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5h |
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Coherent $U(N)$ intertwiners in Loop Quantum Gravity (LQG) and a measure on the Grassmannian @levitopher cont. In fact, the procedure adopted by Qmechanic is a part of the "Faddeev-Popov" procedure for reducing gauge symmetries, and in principle it can be completed to get a BRST invariant integrand. |
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5h |
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Coherent $U(N)$ intertwiners in Loop Quantum Gravity (LQG) and a measure on the Grassmannian @levitopher Yes, the $U(2)$ volume in this case is a component of the integration over $GL(2)$. In fact $Z$, or ( $u_1$ and $u_2$) are redundant for the Grassmannian, even with the constraints because they sum up to $2N-4$ coordinates, while the dimension of the Grassmannian is $2N-8$. Another way to look at it is that the constraints are gauge fixing conditions and $U(2)$ represents the gauge symmetry. This is exactly the meaning of equation 82 in the article. |
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1d |
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Coherent $U(N)$ intertwiners in Loop Quantum Gravity (LQG) and a measure on the Grassmannian deleted 1 characters in body |
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1d |
answered | Coherent $U(N)$ intertwiners in Loop Quantum Gravity (LQG) and a measure on the Grassmannian |
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May 13 |
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Parametrization of $U(N)$ non-linear sigma model deleted 2 characters in body |
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May 13 |
answered | Parametrization of $U(N)$ non-linear sigma model |
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May 9 |
awarded | Nice Answer |
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May 7 |
awarded | Enlightened |
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May 7 |
awarded | Nice Answer |
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Apr 29 |
revised |
What does the sum of two qubits tell about their correlations? added 22 characters in body |
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Apr 29 |
answered | What does the sum of two qubits tell about their correlations? |
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Apr 25 |
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What does the sum of two qubits tell about their correlations? Cont. What is important is that on each manifold type there is an explicit parameterization of the 4-dimensional state vector (in terms of 3, 4, and 5 parameters respectively). (The generic case is not written explicitly in the article but can be easily worked out). Given the state vector, the $S_z$ probability distribution can be explicitly computed for each type of orbit and compared to the experimental distribution |
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Apr 25 |
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What does the sum of two qubits tell about their correlations? The idea is to use the classification given in arxiv.org/abs/quant-ph/0006068 (Geometry of entangled states) by Marek Kuś Karol Žyczkowski. The manifolds of equal entanglement of the pure two-qubit system fall into three strata parameterized by a single parameter $\theta$. The nongeneric orbits are $\mathbb{R}P^3$ and $S^2 \times S^2 $ correspond to $\theta = \frac{\pi}{2}, 0 $ respectively. The generic orbits correspond to $0 < \theta <\frac{\pi}{2}$ are five dimensional twisted $S^2$ bundles over $\mathbb{R}P^3$. |
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Apr 25 |
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What does the sum of two qubits tell about their correlations? Will the special case of a pure (but otherwise arbitrarily entangled) composite system be of interest to you. |
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Apr 22 |
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Aharonov-Bohm Effect and Integer Quantum Hall Effect @ramanujan_dirac, Yes your explanation is correct and conforms with Laughlin's original argument. |
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Apr 22 |
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Aharonov-Bohm Effect and Integer Quantum Hall Effect @ramanujan_dirac: (cont.) The gauge transformation $\phi$ must be a true function on the circle, i.e. $\phi(0) = \phi(2\pi)$. Thus we can choose a (large i.e., nonhomotopic to the identity) gauge transformation $\phi = n \theta$ to remove the integer part of the flux, such that only its fractional part will enter the Schroedinger equation. |
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Apr 22 |
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Aharonov-Bohm Effect and Integer Quantum Hall Effect @ramanujan_dirac: The Schroedinger equation a particle moving on a circular ring is invariant under the gauge transformation $\psi\rightarrow e^{i\phi} \psi$, $ A_{\theta }= A_{\theta } + \frac{c \hbar}{eR} \partial_{\theta}\phi$. |
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Apr 17 |
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When can a global symmetry be gauged? @Tomáš Brauner: Yes, the obstruction to gauging can be understood intuitively by means of the Dirac's constraint theory. If the theory can be gauged, the currents couple to the gauge field. Since the time component of the gauge field is non-dynamical, the charge densities of the symmetry currents will become constraints, i.e., must become zero on the gauge surface which can be thought as a reformulation of the theory with gauge invariant fields. But, then how can the bracket of two vanishing quantities give a nonzero constant. |
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Apr 17 |
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When can a global symmetry be gauged? Tomáš Brauner: Regarding the remark by drake, Please see section 7.5 of the lecture notes by: Riccardo Rattazzi: itp.epfl.ch/files/content/sites/itp/files/groups/ITP-unit/…. @drake: The scalar Stueckelberg field is real and so is its shifting transformation. The extension to complex values, together with the fact that the Schrodinger Lagrangian is linear in the time derivatives make the shifting transformation anomalous. |
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Apr 17 |
answered | When can a global symmetry be gauged? |
