# “Crystal electric field”, zero temperature, and the idea that the crystal electric field will have a “symmetry reflecting that of the crystal lattice”

I was searching for information on "phonon broadening" in the context of lasers. I found the document Paper BIII: Diatomic Molecules & Laser Physics by Prof. Simon Hooker, where chapter 5.2.3 Phonon broadening says the following:

In many solid-state lasers the active species are ions doped into a crystalline host. The energy level structure of ions in a crystalline environment is generally different (sometimes very different) than that of an isolated ion owing to interactions with the surrounding ions of the lattice. These interactions can be described in terms of a crystal electric field which, for a perfect crystal at zero temperature, will have a symmetry reflecting that of the crystal lattice.

It isn't totally clear to me what is meant by this part:

These interactions can be described in terms of a crystal electric field which, for a perfect crystal at zero temperature, will have a symmetry reflecting that of the crystal lattice.

Can someone please provide a clearer and more elaborate description of what is being described here (perhaps including illustrations)? For instance, what precisely is meant by the "crystal electric field", why is zero temperature relevant for this explanation, and what is meant by the idea that the crystal electric field will have a "symmetry reflecting that of the crystal lattice"?

This is really a group theoretical question. The best references for group theory in solid state are (in my opinion) "The Mathematical Theory of Symmetry in Solids: Representation Theory for Point Groups and Space Groups" by Bradley and Cracknell and "Group Theory: Application to the Physics of Condensed Matter" by Dresselhaus. The latter is available for free as lecture notes online here.

The basic idea of crystal field splitting is as follows: A free atomic orbital has full rotational symmetry and therefore lives in a representation of a general rotation group of the SO-type. Once you put this atom into a lattice, however, the lattice imposes constraints on the rotational symmetry - not every angle is allowed (lattices can in general have twofold, threefold, fourfold or sixfold rotational symmetries). This changes the degeneracy of your atomic energy levels.

The question of how you go from the full rotational symmetry to the reduced symmetry is answered by the method of group subduction. This is a mathematical procedure which determines which representations of the smaller symmetry group your full atomic orbital splits into.

Physically, the idea is the following: the fact that your atom lives in a lattice puts it into a potential from the surrounding atoms (the crystal field). This potential splits the level degeneracy. How it splits is what group subduction tells you. At zero temperature (barring quantum zero-point motion, which can be important for some light atoms!) you think of the atoms as being in a fixed position. Thus, the atoms respect the symmetry of your underlying lattice, and so does the potential (crystal field) which results from them. At higher temperature, your atoms only respect the symmetry "on average" (because of the vibrations/phonon modes).

So what exactly is a crystal field? You can think of it classically as a result of the Coloumb field from all of the atoms surrounding you. Looking at a specific atom in the crystal, it will feel some Coloumb effect from every other atom in the whole lattice. Luckily, these will tend to cancel out, so the most important contribution is from the nearest neighbors. So you can think of it as the potential you would feel if you had charges surrounding you, in the geometry of the crystal. Thus, if you have an octahedral crystal (every atom has six neighbors), it's the effective Coloumb field from six charges symmetrically located around the origin in an octahedral geometry. This is why the crystal field has the symmetry of the lattice. There is an explicit example of how to find these fields in section 6.2 of the lecture notes I linked above.

I have included a picture from chapter 6 of the lecture notes version of the Dresselhaus book, which are freely available here. What you see here is how the energy levels split as you decrease symmetry. On the left, you start with five free d-orbitals ($$Y_{2,m}$$ for m in $$\{-2,-1,0,1,2\}$$), which all have full rotational symmetry and are perfectly degenerate. Then you put these d-orbitals into an octahedral lattice (six nearest neighbors). This gives the crystal field octahedral symmetry, and splits some of the degeneracy of your d-orbitals. Then you decrease the symmetry further, going to a tetragonal phase (four neighbors), which leads to more splitting. The labels in the figure refer to representations of the symmetry group.

• Your answer is very informative. It seems that I underestimated the complexity involved here, so some of this is beyond my current physics knowledge (although, I know enough to understand the main points you're making). One thing, though: I don't think you directly explained precisely what a "crystal electric field" is supposed to be. This is probably an elementary question, so you might have assumed this knowledge and skipped over it. – The Pointer Dec 23 '20 at 10:26
• I have edited my answer. You can safely ignore the group theoretical comments I made if you are not comfortable with group theory. I think the intuitive picture is quite straightforward, and I hope I have not obscured it. The idea is just that if you put your atom in an external potential (in this case originating from the lattice), the degenerate levels will split. But the actual calculations can get quite ugly. – G.Lang Dec 23 '20 at 10:43
• Your explanations are clear and informative. Thank you for the answer! – The Pointer Dec 23 '20 at 10:46

a rough addition to the above answer. think about a ion, for example Fe$$^{3+}$$; it has 5 electron (negative charges) distributed in 5 different orbitals, characterized by different shapes and orientations. Obviously this 5 orbitals will host negative charges, but in the free space they will have the same energy.

Now put this Fe$$^{3+}$$ at the centre of an octahedron with O$$_{2}^-$$ ions at vertexes (this ion arrangment will produce a crystal field). You will have $$d_{x^2-y^2}$$ and $$d_{z^2}$$ orbitals (that are negatively charged because an electron is delocalized inside them) pointing towards the negatice O$$_2^-$$ ions at vertexes, whereas the $$d_{xy}$$, $$d_{xz}$$ and d$$_{yz}$$ orbitals (negtively charged too) will point towards less negatively charged region of the space (edges or faces of the octahedron). As a consequence of the electrostatic repulsion, the $$d_{x^2-y^2}$$ and $$d_{z^2}$$ orbitals will be found at the 2-fold higher energy level (see the picture above), whereas the remaining $$d_{xy}$$, $$d_{xz}$$ and $$d_{yz}$$ orbitals at the 3-fold lower energy level.

If Fe$$^{3+}$$ would be in tetrahedral coordination, the energy levels would be inverted.

• I was trying to edit your answer to clean up the formatting, but it isn't clear to me what d_x^2_y^2 is supposed to be. Please clarify this. – The Pointer Dec 23 '20 at 10:36
• these are the different d orbitals: dx2-y2, dz2, dxy, .... apex and subscript characters would be needed – gryphys Dec 23 '20 at 10:42
• It seems that gryphys fixed it. What do you mean by "in the free space they will have the same energy."? – The Pointer Dec 23 '20 at 10:52
• the 5 orbitals are degenerate in the free space (isolated ion) because there is no charge around the Fe ion lifting the energy levels of the d orbitals – gryphys Dec 23 '20 at 10:58