| bio | website | mat.univie.ac.at/~neum |
|---|---|---|
| location | Vienna, Austria | |
| age | ||
| visits | member for | 1 year, 2 months |
| seen | Dec 8 '12 at 10:35 | |
| stats | profile views | 3,259 |
Some physics material I wrote:
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Dec 3 |
reviewed | Approve suggested edit on Stiffness tensor |
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Dec 3 |
answered | Why the Hamiltonian and the Lagrangian are used interchangeably in QFT perturbation calculations |
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Dec 3 |
answered | Time evolution of a reduced density matrix |
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Dec 3 |
awarded | Nice Question |
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Dec 3 |
reviewed | Approve suggested edit on Change in time period of pendulum |
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Dec 3 |
reviewed | Reject suggested edit on How does a comet form? |
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Dec 3 |
reviewed | Edit suggested edit on momentum four vector and dirac matrices |
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Dec 3 |
revised |
momentum four vector and dirac matrices I corrected your brackets. |
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Dec 3 |
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Relativistic Hamiltonian Formulations I only claimed that it does not reproduce the standard results. |
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Dec 3 |
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Is the momentum operator well-defined in the basis of standing waves? The Hilbert space of states for the square well problem only consists of the L^2 wave functions that vanish at the boundary (and outside). |
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Dec 3 |
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Is the momentum operator well-defined in the basis of standing waves? I had meant: Thus $(\hat p\psi)(x)$ is generally nonzero. Thus $\hat p$ does not preserve the boundary condition. |
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Dec 3 |
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Is the momentum operator well-defined in the basis of standing waves? @VladimirKalitvianski: I expressed it globally rather than locally; momentum is infinitesimal translation. if $psi(x)=0$ it doesnt always imply that $\psi'(x)=0$. Thus $\hat p(x)$ is generally nonzero at the boundary. |
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Dec 3 |
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Relativistic Hamiltonian Formulations @juanrga: e.g., hep-th/9601021 ''calculate certain elementary processes, including Compton scattering and Moeller scattering. These calculations lead to qualitative deviations from the usual scattering cross-sections'' |
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Dec 3 |
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Is the momentum operator well-defined in the basis of standing waves? @VladimirKalitvianski: There are two walls. If you translate to the right, the right boundary condition will be violated. |
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Dec 3 |
answered | What makes an equation an 'equation of motion'? |
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Dec 2 |
revised |
What is the meaning of the word “particle” in particle physics? added 16 characters in body |
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Dec 2 |
revised |
Is the momentum operator well-defined in the basis of standing waves? added remark on the lack of proerties of the momentum operator. |
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Dec 2 |
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What is the meaning of the word “particle” in particle physics? ... just completed. |
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Dec 2 |
answered | What is the meaning of the word “particle” in particle physics? |
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Dec 2 |
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What is the meaning of the word “particle” in particle physics? I read the article, and now understand how he is using the terminology. It is quite uncommon to phrase things the way he does, but with the proper interpretation, what he says corresponds to something correct. I'll write my own answer explaining this. |