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I am trying to grasp the basics of rigorous quantum field theory. Let me summise how the setup of non-interacting quantum field theories look like to me.

Let $\mathcal{H}$ be a Hilbert space in which the state of a single particle of some type resides. For a particle moving in $\mathbb{R}^d$ with some spin $k$ this will be $\mathcal{H} = \operatorname{L}^2(\mathbb{R}^d) \otimes \mathbb{C}^{2k+1}$. Then to build a theory of arbitrarily many indistinguishable particles of that type we pass over to either the fermionic (antisymmetric) or bosonic (symmetric) Fock space. To simplify notation, let us assume that we the particles in question are bosons so we get the symmetric Fock space $\operatorname{Sym}(\mathcal{H})$ which is a quotient of the tensor algebra $\bigoplus_{n \in \mathbb{N}} \mathcal{H}^{\otimes n}$.

Now in single particle quantum mechanics a state is a unit vector $\psi \in \mathcal{H}$, an observable is a self-adjoint unbounded operator $T$ on $\mathcal{H}$ and the expecation value of $T$ in the state $\psi$ is given by the number $\langle T \psi, \psi \rangle \in \mathbb{R}$.

In the many particle case this should be the same: a state is an element of $\operatorname{Sym}(\mathcal{H})$ and observables are self-adjoint unbounded operators on the Hilbert space $\operatorname{Sym}(\mathcal{H})$. For example if $\psi$ is some single particle state, the element $\frac{1}{\sqrt{2}}\psi \otimes \psi \in \operatorname{Sym}(\mathcal{H})$ will be the two-particle state corresponding to two particles being in the state $\psi$. So far so good.

But now in quantum field theory one does not focus on unit vectors and self-adjoint operators, but rather on operator-valued distributions on $\operatorname{Sym}(\mathcal{H})$. A quantum field $\Phi$ then returns for every suitable test function $f$ on $\mathbb{R}^d$ an operator $\Phi(f)$ on $\operatorname{Sym}(\mathcal{H})$. In general this operator need not be self-adjoint if one has a “charged field”.

Now to my question: Why are quantum fields now suddenly these complicated operator-valued distributions and not just states on $\operatorname{Sym}(\mathcal{H})$? Since states on $\operatorname{Sym}(\mathcal{H})$ still make sense and correspond to so-and-so many particles being in specific states, how do these connect to the quantum field $\Phi$? In particular given a state $\psi \in \operatorname{Sym}(\mathcal{H})$ one can form the quantity $\phi(f)\psi$. How should one interpret this new state?

Since most of the times $\Phi$ seems to be built out of creation and annihilation operators on the Fock space, $\phi(f)$ creates and destroys particles inside of $\operatorname{supp}(f) \subseteq \mathbb{R}^d$. What is the meaning in passing from the particles $\psi$ to the particles $\phi(f)\psi$? I guess the question lurking behind this vague uncomfortableness of the deviation of quantum field theory from the standard quantum mechanics axioms is the question why it does not suffice to just build $\operatorname{Sym}(\mathcal{H})$, its operator algebra and the states on that operator algebra. Why operator valued distributions?

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  • $\begingroup$ Maybe related: physics.stackexchange.com/questions/563106/… and physics.stackexchange.com/questions/337423/… $\endgroup$
    – Plop
    Commented Mar 1, 2023 at 9:16
  • $\begingroup$ In ordinary quantum mechanics, what is the mathematical content of the quantum description of a spin-$\frac{1}{2}$ particle. Just a unit vector in $\mathbb{C}^2$? Or is the set of Pauli matrices? The states on the HIlbert space are still representing "states of spacetime", but the quantum description of the fields is much more than that. $\endgroup$
    – Plop
    Commented Mar 1, 2023 at 9:22
  • $\begingroup$ Finally, I'm not sure about it, but considering operator-valued distribution instead of operator-valued functions is a technical problem, not really a philosophical matter. $\endgroup$
    – Plop
    Commented Mar 1, 2023 at 9:25
  • $\begingroup$ See "PCT, Spin and Statistics, and All That" from Streater and Whiteman. $\endgroup$
    – jjcale
    Commented Mar 2, 2023 at 18:31

4 Answers 4

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tl;dr: The reason for operator-valued distributions is because the physically meaningful "measurements" in QFT are things that preserve locality and that can be measured at any location.


