get back something

$$\phi(p_1) =\int d^3x\;e^{i\color{red}{p_1}x} \phi(x) \qquad \phi(p_2) = \int d^3x\;e^{i\color{red}{p_2}x} \phi(x)$$

while the position $x$ is summed over, so you have every possible position. In the same way the position in $\phi(x)$ is just a variable that can have only one definite value.

So you need to distinguish the two momentum when plugging it in the fourier transform because if you don't do it, you let the exponential in the fourier transform have every possible value of $p$ and so loose the dependence on the momentum variable in $\phi(p)$.

If you find this confusing, being the momentum in $\phi(x)$ just a dummy integration variable, you can just write down the following

$$ \phi(x) = \int\frac{d^3 k}{(2\pi)^3}\frac{1}{\sqrt{2E_k}} [a_ke^{ikx}+a_k^\dagger e^{-ikx}]$$

so that you won't have any confusion when plugging it in the fourier transform.