Recent content by RadonX
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Brillouin Zone Size: Debate Settled
Daveyrocket! Thank you very much! That's a great help. Just checking through our stuff to see if we get your (2pi)^3 result when multiplying the reciprocal cell by the real space cell. We get (2Pi)^2 in our case which makes sense for a 2d lattice area we think?- RadonX
- Post #3
- Forum: Advanced Physics Homework Help
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Brillouin Zone Size: Debate Settled
Hi guys, Me and a few of my coursemates are revising for a solid state exam but have hit a problem. Is the area of every 2D brillouin zone (independent of lattice type) (2Pi/a)^2? For a square lattice in 2D real space with lattice constant, a, the reciprocal lattice vectors can easily be found...- RadonX
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- Brillouin Brillouin zone
- Replies: 3
- Forum: Advanced Physics Homework Help
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What is the density of states in k-space?
I THINK I have it. For each axis we have a state of a different k value at every; \frac{2 \pi}{a}n So rearranging for n and summing up (integrating) over each axis we end up with; (\frac{a}{2 \pi})^3 \int d^{3} k The factor of 2 difference between this answer and the one quoted in my notes is...- RadonX
- Post #2
- Forum: Advanced Physics Homework Help
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What is the density of states in k-space?
Homework Statement I'm having a dumb moment in proving why the following is true: D(k)d3k=\frac{V}{4\pi^{3}}4\pi k^{2}dk Homework Equations NA The Attempt at a Solution I realize that the second part 4\pi k^{2}dk is an integration element d3k. But the density of states I can't...- RadonX
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- Density Density of states States
- Replies: 1
- Forum: Advanced Physics Homework Help
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Quick Q: First order perturbation theory derivation
Thank you very much! After your response I made a little headway on it late last night but in my tiredness began to worry myself over a problem that didn't exist! For the first term, although I saw where operating H_0 on \langle\phi_m| would give me my E_m^{(0)} term, I mistakenly kept trying...- RadonX
- Post #5
- Forum: Advanced Physics Homework Help
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Quick Q: First order perturbation theory derivation
Yep, I think so. I think they could be expressed |\psi_n^{(0)}> (That's if I'm understanding the perturbation theory correctly? That'd be the zeroth order approximation of the perturbed Hamiltonian, ie. UNperturbed yes?- RadonX
- Post #3
- Forum: Advanced Physics Homework Help
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Quick Q: First order perturbation theory derivation
Homework Statement Going over and over the perturbation theory in various textbooks, I feel that I've NEARLY cracked it. However, in following a particular derivation I fail to understand a particular step. Could anyone enlighten me on the following? Multiply |\psi^{1)_{n}>...- RadonX
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- Derivation First order Perturbation Perturbation theory Theory
- Replies: 4
- Forum: Advanced Physics Homework Help