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In Raman scattering, the molecule absorbs the photon into a virtual state, which doesn't actually exist. Unlike an excited state, the molecule can't stay in that state for longer than a time $\Delta t$ where $\Delta t \Delta E \leq \hbar/2$ - the Heisenberg uncertainty relation. A virtual state can have any energy level, though, and that's the reason for ...

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Make sure when you vary your first equation that you use a different index instead of $n$. Do the following: $$\frac{\delta U^{harm}}{\delta u(ma)}$$ this gives $$\frac{1}{2}K \sum_{n} 2 [u(na) - u([n+1]a)]\left[\frac{\delta u(na)}{\delta u(ma)} - \frac{\delta u([n+1]a)}{\delta u(ma)} \right]$$ by the chain rule. But $$\frac{\delta u(na)}{\delta ... 0 I think the difference between luminescence and Raman scattering lies in whether or not the mixed photon-molecule state maintains coherence with the exciting radiation. In Raman scattering, we imagine that coherence is maintained. In luminescence we imagine that coherence is disturbed by any one of a variety of interactions: for example, collisions with ... 0 To kick things off: you're right. Molecular vibrational excitations are exactly the same as phonon modes. We don't use that language very much because the system is too small (so we can't have things like travelling waves which have momentum, and we need to work only with stationary waves) but the analogy is indeed exact. Now, as to what exactly is ... 0 In this context, vibration usually refers to the relative motion of the nuclei. In a periodic solid (crystal), vibrational modes are called phonons. We can say, for example, that a normal vibrational mode of a branch s with wavevector \mathbf{k} is in its nth excited state, or equivalently, that there are n_{\mathbf{k}s} phonons of branch s with ... 4 In a crystal, each electron polarizes its vicinity, meaning positive charges prefer to stay nearer to the electron while negative charges move away from it. As the electron traverses through the crystal, it has to drag the polarization cloud with it, which can effectively increase or decrease its mass. Actually, the system "electron+polarization cloud" can ... 1 As far as I understand. The effective mass is given by a lattice. There is a periodic field that makes electrons move differently in comparison to free electrons. 0 It is easiest if you consider a rectangular box in 3D, with length L_x in the x-y plane, and length L_z \ll L_x in the z direction. Now, L_x is very large, and you are only interested in the physics deep within the bulk, far from the boundaries. Therefore it is of little importance exactly which boundary conditions you choose. It is often easier to ... 2 DFT is based on two important theorems: (1) Hohenberg & Kohn: the potential and the density are connected by a one-to-one map (2) Kohn & Sham: there is always a non-interacting reference system (map: V_xc: non-interacting <-> interacting problem) having the same density as the interacting one. In a nutshell: the potential and the density of the ... 0 I'm not sure if it's correct. However assuming that it is, keep in mind atoms in a metal are positively charged relative to the "electron sea". That being the case, intuition says if you have more atoms near eachother in a metal, you have larger positive charge density, which diminishes the electrostatic attraction (potential due to separation of charges) ... 0 It is the total H = \sum_k H_k invariant under time reversal 1 The mathematics is okay but how does this makes sense from the physical point of view? How does the dimension of the system dictate that in 3D we have more states with high energy where as in a 1D the system we get fewer and fewer states for increasing the energy? The density of states just comes from counting the states with energy less that E ... 1 Assuming that the expressions given for a 2D electron gas here are valid, there really is no contradiction. Remember that there is an implicit sum in a repeated index (Einstein summation convention). Equivalently, an inverse matrix need not be the component-by-component inverse of the original matrix. For example, we get:$$[\sigma\cdot\rho]_{xx} = ...

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Superconductors have both a critical temperature, at which they transition to the normal phase, and a critical applied magnetic field value. Once the applied magnetic field is at the critical value, a transition to normal occurs, regardless of the fact that the superconductor is below its critical temperature. The critical value of the applied magnetic ...

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Imagine you have a string of irregularly placed (that is, not evenly spaced) christmas lights, and you want to calculate the total power emitted by all of them. There are two ways you can do this. One way is to sum over all lights the power emitted by a light. Another way would be divide the string of lights into segments of length $\Delta x$, and then sum ...

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Ok, finally solved it in a very simple geometrical way. IF we take a square slanted lattice in the hexagonal lattice, like in the image, which is $N$ particles along each side. Then the number of particles inside is $N^2$. The volume of that area is just $V=(Na)^2\cos(30)$, and so: $$\rho a^2=\frac{2}{\sqrt 3}$$. I'm sorry for asking, I was frustrating ...

