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Mathematical methods in classical mechanics, classical and quantum field theory, quantum mechanics, statistical mechanics, condensed matter, nuclear and atomic physics.
4
votes
1
answer
227
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Computational trick used in QFT and the Jones Polynomial
Given a compact, simple Lie group $G$, and a compact, oriented three manifold $M$, we can consider the following smooth homotopy groups:
$(C^\infty(M;G)/{\sim},\cdot )$: Here, $\sim$ is smooth …
1
vote
0
answers
83
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Characterization of the Subspace of Quasifree States of the CAR Algebra
Consider $\mathfrak U(\mathfrak H)$, the CAR algebra over a separable Hilbert space $\mathfrak H$. The states $E_{\mathfrak U}$ over this algebra are defined to be positive linear functionals of norm …
2
votes
1
answer
272
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Hessians on Kahler Manifolds
This is primarily a linear algebra question, but for motivation I want to state this question in its natural, global context. Whenever we have a non-relativistic quantum field theory (renormalized, o …
2
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0
answers
79
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Algebraic machinery for boundary conditions: may spectral data be "lifted" via the Toeplitz ...
Let $\tilde{\mathcal H}$ be a Hilbert space, and let $L(\tilde{\mathcal H})$ denote the corresponding space of linear operators. By fixing a basis, we can, via Fourier transform, identify an important …
3
votes
1
answer
125
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Classification of $2k$-vectors modulo orthogonal transformations
Consider the following chain $\{A_1,A_2,A_3,\cdots,A_{n}\}$ of orbit spaces of even-rank anti-symmetric tensors, where
$$A_k:=\frac{\Lambda^{2k}(\mathbb{R}^{2n})}{e_{i_1}\wedge \cdots \wedge e_{i_{2k} …
4
votes
1
answer
314
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Spurious length-scale cutoff emerges in propagator defined in Costello's "Renormalization an...
In page 9 of the introductory chaper of Renormalization and Effective Field Theory (the introductory chapter is available free here), Kevin Costello defines a propagator $P$ for the Laplace operator $ …
3
votes
Accepted
Why does the Bogolyubov transformation work? - In language of Clifford Algebras?
There is an elegant formulation of the Bogolyubov transformation in terms of Clifford algebras. Note that a quadratic Hamiltonian (noted by a hat), is a hermitian element of the representation of a Cl …
4
votes
1
answer
503
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Why does the Bogolyubov transformation work? - In language of Clifford Algebras?
Letting the standard Clifford algebra of dimension $2k$ be denoted by $Cl_{2k}$, let's denote the corresponding complex Clifford algebra via $$\mathbb{C}l_{2k}\equiv Cl_{2k}\otimes_{\mathbb{R}}\mathbb …