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## Hot answers tagged density-of-states

7

The "modes" in this case refer to the standing waves that can exist in a cavity. A very nice diagram / explanation is given at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html To summarize: if you consider a cavity of dimension $L$, the modes that fit inside the cavity have wave numbers $n_1$, $n_2$, $n_3$ such that $$n_1^2 + n_2^2 + n_3^3 = \... 7 You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is G_R(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2} You can ... 5 Maybe a picture like this would help? It shows the density of states along with the band structure. You get a spike in the DOS most of the time when one of the dispersion curves goes flat. I don't think that focusing on the origin is that instructive since that's a special (albeit important) case. Whenever the dispersion curve goes flat, you have "a lot" of ... 5 Suppose we have a drum. When you bang on the drum it will vibrate. When you look at any point on the surface of the drum head, you will see it go up and down, similar to if to take a mass hanging from a spring attached to the ceiling and watch it bob up and down. However, there is also an important difference between these two situations. In the situation ... 5 Essentially, I would like to prove$$ \sum_k f(k) \to \int f(k) \rho dE \tag{1}$$where$$ \rho = \frac{dk}{dE} \tag{2}$$is the density of states and k \to \infty. As mentioned in the comments, you need to introduce a measure on the LRS to get the dimensions to work out. To put it another way, your f(k) on the LHS can't be the same as your f(k)... 5 For the absolutely continuous part of the spectrum of a self-adjoint operator H, the "density of states" is provided by the Radon-Nikódym derivative of the spectral measure of HP_{ac} with respect to Lebesgue measure, where P_{ac} is the orthogonal projection onto the absolutely continuous subspace of the domain of H. This formula is well defined ... 4 In general, the density of states can be computed as follows: Find eigenstates of the Hamiltonian, \psi_{s}, so that H\psi_s=E_s\psi_s (except in special cases, this is usually the hardest part). Compute N(\epsilon), defined as the number of states \psi_s with energy E_s<\epsilon. D(\epsilon)=\frac{dN}{d\epsilon} is the density of states. ... 4 I am not sure I understand perfectly you question but formally in the canonical ensemble we can write the partition function Q(\beta) as being: $$Q(\beta) = \int\cdot \cdot \int d\mu(x)\: e^{-\beta H(x)} = \int_0^{+\infty} dE \: \rho(E)e^{-\beta E}$$ where d\mu(x) is the volume measure for the micro states in the system, H(... 4 Edited and simplified on behalf of the crowd (useless at this point of other very good answers): Consider a cube of edge length L in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength λ only if an integral number of half-wave cycles fit into an interval in the cube. In other words, ... 4 The continuum states are different in several aspects. First, there are a countably infinite number of bound molecular states vs an uncountably infinite number of states in any finite range of the continuum. Thus, the ratio of total number of bound states over the number of states in even a small range at the beginning of the continuum is effectively ... 3 The relationship between the density of states (DOS) and the dispersion can be more easily visualized if you imagine a discrete dispersion rather than a continuous one (which is exaggerated here for visualization). The image on the left shows the dispersion E(k) for a 1D parabolic band. If we wanted to visualize the density of states, what you would do ... 3 The partition function is not in general equal to one. It is a normalization constant, i.e. the probability of being in a configuration \{\sigma\} is$$ P(\{\sigma\}) = \frac{e^{-H(\{\sigma\})/k_B T}}{Z} $$where$$ Z(T) = \sum_{\{\sigma\}} e^{-H(\{\sigma\})/k_B T} $$Since multiple configuration can have the same energy, you can define (with a little ... 3 Calculating DOSs can be tricky business. It sounds like for your purposes, you don't have to be super accurate. So, I suggest using a quick and dirty way of calculating the DOS: Choose a bunch (at least a million, but feel free to go wild) \mathbf{q} points at random. For each one, find \omega\left(\mathbf{q}\right) Make a histogram of the \omega ... 2 I will present the result for the one-dimensional case only. It is readily generalized to your three-dimensional situation. Knowing the phonon dispersion \omega(q), the DOS can be defined as$$D(\omega) = \frac{1}{2\pi}\int\mathrm{d}q\,\delta(\omega-\omega(q)).$$Using the formal identity$$\delta(g(x)) = \sum_{i}\frac{\delta(x-x_i)}{|g'(x_i)|}$$where ... 2 Your confusion arises from the fact that you are confusing scalars and vectors. Scalars, are like numbers, and they have only magnitude. Vectors on the other hand have direction in addition to magnitude. In your question, you mention the wave vector, which, as its name suggests, is a vector. Typically vectors are written in bold or with an arrow over them; ... 2 The energy change of a dipole in a magnetic field is 2\mu H. So the larger this energy is, the fewer the number of states within a certain energy range. 2 It seems like the problem is just a misconception. OP's Eq. (1) really is not the density of one-particle states, but instead the density of one-particle modes. To obtain the real density of states, one indeed has to multiply the formula by 2. In rate calculations using this density in Fermi's golden rule, many common textbooks (like Sakurai's "Advanced ... 2 The formulas in Griffiths are correct, but the explanation is pretty clumsy, because he's basically done the derivation 'in reverse'. For simplicity I'll just talk about the distinguishable particle case, but the others are similar. The derivation in the forward direction looks like this: the Maxwell-Boltzmann distribution is the distribution that maximizes ... 2 To understand how to compute the density of states, you must first understand where it comes from. In statistical physics, we often have sums which look like this:$$\sum_{s}\sum_{\vec n}$$where s labels spin and \vec n is a vector of natural numbers labelling the quantum states, and it has as many components as the number of dimensions. We could ... 2 The density of states of one particle in 2D is$$ G(\varepsilon) = \int\!\!\int d^2\vec{p}\ d^2\vec{q}\ \delta\left(\varepsilon - H(\vec{p},\vec{q})\right) $$For the given Hamiltonian function this integral can be transformed to the following expression$$ G(\varepsilon) = \frac{2\pi^2m}{A}\int_0^\infty\!\! d\xi \int_0^{AR^2}\!\! d\eta\ \delta(\varepsilon -...