In quantum mechanics modelled by a Hilbert space $\mathcal{H}$, it was never really the case that "a state is a unit vector $\psi \in \mathcal{H}$". What is actually correct is: a pure state is a point in the projective space $\mathbb{P}\mathcal{H}$, i.e. a line in $\mathcal{H}$, i.e. a unit vector up to phase factor. The best definition is: there is a C* algebra $\mathcal{A}$ of bounded operators; a state is a linear functional $\langle - \rangle : \mathcal{A} \to \mathbb{C}$ such that $\langle 1 \rangle = 1$ and $\langle AA^* \rangle \in \mathbb{R}_{\geq 0}$ for any $A \in \mathcal{A}$. For this definition, it is manifest that any convex combination of states is again a state, and so it makes sense to talk about "pure" states, and then it is a nontrivial theorem that when $\mathcal{A}$ is the bounded operators on $\mathcal{H}$, the pure states are precisely the states of the form $\langle \psi | - | \psi \rangle$.

Indeed, the bigger point is that the Hilbert space itself is not physical. What's physical is the algebra of observables, and everything else should be derived from it. You can recover the Hilbert space up to isomorphism, and that isomorphism is unique up to an overall phase ambiguity, from the algebra of observables. Conversely, Hilbert spaces supply models.

In quantum field theory, as you said, the Hilbert space is (perhaps, depending on the model) $\mathcal{H}_{\text{many-particle}} = \mathrm{Sym}(\mathcal{H}_{\text{1-particle}})$; for example, in QFT in dimension $d+1$, the Hilbert space might be $\mathrm{Sym}(L^2(\mathbb{R}^d) \otimes \mathbb{C}^{2k+1})$. But the natural physical class of "operators" are not all (bounded) operators on this space, but rather the ones that preserve some "locality". Which locality? Maybe you want "operators" that do not increase support; maybe you want ones that increase support only by $\epsilon$; maybe you allow to create tails that vanish exponentially. Going between different notions of locality is a question for the functional analysts. We also should be particularly interested in "operators" on $\mathcal{H}_{\text{many-particle}}$ which make sense at any location in space, just like how in QM you can make measurements at any time.

Well, suppose you have an observable who can be inserted at any point in space. A typical example if you had just a one-particle Hilbert space: the operator who evaluates the value of your field. A typical example in the full Hilbert space: the operator who creates or annihilates a field. How does this operator look functional analytically? It is some sort of "function" $\mathbb{R}^d \to \text{bounded operators on }\mathcal{H}_{\text{many-particle}}$, I mean the function that inputs your choice of where to do the measurement and outputs the operator in the sense of QM. Which sort of function? As another answerer has answered, you are forced to use distributions (which aren't really functions, but are morally functions) so that uncertainty/commutator relations can be $\delta$-valued.

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Q: Why do we need operator-valued distributions instead of operator-valued functions?

Consider the commutator of two canonically conjugate quantum fields, for example, for $\mathbf{p},\mathbf{p}'\in\mathbb{R}^3$ consider the commutator of the creation and annihilation operators, $$[a(\mathbf{p}), \mathbf{a}^\dagger(\mathbf{p}')]=(2\pi)^3\delta^3(\mathbf{p}-\mathbf{p}')I.$$ The right-hand-side is a distribution, so the fields must be distributions as well. The same applies to the space-time fields created by Fourier transformation of the creation and annihilation operators. Their equal-times correlator is a delta function, a distribution, so the space-time fields must be operator-valued distributions.

Further reading: Introduction to Quantum Field Theory for Mathematicians.

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  • $\begingroup$ Let $F$ be the space of generalized functions that the fields live in. It's clear that the elements of $F$ can’t be pointwise-defined. If the question is why $F$ must be specifically the space of distributions, rather than some other generalized function space, the reasoning here doesn’t actually establish this claim. We don’t want to simply assume the Schwartz kernel theorem holds for $F$. The commutator on the left-hand side should be interpreted as a bilinear form on test functions. So the right-hand side shouldn’t be interpreted as an element of $F$, but as the corresponding bilinear form. $\endgroup$
    – ldrinehart
    Commented Feb 28, 2023 at 16:24
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I want to address the core question here:

Why are quantum fields now suddenly these complicated operator-valued distributions and not just states in Sym(H)?

An element of $\operatorname{Sym}{\mathcal{H}}$ is a state. But quantum fields are not states. They are families of observables, representing measurements that we can make. The notation around this point can be very confusing, so let me spell it out in a simpler setting, classical field theory.

