# Operators, Distributions and States in QFT

First of all, I will mention what I understand (pls. correct if wrong):

• States are vectors in the Hilbert space, to include continuous spectrum (and thus distributions), we expand this space to Rigged Hilbert space, then, distributions, same as usual states, are vectors (in some representation) in this space.
• Operators are just a map from vector space, to another, this includes Rigged Hilbert space.
• The $\hat{a},\hat{a}^{\dagger}$ (usual creation/annihilation operators) are operators on Fock space which build up from a set of Rigged Hilbert spaces (each of different number of particles).

So my questions are:

1. Dose: $$\hat{a}\left(x\right)\delta\left(\vec{x}-\vec{y}\right)\stackrel{?}{=}\delta\left(\vec{x}-\vec{y}\right)\hat{a}\left(x\right)$$ from above, those operators should act on this delta distribution, same as they act on states, aren't they?

2. If above is true, then I can replace $\hat{a}$ by any other operator, like $d/dx$ , then we have derivative of Delta distribution, and above relation is not valid. Some told me that previous differential operator and $\hat{a}$ are acting on “different spaces”, thus it above equality is true for $\hat{a}$ , but not for the differential operator. However, they could not explain for me why they acts on different spaces, even so both of them just operators, (I understand that $d/dx$ is already written in some particular representation, but we can always change it.), this even becomes stranger if I boil down QFT to one degree of freedom and will get simple quantum harmonic oscillators, in which $\hat{a}$ is defined by above differential operator (the momentum).

3. In general, for arbitary operators, we have obviously:$$\frac{d}{dx}\hat{A}\hat{B}\neq\frac{d}{dx}\hat{A}\cdot\hat{B}+\hat{A}\cdot\frac{d}{dx}\hat{B}$$ so how in QFT we can use “integrating by parts” to write something as:$$\int d^{3}x\,\hat{a}^{\dagger}\nabla^{2}\hat{a}=\int d^{3}x\,\nabla^{2}\hat{a}^{\dagger}\hat{a}$$ (such expressions can be found in non-interacting scalar field formulation), unless we consider something like $\left[\hat{a}^{\dagger}(x),\nabla_{x}^{2}\right]=0$ , which has no sense if we consider that $\nabla^{2}$ and $\hat{a}^{\dagger}$ acting on “different spaces”, and then $\nabla^{2}$ can be moved around as some “constant” (as been suggested to me).

Thank you in advance for clarifying those confusions to me.

• 1) yes: $a$ is an operator-valued distribution (q-number), and $\delta$ is a distribution (c-number): they commute. 2) no: $a$ and $d/dx$ act on different spaces: they are not interchangeable (also, $\hat p\neq -id/dx$, because $p$ acts on $\mathcal H$ and $d/dx$ acts on $L^p$). The fact that both $a$ and $d/dx$ are operators doesnt mean they act on the same space: think of the identity matrix in 2 and 3 dimensions: they are both operators, but they are not interchangeable 3) what are $A$ and $B$, and why is this so obvious? Feb 6 '16 at 13:01
• 1)As I found by googling, c,q-numbers are old Dirac notations, is there more modern rigor reason? and is it considered as Axiom? 2) Why I can not consider $d/dx$ acting on Hilbert space? is being complete not enough to be square integrable? at the end both should act on state vectors, if not what nabla "do" in third question? 3) $A,B$ arbitrary operators, if I act them on some function, that relation obviously wrong in general, is the case different for $a,a^{\dagger}$?
– TMS
Feb 6 '16 at 13:24

## 1 Answer

1. The $a,a^\dagger$ act on the Fock space. If you write a random $\delta(\vec x - \vec y)$ is it neither an element of a Fock space nor an operator on it - the equation doesn't really make sense without further context. However, $\delta$ is just a distribution on functions of spacetime and not operator-valued itself, so the meaning of $a(x)\delta(x-y)$ is obviously that it gives $a(y)$ when integrated over $x$, no matter the order.

2. In QFT, $a(x)$ is an operator on the space of states, but $\frac{\mathrm{d}}{\mathrm{d}x}$ is not. States are not functions of spacetime as they are in the wave mechanics of usual quantum mechanics, they are functionals of field configurations, so $\frac{\delta}{\delta\phi}$ is an operator on the space of states in a certain representation, but the spacetime derivative never is. If you reduce to the harmonic oscillator, you do not have $a(x)$ anymore - you just have $a$, and then your Fock space is just the usual $L^2(\mathbb{R}^3)$. The Fock space of QFT is not that, it is expressions in the fields, not in the coordinates of spacetime.

3. Your "obviously" is obviously non-sensical. If $A,B$ are not operator-valued functions, then writing a spacetime derivative in front of them does not make any sense. If they are operator-valued functions, then the $\frac{\mathrm{d}}{\mathrm{d}x}$ is not an operator on the spaec of states itself and the product rule is obviously true.

Your basic confusion seems to be what the spacetime derivative acts on. It acts on functions of spacetime, which may be operator-valued, but it doesn't know about that, since it is not an operator on the space of states itself - QFT is not quantum mechanics where you have wavefunctions.

• Thx, it seems litle bit clearer, but I am not familiar with object like "Operator-valued functions" you speaking about, maybe you can refer to some book with rigor introduction to QM/QFT (but not too rigor), or maybe explain those concepts little bit, because classical books does not touch this issues what only confuses me,thx.
– TMS
Feb 6 '16 at 16:14
• @TMS: An operator-valued function is exactly what it says on the tin: It's a function $\mathbb{R}^n\to \mathcal{O}(\mathcal{H})$ where $\mathcal{O}(\mathcal{H})$ is the algebra of operators on your space of states. There's not much to be "familiar with", it's exactly like the time-evolution operator $U(t)$ which you could also see as an operator-valued function $t\mapsto U(t)$, just instead of $t$ you have $x$ as the argument. Feb 6 '16 at 16:48