2

In step one, the angular part of the integral is $4\pi$ because the $\vec{k}$ integral is an integral in 3 dimensions, so you're integrating over a sphere: $$\int d\vec{k} = \underbrace{\int_{0}^{2\pi} \,d\phi \int_{0}^\pi \sin\theta \,d\theta }_{=\,4\pi} \int_0^\infty k^2 \,dk \,.$$ In step two, the lower bound of the integral can be extended to $-\infty$, ...

1

But in reality, when calculating the $N(\epsilon)$, we just divide the whole energy range into small steps with equal size. Then count the number of electrons whose energy lies in each small steps. After further scaling we can get the $N(\epsilon)$. I just want to know why is this holding. You are right. But then, if we want to know the number of electrons ...

1

$f(E)$ is the probability that a quantum state of energy $E$ is occupied. There are two quantum states (for two spin states) at each energy. The probability cannot be doubled, since that could then exceed 1. All that happens for a spin $1/2$ particle is that the number of available quantum states is doubled.

1

First off, let's answer the question: what part of that equation would change? $f\left(\epsilon\right)$ is always the same if you're in equilibrium. Scattering won't change that, because scattering alone won't move you out of equilibrium. If anything, scattering has the opposite effect. What can change is $Z\left(\epsilon\right)$. How it would change ...

1

The Fermi energy is the energy of the highest occupied state at absolute zero of a system of non-interacting (or mean-field interacting) fermions. So for a band structure, the Fermi energy is the energy of the highest-occupied level after you have filled al your bands with electrons. Note that each (spin) band can hold $N_{\textrm{cells}}$ electrons, where $... 1 This is all pretty standard stuff that you can find elsewhere on this site, but it is not always clearly presented and I feel like working through it again: Intuitively, anything that changes the dispersion relation$E(\vec{p})$also changes the density of states. This is just because we typically label the states by their momentum and assume there are two ... 1 I am not sure why you consider the maximum amplitude of any variable as a sensible measure of the number of states. Normally, the number of states is proportional to the energy of the system, here the area of the ellipse in phase space, so proportional to the product of the position and momentum maximum amplitudes. Absorb the units into x and p to make the ... 1 Absorbing the irrelevant ħω constants into the normalization of the suitable quantities, for the 3D isotropic oscillator,$\epsilon=n+3/2$, while for each n the degeneracy is$(n+1)(n+2)/2$; (see SE ). Scoping the power behavior of a large quasi-continuous n, leads you to the answer. The number of states then goes like$N\propto n^3 \propto \epsilon^3\$, ...

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