In non-interacting relativistic classical scalar field theory, a state is a solution $\phi_c: \mathbb{R}^4 \to \mathbb{R}$ of the equations of motion, which have the form $(\Delta + m^2)\phi_c = 0$. The fundamental observables of this theory are the evaluation observables: $\operatorname{ev}_x$, which are the functions on the set of solutions that send $\phi_c\mapsto \phi_c(x)$. Usually people don't give these evaluation functions explicit names, and instead just write their values $\phi_c(x)$. One can also smear out these observables, obtaining $\operatorname{ev}_f = \int \phi_c(x) f(x) dx$.

In the quantum version of this theory, which is what you're considering, the smeared quantum fields are analogous to the smeared evaluation observables $\operatorname{ev}_f$, not to the states $\phi_c$. So they are represented by observables acting on the Hilbert space, not by functions which are elements of the Hilbert space. Thanks to hysteresis and inertia, people still write these observables as $\hat{\phi}(f)$, but really, they should be called $\widehat{\operatorname{ev}_f}$.

In the classical theory, we have one observable $\operatorname{ev}_x$ for each point $x$ in spacetime, and these observables can reasonably be thought of as smoothly varying functions of $x$, because the states are represented by smoothly varying functions of $x$. But in the quantum theory, this just isn't the case any more: the results of measurements have some regularity, but not much. They don't vary smoothly or even continuously in $x$. The best we can do is say that the various smeared observables $\widehat{\operatorname{ev}_f}$ define an operator-valued distribution.

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This is a personal opinion which is perhaps tangential to the question but which I hope might shed some light on this topic (if otherwise, it can be ignored). It is arguable that the important question here is not whether observable valued distributions are the appropriate mathematical tool for QFT but rather, how such objects can be precisely defined mathematically. A similar theme is that of the holomorphicity of functions between spaces of observables. In most expositions in the literature, it is assumed that these concepts are self-evident (which ignores the subtleties of the algebraic and topological structure of the space of observables). The ones which do address this problem assume that the operators involved have the same domain of definition or at least are all defined on a common dense subspace. It easy to give examples which display the inadequacies of these formulations (for examples the sequence $(z^n)$ can be regarded as an unbounded operator on $\ell^2$ and $z\mapsto (z^n)$ should clearly be regarded as holomorphic in $z$. However,the domains of definitions vary wildly. Even worse is $z T+ I$ which is even linear but not holomorphic for this definition). As for the weaker definition, it is hopeless since if you know an observable on a dense subspace you know nothing. Just think of the Laplace operator on the test functions.

However, there is a remedy: if we assume, as we shall, that the Hilbert space is separable, then there is a natural topology on the space $O$ of observables under which the latter is a polish space. We then define the space of holomorphic functions resp. tempered distributions with values in $O$ to be the space of continuous functions from $H(U)´$ resp. $\cal S$ to $O$ which are linear in the special sense that if $f$ and $g$ are elements of the given lcs´s such that the sum of $Tf$ and $Tg$ exists in the usual sense of addition for unbounded s.a. or normal operators, then we have the equality that one would expect.

Here we are building on the pioneering work of the functional analysts of the 50-s (Sebastiao e Silva, Köthe, Schwartz, Grothendieck) on spaces of test functions and holomorphic functions, their duality and the representation of operator valued functions or distributions as spaces of suitable continuous linear mappings thereon). ($\cal S$ and $H(U)$ are the (nuclear Frechet) spaces of smooth functions of rapid decrease, resp of holomorphic functions).

Using these definitions it is fairly routine to show that the spaces have the properties which one would expect of such concepts--natural relationship with the classical versions, i.e. for the case of bounded observables, functoriality (so that one discuss, e.g., group actions thereon), the fact that these objects are differentiable, have series representations (Hermite or Taylor), few of which are known (or even true) for the versions in the literature. It is also relatively easy to construct non-trivial and interesting examples so that they are natural candidates as a framework for a rigorous treatment of analysis in the space of observables.

The above remarks are purely mathematical in nature, simply a modest attempt to give a rigorous approach to the concepts of analytic functions and distributions in the context of analysis for observables, .i.e., unbounded, self-adjoint of normal operators on a separable Hilbert space. Clearly, in the hope that could be of use in the mathematical formulation of QFT.

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