It seems to me that the definition of "nematic phase" is quite chamaleonic.

Quoting from its first definition [Nature 393, 550 (1998)] for a 2-dimensional square lattice "The nematic phase breaks the four-fold rotation symmetry of the lattice, but leaves both translation and reflection symmetries unbroken". This is clearly not possible; if you don't eliminate even reflection symmetries, you cannot break the 4-fold rotation symmetry (mirror planes act as generators of the 4-fold rotation symmetry). Then it is not clear if the therm 'lattice' refers to a true mathematical 'lattice' or to the crystal structure (that is a convolution of the lattice with the basis).

Quoting from [Phys. Rev B77, 224509 (2008)] in iron superconductors "... a closely associated structural transition, which we wish to identify as the transition to an “electron nematic phase”". But the structural transition is from the tetragonal to the orthorhombic system and actually breaks both mirror and translation symmetries. It seems to me that this statement is in contradiction with the above definition.

In another paper it is stated "...the distortion is driven by electronic nematicity rather than a lattice (phonon) instability" [Nature Physics 11, 959–963(2015)].

So, is it a nematic transition a structural transition simply not driven by lattice (structural) degrees of freedom?


1 Answer 1


Yes, nematic order is an alignment of rotational degrees of freedom without spatial structure.

One good example of a nematic phase occurs in a liquid crystal composed of elongated molecules. The molecules can align their long axes (breaking the rotational symmetry) without forming any lattice or long-range correlations in their spatial positions. If you were to just look at the center-of-mass positions of the molecules, there would be no correlations.

This term is often poorly defined and is used in many different ways, which can make it really difficult to know exactly what it means in a given situation.

  • $\begingroup$ yes, but it seems to me that the concept of nematicity developed for superconductivity has much more complications, since that in this case the nematicity is intrinsecally formed inside a crystal structure. in liquid crystals the alignement is provided by an external field; in superconductors what is called nematicity fundamentally appears by decreasing temperature. so the same term apparently defines very different phenomena (at least for what I can understand). $\endgroup$
    – gryphys
    Dec 29, 2020 at 13:19
  • $\begingroup$ So in the context of superconducting materials the concept of "without spatial structure" seems to me physically meaningless, because nematicity actually develops within a crystalline structure. $\endgroup$
    – gryphys
    Dec 29, 2020 at 13:22
  • $\begingroup$ I haven't been able to read your references in detail, but in that case they may be using "nematic phase" to refer to the development of some orientational order without positional order, either because the electron positions are still disordered (despite the lattice) or to emphasize that the positional order provided by the lattice already existed and did not arise at the same time as the orientational order. $\endgroup$ Dec 29, 2020 at 14:15
  • $\begingroup$ At first, as far as I understood, it seemed to me that nematicity was meant has a decoupling between the symmetry of the electronic structure and that of the underlying crystal structure. For example an orthorhombic symmetry of the electronic structure (with 2-fold rotation symmetry) hosted in a tetragonal crystal structure (with a 4-fold rotation symmetry). But then the same author (and other authors as well) assumed nematicity in an orthorhombic crystal structure; so I don't see this decoupling any more. It seems to me a rather obscure definition. $\endgroup$
    – gryphys
    Dec 29, 2020 at 15:04
  • $\begingroup$ From my experience, the term nematic is often used in a confusing way like that, haha. It might be worthwhile to directly reach out to one of the authors. Professors can be difficult to get ahold of, but students and postdocs are often quite happy to take the time to answer your questions. $\endgroup$ Dec 30, 2020 at 11:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .