# Is there any theorem that suggests that QM+SR has to be an operator theory?

## UPDATE

To make my question more precise, I'll define what I mean by an operator theory:

• An operator theory is a theory in which the dynamical objects are operators, i.e., the equations of motion are imposed on operators.

• A wave function theory, on the other hand, is a theory in which the dynamical objects are functions of space-time, i.e., the equations of motion are imposed on functions.

In wave-function theories we have differential operators (e.g., $-i\nabla$), but these aren't dynamical, so there is a clear distinction between an operator and a differential operator. To make the distinction more evident, I'll place hats $\hat{\square}$ over dynamical operators.

At the first stages of the old (non-relativistic) quantum theory Heisenberg developed a theory in which the fundamental objects were operators, namely, the position and momentum operators: \begin{align} i\frac{\mathrm d}{\mathrm dt} \hat X(t)&=[\hat X,\hat H] \\ i\frac{\mathrm d}{\mathrm dt} \hat P(t)&=[\hat P,\hat H] \end{align} \tag{QM.1}

More or less one year later Schrödinger published a new theory in which the fundamental object is just a function of space-time: $$-i\partial_t\psi(x)=-\frac{1}{2m}\Delta\psi(x)+V(x)\psi(x) \tag{QM.2}$$

Later on, Schrödinger proved that the two formulations are actually equivalent.

Nowadays we have QFT, which is an operator theory, because the equations of motion are imposed on fields, i.e., operators. For example, a scalar field $\hat\phi$ evolves through \begin{align} i\frac{\mathrm d}{\mathrm dt}\hat\phi(x)=[\hat\phi,\hat H]\\ i\frac{\mathrm d}{\mathrm dt}\hat\pi(x)=[\hat\pi,\hat H] \end{align} \tag{QFT.1} where $\hat \pi$ is the field conjugate to $\hat\phi$.

To me, it seems more or less natural to ask about a possible $(\mathrm{QFT.2})$, i.e., a formulation of QFT as a wave-function theory: $$\begin{array} {}&\text{non-relativistic} & \text{relativistic}\\ \text{operator theory} & (\mathrm{QM.1}) & (\mathrm{QFT.1}) \\ \text{wave-function theory} & (\mathrm{QM.2}) & \quad ?? \end{array}$$

IMHO functional methods (i.e., path integrals) belong to an operator point of view. I believe it is not possible to use path integrals to calculate, for example, scattering amplitudes without using operators sooner or later, which means that path integrals don't actually fit in the $??$ slot.

## My question(s):

• Has someone done anything similar to Schrödinger, in the sense of reformulating QFT using solely functions of space-time and differential operators? Is there any theory of relativistic quantum mechanics in which the formalisms consists of PDE's?

• If the answer to the first question is that no wave-function theory exists as for today, then is there any reason to expect that there will never be? I mean: is there a no-go theorem or an argument that suggests that it is impossible to formulate QFT as a wave-function theory?

## My thoughts on this

Any relativistic theory of interactions must be able to describe creation/annihilation phenomena, which operators easily do (through $a,a^\dagger$). But a single wave-function must have a fixed number of space-time arguments, so the number of particles must be fixed. This means that one wave-function will not suffice.

A wave-function theory of interactions must, therefore, consist in an infinite number of wave-functions, each with a different number of space-time arguments: \begin{align} &\psi(x_1)\\ &\psi(x_1,x_2)\\ &\psi(x_1,x_2,x_3)\\ &\cdots \end{align} which means that we must have an infinite number of PDE's, one for each $\psi$. And the solutions must be consistent with the operator formulation of QFT, i.e., we must prove that the new formulation is equivalent to the old one.

I believe that the easier way to archive this is to use the correlation functions of QFT (or $n$-point functions) \begin{align} \psi(x_1)&=\langle\hat\phi(x_1)\rangle\\ \psi(x_1,x_2)&=\langle\hat\phi(x_1)\hat\phi(x_2)\rangle\\ \psi(x_1,x_2,x_3)&=\langle\hat\phi(x_1)\hat\phi(x_2)\hat\phi(x_3)\rangle\\ &\cdots \end{align}

With this, it should be possible in principle to find the PDE's for the $\psi$'s in terms of the PDE's for $\hat\phi$. This means, we should use $(\partial^2+m^2)\hat\phi=\hat j$ to get \begin{align} \mathcal O_1\ \psi(x_1)&=\mathcal J(x_1)\\ \mathcal O_2\ \psi(x_1,x_2)&=\mathcal J(x_1,x_2)\\ \mathcal O_3\ \psi(x_1,x_2,x_3)&=\mathcal J(x_1,x_2,x_3)\\ &\cdots \end{align} for some differential operators $\mathcal O_i$ and some sources $\mathcal J$. Once we find the appropriate $\mathcal O_i,\mathcal J$, there will be no explicit reference to any operator, and we only use wave-functions (of course, these wave-functions don't have the same probabilistic interpretation of non-relativistic QM, but this is irrelevant: the point is to find an equivalent formulation of QFT using just functions of spacetime).

I don't know if this makes much sense, or whether there is a better approach. Any comment will be appreciated.

