2
$\begingroup$

In Reed & Simon's Methods of Mathematical Physics Volume II, they define a (Hermitian scalar) quantum field theory to be the quadruple $\langle \mathcal{H}, U, \varphi, D\rangle$ that satisfies the Wightman axioms. Here $\mathcal{H}$ is a separable Hilbert space, $U$ is a strongly continuous unitary representation of $\mathcal{H}$ of the restricted Poincare group, $D \subset \mathcal{H}$ is dense, and $\varphi: \mathscr{S}(\mathbb{R}^4) \rightarrow T(\mathcal{H})$ where $\mathscr{S}$ is the Schwartz space and $T(\mathcal{H})$ is the set of all (unbounded) operators on $\mathcal{H}$.

They define the Wightman functions as the operator valued distributions $\{\mathcal{W}_n\}$ where $$\mathcal{W}_n (f_1, \ldots, f_n) = \big(\psi_0, \varphi(f_1), \cdots \varphi(f_n)\psi_0\big)$$ and $\psi_0$ is the vacuum state in $\mathcal{H}$.

From what I have read, Schwinger functions $\{\mathcal{S}_n\}$ are defined are just the analytic continuation of $\{\mathcal{W}_n\}$ to imaginary time. Most resources present both sets of distributions as functions on $\mathbb{R}^4$ so the Wick rotation is easy to visualize, but this is less obvious when the domain is a function space and not space-time.

Based on the definition given by Reed and Simon it seems we are often given $\varphi$ so $\{\mathcal{W}_n\}$ is easily defined. When I first started studying axiomatic QFT my impression was that the axioms are used to construct the quantum field $\varphi$ from scratch. However, looking at these axioms, my impression now is that one instead constructs $\varphi$ using traditional means (for example, from a Lagrangian), and then checks if all the axioms are satisfied.

My questions can be summarized as follows:

  1. Given that both $\mathcal{W}_n$ and $\mathcal{S}_n$ are distributions, how does one Wick rotate between them?

  2. In reference to the last paragraph above, how does this work in practice? Where does $\varphi$ come from?

  3. Lastly, is there some sort of uniqueness result for the Wightman functions? That is, if I have two sets of Wightman functions $\{\mathcal{W}_n\}$ and $\{\tilde{\mathcal{W}}_n\}$, do they correspond to the same scalar field (up to unitary equivalence)? And thus does are Wightman functions only applicable to scalar fields?

$\endgroup$
1

1 Answer 1

3
$\begingroup$

1. Let's write $x=(\mathbf{x},t)$ for spacetime points in $\mathbb{R}^d$ so the first $d-1$ coordinates $(x_1,\ldots,x_{d-1})=\mathbf{x}$ pertain to space and the last coordinate $x_d=t$ pertains to time. Then the $n$-point Wightman function is $$ W_n(\mathbf{x}_1,t_1;\ldots;\mathbf{x}_n,t_n):= \langle\Omega|e^{it_1 H}\phi(\mathbf{x}_1,0) e^{-i(t_1-t_2)H}\phi(\mathbf{x}_2,0)e^{-i(t_2-t_3)H}\cdots e^{-i(t_{n-1}-t_n)H}\phi(\mathbf{x}_n,0)e^{-it_n H}|\Omega\rangle $$ $$ =\langle\Omega|\phi(\mathbf{x}_1,0) e^{-i(t_1-t_2)H}\phi(\mathbf{x}_2,0)e^{-i(t_2-t_3)H}\cdots e^{-i(t_{n-1}-t_n)H}\phi(\mathbf{x}_n,0)|\Omega\rangle $$ where $H$ is the Hamiltonian and $|\Omega\rangle$ is the vacuum state, of the possibly interacting theory. The exponentials near the vacuum disappeared at both ends because $\Omega$ is an eigenvector of $H$ for the eigenvalue zero.

On the other hand, and assuming now that $\tau_1<\cdots<\tau_n$, the Schwinger functions are $$ S_n(\mathbf{x}_1,\tau_1;\ldots;\mathbf{x}_n,\tau_n):=W_n(\mathbf{x}_1,i\tau_1;\ldots;\mathbf{x}_n,i\tau_n) $$ $$ =\langle\Omega|\phi(\mathbf{x}_1,0) e^{-(\tau_2-\tau_1)H}\phi(\mathbf{x}_2,0)e^{-(\tau_3-\tau_2)H}\cdots e^{-(\tau_{n}-\tau_{n-1})H}\phi(\mathbf{x}_n,0)|\Omega\rangle $$ In general, analytic continuation of functions in the complex domain is one of those riddles wrapped in a mystery etc. See the Voronin Universality Theorem for a cautionary example. To do analytic continuation one needs some very strong structural property of the function at hand. More often than not, this is a positivity property. For example if $X$ is a random variable with moments growing as $C^n n!$ then the characteristic function $\mathbb{E}(e^{izX})$ is obviously analytic in a disc $|z|<\varepsilon$ but what is remarkable is that it is then automatically analytic in the strip $|\Im z|<\varepsilon$. The reason is the elementary inequality $$ \left|\int f(\omega)\ {\rm d}\mu(\omega)\right|\le \int |f(\omega)|\ {\rm d}\mu(\omega) $$ which holds because the measure $\mu$ is positive. Here the needed positivity property is that of the self-adjoint operator $H$.

The above is just a quick sketch. For a serious introduction see for example Ch. 6 and 19 of the book "Quantum Physics: a Functional Integral Point of View" by Glimm and Jaffe. See also: P. Kravchuk, J. Qiao, and S. Rychkov, "Distributions in CFT. Part II. Minkowski space". JHEP 2021, Article number: 94 (2021).

2. What you said in that last paragraph is correct. Axiomatic QFT gives you zero information as to how to produce/construct the interacting field. You need to do that with completely different methods. The area of mathematics with tackles this problem is Constructive QFT. A noteworthy example where one has a "construction of the field" $\varphi$ is the 2d Ising CFT. There the Schwinger functions are given explicitly by $$ S_n(x_1,\ldots,x_n)=\sqrt{\sum_{q} \prod_{1\le i<j\le n}|x_i-x_j|^{\frac{q_i q_j}{2}}} $$ where $q=(q_1,\ldots,q_n)$ is summed over all "charge-neutral" elements of $\{-1,1\}^{n}$, i.e., those satisfying $\sum_{i=1}^{n}q_i=0$. As far as I know, nobody knows how to prove the Axioms, specifically Osterwalder-Schrader positivity, from the above formula.

3. No. Once again the axioms cannot be used for the purpose of defining/constructing QFTs, because they do not give rise to a mathematically well- posed problem à la Hadamard. There is no uniqueness. One would need to add a lot more axioms to get uniqueness. In a sense, this is the philosophy in Conformal QFT where one needs the OPE, its convergence, the list of scaling dimensions, OPE coefficients.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for this wonderful answer, I found it very helpful! $\endgroup$
    – CBBAM
    Commented Feb 22, 2023 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.