I have been thinking lately about the following question. In quantum field theory, we define spectral density function $\rho(\mu^2)$ using two-point function as follows (Källén–Lehmann formula) $$ \langle 0|T \phi(x) \phi(y)|0\rangle = \int \limits_0^\infty d\mu^2 \rho(\mu^2) D(x-y,\mu^2) $$ where $D$ is free field propagator. The function $\rho(\mu^2)$ is said to contain information about all possible excitations in a system (single particle states are its poles, multiparticle ones correspond to branch cuts, etc,etc)

The question is as follows: is knowing the spectral density in QFT enough to specify the theory? Or, in other words, say we have two QFTs with the same spectral density function; is it possible to identify them? Say, in quantum mechanics (given boundary conditions) there is such a thing as inverse scattering problem. In a good enough situation one can reconstruct the potential if knowing the spectral measure - energy levels and their degenerateness (I am not sure the analogy here is correct, but this analogy is one of the things that make me confused about this question). Yet it seems that in QFT only knowing two-point functions are not enough; we kind of only have information about "one-particle" excitations this way. Yet, it still contains a lot of information about interactions such as bound states, energy corrections for multiparticle states and so on. Is this information enough in QFT?

I would appreciate any discussion or known result connected with this problem.


1 Answer 1


The two-point function is not nearly enough to determine a general QFT. You need the whole set of correlators, which is infinite. The formal statement is Wightman's reconstruction theorem, namely that given the Wightman functions $$ \{\langle 0|\phi(x_1)\cdots\phi(x_n)|0\rangle\}_{n\in\mathbb N} $$ one can reconstruct the Hilbert space of the theory, uniquely. But one needs all these functions.

Two remarks are in order:

  • The theorem above is stated in terms of non-time-ordered correlators, but a similar statement holds for time-ordered correlators as well (cf. this PSE post).

  • In some specific theories, such as free theories or CFTs, a finite number of correlators does in fact uniquely determine the theory. For free theories the two-point function is enough (it knows the spin and the mass, which is all you need), and for CFTs1 you also need the three-point function (OPE coefficients).

The Källén-Lehmann density knows a lot about the theory, but it is a "finite lot"; a QFT requires an "infinite lot" – no finite set of functions can be enough. The intuition is that QFT describes creation and annihilation phenomena, i.e., the number of particles can take any value. Thus, you essentially need an infinite number of wave functions to describe the dynamics; no truncation can contain the information for all these infinitely many possible processes.

The analogy with standard QM breaks down because such theories have a finite number of particles, and so a finite number of functions suffice to specify the theory. For example, a single (multiparticle) wave function is enough. If one is lucky, other functions may work as well, such as say the interparticle potential. But in QFTs, you would need an infinite number of functions, because you have an infinite number of degrees of freedom.

1: In fact, CFTs are the easiest counterexample to OPs intention: by properly choosing the fields of the theory, one can always take the two-point function of any CFT to be $\langle \phi_i\phi_j\rangle=\delta_{ij}/x^{\bullet}$ for some power that depends on e.g. the dimension of spacetime. It is clear that knowing this function is mostly irrelevant for identifying the theory: all CFTs look like this!

  • $\begingroup$ Thank you very much for a detailed answer! Indeed, the CFT counterexample somehow did not occur to me when I was thinking about this statement. $\endgroup$ Commented Apr 28, 2020 at 21:32

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