Lattice constant from first principles

[Demystifier]

I would like to know how to estimate the value (length) of the lattice constant in crystals in terms of fundamental parameters such as electron charge, electron mass, proton mass and Planck constant. (Alternatively, the formula may contain the Bohr radius, since I know how to calculate Bohr radius from the fundamental parameters). I am not interested in an accurate calculation, by DFT or whatever. All I need is a simple formula that estimates the order of magnitude.

If the formula can be found in a standard textbook, please tell me which textbook and a precise location (page number, equation number, etc.).

Thanks!

 

[M Quack]

Chemists have tables of ionic radii. You can combine that with space filling factors for different packings of spheres to get a very rough estimate of the unit cell volume. After that you have to make assumptions about the symmetry of the cell (cubic, hexagonal, ...). Note that the results will be a very very crude approximation.

From Wikipedia, the densities of graphite and diamond:

diamond: 3.515 g·cm−3
graphite: 2.267 g·cm−3

This gives you a taste of what kind of accuracy you can expect from such a formula.

 

[Demystifier]

You take the measured densities as an input, but that's not what I want. In a formula, I don't want any non-fundamental parameters such as those measured densities.

A kind of formula I would accept is something like
$$a=4\pi a_{Bohr}$$
where $a_{Bohr}$ is the well known expression for the Bohr radius.

 

[f95toli]

I doubt there is a "simple" formula for that. Unless I am missing something this is a complicated optimization problem which it is always done numerically. You don't define what you mean by "crystal", but the crystallography can get very complicated even if you only consider pure elements (carbon would be the obvious example); and unless you know the crystal structure you can't calculate the lattice constants.

It might be possible if you consider some "simple" element and you assume a simple cubic structure, but I don't remember ever seeing such a formula.

 

[Demystifier]

f95toli, can one, at least, make a qualitative theoretical argument that $a$ cannot be much larger than $a_{Bohr}$? If such an argument can be made (without invoking experimental facts on $a$ for different materials), that would satisfy me.

 

[TeethWhitener]

What do you mean by "much" larger? If you're talking about, say, protein crystals that are grown in molecular biology research labs, $a$ can be hundreds of times larger than $a_{bohr}$. Furthermore, the lattice constants are simply the repeat lengths of a crystal, so if you were to, for example, create regularly spaced defects in a diamond that were a distance of 1mm apart, the lattice constant would technically be very close to 1mm, even though the lattice constant of pure diamond is only a few angstroms.

 

[DrClaude]

f95toli, can one, at least, make a qualitative theoretical argument that $a$ cannot be much larger than $a_{Bohr}$? If such an argument can be made (without invoking experimental facts on $a$ for different materials), that would satisfy me.

Asuming you are talking about a crystal of atoms, and you interested in the average interatomic distance (see TeethWhitener's comment), then the only way I can think of is to start from a diatomic molecule. The only case that can be solved exactly is H₂⁺ in the Born-Oppenheimer approximation. You then get that the minimum energy is at about $2 a_{Bohr}$. You then need to extrapolate by saying that this distance is (possibly) not much affect by the presence of other atoms, including when an atom in bounded to many other, such that in a solid the atoms will be found this order of magnitude apart one from another.

 

[DrDu]

You might give W. Thirring,
Quantum Mathematical Physics: Atoms, Molecules and Large Systems
a try. He derives some bounds on the stability of extended matter.

 

[Demystifier]

What do you mean by "much" larger? If you're talking about, say, protein crystals that are grown in molecular biology research labs, $a$ can be hundreds of times larger than $a_{bohr}$. Furthermore, the lattice constants are simply the repeat lengths of a crystal, so if you were to, for example, create regularly spaced defects in a diamond that were a distance of 1mm apart, the lattice constant would technically be very close to 1mm, even though the lattice constant of pure diamond is only a few angstroms.

I, of course, meant the distance between the nearest neighbor atoms. Otherwise, quasicrystals would have infinite $a$.

 

[Demystifier]

Asuming you are talking about a crystal of atoms, and you interested in the average interatomic distance (see TeethWhitener's comment), then the only way I can think of is to start from a diatomic molecule. The only case that can be solved exactly is H₂⁺ in the Born-Oppenheimer approximation. You then get that the minimum energy is at about $2 a_{Bohr}$. You then need to extrapolate by saying that this distance is (possibly) not much affect by the presence of other atoms, including when an atom in bounded to many other, such that in a solid the atoms will be found this order of magnitude apart one from another.

That's close to what I wanted. But in the meantime, I have found my own intuitive qualitative answer which I would like to share.