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I would just like to add this: in ordinary substances, we have the system in thermal equilibrium with the environment, and with a fixed number of particles. However we go to the grand canonical system for efficiency, where we do not have to restrict ourselves to a fixed particle number $N_{0}$, and that simplifies calculations enormously. In this, the system ...

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In real life most fractures occur at defects. Even such everyday materials as cement can have their strength increased many times by reducing the defect density within them. You'll often see claims for the incredible strength of nanostructures, but the strength is just due to the fact that these structures are free of defects. It's a lot easier to make a ...

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A very common approach for modelling the recombination dynamics in semiconductors is, $$R = An + Bnp$$ This equation assumes, Monomolecular recombination of electrons dominated over that of holes, with a rate $(s^{-1})$ given by the first term. Bimolecular recombination requires and electron and a hole. Clearly these assumptions will be material ...

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Start from the general definition, how this thing is set up in practice: You start from the mean energy of a system in contact with a thermal reservoir. The systems of the representative statistical ensemble are distributed over the entire number of possibilities, in accordance with the canonical ensemble $$P_i = C \exp(-\beta E_i) = \frac{\exp(-\beta ... 0 But what is the physical interpretation of the partition function and it's significance to Thermodynamics? I'm seeking a simple yet understandable intuition. The partition function has one simple physical interpretation in terms of Thermodynamic functions: Its natural log is proportional to the Free Energy (the proportionality constant is the ... -1 The significance is actually quite easy to explain at an elementary level. Namely, the partition function contains roughly all the information the thermodynamic system carries. For example, the expected energy of a system:$$\langle E\rangle =\sum E_i p_i=\sum_i\frac{E_ie^{-E_i/k_BT}}{e^{-E_i/k_BT}}=\frac{-1}{Z}\frac{\partial Z}{\partial (1/k_BT)}$$. The ... 1 Partition functions are a measure of the allowed volume in (microscopic-)configuration space for the system, and as such they are the normalizing function for probabilities expressed as volumes in configuration space (and assuming the applicability of the ergodic hypothesis). I know that this is very abstract, but it is also very general. 0 First I think that the hamiltonian is \hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right)^2, not \hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right). Then the remaining is just calculation. Using [\hat{p},A]=0, expanding hamiltonian, we get$$\hat{H}=\frac{1}{2m}(-\hbar^2 \frac{\partial^2}{\partial x^2}+\frac{e^2B^2}{c^2}x^2+\frac{2 \hbar e ...

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$$\newcommand{\ket}[1]{| #1 \rangle}$$ I'll try to answer the last two. with an arbitrary superposition, the probability density for the electron could be anything - can we actually find the coefficients of the superposition an electron actually is in? I'm a little confused about what you mean here. If we are given $\ket{\psi}$ as a combination of, ...

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Regarding your first two points: The symmetry axis of an orbital is free for a free atom. If it's bound to some other atom through one of these one-dimensionally elongated orbitals, the orientation of one orbital is fixed. If you take e.g. carbon, silicon or germanium, you have one s orbital and three p orbitals, which are oriented perpendicular to each ...

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Your second figure is a simplification of the first one, usually in the $\Gamma$ point, but it could be any other as well. Regarding your questions: There are multiple lines in valence and conduction band because there are several allowed bands or energy eigen states. Technically there is even an infinite number of allowed bands, but usually you would ...

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In any case, for undoped graphene, the Fermi level of electrons and holes are symmetric so the Fermi Energy lies at the Dirac Point, so these two definitions would be equivalent.

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Lets get something out of the way first: The threshold, or turn-on voltage, is not really an intrinsic device property per se. It originates more from a desire by circuit designers to have a rule of thumb about how much a diode has to be forward biased to get it into conduction mode. As such, one takes the inherently non-linear current vs voltage response ...

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Let us understand Brownian motion in liquids before we look at the motion in solids. If you observe a glass of water at rest on a table, it "appears" to be motionless. However, all we need is a magnifying glass to observe the random, incessant motion of water on the surface. This random motion is a manifestation of heat. The same thing happens in a solid. A ...

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Since there is no interaction between the parts of your lattice, the response of the total system is sum of responses of the individual parts. These parts are harmonic oscillators, whose response function is well-known. If these have many different frequencies, the total system will have response function with peaks at these frequencies. No need for boson ...

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