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We have the usual commutation relations for creation/annihilation operators in QFT, $$\left[\hat a(\vec p),\hat a(\vec q)\right]=\left[\hat a(\vec p)^\dagger,\hat a(\vec q)^\dagger\right]=0\tag{1}$$ $$\left[\hat a(\vec p),\hat a(\vec q)^\dagger\right]=(2\pi)^3 2E(\vec p)\delta^{(3)}\left(\vec p- \vec q\right)\tag{2}$$

I'm trying to understand the last two equalities of the following proof:

$$\left[\hat a(\vec p)^\dagger\hat a(\vec p), \hat a(\vec q)^\dagger\hat a(\vec q)\right]$$

$$=\hat a(\vec p)^\dagger\left[\hat a(\vec p),\hat a(\vec q)^\dagger\hat a(\vec q)\right]+\left[\hat a(\vec p)^\dagger,\hat a(\vec q)^\dagger\hat a(\vec q)\right]\hat a(\vec p)$$

$$=\hat a(\vec p)^\dagger\Big(\hat a(\vec q)^\dagger\left[\hat a(\vec p),\hat a(\vec q)\right]+\left[\hat a(\vec p),\hat a(\vec q)^\dagger\right]\hat a(\vec q)\Big)$$

$$+\Big(\hat a(\vec q)^\dagger\left[\hat a(\vec p)^\dagger,\hat a(\vec q)\right]+\left[\hat a(\vec p)^\dagger,\hat a(\vec q)^\dagger\right]\hat a(\vec q)\Big)\hat a(\vec p)$$

$$=\hat a(\vec p)^\dagger\left[\hat a(\vec p), \hat a(\vec q)^\dagger\right]\hat a(\vec q)-\hat a(\vec q)^\dagger\left[\hat a(\vec q), \hat a(\vec p)^\dagger\right]\hat a(\vec p)$$

$$\color{red}{=}(2 \pi)^32E(\vec p)\delta^{(3)}\left(\vec p- \vec q\right)\Big(\hat a(\vec p)^\dagger \hat a(\vec q)-\hat a(\vec q)^\dagger \hat a(\vec p)\Big)\color{red}{=}0$$

Taking these equalities (highlighted red) one in turn, starting with the first,

$$\hat a(\vec p)^\dagger\left[\hat a(\vec p), \hat a(\vec q)^\dagger\right]\hat a(\vec q)-\hat a(\vec q)^\dagger\left[\hat a(\vec q), \hat a(\vec p)^\dagger\right]\hat a(\vec p)$$ $$=(2 \pi)^32E(\vec p)\delta^{(3)}\left(\vec p- \vec q\right)\Big(\hat a(\vec p)^\dagger \hat a(\vec q)-\hat a(\vec q)^\dagger \hat a(\vec p)\Big)\tag{3}$$

Taking eqn. $(2)$, $$\left[\hat a(\vec p),\hat a(\vec q)^\dagger\right]=(2\pi)^3 2E(\vec p)\delta^{(3)}\left(\vec p- \vec q\right)$$ and switching the momenta labels, $\vec p \to \vec q$ and $\vec q \to \vec p$ should result in a similar equation to $(2)$, namely, $$\left[\hat a(\vec q),\hat a(\vec p)^\dagger\right]=(2\pi)^3 2E(\vec q)\delta^{(3)}\left(\vec q- \vec p\right)\tag{4}$$

Okay, so the $\delta$ 'function' is even, so $\delta^{(3)}\left(\vec p- \vec q\right)=\delta^{(3)}\left(\vec q- \vec p\right)$.

But eqn. $(3)$ can only be true if $(2)$ and $(4)$ are equal, and this can only happen if $E(\vec p)=E(\vec q)$. But why should this be the case?

For the final equality, $$(2 \pi)^32E(\vec p)\delta^{(3)}\left(\vec p- \vec q\right)\Big(\hat a(\vec p)^\dagger \hat a(\vec q)-\hat a(\vec q)^\dagger \hat a(\vec p)\Big)=0\tag{5}$$

I fail to see how eqn. $(5)$ holds as to invoke the 'sifting/sampling property' of the $\delta$ function one would need to be integrate over $\vec q$ (or $\vec p$) so that the term in parentheses is zero, or $$\hat a(\vec p)^\dagger \hat a(\vec q)=\hat a(\vec q)^\dagger \hat a(\vec p)$$ when $\vec p = \vec q$.

But no such integration is being performed, so why is eqn. $(5)$ equal to zero?

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    These equalities are to be interpreted in the sense of distributions. Roughly speaking, it does not have meaning unless it is being integrated out against some suitable function. But when you integrate such a suitable function against the delta $\delta(p-q)$, one evaluates the function at $p = q$ and the terms inside the parenthesis in (3) and (5) become zero. – JustWannaKnow Jan 06 '24 at 17:04
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    For any continuous function, $f(x),\delta(x-y) = f(y),\delta(x-y)$ in the sense of measures (or distributions of order $0$) – LL 3.14 Jan 06 '24 at 23:52
  • @LL3.14 Thanks for your comment. That equality in your comment only makes sense to me if it is integrated over. I am interested as to how that equality holds without integrating. Do you have a source or book recommendation of this property in the sense of measures or distributions? I am not familiar with these concepts, but as a beginner I would like to learn about this. –  Jan 07 '24 at 06:28
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    That's exactly what "equality in the sense of measures" means. The Dirac delta is not a function, there is no meaning to $\delta(x)$ at a given point $x$. Distributions are defined as linear form acting on test functions, in a way that is identified with integrating. Measures are functional on measurable sets, and one can define the concept of integrating against a measure. I did a summary here https://math.stackexchange.com/questions/4538645/prove-properties-of-dirac-delta-from-the-definition-as-a-distribution/4539654#4539654 – LL 3.14 Jan 14 '24 at 16:45

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