What fundamental force binds the atoms in a crystal together? This, of course, bust be some manifestation of the electric force. But atoms are electrically neutral, so the force cannot be the electric monopole force. So it must be some electric multipole force, originating from a non-trivial distribution of charge in the atoms. But the atom (and its charge distribution) looks point-like when seen from a distance much larger than the atom size. Consequently, the multipole force must be negligible at such large distances. Therefore the distance $a$ between the neighboring atoms which act on each other by a mutipole force cannot be much larger than the size of the atom. Since the size of atom is of the order of $a_{Bohr}$, it implies that $a$ cannot be much larger than $a_{Bohr}$. Q.E.D.

 

[DrClaude]

But atoms are electrically neutral, so the force cannot be the electric monopole force.

That completely neglects all the QM processes that lead to stable bonds, and throws under the carpet the fact that electrons do not stay "attached" to one particular atom.

So it must be some electric multipole force, originating from a non-trivial distribution of charge in the atoms.

See van der Waals interaction.

But the atom (and its charge distribution) looks point-like when seen from a distance much larger than the atom size.

That is not true: the induced-dipole - induced-dipole attraction that is the van der Waals interaction takes place on length scales much greater than the size of the atoms.

Consequently, the multipole force must be negligible at such large distances. Therefore the distance $a$ between the neighboring atoms which act on each other by a mutipole force cannot be much larger than the size of the atom. Since the size of atom is of the order of $a_{Bohr}$, it implies that $a$ cannot be much larger than $a_{Bohr}$. Q.E.D.

What I like least about this "demonstration" is that you have just proved that gases (and even liquids) have no cohesion, since the atoms have then much greater mean separations than in solids.

 

[DrClaude]

DrClaude, thank you for your notes. So, given the existence of van der Waals interaction, what is the typical distance at which it is not negligible, and how would you express this distance in terms of fundamental parameters?

It goes as $r^{-6}$. It's hard to say at what point it becomes negligible: ngligible with respect to what? Calculating its magnitude requires knowing the polarizability of the atoms, which is not something you get directly from fundamental constants.

If you do not want to introduce chemical bonding, then I think the simplest answer is just ~2 a_Bohr, which is what you get when packing atoms-as-spheres together.

 

[ShayanJ]

DrClaude, thank you for your notes. So, given the existence of van der Waals interaction, what is the typical distance at which it is not negligible, and how would you express this distance in terms of fundamental parameters?

I think temperature becomes important here. Because whatever the energy of the interaction is, if it is less than the thermal vibrations, no bond can form. So here "not negligible" has something to do with temperature.
But if you assume temperatures very close to absolute zero, then of course there will be a bond with any interaction energy, only the lattice constant should be determined. I think considering a first-neighbour approximation can give a good starting point.
But such considerations will be classical, so I think its better to be careful about them at very low temperatures.

 

[DrDu]

Your argument should somehow hold true for nonpolar atomic and molecular solids. VdW interaction is not simple. There is a long range part with gives rise to surface tension and other effects. You can probably separate the long range forces off somehow. For the short range, you have Pauli repulsion which decays exponentially for distances larger than the Bohr radius and the VdW attraction. So you end up with some Lennard Jones intermolecular potential. The only length scale entering is the Bohr radius.

 

[Demystifier]

It goes as ##r^{-6}##. It's hard to say at what point it becomes negligible: ngligible with respect to what?

Well, to get something which has dimension of force, the $r^{-6}$ factor must be multiplied with some constant dimensional factor. This dimensional constant contains something with a dimension of length (with an appropriate power). So what this parameter with a dimension of length can be, if not something of the order of $a_{Bohr}$?

I suspect that the length scale is still proportional to $a_{Bohr}$, but multiplied with something relatively big such as $(4\pi)^4$.

 

[Demystifier]

The only length scale entering is the Bohr radius.

But for long-range interactions, it must be multiplied with something relatively big as in my post above, right?

 

[DrDu]

For me, 4$\pi$ is still $O(a^0)$, so it is neither big nor small.

 

[Demystifier]

For me, 4$\pi$ is still $O(a^0)$, so it is neither big nor small.

For me, $(4\pi)^4\approx 24936\gg1$, so it's big.

 

[DrDu]

Not in comparison with the inverse bohrs radius. Where do you take the 4th power from ?

 

[Demystifier]

Not in comparison with the inverse bohrs radius. Where do you take the 4th power from ?

I have not derived the 4th power, but perhaps something like that might originate from the 6th power in $r^{-6}$ compared to the 2nd power in Coulomb law.

Anyway, it is certainly true that $(4\pi)^4 a_{Bohr} \gg a_{Bohr}$. My idea is that some relation of this form might be the reason why van der Waals is effective at distances much larger than $a_{Bohr}$. That's just a conjecture, I don't have a proof.

