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Apr
10
comment Relation between Borchers class and the LSZ formula on S-matrix equivalence
@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 $m$.
Apr
10
comment Relation between Borchers class and the LSZ formula on S-matrix equivalence
@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.
Apr
10
comment Relation between Borchers class and the LSZ formula on S-matrix equivalence
@AccidentalFourierTransform I would say that's dependence on the field type, i.e. the spin of the field, not so much the Lagrangian itself.
Apr
8
comment Is the Noether charge always a Hermitian operator?
Thanks for the updated answer. I agree for the most part, but it still doesn't seem satisfying. If we have a symmetry of the Lagrangian, we don't get to pick the charge that comes out. The associated symmetry $V(\theta)=e^{i\theta Q}$ will either be unitary or not, and this is not a condition we can set. Of course, we can reject any symmetries of the Lagrangian which generates such a non-unitary $V$ as non-physical, which is what I think you're suggesting. But the underlying mathematical problem is whether such non-physical symmetries exist in the first place, and that hasn't been resolved.
Apr
8
comment Is the Noether charge always a Hermitian operator?
Sorry for being obtuse, but the argument still seems circular to me. Let me see if I understand you correctly. Let's use a complex scalar field as example. The Lagrangian is invariant under $\phi \mapsto e^{i\theta}\phi$, which generates a charge $Q$. This charge acts on the Hilbert space, with the symmetry implemented via $V(\theta) = e^{i\theta Q}$. So far, I agree. But if we do not know a priori that $Q$ is Hermitian, how do we know that we have $|\langle\xi|V^{\dagger} V|\psi\rangle| = |\langle\xi|\psi\rangle|$, which is the condition for Wigner's theorem to apply?
Apr
8
comment Is the Noether charge always a Hermitian operator?
Wigner's theorem is a statement about symmetries acting on the underlying Hilbert space. Noether's theorem concerns symmetries internal to the Lagrangian. I'm not too sure how the two are related. In particular, I'm not entirely convinced that every symmetry for the Lagrangian needs to be unitary.
Apr
8
asked Is the Noether charge always a Hermitian operator?
Apr
4
comment Relation between Borchers class and the LSZ formula on S-matrix equivalence
...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.
Apr
4
comment Relation between Borchers class and the LSZ formula on S-matrix equivalence
@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.
Apr
2
asked Relation between Borchers class and the LSZ formula on S-matrix equivalence
May
23
awarded  Good Question
Dec
30
awarded  Benefactor
Dec
29
accepted Independence of thermodynamic variables
Dec
23
comment Independence of thermodynamic variables
@Christoph Thank you Christoph. If you convert your comments to an answer, I'd be happy to award you the bounty.
Dec
23
comment Independence of thermodynamic variables
cont.. This is because for any partition of extensive/intensive variables that we choose, there is some thermodynamic potential for which the variables serve as natural variables and hence is enough to determine the entire thermodynamic state. So I guess this is a sufficient requirement. Am I correct in assuming that this is also sufficient? I.e. the inclusion of a conjugate pair of variables means that the state will not be uniquely determined. So for a simple system with variables $\{U,S,V,T,p\}$, we cannot use $\{S,T,p\}$ or $\{V,p,T\}$ as coordinates for example.
Dec
23
comment Independence of thermodynamic variables
@Christoph Thank you for your response. My current understanding is this: Let $\{X_i\}_{i=1}^k$ denote the extensive parameters of the system and let $\{Y_j\}$ denote the conjugate intensive parameters. Let $I\cup J = {1,2,\cdots,k}$ denote a partition of the indices. Then $\{X_i\}_{i\in I} \cup \{Y_j\}_{j\in J}$ will serve as a valid coordinate system for the thermodynamic states so long as we include at least one extensive parameter.
Dec
23
awarded  Promoter
Dec
21
asked Independence of thermodynamic variables
Oct
24
accepted A vector function of a vector $\mathbf{S}$ must be given by a multiple of $\mathbf{S}$?
Oct
15
awarded  Curious