One of the active research areas in present is Strong interacting, Strong Coupling, Strong Correlated regime of the phases of matters.

It seems to me that some physicists in the fields often mix the usages of these twos: Strong Coupling, Strong Correlated.

However, in my viewpoint, they are NOT the exactly same, I regard that

$\bullet$ Strong Coupling: implies the large coupling of interactions comparing to the free part of theory. Say, suppose there is a Lagrangian description, then the action $S$ $$ S=S_{free} +g S_{interact} $$ the Strong Coupling means $g >>1$. So this can be the confined phases of QCD, where coupling $g$ of quarks and gluons runs large.


$\bullet$ Strong Correlated: in my view, usually implies the fractionalization of the elementary particles into fractional quantum numbers. For example, this happens at 1+1D Luttinger liquids, where spin and charge can separated their degree of freedom from the elementary constituents(electrons), but the system needs NOT to be Strong Coupling. i.e. this example is Strong Correlated but NOT Strong Coupling. This is about the fields of Strong Correlated Electron on arXiv.


My question, so what are other examples of systems that are:

1. YES Strong Coupling and YES Strong Correlated

2. YES Strong Coupling but NOT Strong Correlated

3. NOT Strong Coupling but YES Strong Correlated

See also this relevant post.

  • 1
    $\begingroup$ You're right they are not the same. The definition of strong coupling is pretty obvious, but that of strong correlation is much less so. For some people, it means "large correlation length", with the extreme example of critical points (classical or quantum). For other, it just means "a naive/mean-field expansion (weak or strong coupling)" does not work (examples for this: BCS, Bose Mott phases are not strongly correlated). For some (like you?), it means that quasi-particles do not exist. So it might mostly be a matter of definition... $\endgroup$
    – Adam
    Jun 7, 2014 at 4:12
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    $\begingroup$ A blog post about this question: condensedconcepts.blogspot.com/2014/02/… $\endgroup$
    – Adam
    Jun 7, 2014 at 4:16
  • $\begingroup$ And what about the difference between coupling and interaction? it follows from your post that there's one but you don't comment about it, could you clarify it, please? Also, what it means the $S_{free}$? it includes the kinetic part? $\endgroup$
    – cla
    Jan 5, 2017 at 16:01
  • $\begingroup$ @cla, in a regular day basis, interactions tend to be regarded by physicists as one of the "Big 4": nuclear(weak or strong), electromagnetic or gravity. The mathematical understanding of all but gravity - up until today - require the exchange of gauge bosons. The strength of interactions depend on the relative values of each characteristic coupling (e.g. the fine structure constant for QED being larger than its gravitational relative en.wikipedia.org/wiki/Gravitational_coupling_constant). $\endgroup$
    – daydreamer
    Oct 19, 2020 at 1:42

1 Answer 1


From the point of view of the condensed matter physics, the distinction between strong coupling and strongly correlated seems rather evident, and largely agrees with the point of view expressed in the question. First of all, this is because in the condensed matter physics the coupling strength is not fixed by the world constants but material-dependent, which allows exploring regimes with different coupling. Moreover, strongly correlated systems are often treated as a subfield of condensed matter, rather distinct from the rest. Let me give a few examples.

The polaron problem
The problem is about the kinetic and thermodynamic properties of an electron dressed by a phonon cloud, rather similar in spirit to many QFT problems $$ \hat{H} = \frac{\mathbf{p}^2}{2m} + \sum_\mathbf{q}\hbar\omega_\mathbf{q}a_\mathbf{q}^\dagger a_\mathbf{q} + \sum_\mathbf{q}g_\mathbf{q}\left(a_\mathbf{q}e^{i\mathbf{qr}} + a_\mathbf{q}^\dagger e^{-i\mathbf{qr}}\right)$$ It has been extensively studied both in weak-coupling and strong-coupling regimes, either of which is realizable. We literally talking here about the magnitude of the coupling constant.

Ising model
Ising model is the main toy model in the field of strongly-correlated systems: $$ H = -J\sum_{\langle i, j\rangle}S_i S_j - H\sum_i S_i $$ Speaking of strong correlations one means that spatial and temporal correlations between parts of the system are decaying slower than exponentially (typically as power laws). Big part of this field is about studying phase transitions: the exchange coupling $J$ in the Ising model can be small, yet below the transition temperature the model will exhibit long-range correlations (in two or more dimensions). One must note that the transition temperature is dependent on the size of coupling, so it is hard to call it strong or weak - it is not clear what it must be compared with.

Kondo model
The examples above actually demonstrate that terms strong coupling and strong correlations are often applied in different contexts: the former often deals with a small system coupled to a bath (e.g., electrons and phonons) where coupling strength mediates the interaction between the two, the latter deals with collections of many identical interacting entities, where coupling strength is hard to identify.

An example at the intersection of these two is the Kondo problem of an impurity spin coupling via exchange interaction to the electron sea: $$ \hat{H} = \sum_{\mathbf{k}\sigma}\varepsilon_{\mathbf{k}\sigma}c_{\mathbf{k}\sigma}^\dagger c_{\mathbf{k}\sigma} - \mu_B\mathbf{H}\cdot\mathbf{S} - \sum_{\mathbf{k}_1\sigma_1,\mathbf{k}_1\sigma_2}\mathbf{S}\cdot\mathbf{s}_{\sigma_1\sigma_2}c_{\mathbf{k}_1\sigma_1}^\dagger c_{\mathbf{k}_2\sigma_2} $$ It can be characterized in terms of a coupling strength, yet it exhibits many features of the strongly correlated systems. Again, the transition to the strongly correlated regime is determined by the Kondo temperature $T_K$, which depends on the coupling strength $J$.

  • 1
    $\begingroup$ This should be the accepted answer: Ising Model alone exhibit the whole 3 features OP asked for. We just need to walk over the phase diagram and tinker with J values. $\endgroup$
    – daydreamer
    Oct 19, 2020 at 1:38

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