 

[TeethWhitener]

I have not derived the 4th power, but perhaps something like that might originate from the 6th power in $r^{-6}$ compared to the 2nd power in Coulomb law.

Anyway, it is certainly true that $(4\pi)^4 a_{Bohr} \gg a_{Bohr}$. My idea is that some relation of this form might be the reason why van der Waals is effective at distances much larger than $a_{Bohr}$. That's just a conjecture, I don't have a proof.

First, $r^{-2}$ dies off much more slowly than $r^{-6}$. But that's not really relevant. The $r^{-6}$ dependence of the van der Waals energy comes from adding a dipole-dipole interaction at 2nd order as a perturbation in the Hamiltonian. An individual atom has an instantaneous dipole moment, and the interaction of 2 dipole moments from separate atoms goes as $r^{-3}$. So you have a perturbative term in the Hamiltonian $V_{pert}\propto\frac{1}{r^3}$. Expanding the perturbation matrix in the standard fashion:
$$E_{pert}=\langle \varphi_0 | V_{pert} | \varphi_0 \rangle + \sum\limits_{k} \frac{|\langle \varphi_{k} | V_{pert} | \varphi_{0} \rangle|^2}{E_0 - E_k}+...$$
The first order term disappears because $V_{pert}$ has instances of the position operator in the numerator, so the dominant contribution to the perturbation energy goes as $V_{pert}^2 = \frac{1}{r^6}$.

 

[Demystifier]

First, $r^{-2}$ dies off much more slowly than $r^{-6}$. But that's not really relevant. The $r^{-6}$ dependence of the van der Waals energy comes from adding a dipole-dipole interaction at 2nd order as a perturbation in the Hamiltonian. An individual atom has an instantaneous dipole moment, and the interaction of 2 dipole moments from separate atoms goes as $r^{-3}$. So you have a perturbative term in the Hamiltonian $V_{pert}\propto\frac{1}{r^3}$. Expanding the perturbation matrix in the standard fashion:
$$E_{pert}=\langle \varphi_0 | V_{pert} | \varphi_0 \rangle + \sum\limits_{k} \frac{|\langle \varphi_{k} | V_{pert} | \varphi_{0} \rangle|^2}{E_0 - E_k}+...$$
The first order term disappears because $V_{pert}$ has instances of the position operator in the numerator, so the dominant contribution to the perturbation energy goes as $V_{pert}^2 = \frac{1}{r^6}$.

What your analysis misses is the central thing I am interested about: dimensional constants which give the potential the correct dimension.

 

[TeethWhitener]

The interaction energy between two dipoles can be found in any decent EM book. It's derived by looking at the energy of one dipole in the field of another. I've reproduced the formula here for easy access:
$$V_{int}=- \frac {p_{1} \cdot p_{2} - 3(p_{1} \cdot \hat{r}_{12}) (\hat{r}_{12}\cdot p_{2})}{4 \pi \epsilon r_{12}^3}$$
where $p_i$ is the dipole moment created by the ith electron-nucleus pair and $r_{12}$ is the distance vector connecting the two point dipoles.

Here's the derivation: http://depts.washington.edu/chemcrs/bulkdisk/chem551A_win06/notes_van_der_Waals_Interaction.pdf
It's a scanned pdf (I think of Messiah's QM book?). It also has a discussion of the approximate magnitude of these interactions in terms of Bohr radii. It gives
$$V_{vdW}=-\frac{C}{r^6}\approx -\frac{6e^2}{r} \left(\frac{a_0}{r}\right)^5$$

[EDIT]: This equation is only valid where the wavefunction overlap of the two atoms is negligible. So it wouldn't really apply to crystals.

 

[Demystifier]

Thanks TeethWhitener!

The scanned pdf is from the book by Cohen-Tannoudji, Dui and Laloe. It's nice to have the final formula for the potential (that's exactly what I wanted), but now I am confused. This formula predicts that the VdW force is not significant at distances much larger than the Bohr radius $a_0$. It is in agreement with my intuitive argument in post #10, but directly contradicts the claim by DrClaude in post #11. oo)

 

[TeethWhitener]

The scanned pdf is from the book by Cohen-Tannoudji, Dui and Laloe. It's nice to have the final formula for the potential (that's exactly what I wanted), but now I mam confused. This formula predicts that the VdW force is not significant at distances much larger than the Bohr radius $a_o$, contrary to the claim in post # .oo)

Right, as I mentioned before, the vdW potential falls off as $r^{-6}$, so it becomes negligible very quickly (w/in a few angstroms), much more so than a $\frac{1}{r}$ monopole potential. However, the vdW interaction acts as a potential between neutral atoms, which I think is what you were looking for.

