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  Axioms for Euclidean Green's functions's paper by Osterwalder and Schrader

+ 4 like - 0 dislike

Hi, I am refering to the following paper http://link.springer.com/article/10.1007%2FBF01645738 (free version from Project Euclid: https://projecteuclid.org/euclid.cmp/1103858969)

on page 91, I don't see how "by (E1), for $f,g\in \mathcal{L}_+$, $(f,U_s(a)g)=(U_s(-a)f,g)$" does this follow from (E1)? where (E1) appears on page 88:(E1) $\mathcal{\sigma}_n(f)=\mathcal{\sigma}_n(f_{(a,R)})$, where $R\in SO(4) , a\in \mathbb{R}^4$.

I appreciate your help.

asked Jun 22, 2015 in Mathematics by MathematicalPhysicist (200 points) [ revision history ]
edited Jun 22, 2015 by Jia Yiyang

2 Answers

+ 3 like - 0 dislike

As @ArnoldNeumaier stated you need to use the definition of (4.3) and (E1). Where (4.3) is

\[(f,g) = \sum_{n,m} G_{n+m}(\theta f_n^* \times g_m)\]

and (E1) states that the Euclidean Green function is invariant under SO(4) rotations and translations. \(G_n(f) = G_n(f_{(\underline{a}, \mathbf{R})})\) where \(f_{(\underline{a}, \mathbf{R})}(\underline{x_1},...,\underline{x_n})= f(\mathbf{R}\underline{x_1} + \underline a,...,\mathbf{R}\underline{x_n} + \underline a)\).

While \(\hat U_s(\vec a)\) generates purely spatial translations on the functions it acts on. Now we can see how the statement follows. 

\[(f,\hat U_s(\vec a)g) = \sum_{n,m} G_{n+m}(\theta f_n^* \times (\hat U_s(\vec a)g)_m) = \sum_{n,m} G_{n+m}(\theta f_n^* \times g_{(-\vec {a}, 1)m}) \]

Then we use (E1) to shift the whole Green function by a value \(\vec a\), where we don't need to worry about \(\theta\) for it is just temporal inversion, which commutes with spatial shifts. Thus we find

\[\sum_{n,m} G_{n+m}(\theta f_n^* \times g_{(-\vec a, 1)m}) = \sum_{n,m} G_{n+m}(\theta f_{(\vec a, 1)n}^* \times g_{m})= \]

\(= \sum_{n,m} G_{n+m}(\theta (\hat U_s(-\vec a)f_{n})^* \times g_{m}) =(\hat U_s(-\vec a)f , g)\)

answered Jun 23, 2015 by Peter Anderson (205 points) [ revision history ]
edited Jul 28, 2015 by Peter Anderson
+ 0 like - 0 dislike

You need to use the definition (4.3) and $U_s(a)^*=U_s(-a)$. (E1) is then applied to the resulting expression.

answered Jun 23, 2015 by Arnold Neumaier (15,787 points) [ no revision ]

Looking how they define \(\hat U_s(a)\) above the statement in question, we should find \(\hat U_s(a)\hat U_s(-a) = 1\) by definition. With the addition of (E1) on the definition (4.3) you then find that it will be Unitary. It would be circular to require from the start that \(U_s(a)^\dagger=U_s(-a)\)

One needs both conditions to derive the desired equation. How one gets the unitarity condition is immaterial. (By the way, there is no circularity if one doesn't use a result to prove thes result itself.)

The whole point of the statements in question is to prove unitary, I'd be very surprised if the required it (as is commented in the paragraph afterwards in the article). I posted how I interpreted it and why I don't think you need that condition. Please correct me if I'm wrong =)

@PeterAnderson: There are two different kinds of unitarity involved. Unitarity of the representation of the group on the classical $N$-particle test functions is assumed (and indeed trivial for scalar fields). Unitarity of the group in the Hilbert space of the field theory is the goal. One must make use of the former to get (ultimately) the latter.

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