# About symmetry, and about electron density in crystals in particular

The book Introduction to Solid State Physics by Kittel says:

"We have seen that a crystal is invariant under any translation of the form T [...]. Any local physical property of the crystal, such as the charge concentration, electron number density, or magnetic moment density is invariant under T. [...] n(r+T) = n(r)"

Why is this the case? Even though the underlying atoms are invariant under translations in T, why should the electron density also be invariant under the same transformation? Couldn't for example if you have this 1D crystal:

o   o   o   o   o


This is invariant under transformations $T = {...,-1,0,1,2,3,...}$. Why couldn't the electron densities be like this:

o h o l o h o l o h o


Where h is a high density and l is a low density. So they would only be invariant under $T = {...,-2,0,2,4,...}$.

This is just an example: why do the local physical properties have to be invariant under T?

I suppose this question could be phrased even more generally: if the causes of some phenomenon have a symmetry, does the phenomenon also always have the same symmetry? If yes, why, and if no, what are the conditions under which the phenomenon does have the same symmetry?

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What you are describing is called a “charge density wave” (try to google that). And yes, it does arise in some exotic materials, from spontaneous symmetry breaking. –  Edgar Bonet Jul 8 '11 at 10:06
Mmm... I forgot to mention: a charge density wave has some electrostatic effect on the underlying lattice a thus causes a slight periodic compression/extension of the lattice constant. Thus, to be fair, when the wave arises (below a transition temperature), both the electron density and the lattice have their translation symmetry broken. –  Edgar Bonet Jul 8 '11 at 10:22

Of course you can have configurations which do not respect the underlying symmetry of they dynamics, the issue is that those will not be eigenstates of the Hamiltonian. Typically when talking solid state people are interested in energy bands which are eigenstates of the Hamiltonian, and these must also be eigenstates of the lattice translations. This follows from elementary operator algebra. The non-symmetric configurations would correspond to superpositions of different energy bands and would not be stationary in time.

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I see. I was not able to see why the lowest state has to be symmetric under T (in fact I do not even know how to start proving this). Why is this true? –  Jules Jul 7 '11 at 21:27
@Jules: it need not be. For the sake of simplicity, consider the Ising model (a lattice with just values $\{+1, -1\}$ at every point and interaction liking the neighboring sites to have the same value). The ground state is either all $+1$ or all $-1$. Both of these states break the spin-flip symmetry of the Hamiltonian. In fact, configuractions (whether thermal or energy eigenstates) of most of the interesting systems at low energies do break the symmetry of the theory. The phenomenon of spontaneous symmetry breaking is the daily bread of many fields of modern physics. –  Marek Jul 7 '11 at 21:51
For simplicity, assume a unique ground state |0> with energy e0, s.t. H|0> = e0|0>. If [H,T] = 0, then HT|0> - TH|0> = HT|0> - e0 T|0> = (H-e0)T|0> = 0, so it must be that T|0> = constant * |0> since the ground state is unique. Now the constant need not be equal to 1, but it usually is. –  user1631 Jul 7 '11 at 21:53
@user1631: 1) Try to prove that the ground state is unique. ;-) 2) Even if it is indeed unique, the energy difference between the ground state and the first exited state is in some cases small enough to be irrelevant. 3) Spin glasses have such a large state density around the ground state that they keep finite entropy even at $T \to 0$. –  Edgar Bonet Jul 8 '11 at 10:13

my guess is that this is just an approximation.

The electrons in the symmetric grid of the crystal are acting more like a wave than as cloud of point-like charges. The lowest energy modes for the electron standing-waves are those with long(er) wavelengths and are approximately uniform in the lattice.

So, wavefunctions that vary significantly between nearby lattice cells can be interpreted as representing high-energy modes, which might occur only far from the equilibrium. However i'm not familiar with the condensed matter specifics of crystals enough to tell you under what circumstances that approximation will hold

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This is actually a very deep question. Until very recently scientists believed that yes, if a problem has a certain symmetry, then the solution will share it as well. But then the concept of a spontaneously broken symmetry arose:

Wikipedia link for more advanced discussion and examples

I can provide a very simple explanation, since I haven't learned the mathematical aspect of it yet (usually people apply it to particle physics and the Standard Model). Look at the graph on the tab of your web browser, next to "solid state physics - About ..." It's called a Mexican Hat potential, funnily enough, and it's important in dealing with the Higgs Mechanism. Now if you place a ball at the top of the hat, there is rotational symmetry about the axis of the hat. But this doesn't mean that this symmetry exists for all future times; as soon as the ball rolls down, it picks a direction, and we say that the symmetry is broken. Basically if a certain "direction" has to be picked, but it doesn't matter which one is picked, spontaneous symmetry breaking has occured.

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This makes a lot of sense for the general case. So Kittel is making the (unproven) claim here that the state where the electrons obey the symmetry is in fact a local (or global?) minimum of energy. Why is this true? –  Jules Jul 7 '11 at 21:25
I'm not sure if you can prove this using more general arguments, but for this case I'd say the fact that the crystal is invariant under T means that the background is 'smooth' or 'flat', so that there isn't a particular place electrons want to go or don't want to go. And since the electrons minimize their interaction energy by staying as far away from each other as possible, on this 'flat' background they'll spread out into a more or less uniform distribution, which happens to be invariant under T. –  Arun Nanduri Jul 7 '11 at 21:40