# Is the charge-conjugation symmetry in cond-mat physics different from that in QFT?

In condensed matter physics, the terms "particle-hole symmetry" and "charge-conjugation symmetry" are often used interchangeably. As far as I understand, they refer to the switching of nothing other than a unit charge. For e.g., in a condensed matter system, introducing a particle of positive unit charge is equivalent to removing a particle of negative unit charge such as an electron. Doing either of the two would apply the charge-conjugation transformation to the system.

In quantum field theory however, charge-conjugation means much more than what the name implies $$-$$ conjugation of not just the charge but the replacement of a particle by its antiparticle. An antiparticle differs from the particle not just by having the opposite charge but also opposite values of other properties, such as helicity, parity, etc. In my understanding, in QFT we study fundamental particles in isolation such that charge-conjugation means replacing a fundamental particle by its antiparticle. For this reason charge-conjugation in QFT cannot be achieved by simply adding/removing a particle as can be done in particle-hole symmetry transformation of condensed matter systems. Is my understanding on this topic correct or is there a different way to look at it?

In condensed matter physics one can define holes as simple charge conjugation. E.g., if we define the ground state as $$a_{\mathbf{k},\sigma}|0\rangle=0,\text{ if } k>k_F,\\ a^\dagger_{\mathbf{k},\sigma}|0\rangle=0,\text{ if } k then the hole annihilation operator can be defined as $$h_{\mathbf{k},\sigma}=a^\dagger_{\mathbf{k},\sigma},$$ and many calculations in condensed matter would work out just fine.

Nevertheless, one usually uses a more elaborate definition: $$h_{-\mathbf{k},-\sigma}=a^\dagger_{\mathbf{k},\sigma},$$ i.e., switching directions of momentum and spin (very similar to the QFT.)

From the condensed matter point of view, this is motivated by physical considerations: creating a hole means removing a spin from a filled band, i.e., reducing the total spin of the band. Similarly, creating a hole means reducing the total momentum of the band, which is reduce by the value of the momentum of the removed electron: if the full band had net momentum $$0$$, the new state, formed by removing an electron of momentum $$\mathbf{k}$$, has momentum $$-\mathbf{k}$$.

As stated above, many mathematical treatments would be unaffected by this - notably those where one considers the properties of the band itself, mostly related to electron-electron interactions. However, there are situations where the correct definition is important - e.g., when discussing the selection rules for photon absorption.

Somewhat aside, the important difference between the condensed matter and the QFT is that conduction electrons and holes live in different bands with different dispersion laws. Moreover, it is not uncommon to have different numbers of bands: e.g., for typical semiconductor materials (Si, Ge, GaAs/AlGaAs) one usually considers the conduction band and three valence bands (heavy holes, light holes, and split-off band.) In this sense holes in the valence bands are not antiparticles of the electrons in the conduction band. Again, these are nuances usually overlooked in QFT for condensed matter type of texts, but which are essential once one does actual physics.
(image source) 