DOS Formulas: Derivation & Comparison

[Douasing]

Hi,everyone,there is an important formula of DOS as follows,
D_{n}(E)=\frac{Ω}{(2\pi)^{3}}\int_{BZ}dk\delta[E-\epsilon_{n}(k)] (1)
On the other hand, another formula of DOS is often mentioned as follows,
D_{n}(E)=\frac{Ω}{(2\pi)^{3}}\int_{S}\frac{dS}{\nabla_{k}E(k)} (2)
But,how to derive the first formula in a simple and accesible way ?
Another question is how to derive the second formula according to the first one ?

 

[Simon Bridge]

1. what do you mean by "simple and accessible?" - suspect the answer to this question is "you can't".
2. To derive one formula from the other you need to know how they are related. Start with what each of the symbols means, and check that they are being used in equivalent circumstances.

 

[Douasing]

1. what do you mean by "simple and accessible?" - suspect the answer to this question is "you can't".
2. To derive one formula from the other you need to know how they are related. Start with what each of the symbols means, and check that they are being used in equivalent circumstances.

Thanks for your reply.

(1) Here is a discussion about "The definition of Density of States",which comes from "Physics Stack Exchange",and the link is
http://physics.stackexchange.com/questions/55243/the-definition-of-density-of-states
But I think the discussion is not very clear, and also, lack of some important details.

(2) Generally,the first formula is thought to be the definition of DOS, a more standard form is
D_{n}(E)=\frac{Ω}{(2\pi)^{3}}\int_{BZ}f(k)\delta[E-\epsilon_{n}(k)]dk
So I think the second one should be derived from the difinition of DOS, but I found a big difficulty. Maybe... my understanding is not very reasonable ?

 

[Simon Bridge]

So an accessible and simple description would be one that is incomplete?
I think my suspicion is born out.

 

[M Quack]

The density of states is basically the answer to the question "How many states are there with energy E?"

The formula more or less reads like this:

You go through your Brillouin zone (you vary k throughout the BZ), and each time you find a band that has energy E, you increase the density of states. (delta(E-epsilon(k)).

To this basic idea you have to add details like normalization, "going around" becomes and integral, etc.

I suppose this qualifies as an incomplete description :-)

 

[Simon Bridge]

Yah - the definiton of the density of states is the number of states with energy between E and E+dE
The exact calculation will depend on the setup.
That's what's missing from the two integrals in post #1 - what are they intended to be for.
OP needs to look more closely or refine the problem description to make headway.

 

[t!m]

I don't understand why some are being so pedantic about the OP's questions, which are entirely reasonable.

The first expression as written is indeed close to the definition of the density of states. M Quack gave a good physical interpretation in terms of integrating over k-space and adding to the DOS at each energy level. The prefactors come from taking the thermodynamic limit and converting a k-space sum into an integral.

The second expression essentially follows from the first by vector calculus. In one dimension, there is a useful relation for delta functions with functions as arguments,
\int dx\ \delta(f(x)) = \sum_{x_0}\frac{1}{\left|f^\prime(x_0)\right|}
where the sum is over all points x_0 satisfying f(x_0) = 0. Generalizing this to the case here, observe that all the points k_0 satisfying E-E_n(k_0)=0 define a continuous, constant energy surface in k-space. Calling that surface S, we have
D_n(E) = \frac{\Omega}{(2\pi)^3} \int d^3k \delta(E-E_n(k)) = \frac{\Omega}{(2\pi)^3} \int_S dS \frac{1}{|\nabla_k E_n(k)|}

More discussion can be found in Ashcroft and Mermin, and the latter definition in particular when discussing "van Hove singularities."

 

[Simon Bridge]

I don't understand why some are being so pedantic about the OP's questions, which are entirely reasonable.

... the idea is to guide OP into a better understanding of what has been written down in post #1 through Op's own efforts.

The question is "how do you get from one equation to the other?"

OP has also asked that the reply be "accessible and simple" ... the example of what this means was criticized as lacking "some (unspecified) important details" ... so it is reasonable to ask OP to refine the question further.

To best answer - we, at least, need some idea of the OPs understanding of the equations - since this has not been forthcoming, fair enough to redirect to a resource like a textbook ;)

 

[t!m]

Completely agree, well-formed questions with knowledge of the OPs background is ideal, but I felt there was enough in the post to start discussing specifics.

To derive one formula from the other you need to know how they are related. Start with what each of the symbols means, and check that they are being used in equivalent circumstances.

could apply to any question about anything. And since this isn't the Homework forum, I assume there's nothing wrong with hopefully pointing the OP towards the answer he/she seeks. But best not derail the post; hopefully the OP returns with more questions if I've mis-judged his/her background.

 

[Douasing]

You go through your Brillouin zone (you vary k throughout the BZ), and each time you find a band that has energy E, you increase the density of states. (delta(E-epsilon(k)).

I suppose "these qualifies" just as a specific description of my so-called "accessible and simple way" ,not only "as an incomplete description" .

Many a time,"derivation" may be as a clear physical description other than "onerous formulas".

The second expression essentially follows from the first by vector calculus.

But, tim's method of derivation reminds me that the "onerous formulas" could not be a difficult thing.


By the way,several times, when I encountered that " first interpolating E and fon a finer k-point grid using the trilinear method.Then for each E_{n}(k) on the finer grid the nearest E is found and f(k) is accumulated in D_{n}(E). ",I always wanted to suspect which formula is the best choice.Yeah,that was indeed my original meaning,but actually,I gain more than that.Thanks to both of you.