The cohesive energy of a solid is the (average) energy required to isolate the atoms of a solid, which means it's given by the difference between by the energy of solid and the energy of the isolated atoms. I assume in this context by "energy of the atoms" we mean the total binding energy of the electrons to the nuclei, while "energy of the solid" is the sum of the kinetic energy of the electrons, the binding energy of the electrons to the lattice, the energy of lattice vibrations/phonons and the binding energy of the atoms with respect to other atoms.

I have a few questions on the definition of cohesive energy itself: since the (average) kinetic energy of electrons depends on the temperature of the solid, does this mean cohesive energy depends on temperature as well?

How does one compute the contribution to the cohesive energy knowing only the electron hamiltonian and the primitive vectors of the lattice?

This last question is part of an exercise I've been trying to solve without much success. In this exercise we consider a tight-binding hamiltonian with atomic binding energy $E_0$ e hopping energy $V$ such that:

$ H = -E_0\sum_n|n><n| - V\sum_n|n+1><n| + |n><n+1|$

and $a$ is the distance between an atom and the next.

One then solves the Schrodinger equation using Block sums as ansats to find that there is one band:

$E(k) = -E_0 - 2V\cos(ka)$

From which one derives the density of states:

$ D(E) = 2\sum_k\delta(E(k) - E) \propto \frac{1}{\sqrt{1 - \big(\frac{E + E_0}{2V}\big)^2}} $

and the Fermi Energy and wavevector:

$ k_F = \frac{\pi}{2a} \qquad E_F = E(k_F) = -E_0 $

(Is it then correct to say that the Fermi temperature is $T_F = E_0/k_B$ or is it not well defined in this case? Should the Fermi energy be calculated with respect to $-2V$? I've only seen this temperature calculated in the free electron model and it doesn't seem to make as much sense in this case.)

I've looked up various exercises with solutions to see how to approach the problem using the given information but I've trouble making sense of it all. One approach I've seen used Lennard-Jones-type potentials, which requires the equilibrium distance $r_0$, which is not given to me in the exercise and I believe is an appropriate approach only in Ionic Crystals (correct if I'm wrong).

Another approach calculated the cohesive energy (again, the contribution due to electrons) as the average energy at $0K$ (the exercise did not explicit say this, but I thought this part was clear based on the expression alone, maybe it can be interpreted differently).

$ (E_{coh})_{el} = \int_{-E_0 - 2V}^{E_F} ED(E) dE $

However this brings me back to my first question, shouldn't cohesive energy depend on the temperature since it should take into account the kinetic energy of the electrons? Or is this energy excluded?

I'm sorry if the text was a bit confusing, I'm not a native speaker, if anything is not clear I will try to explain it better. Thanks for any help, even a resource (book, pdf) where this is explained in detail would be helpful.


1 Answer 1


I think you're right, in the general case we should consider temperatures, but that is a matter of definition and I think cohesive energy is defined for zero temperature. In this particular case, I will try to answer your questions for $T=0$.

We will do this by neglecting the kinetic energy of the nuclei (phonons). What essentially happens is at $T=0$ we go to the ground state of the system, and the ground state should not have any phonons. The general process goes like this - calculate the energy of an electron in an Isolated atom and multiply it by $N$, which is the number of atoms. Then you compare this with the energy of the ground state of the electron Hamiltonian in the lattice. The ground state (neglecting interactions) will look like the electrons occupying the $N$ lowest energy states. So the energy of the lattice will be the sum of the energies of all these states, which will generally be lower than the energy of $N$ isolated atoms (otherwise we wouldn't have solids!). This difference will give us the cohesive energy.

Now let's try to do the tight binding problem. \begin{equation} H = -E_0\sum_n |n\rangle\langle n|-V\sum_n (|n\rangle\langle n+1| +|n+1\rangle\langle n|) \end{equation} This is an approximation to a system of non-interacting electrons in a 1D lattice. $-E_0$ is like the energy of an electron in an isolated atom and when you bring these atoms together the $V$ terms lets the electron `hop' between neighboring sites. But this changes the system drastically. Like you mentioned, we get the dispersion \begin{equation} E(k) = -E_0 - 2V \cos (ka) \end{equation} where $a$ is the lattice constant. This gives us the density of states $$ g(E) \sim \frac{1}{\sqrt{1-(\frac{E+E_0}{2V})^2}} $$ An understandable assumption here would be that each atom donates one electron to the system, so we have a total of $N$ electrons and $N$ atoms. But this means that the whole band is filled, and if you integrate to calculate the energy of this system, you will get exactly $-NE_0$, which is equal to the energy of $N$ isolated atoms. So the cohesive energy is zero, which means such a crystal will not form due to electron delocalization. Note: if you have learnt some Molecular Orbital theory, this is exactly why the molecule $He_2$ does not exist!

The idea of Fermi temperature is not very useful here since there are no electronic excitations possible but if one were to define such a quantity , it would be $T_f = \frac{2V}{k_B}$ since it should not depend on $E_0$, which simply shifts the energy of all states. So we choose it such that the lowest energy state is at $T=0$.

Why do we have metals, then? What did we miss? The thing we missed is that metals generally have free orbitals at energies close enough to the energy of the orbital (don't forget spin degeneracy!) the electron is in ($-E_0$), so we have additional bands where the electrons go in and reduce their energy.

One final comment on temperature - I am not sure about the exact definition of cohesive energy, but if you wanted to use cohesive energy to see if a solid forms at any finite temperature, you have to take entropy into account. The state that minimises the Helmholtz/Gibbs free energy will be observed.

Some nice textbooks for this would be The Oxford Solid State Basics by Steven Simon and the classic textbook by Ashcroft and Mermin.

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    $\begingroup$ If one takes into account spin shouldn't the band be half-full, since there are N states due to boundary conditions? (As I said implicitly in my question) I've consulted both texts but didn't arrive at a clear answer, I'll try to read more carefully $\endgroup$ Jun 19, 2020 at 21:44
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    $\begingroup$ Ahh, sorry. I thought you were doing the problem without spins. You're right. There will be two overlapping bands with the same dispersions and both the bands will be half filled in this case, if each atom contributes on electron. $\endgroup$ Jun 20, 2020 at 5:51

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