Come to think of it, I'm not really sure what you're looking for. As far as I can tell, the case you're concerned about is one in which the crystal has cubic symmetry (so all lattice constants are equal), and the unit cell has one atom (so that there is no electron density transfer between atoms--IOW, all atoms in the crystal are equivalent). Moreover, the lattice would have to be simple cubic in order for the lattice constant to equal the nearest neighbor distance (in an fcc closest packed crystal, the nearest neighbor distance is $\frac{\sqrt{2}}{2}a$, where $a$ is the lattice constant). And from these assumptions, you want to know if there's a simple relationship between fundamental constants and an estimate for the lattice constant. Is this roughly the right idea?

 

[Demystifier]

Come to think of it, I'm not really sure what you're looking for.

Well, you explained a lot and almost everything is now clear to me. The only thing which is still not clear to me is how can all this be compatible with the claim by DrClaude in post #11. Can we conclude that he was wrong?

 

[TeethWhitener]

Well, you explained a lot and almost everything is now clear to me. The only thing which is still not clear to me is how can all this be compatible with the claim by DrClaude in post #11. Can we conclude that he was wrong?

Well, I'd say he likely was unclear on what you were asking. (Who knows, maybe his full name is Dr. Claude Cohen-Tannoudji?). What is true is that for two uncharged species with no permanent dipole moments, the vdW (induced dipole-induced dipole) interaction is really the only meaningful interaction that occurs further out than the immediate vicinity of the atom. So in that sense, chemists and physicists typically refer to vdW as a long-range interaction.

 

[ehild]

That's close to what I wanted. But in the meantime, I have found my own intuitive qualitative answer which I would like to share.

What fundamental force binds the atoms in a crystal together? This, of course, bust be some manifestation of the electric force. But atoms are electrically neutral, so the force cannot be the electric monopole force. So it must be some electric multipole force, originating from a non-trivial distribution of charge in the atoms. But the atom (and its charge distribution) looks point-like when seen from a distance much larger than the atom size. Consequently, the multipole force must be negligible at such large distances. Therefore the distance $a$ between the neighboring atoms which act on each other by a mutipole force cannot be much larger than the size of the atom. Since the size of atom is of the order of $a_{Bohr}$, it implies that $a$ cannot be much larger than $a_{Bohr}$. Q.E.D.

A crystal can be made of atoms, ions or molecules, these units repeating periodically is the space. The distance between the closest repeating units are the lattice parameter. The repeating units can be made of quite a lot of atoms, so the lattice parameter can be much larger than the Bohr radius.

The forces between the repeated units can be of different nature from the forces between the atoms.

Your argument would be true for atomic crystals, consisting of single neutral atoms. Very few crystals are like that!

The simplest crystals like NaCl are ionic: they consist of positive and negative ions, kept together by Coulomb forces. The Coulomb force is a long-range force, being inversely proportional to r².

In metals, the repeated units are positive ions held on their place by the free electrons shared by each of these ions.

In such mono-atomic covalent crystals like diamond or silicon, the atoms are kept together by their chemical bonds.These bonds can not be explained by pure electric interaction. The bonds form of pair of electrons which are negative, repulsing each other electrically, but still forming strong bonds in the covalent crystal.Those crystals are like a single molecule.
In the previous cases, you get the lattice parameter as the distance between the closest ions, and you can say that it is comparable to the Bohr radius.

But crytals can be made also of molecules. The atoms in the molecule are kept together by chemical bonds, which are not purely electronic. The molecules in the crystal are kept together by dipole and multipole interaction between the molecules. Molecules can be of any shape and quite large. The interaction between the molecules are not so strong as the chemical bonds between the atoms in the molecule. You can not say that the units of the crystal interact with the same kind of force as the force keeping the atoms together in a molecule.

 

[DrClaude]

Who knows, maybe his full name is Dr. Claude Cohen-Tannoudji?

I wish!

Well, you explained a lot and almost everything is now clear to me. The only thing which is still not clear to me is how can all this be compatible with the claim by DrClaude in post #11. Can we conclude that he was wrong?

I might well be wrong, but I think there is a misunderstanding be what is meant by "long-range." My point is that there is cohesion between atoms even when the mean separation is much greater than the Bohr radius (although in that case we don't have a solid, but a gas).

 

[Demystifier]

I might well be wrong, but I think there is a misunderstanding be what is meant by "long-range." My point is that there is cohesion between atoms even when the mean separation is much greater than the Bohr radius (although in that case we don't have a solid, but a gas).

I believe now everything is clear and consistent. :)

VdW is important at large distances not because it is strong there, but because other forces (if any) are even weaker. For example, "quantum forces" due to the overlap of wave functions of different atoms fall of exponentially with $r$.