Timeline for Curvature of Hilbert space
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2018 at 13:28 | comment | added | Ernesto Lopez Fune | Hilbert spaces are isometrically-isomorphic to the space $\mathcal{l}^{2}(\mathds{C}).$ Each element $\psi$ of a Hilbert space has a natural coordinate representation $x_{i}$ in $\mathcal{l}^{2}(\mathds{C})$ once provided a basis $\{\phi_{i}\}:$ $x_{i}=\langle \psi,\phi_{i}\rangle.$ So, since the metric in $\mathcal{l}^{2}(\mathds{C})$ is similar to the Euclidian metric (except adding infinite number of terms), the curvature should be zero, however, this is just an estimate since one needs to study carefully the convergence in such infinite dimensional spaces. | |
| Jun 28, 2016 at 20:17 | comment | added | Peter R | How would you interpret a curved Hilbert space? It's a mathematical tool for quantum states. I don't see the geometric extension of the idea. | |
| Jun 28, 2016 at 17:42 | comment | added | Les Adieux | Since nobody wants to pay for it, I hacked it. | |
| Jun 28, 2016 at 17:17 | comment | added | vbj | Here's the link of the paper, worldscientific.com/doi/pdf/10.1142/S0217732393001148 | |
| Jun 28, 2016 at 17:11 | comment | added | vbj | I read a paper which attempts to define the curvature as a Gaussian curvature using Heisenberg-Weyl representation. It states that "The metric can be defined by considering a distance function as $D^2(|\psi_i\rangle,|\psi_j\rangle)=inf_{(\delta,\gamma)}|||\psi_i\rangle e^{\mathrm{i}\delta}-|\psi_j\rangle e^{\mathrm{i}\gamma}||$ and then considering a coherent space representation.". I am looking for the link of the paper. | |
| Jun 28, 2016 at 16:55 | comment | added | John Rennie | It would be interesting to see this fleshed out. e.g. what is the metric and how would we calculate the curvature from it? | |
| Jun 28, 2016 at 16:23 | history | answered | Quantumwhisp | CC BY-SA 3.0 |