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## Hot answers tagged density-of-states

8

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 ...

7

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 ...

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 $$n_1^2 + n_2^2 + n_3^3 = \... 6 OP's equality involving a delta function is probably easier to appreciate in its equivalent integrated form$$\begin{align}\int \!d\Omega(E) ~f(E) ~=~ &\int \!dE~D(E) ~f(E)\cr ~=~&\sum_{n} \int \frac{d^3k}{(2\pi)^3}f(\epsilon_n(\mathbf{k})),\end{align} $$where f(E) is an arbitrary function. 5 Essentially, I would like to prove$$ \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 ... 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 ... 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 ... 4 We can use both definitions to determine the density of state, D(E'), for a free particle of mass m confined to a volume V with energy E'. The free particle has wave vector \vec{k}' and energy E'=\frac{\hbar^2k'^2}{2m}. \Omega(E') is defined to be the number of states with energy E<E' over the total position-space volume, L^3. In k-... 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: \begin{equation} Q(\beta) = \int\cdot \cdot \int d\mu(x)\: e^{-\beta H(x)} = \int_0^{+\infty} dE \: \rho(E)e^{-\beta E} \end{equation} where d\mu(x) is the volume measure for the micro states in the system, H(... 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 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 ... 4 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}$$where s labels spin and \vec n is a vector of natural numbers labelling the quantum states, and it has as many components as the number of dimensions. We could ... 4 Explicitly computing the density of states for an otherwise free particle confined to an arbitrary volume V is not possible, simply because the spectrum of the Hamiltonian cannot be determined without more information about V. For example, if V is a rectilinear box the allowed energies would be different than if V were a sphere. However, this is not ... 3 I will present the result for the one-dimensional case only. It is readily generalized to your three-dimensional situation. Knowing the phonon dispersion \omega(q), the DOS can be defined as$$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)|}$$where ... 3 The partition function is not in general equal to one. It is a normalization constant, i.e. the probability of being in a configuration \{\sigma\} is$$ P(\{\sigma\}) = \frac{e^{-H(\{\sigma\})/k_B T}}{Z} $$where$$ Z(T) = \sum_{\{\sigma\}} e^{-H(\{\sigma\})/k_B T} $$Since multiple configuration can have the same energy, you can define (with a little ... 3 The relationship between the density of states (DOS) and the dispersion can be more easily visualized if you imagine a discrete dispersion rather than a continuous one (which is exaggerated here for visualization). The image on the left shows the dispersion E(k) for a 1D parabolic band. If we wanted to visualize the density of states, what you would do ... 3 In step one, the angular part of the integral is 4\pi because the \vec{k} integral is an integral in 3 dimensions, so you're integrating over a sphere:$$\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 \,.$$In step two, the lower bound of the integral can be extended to -\infty, ... 3 You can use the general formula for denisty-of-states$$N(E)=2\sum_\alpha\delta(E-E_\alpha)Now using our energy dispersion relation we get \begin{align*} N(E) &= 2\sum_{\alpha=n,k_x,k_y}\delta(E-E_\alpha)\\ &\approx 2\frac{L_xL_y}{(2\pi)^2}\int_{-\infty}^\infty dk_x \int_{-\infty}^\infty dk_y\sum_n\delta(E-E_n(k))\\ &= 2\frac{L_xL_y}{(2\pi)^2}\... 3 The density-of-states is given byD(E) \propto \int dk_x dk_y dk_z \delta\left[E-\epsilon(\mathbf{k})\right],$$where I omitted the prefactor for simplicity. The result of this integration depends on the form of the dispersion law \epsilon(\mathbf{k}). Assuming that the dispersion is isotropic (which is really rarely the case in real materials) we have ... 3 The internal energy if water is 80 calories per gram higher than that of ice, which represents a finite but incredibly small mass increase, as is clear from Einstein's relation E=mc^2. Otherwise the mass is constant. The weight depends also on the gravitational field, which you can assume to be constant over the volume of ice and water. All considered the ... 2 Black-body radiation is caused by electronic degrees of freedom in the black body radiator. An ideal black body has eDoFs resonating at any frequency relevant to the temperature range that you consider. Your conductor certainly is far from an ideal black body at low temperatures but maybe quite close at high temperatures. For example Tungsten wires behave ... 2 Absorbing the irrelevant ħω constants into the normalization of the suitable quantities, for the 3D isotropic oscillator, \epsilon=n+3/2, while for each n the degeneracy is (n+1)(n+2)/2; (see SE ). Scoping the power behavior of a large quasi-continuous n, leads you to the answer. The number of states then goes like N\propto n^3 \propto \epsilon^3, ... 2 Your confusion arises from the fact that you are confusing scalars and vectors. Scalars, are like numbers, and they have only magnitude. Vectors on the other hand have direction in addition to magnitude. In your question, you mention the wave vector, which, as its name suggests, is a vector. Typically vectors are written in bold or with an arrow over them; ... 2 You can't use$$Z = \int e^{-\beta E} dE$$because that integral weights every energy equally, when you really want to be weighting each energy by the number of microstates at tha energy - the "density of states'' \rho(E). You can think of \rho(E) as being like dN/dE, where dN is the number of states in the window [E, E + dE], so$$Z = \int e^{-...

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 ...

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