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  No Lagrangian description v.s. No quasi-particle description

+ 4 like - 0 dislike

This post is aimed to stimulate some discussions.

We are familiar with many physical descriptions and theories of the (many-body quantum) system, with both quasi-particle description and Lagrangian description. For example:

the Landau Fermi-liquids theory.

Here quasi-particle description are simply a way to find effective excitations for the (many-body quantum) system. The effective excitations may not be the original elementary constituents or elementary particles/spins of the system. These effective excitations contain quasi-particle, quasi-string, quasi-brane excitations, etc.

$\bullet$ My question is that:

  1. what are systems with No quasi-particle description but with Yes Lagrangian description.

  2. what are systems with Yes quasi-particle description but with No Lagrangian description.

  3. what are systems with No quasi-particle description and with No Lagrangian description.

NOTE: For instance, I suppose that,

1+1-dimensional Luttinger liquids are examples of 1. No quasi-particle description but with Yes Lagrangian description.

See, e.g. T. Giamarchi, "Quantum Physics in One Dimension" Chap 2.

enter image description here

Fig. 2.4. The occupation factor $n(k)$ of 1+1D Luttinger liquids. Instead of the usual discontinuity at $k_F$ for a Fermi liquid, it has a power law essential singularity. This is the signaturethat fermionic quasiparticles do not exist in one dimension. Note that the position of the singularity is still at k_F. This is a consequence of Luttinger's theorem.

On the other hand, it is likely that

examples of 2 or examples of 3 happens in RR fields or D-branes of string theory, which has Yes/No quasi-particle description, with NO Lagrangian description.

Eg. see this Ref:Stability of Fermi Surfaces and K-Theory by Horava, see page 1 right column: This implies that the RR fields are also objects in K theory, and not differential forms [9], making the low energy description of string theory on manifold Y in terms of Lagrangian (1) questionable. ... once the p-form $C_p$ are reinterpreted as K-theory objects, it is not clear how to even define Lagrangian (1). This crisis of the Lagrangian formulation of low-energy string theory is further supported by the discovery [10] of apparently non-Lagrangian phases in the partition functions of various string and M-theory vacua. Perhaps this means that the Lagrangian framework currently available is insufficient for RR fields but its suitable generalization awaits to be discovered. (Important steps in this direction have been taken [11].) Alternatively, the theory may require a non-Lagrangian formulation. (This may already be suggested by the presence of a self-dual RR field strength in Type IIB theory). Before we settle on either of these two alternatives, however, we should consider a third possibility. The subtle K-theory features of string theory could be an emergent phenomenon, with D-branes and RR fields emerging as composites of some more elementary degrees of freedom that admit a conventional Lagrangian description.

What else physical systems and theories are examples of 1. 2. 3.?

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user wonderich
asked Jun 7, 2014 in Theoretical Physics by wonderich (1,500 points) [ no revision ]
I don't really understand your question. Since the Hamiltonian is the generator of time translations, it always exists. Barring some pathologies, I guess one should be able to perform a Legendre transformation in order to obtain an Lagrangian. It may happen that the transformation is non-trivial by virtue of existence of constraints, but Dirac's theory should tackle this. I'm not aware of any other problem in this regard. The paper you quote seems to state that there is no Lagrangian in terms of those degrees of freedom, not non-existence of Lagrangian description at all (...)

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user cesaruliana
(...) Therefore the statement in the paper appears rather trivial, just that the particular system of interest is a effective theory of a more fundamental Lagrangian. Considering this, could you clarify if you're interested in effective theories without an effective Lagrangian, or in absence of the Lagrangian at all? And why do you think this should be related to quasi-particles, which I always though to be just a reflection of some property of the spectral function? As in TwoBs' answer, even your example in Luttinger liquids is not clear

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user cesaruliana
@ cesaruliana, well, I can imagine that there are certainly non-Lagrangian description of some theory, as cited in the Ref given. Besides, possibly hinted from AdS/CFT, we can have some description of unknown CFT which Lagrangian is not clearly-known, but we can compute its property from the bulk gravity AdS side.

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user wonderich
On the other hand, those strongly-coupled CFT may NOT have quasi-particle descriptions. An AdS/CFT example can be: arxiv.org/abs/1301.1986. Other condensed matter example, can be found in arxiv.org/abs/0803.4009, etc.

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user wonderich
Great, I think that clarifies your question as related to theories with unknown Lagrangian description (via AdS/CFT for instance) and it's possible relation to quasi-particle states.

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user cesaruliana
@cesaruliana - your reasoning about "Legendre transform" is just plain naive. Legendre transform is an operation done with classical fields. So the Hamiltonian entering the Legendre transform is a classical function. On the other hand, in quantum mechanics, the Hamiltonian is a general operator made of non-commuting stuff. Even in quantum mechanics, Lagrangian is a classical function - one Feynman path-integrates over classical histories. So Lagrangian, Hamiltonian are in no way "analogous objects" in quantum mechanics and they can't be related by any symmetric transform that works in QM.

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user Luboš Motl

1 Answer

+ 2 like - 0 dislike

As for the third class, I think that most of strongly coupled CFTs do not admit a lagrangian description and have no particle like excitations.

In the second class, you can put perhaps strongly coupled model (again, say a strongly coupled CFT, or close to conformality) with a global symmetry that is broken spontaneously. The theory will contain Goldstone bosons that are particles, although the UV theory does not necessarily admit a lagrangian description. Actually, if the CFT is broken spontaneously a dilaton emerges too, and you don't need extra global symmetries .

Finally, I don't think I agree with your first example on Luttinger liquids. In fact, they admit a particle description in terms of free massless bosons. In this case I would put Luttinger liquids in the class of theories that admit both Lagrangian description and particle description. Moreover, Luttinger liquids show a strong-weak duality which is the reason behind the fact that even if strongly coupled in terms of fermionic degrees of freedom the theory can be recast as a weakly coupled one in other (bosonic, in this case) variables.

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user TwoBs
answered Jun 7, 2014 by TwoBs (315 points) [ no revision ]
+1, thanks TwoBs. I will think over for Luttinger liquids. What you evoke is bosonization. But in the pure fermion description, the paragraph I wrote above is from the book I referred to.

This post imported from StackExchange Physics at 2017-11-26 22:02 (UTC), posted by SE-user wonderich

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