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New answers tagged many-body

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Yes. Given any Hamiltonian $H'$, let $C_1=1$ be the identity operator and $h_{11}:=H'$ be a $1\times 1$-matrix. (If you want to have $h$ take values in a field, the fact that you propose a discrete sum over states will be insufficient if the Hamiltonian involves operators with a continuous spectrum like the standard momentum operator. I see you also leave ...

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Then, there is the case that such an operator is defined on the full interval I assume that by "full interval" you mean the whole real line. First question: Do we then need any boundary conditions? Yes, as noted by Sam Bader, boundary conditions are part of the Hamiltonian. In my physics lecture we used so-called Born von Karmann boundary ...

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Briefly, no. Charged particle interaction is fundamentally a non-instantaneous photon exchange. The interaction can be written or expanded as the ''naive'' instantaneous interaction ($e^2/r$) plus a photon exchange portion which contains exactly a $-e^2/r$ term and a non-instantaneous (retarded in time) term. The author calls the first term the Coulomb ...

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It's good that you're considering questions like this; I find that this type of questions really forces a student to a deeper understanding of the math involved. Do we then need any boundary conditions? Yes, boundary conditions should be considered as part of the definition of the Hamiltonian and its domain. Different boundary conditions can result in ...

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any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true? No!! The eigenfunctions are Bloch waves $\psi(x) = u(x)e^{ikx}$, where $u$ is periodic (with the period of the lattice). But the product $\psi$ is not periodic (with the period of the lattice) unless $k=0$. I put up an example on Wikipedia ...

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I can offer a few comments on your project, although without solving your specific problem. I believe my outlook is rather different from yours. Historically, statistical mechanics arose from macroscopic thermodynamics; you cannot divorce the two. We can take Gibbs and Boltzmann as the creators of statistical mechanics, with a hat tip to Maxwell. For ...

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It means that every state (wavefunction) that has the same total energy has also the same probability, that is, if you have the system prepared in the same way many times, and measure the state, the probability of measuring a specific state is not a function of the energy. Is it saying all the wave functions are 'equal' to one another in the sense ...

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The bigger the crystal is, the closer-together those discrete values of k are. A real crystal might be 1cm long, or 100 million atoms. Then there would be 100 million little dots between the vertical bars on your figure. Those dots would be equally spaced with respect to the horizontal axis. It's too many dots to see, they blend together to look just like a ...

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$k$ is the wave vector of the valence band that the hole resides. $m_V$ is usually denoted as an effective mass of the hole. The effective mass is described in the band structure of the semiconductor. $E_{0,v}$ is most likely a first-approximation of the electrostatic energy of the lattice. In the first approximation, it just shifts the allowed quantized ...

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The Hilbert space of the states in this case is the Fock space. It is a linear space "constructed" by acting by the creation operators $a^\dagger_k$ on the vacuum state $|0\rangle$, which has the property $a_k|0\rangle=0$. All other states are related so that the commutation relations between $a,a^\dagger$ are satisfied. The individual states like ...

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Usually in many body theory these operators create and annihilate particles. There are different annihilation and creation operators for fermions and bosons (they obey different commutation relations). The states they act upon and the states "created" by them respect the required symmetries (antisymmetric for fermions, symmetric for bosons). The operators ...

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"Trivial" is used because the equation reduces to something a bit simpler. Take the equation $E_{\pm}=\pm \sqrt{\left(\frac{p^2}{2m}-\mu\right)^2+|\Delta|^2p^2}$ Setting $\Delta$ to $0$ gets rid of the last term, and making $\mu<0$ means that the first term will reduce to $$E_1=\pm \left(\frac{p^2}{2m}-\mu\right)$$ which is pretty simple. That's what ...

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