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Buzz
  • Member for 4 years
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  • The American South
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What is the relationship between the Dirac algebra and the Clifford algebra?
It is not that hard to show, by explicit computation, that a finite-dimensional irrep of a Clifford algebra with $n$ generators $\gamma^{\mu}$ (in the Dirac theory, $n$ is the dimensionality of spacetime) must be $m$-dimensional, where $m=2^{[n/2]}$. This is mostly a matter of showing that the fundamental anticommutation relations imply that the algebra contains at least $m^2$ linearly independent elements. That a $m\times m$ matrix irrep exists is then easy. (Dan Freedman had some good lecture notes on this.) This basically covers the universality for finite-dimensional representations.
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Why does Riesz's Representation Theorem apply in quantum mechanics?
Why do you think that the collection of all positive linear functions on $A$ is the actual object of interest?
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Rigorous Euler-Lagrange equations for fields
I don't think there is a rigorous derivation. For one thing, it is possible to construct systems in which the physically-correct time evolution does not correspond to a minimization of the action. It could be a maximum of the action or—more crucially—a saddle point of the action. But what does it mean, precisely speaking, to be a saddle point of the action? Well, the fields obey the Euler-Lagrange equations....
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Independence result where probabilistic intuition predicts the wrong answer?
@TimothyChow No, it's not quite the same, you're right. It does seem like a relevant point of comparison though, since it expressed a converse situation where what looks like a "probability 1" situation is not (necessarily) generic.
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