Why is the density of states required conceptually? Should it be seen as a mathematical trick related to Fourier series?

Question

[edit]: My misunderstanding is more precisely asked here: Density of states and boundary conditions: how the density of states is physical if it depends on box size :it was suggested to open a new post as the details of the question changed too much from this one.

My question is about the motivation of defining density of states. This is a notion I did not use since a while so I might be saying stupid mistakes here.

I take a simple example: density of states of $1D$ gas of electron.

If I want to describe a function defined in $[0,L]$ I can describe use Fourier series, considering the wavevectors $k_n=n \frac{2 \pi}{L}, n \in \mathbb{Z}$. My function will then be $L$-periodic, but as soon as the physics I am interested in is within $[0,L]$, I can encode an "arbitrary" shape playing with the Fourier coefficients.

Now, the energy of a free electron of wavelength $k$ is $E=\frac{\hbar^2 k^2}{2m}$.

The density of states in the $|k|$ space is: $L/(2\pi)*2*2$ where the first $*2$ is here to take in account the two possible spins for the electron, the second one is to take in account the two direction of propagations (as it is the modulus of $|k|$ space)

From this, I can deduce the density of states in the $E$ space:

$$\rho(E)=\frac{d|k|}{dE} \rho(k(E))=\sqrt{\frac{m}{2E}}\frac{1}{\hbar}*L/(2\pi)*4=\sqrt{\frac{m}{2 E \hbar^2}}\frac{2 L}{\pi}$$

Then we use this density of state $\rho(E)$ in order to compute various quantities depending on $E$ by the mean of an integral:

$$\int_0^{+\infty} dE\rho(E) f(E)$$

My questions

When using the density of states to compute quantities, an approximation is being done. Indeed we approximate a sum by an integral. This will be more and more correct the bigger the length $L$ is. But for any fixed length $L$ we are doing a "mistake" by integrating. Is that correct ?

The starting point of all this reasonning is a "mathematical trick". We want to use Fourier series to describe the physics. This comes at the cost of desribing the physics in a finite length space $[0,L]$ but this has the advantage to be easier to handle: we can easily find the density of states that then allow us to go into the continuum. Is that correct as well ?

Can we then say that the notion of density of state is just a mathematical tool that is introduced to go from an easy physical situation (the physics is discrete) to a more complicated scenario (continuum). But someone could come and start by a continuum description right from the beginning. Then, he would never need to use any density of states.

Answer 1

But for any fixed length L we are doing a "mistake" by integrating. Is that correct?

The word mistake suggests we don't mean to do it. We are introducing a useful (and often very, very good) approximation.

I don't really understand your second question. It tends to be easier to work with compact spaces rather than infinite ones and then take a limit when necessary, but in principle this is not necessary.

Can we then say that the notion of density of state is just a mathematical tool that is introduced to go from an easy physical situation (the physics is discrete) to a more complicated scenario (continuum).

First, I see this as being backward. Integration is far easier than summation from an operational point of view, and having a smoothly varying density of states is far simpler than a vast array of discrete points with infinitesimal spacing between them.

Second, the density of states is a deeply physical quantity which tells you how many states exist in a given range of energy (or momentum). It's not a purely mathematical trick by any means. As an example, when you perform scanning tunneling spectroscopy measurements, you directly observe the local density of states of a material by observing how rapidly the electron tunneling current $I$ changes respect to the applied voltage $V$; a large density of empty states $E>E_F$ which electrons can tunnel to (or a large density of filled states $E<E_f$ which electrons can tunnel from) means a large $dI/dV$.

Answer 2

TL; DR: DOS is converting summation over (multiple) quantum numbers into the (single) integral over energies

Normalization in a box
Normalization in a box $[0,L]$ is indeed a mathematical trick, but in the end one usually takes the limit $L\longrightarrow+\infty$. The confusion here may arise due to the fact that taking this limit is rarely done explicitly, as $L$ often cancels out from the final result. One could perform calculations without using the mathematical trick, working directly with the states normalized to delta-function, and defining the density-of-states as an integral: $$ D(E)=\int dk \delta\left(E-\epsilon(k)\right) $$

DOS is a physical quantity
As the answer by @JMurray correctly points out, density-of-states is a measurable physical quantity. In particular, it is measured on many kinds of spectroscopic measurements: those using electron spectroscopy, light spectroscopy or conductance measurements.

DOS as an object of study
Density-of-states is the central object of study in some physical theories, since it pretty much summarizes all the physics of the relevant phenomena. Notably, studies of localization and weak localization are often reduced to calculating DOS.

