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In QFT, we usually write creation and annihilation operators in the following way: ${a^s_{\textbf{p}}}^\dagger$, $a^r_{\textbf{q}}$, where $r,s$ denote the spins and $p,q$ the three-momenta of particles annihilated or created, as if to say that these operators are associated with a certain momentum and spin but have not earned the notation of a functional dependence. I have now stumbled upon an exercise where one needs to integrate over the following expression to show that it vanishes after a change of variable:

$$\int d^3p\frac{1}{2E_p}(p\cdot\mathbf{\sigma})a^r_p(\sigma^2)_{rs}a^s_{-p}.$$

The author of this solution (page 12) explains that since the integrand is odd in $\mathbf{p}$, we can show that the integral vanishes. But is the integrand really odd? How do I know that the creation and annihilation operators don't make the integrand even in the momentum? In fact, are they even functions of the momentum (or spin for that matter)?

When we do the next step, we see indeed that the momentum of the creation/annihilation operators also change sign but the author just easily flips the sign back in the next step, as if it doesn't matter which spin is associated with which momentum:

enter image description here

What exactly is going on here? I feel like there is some subtelety about the relationship of momentum and creation and annihilation operators that I am missing. I know for sure that creation and annihilation operators in in non-relativistic QM were linear in momentum but does this hold in (Dirac) QFT?

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    $\begingroup$ Whether or not that solution you found on the internet is correct has nothing to do with your question about whether or not you can treat the operators as functions. Yes, they are functions. They are operator-valued functions. You should get used to this since you are doing field theory. (Fields are what physicists call functions of space time (or momentum).) $\endgroup$
    – hft
    Dec 7, 2023 at 19:52
  • $\begingroup$ When you write something like $a_i$ and do a sum, like $\sum_{j=1}^{\infty}a_i$, th $a_i$ is a function that just as well have been written as $a(i)$, because it's an assignment $i\to a_i$, where the domain (in this case) is the positive integers. The notation $a_p$ vs $a(p)$ is one of convention and personal taste. I like to use subscripts for discrete variables and function notation for continuous variables, but I don't stick to this convention exclusively. $\endgroup$
    – march
    Dec 8, 2023 at 6:02
  • $\begingroup$ Please use MathJax and not images/screenshots for equations. $\endgroup$ Dec 8, 2023 at 11:21

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Are creation and annihilation operators functions of momentum?

Yes, the creation and annihilation operators you wrote are functions of momentum. It is a matter of notational preference whether you indicate the functional dependence by a subscript like $a_{\vec p}$ or a pair of parenthesis like $a(\vec p)$.

$$\int d^3p\frac{1}{2E_p}(p\cdot\mathbf{\sigma})a^r_p(\sigma^2)_{rs}a^s_{-p}.$$

The author ...explains that since the integrand is odd in $\mathbf{p}$, we can show that the integral vanishes. But is the integrand really odd?

Whether or not the integrand above is odd depends on whether or not the $a_{\vec p}$ commute or anti-commute, since the $\sigma^2_{rs}$ is symmetric, but you have to commute (or anti-commute) the operators past each to put the integrand back in the right order.

The solution you clipped from the internet seems to treat them as commuting... Are they? I would not expect them to be, unless this is a practice problem with commuting spin-1/2 particles...

But, regardless, that has nothing to do with whether or not you can treat the creation and annihilation operators as functions of momentum, which you can.

How do I know that the creation and annihilation operators don't make the integrand even in the momentum?

Maybe they do, depending on whether they commute or anti-commute. You have not told us.

In fact, are they even functions of the momentum (or spin for that matter)?

Yes. They are.

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