• This question is very vague and ill-defined. For there to be a "theorem" you would have to have precise notions of "wave-function theory" and "operator theory". As you say, these two notions are just equivalent expressions of each other in QM. It's not that one has operators and the other doesn't, they just focus on different aspects. Saying "In wave mechanics we don't need operators" is just ridiculous, a PDE is nothing else than a (differential) operator applied to a state in $L^2(\mathbb{R}^n)$. You may avoid the word "operator", but it's there. – ACuriousMind Oct 25 '15 at 17:36
• I didn't specify the precise meaning of "wave-function theory" and "operator theory" because I assume that those who are able to answer this question understand what those terms mean. If you problem with my question relies on the wording, you can exchange "theorem" for "theoretical evidence", or anything like that. – AccidentalFourierTransform Oct 25 '15 at 17:42
• 1. Are you aware of the wavefunction formalism for QFT? 2. The terms do not, in my opinion, have any precise meaning. "operator theory" might deal more with abstract Hilbert spaces while "wave-function theory" might prefer to select a distinguished one like $L^2(\mathbb{R}^n)$, but, well, I really don't see how these two approaches could ever not be equivalent because they just differ in how you do the symbolic manipulations, not in physical content. – ACuriousMind Oct 25 '15 at 17:44
• I am not. This kind of answers is what I was expecting, so thank you. Is there any canonical reference to wavefunction formalism? – AccidentalFourierTransform Oct 25 '15 at 17:48
• Maybe there is evidence that the Hilbert space of QFT is never isomorphic to $L^p$, which would be a perfectly reasonable "proof" that QFT cannot be recast as a wave-function theory. Your answer suggests that this is not the case, though. – AccidentalFourierTransform Oct 25 '15 at 17:50

## 2 Answers

What you call an operator theory is usually called the Heisenberg picture of quantum mechanics. What you call a wave function theory is usually called the Schroedinger picture of quantum mechanics.

It is well-known that for every quantum mechanical model, the Heisenberg picture and the Schroedinger picture are fully equivalent through a dual description, as long as on doesn't consider time correlations.

This therefore also holds for quantum field theories, which are particular cases of quantum theories. In the Schroedinger picture of quantum field theory the ordinary Schroedinger equation takes the form of a functional Schroedinger equation. Here states of a QFT are treated as functionals of the field coordinates in the same way as states in QM are treated as functions of the position coordinates. A thorough discussion of the the functional Schroedinger picture is in the article by Jackiw, Analysis on infinite dimensional manifolds: Schrodinger representation for quantized fields (p.78-143 of the linked document).

On the other hand, the Heisenberg picture is more general than the Schroedinger picture as it allows the discussion of time correlations of obsservable quantities. This is important in statistical mechanics, and essential for the study of quantum field theory at finite time, through the close time path (CTP) formalism. For the latter, see, e.g., Introduction to the nonequilibrium functional renormalization group by Berges.

As mentioned in the comments, the case of 4D relativistic QFT is somewhat peculiar since there are still unresolved foundational problems related to nonperturbative renomalization. As Dirac says, the ''operators'' in the Heisenberg picture don't act anymore on Fock space, so there is no Fock space Schroedinger picture. However, and Dirac does not say this, the Heisenberg picture gives (presumably, proved only in lower dimensions) a valid operator description in a different Hilbert space, and using the time generator of the corresponding unitary representation of the Poincare group, one gets a corresponding Schroedinger picture on this, renormalized, Hilbert space.

• (+1) Thank you very much for your answer. I'm reading your first reference right now, and it looks very nice. Anyway, I'd like to hear your opinion on Dirac's paper Quantum Electrodynamics without Dead Wood, in which he points out that, in fact, Heisenberg & Schrödinger pictures are not equivalent. What do you think about it? – AccidentalFourierTransform Jan 23 '16 at 17:51
• @AccidentalFourierTransform: The Heisenberg picture is the more fundamental one. The Schroedinger picture is not equivalent since it cannot accommodate multiple times. However, for expectations that depend on a single time only, the two pictures are completely equivalent if one assumes a sufficiently general notion of state, by a straightforward duality argument given in many books. – Arnold Neumaier Jan 25 '16 at 6:15
• @AccidentalFourierTransform: Dirac's paper is also concerned with the still unresolved problems of relativistic quantum field theory. The apparent nonequivalence stated for QED is due to ignoring renormalization issues in the Schroedinger version. But this is not a fundamental trait, but a defect of the particular Hamiltonian chosen. – Arnold Neumaier Jan 25 '16 at 10:53
• @AccidentalFourierTransform: In my words, Dirac says that he knows how to renormalize the Heisenberg picture but not how to renormalize the Schroedinger picture (where he presents the naive version only), and concludes somewhat prematurely that only the former can be used for quantum fields. – Arnold Neumaier Jan 25 '16 at 19:07

There is indeed a theory of differential operators applied on wavefunctions (or more accurately wavefunctionals), the Schrodinger functional formalism, though it is not used a lot in most applications. It is defined by a wavefunctional at a time t

\begin{equation} \Psi[\phi(\vec x), t] \end{equation}

Upon which act differential operators, defined by

\begin{eqnarray} \hat\phi \Psi[\phi(\vec x), t] &=& \phi \Psi[\phi(\vec x), t]\\ \hat\pi \Psi[\phi(\vec x), t] &=& -i\hbar\frac{\delta}{\delta\phi(x)}\Psi[\phi(\vec x), t] \end{eqnarray}