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How can particles that exist where quantum mechanics "reigns supreme" be modeled successfully as classical particles? If the model is useful, it gets wide popularity and sticks in physics irrespective of which framework - classical, probabilistic or quantum - it seems closest to. The idea that the quantum theory is some kind of ultimate theory of ...


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Well, $1/2\otimes1/2=0\oplus1$, so a system with two fermions has integer spin. But it is still a two fermion system, and therefore its wavefunction must be antisymmetric, as usual. This is not specific to Cooper pairs, but is basic Quantum Mechanics... [what is specific to Cooper pairs is that their size is $\gg a_0$, which means they are highly ...


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The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it ...


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@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ...


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Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ...


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Do you have access to a precision balance? Then you could weigh the plate, and using the known dimensions of the plate and the density of copper, compute the thickness. For $5\,{\rm cm} \times 5\,{\rm cm} \times 30\,\mu{\rm m}$ the weight would be $0.672\,{\rm g}$ for example. The precision of that measurement depends on how accurately you can measure the ...


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Because by expanding the sinus term into a taylor expansion, you get $\sin(x)\approx x - \frac{x^3}{6} +\cdots$ So, for small values of k you are allowed to take just the linear term.


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$k$ is the incoming beam, $k'$ is the reflected beam, expressed as wave vectors in the reciprocal lattice, which makes $Q=k-k'$ represents a particular plane in reciprocal space. If you assume that the diffracting beam is essentially a plane wave when it elastically scatters off of multiple sites within the crystal, the kinematics of the Laue equation ...


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The Drude model (1900) is based on statistical mechanics plus Maxwell's electrodynamics as updated by Lorentz to include the electron and atoms. It gives good results for some processes, such as conductivity/resistance of metals, but is way off on others, such heat capacity, and totally ignores important features such as band structure. The semi-classical ...


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The higher the dimension, the more phase space you have for your fluctuations to spread. Imagine you make a small local disturbance in your system (caused by e.g. thermal fluctuations). In 1D this disturbance can freely propagate in the system without decaying : it destroys long-range order. In 3D, the disturbance also propagates but it quickly dies out ...


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The force directly acting on particle $s$ in the harmonic approximation is proportional to the displacement from equilibrium, $u$: $$F = - k (x-x_{eq})=-k u$$ If we take into account only nearest-neighbours interactions we will have the total force acting on particle $s$ will depend on the total displacement from equilibrium on the right side ($u_r$) and ...


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In general your relation is $$ \vec{B}(\omega) = (1 + \chi_m(\omega))\vec{B}_0(\omega) $$ or in the time domain $$ \vec{B}(t) =\vec{B}_0(t) + \int\limits_{-\infty}^\infty \chi_m(t,t') \vec{B}_0(t') \;\rm{d}t' $$ Only in the case of instantanous material response, i.e. $\chi_m(t,t') = \chi_{m,0} \cdot \delta(t-t') $, your equation is correct. This already ...


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For any crystal, the First Brillouin Zone is found using the Wigner-Seitz construction for the reciprocal lattice. The high-symmetry points are labeled by certain letters mainly as a convention--like you said Gamma for (0,0,0) etc. The important thing to realize as far as the group theory, is that the group of the wavevector at the Gamma point has the full ...


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The summation given in the OP includes all points within a single unit cell; in this case there are $N$ such points. This looks like the expression for the structure factor, but the weighting for each point is unity, which implies that the crystal is made up of a single type of atom. Lattice sum, as given here, means the sum over all lattice points in the ...


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Both descriptions are correct; some people prefer the geometric description: the lattice of atoms is replace by a collection of planes, with different orientations. This corresponds to the Bragg model of partially reflective mirrors, and the K-vectors give the directions for the reflections which form the diffraction pattern. The description given by ...


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There are often several different methods of synthesizing materials, and in lots of cases they arrive at the same result. Sometimes the experiments you want to do will depend on your growth method, though: for instance, the polycrystalline samples you get from solid state reactions can be good for x-ray or neutron diffraction studies. If you want to ...


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The presence of a ductile to brittle transition temperature implies there are insufficient (ductile) deformation modes at low temperatures to support plastic deformation and therefore fracture occurs to release energy/load. In FCC materials, dislocation slip of both edge and screw dislocations is relatively athermal and due to the number of active slip ...



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