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It seems well known that different quantum fields can give rise to the same $S$-matrix. I know of two ways this is described.

The first is through the Borchers class of relatively local fields, i.e. fields $A(x)$ and $B(x)$ which satisfy $$[A(x),B(y)] = 0,\ \ \ \ \text{$x$ and $y$ space-like separated.}$$ It is then known that all fields within the same Borchers class have the same $S$-matrix.

The second is through the LSZ reduction formula, where the only requirement on the field is that:

  1. It has vanishing vacuum expectation value $\langle 0|\phi(x)|0\rangle = 0$.

  2. It has non-zero overlap with the relevant single particle states $\langle p|\phi(x)|0\rangle \neq 0$.

And any field $\phi(x)$ which satisfies the two above conditions (up to normalization) will give the same $S$-matrix.

My question is a bit vague, but I would like a better understanding as to how the two criteria are related. The two doesn't seem equivalent.

For example, let's consider the simple example of a scalar field $\phi(x)$. We can consider the normal ordered field $:\phi^2(x):$ which lies within the same Borchers class as $\phi(x)$. So we expect the two to give the same $S$-matrix. On the other hand, the overlap of $:\phi^2(x):$ with single particle states is vanishing, so we should expect a different $S$-matrix due to the LSZ formula.

In general, it doesn't seem true that

$$\text{"relatively local" $\implies$ "non-zero single particle overlap",}$$

nor does it seem true that

$$\text{"non-zero single particle overlap" $\implies$ "relatively local".}$$

So how do I reconcile these two (apparently inconsistent) ways of looking at $S$-matrix equivalence?

Qmechanic
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EuYu
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    Wait...are you sure about your statement about LSZ? Surely, a $\phi^3$ and a $\phi^4$-theory do not have the same S-matrices, do they? Or am I misunderstanding what you mean by "any field will give the same S-matrix"? I think the correct statement is that the LSZ formula gives that the S-matrix is field reparametrization/redefinition invariant if both the old and the new field fulfill these conditions, not that any two fields fulfilling these conditions have to have the same S-matrix, and then the question dissolves. – ACuriousMind Apr 04 '16 at 17:28
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    @ACuriousMind Well I'm not entirely sure, but here's what I think. Please correct me if you think I'm wrong. The LSZ formula is established rigorously and non-perturbatively, without any mention of a Lagrangian at all. As far as I know, the only thing that matters is that the $n$-point functions have the same single-particle poles, and this is ensured by the two conditions I've listed. Field redefinitions are just one (easiest) way to get different fields which satisfy these two conditions. – EuYu Apr 04 '16 at 19:06
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    ...cont. As for why $\phi^3$ and $\phi^4$ theories give different $S$-matrices, I understand that to be because they scatter different particles. Afterall, we wouldn't expect the same particle to obey both $\phi^3$ and $\phi^4$ Feynman rules. So while it's confounded by the fact that we call both (distinct) fields $\phi$ and label both (distinct) particles as $|k\rangle$, the single particle overlaps are truly different in this case, so the $S$-matrices are different. – EuYu Apr 04 '16 at 19:08
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    @AccidentalFourierTransform The LSZ reduction formula simply doesn't hold for massless fields (at least not rigorously). From Haag-Ruelle scattering theory, it is well known that LSZ is established with the assumption of a mass-gap. You need to fix the LSZ formula to work for massless fields, so it's not surprising that the formula is different. But again, I don't feel that this has anything to do with my question, or a Lagrangian. – EuYu Apr 10 '16 at 11:04
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    @AccidentalFourierTransform I don't see how the mass depends on the Lagrangian. The bare mass appearing in the Lagrangian isn't even the physical mass anyways. The only input about mass into the derivation of the LSZ formula comes from the fact that the states $|k\rangle$ are eigenvalues of the squared momentum $P^2$ with eigenvalue $m$. – EuYu Apr 10 '16 at 11:11
  • Let us continue this discussion in chat. – AccidentalFourierTransform Apr 10 '16 at 11:13

2 Answers2

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The comment in my book (cited by Diffycue) originated in my own puzzlement when I was a student and reading my own favourite QFT book: that by K. Nishijima, Fields and Particles: Field Theory and Dispersion Relations. In it he seems to use the LSZ technique show that any two fields $\phi_i$ and $\phi_2$ that have matrix elemements between the vacuum and the one particle states of the form $$ \langle k\vert \phi_i(x)\vert 0\rangle = \sqrt Z_i e^{ikx} $$ will have the same $S$-matrix. Since this is a property possesed by $\phi_{\rm in}(x)$ and $\phi_{\rm out}(x)$, which are free, he seemed to have proved that all theories were free. It took me ages and a very careful reading of his argument to realize that he made use of Lorentz transformation properties of time-ordered correlators such as $$ \langle 0\vert T\{ \phi_1(x_1) \phi_2(x_2)\ldots \}\vert0\rangle $$ that are messed up unless $\phi_1$ and $\phi_2$ commute outside the lightcone. Thus Nishijima was tacitly assuming the mutual locality of the various fields.

As a consequence $\phi$ and $\phi+\phi^3$ will have the same $S$ matrix for $\lambda \phi^4$ theories with their ${\mathbb Z}_2$ symmetry, but $\phi$ and $\phi^2$ will not because $\phi^2$ cannot couple the vacuum to the $\phi$ state because it has the wrong ${\mathbb Z}_2$ parity.

mike stone
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  • Thank you for the answer Prof. Stone. Indeed, as you and @Diffycue mention, the assumption that the fields connect the vacuum to particle states is an independent and non-negligible assumption. I had originally thought that it was implied by Borchers' result, but it appears that's not the case. Digging into some older books, I found that this was explicitly mentioned in Bogolubov et al. "General Principles of Quantum Field Theory" (Axiom W.III' on Page 486, and also Corollary 12.2). It is somehow reassuring to know that even experts such as yourself have struggled with these problems :) – EuYu Nov 13 '18 at 07:23
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You seem to have a very good understanding of the LSZ formula; perhaps I am the one who is mistaken here. But I was under the impression that two fields in the same Borchers class were

  1. Mutually local
  2. Connect the vacuum to the same states.

I.e., the statement "the overlap between $: \phi^2 :$ and single-particle states vanishes" would contradict the statement that "$: \phi^2 : $ lies within the same Borchers class as $\phi$."

To me it seems like you are completely correct that locality does not imply nonzero overlap and vice versa, which is why we require condition 2. above.

I'll note also that in my favorite QFT book, The Physics of Quantum Fields by Michael Stone, that the author mentions: "The set of fields which [sic] connect the vacuum to the required states and are mutually local in that they commute at space-like separation is called a Borcher's class." But in the notes my QFT class used, there was no explicit mention of the connectivity property.

Diffycue
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