# Tag Info

5

Lee Smolin doesn't mean that the most fundemental physical theory can have no symmetry. What he means is that symmetry shouldn't be the guiding principle in discerning fundemental physical theories. While symmetry is mathematically useful, it doesn't provide a sufficient reason to accept a theory, this goes back to Leibniz's principle of sufficient reason. ...

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This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization. A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the ...

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There's no real symmetry argument that can explain why the $z$ component is zero. This is because it doesn't have to be: there can always be a uniform magnetic field added to the existing field, without affecting Maxwell's equations. They are all differential equations in the fields, like the following: $$\nabla\times\mathbf{B} = 0$$ which is Ampere's ...

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I think the original source of this claim is the famous unpublished paper of Luescher and Mack. Everyone's citing it. It is more rigorous mathematically and general (they don't assume parity) than Di Francesco. It starts on pages 1-2 of the manuscript. The proof below is basically the same proof, just with added details and a little bit different notation. ...

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I recently read a paper on the possible kinematics: http://scitation.aip.org/content/aip/journal/jmp/9/10/10.1063/1.1664490 It states that under the 3 assumptions they made, there were more then 10 possible Lie-Algebras (while they discarded one by heuristic arguments)

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The mass terms for the $\sigma$ and $\vec{ \pi }$ fields are, $$m _\sigma \sigma \sigma + m _\pi \vec{ \pi } \cdot \vec{ \pi }$$ You have two terms that are going to turn into each other under a symmetry transformation. Thus they need to have the same coefficient in order to remain invariant under the symmetry (feel free ...

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Your own source says Let it be given that the light travels from point A to point B. The whole argument is prefaced with the restriction that we have light emitted from A and received at B. In your example, you want light to start at A and go around B, which is an entirely different thing altogether and has nothing to do with Leibniz. Note also that ...

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There's an argument I think briefly mentioned in Aristotles Physics; where he argues that an object travels in a straight line since to veer would require a cause; he actually says for what reason would it move up or down or to the left? In another sense it's an argument from symmetry; actually this is related to Newtons First Law where cause is interpreted ...

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There is, in a sense, a way to 'guide' oneself to the equations of motion based on the symmetries. The form of mechanics most suitable for this purpose is Hamilton's principle - the system takes a path for which the action has a stationary value for variations with fixed endpoints: $$\delta S=0$$ $S$ is generally expressed as (under some parametrization of ...

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That a Hamiltonian preserves a symmetry means $$[H, C] = 0 \Rightarrow CHC^\dagger = CHC^{-1} = H$$ For a unitary symmetry operator $C$ (or anti-unitary if it is time reversal). The Hamiltonian of a crystalline condensed matter system written in terms of the Bloch matrix is: $$H = \sum_{\vec k} \psi^\dagger(\vec k) H(\vec k) \psi(\vec k)$$ Where ...

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No, you can not write this transformation as $c_{i\uparrow}^\dagger|0\rangle=c_{i\downarrow}^\dagger|0\rangle$, because $c_{i\uparrow}^\dagger|0\rangle$ and $c_{i\downarrow}^\dagger|0\rangle$ are two orthogonal quantum states: they can not be equal. The transformation $c_{i\uparrow}^\dagger|0\rangle\to c_{i\uparrow}|0\rangle$ you start with is also wrong, ...

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