Ballistic transport
Another insight into the meaning of the density-of-state is ballistic transport. Transport through quantum dots often interpreted as measuring the density-of-states in the quantum dot (see, e.g., the classical Meir & Wingreen's paper).

Another important case is the conductance quantization in 1D channels, which results from the exact cancellation of the 1D density-of-states and the group velocity. This phenomenon is also behind the integer quantum Hall effect (which is also often analyzed in terms of DOS).

More general view of DOS
The main point of DOS is converting summation over (multiple) quantum numbers into the (single) integral over energies. This is extremely useful, since many physical quantities either depend only on energy (e.g., partition function) or can be made dependent only on energy via integration (e.g., integrating particle cross-section over angles).

Thus, given quantum numbers $k_1, k_2, ..., k_n, s_1, s_2, ..., s_m$, where $k_j$ are continuous and $s_j$ are discrete, we can write: $$ \int dk_1\int dk_2...\int dk_n\sum_{s_1,s_2,...,s_m}f\left(\epsilon_{s_1, s_2,..., s_m}(k_1, k_2,..., k_n)\right) = \int dE \rho(E)f(E), $$ from which the definition of $\rho(\epsilon)$ immediately follows: $$ \rho(E) = \int dk_1\int dk_2...\int dk_n\sum_{s_1,s_2,...,s_m}\delta\left(E - \epsilon_{s_1, s_2,..., s_m}(k_1, k_2,..., k_n)\right). $$

If we had only one continuous quantum number, $k$ and energy $\epsilon(k)$, the DOS would be simply the Jacobian: $$ \int dk f(\epsilon(k)) = 2\int dE f(E)\rho(E),\\ \rho(\epsilon) = 2\left|\frac{d\epsilon(k)}{dk}|_{\epsilon(k)=E}\right|^{-1}, $$ where factor $2$ appears, if we have symmetry $\epsilon(k)=\epsilon(-k)$ (in case of asymmetric branches, one has to sum over all of them).

Answer 3

One thing that the other answers (which all seem correct at a glance) seem to be missing is that OP seems to be conflating two ideas:

1. Density of states
2. Description in terms of wave-vectors $k$

The density of states is whatever function $g(\epsilon)$ makes the replacement

$$\sum_{\alpha} \mapsto \int d\epsilon \ g(\epsilon)$$

valid, where $\alpha$ runs over all the energy eigenstates. One option for this is of course $g(\epsilon)=\sum_{\alpha}\delta(\epsilon-\epsilon_\alpha)$ but in practice this might be smeared out a little whilst not hurting the approximation (eg convolve with a Gaussian of width $\Delta\ll kT$).

This is an approximation. We expect this approximation to become exact as the level spacing goes to zero under the assumption that the states do not become macroscopically occupied. This fails for instance for a Bose-Einstein condensate, where this replacement holds for all states but the condensate itself.

The above however is completely unrelated to labelling by wave-vectors $k$. The case typically treated is a free particle where the energy eigenstates are conveniently labelled by momentum $k$. Or a particle in a periodic potential where it is labelled by crystal momentum $k$. This isn't really to do with Fourier series per se, but simply a statement of what the good quantum numbers in the problem are.

Answer 4

The previous posts cover all the essentials and I just want to add small remarks.

As mentioned we approximate a sum over the k-states by the integral. In most cases it is a good approximation, since the functions we want to sum are "small". For example we might want to calculate the particle number for fermions by summing the Fermi function over the momenta. The case might be different when we consider bosons e.g. in 3D. The density of states neglects the ground state, since it vanishes as $\sqrt{E}$. But the ground state might be largely occupied due to the singularity of the bose function. Therefore in this case one has to separate the ground state from the sum first before introducing a density of states.

Another point of interest is what happens when interactions are involved. Here the spectral weight might shift and gets broaded. In the equilibrium this is encoded in the spectral function $A_k(\omega)$.

$ A_k(\omega) \propto \int dt e^{i\omega t} < \{c_k^\dagger (t), c_k \} > $

The occupation of a single state is then given as

$ <n_k> = \int d\omega \, f(\omega) A_k(\omega) $

From here we can define an interacting DOS as $\rho(\omega) = \sum_k A_k(\omega) $, so that the particle number is still given by the same formula. The spectral functions for non interacting particles are delta functions and we recover the usual formula for the DOS.

The interpretation of this DOS stays the same for fermions, but might change for bosons depending on the situation. The problem is that spectral functions for bosons are not positive definite and therefore the interacting DOS defined above is also not positive definite. This might happen in the Bose-Hubbard model or when different kinds of bosons interact.