Timeline for decomposition of Hilbert space into tensor product $L^2([0,\tfrac{1}{2}]) \otimes L^2([\tfrac{1}{2},1]) \simeq L^2([0,1])$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2014 at 15:27 | comment | added | blackburne | After your edit, in the last line it should be $A\times B$, not $A\cup B$, no? | |
| Aug 1, 2014 at 13:00 | history | edited | john mangual | CC BY-SA 3.0 |
added 552 characters in body
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| Aug 1, 2014 at 5:03 | answer | added | blackburne | timeline score: 0 | |
| Aug 1, 2014 at 4:46 | answer | added | S. Carnahan♦ | timeline score: 4 | |
| Aug 1, 2014 at 0:42 | comment | added | Yemon Choi | I'm away from references right now, but this kind of "continuous tensor product" decomposition arises in certain models of Fock space as used in quantum probability - perhaps this is closer to what you're after? | |
| Jul 31, 2014 at 23:18 | comment | added | GH from MO | @GeraldEdgar: I agree. So the question, as it stands, does not make sense. | |
| Jul 31, 2014 at 21:39 | comment | added | Gerald Edgar | Well, of course all these spaces are isometric to $l^2$, so the symbol $\simeq$ is not saying much. The question, then is whether there is some "natural" way to identify them. | |
| Jul 31, 2014 at 21:39 | review | Close votes | |||
| Aug 1, 2014 at 16:03 | |||||
| Jul 31, 2014 at 20:55 | comment | added | Christian Remling | The tensor product of $L^2$ spaces $L^2(X)\otimes L^2(Y)$ is naturally identified with the space of square integrable (with respect to product measure) functions $f(x,y)$, and that is probably the structure you want here. | |
| Jul 31, 2014 at 20:50 | comment | added | john mangual | @ChristianRemling can you imagine why they get the tensor product for EE? Usually it is the wavefunction of two different particles, but I have been reading about the entanglement of two regions in space. | |
| Jul 31, 2014 at 20:46 | comment | added | Christian Remling | Since both sides from the equation in the title of your question are separable Hilbert spaces, they are trivially isomorphic. However, the tensor product is not naturally isomorphic to the RHS; you would normally identify $L^2(0,1)$ with the SUM $L^2(0,1/2)\oplus L^2(1/2,1)$. | |
| Jul 31, 2014 at 20:46 | comment | added | john mangual | possibly this is trivial since I can define a function $f$ on $[0,1]$ piecewise and take its Fourier series. | |
| Jul 31, 2014 at 20:41 | history | asked | john mangual | CC BY-SA 3.0 |