Wikipedia:Reference desk/Archives/Mathematics/2018 March 5

= March 5 =

volume in Hilbert space?
Is it customary to speak of the volume of some region of Hilbert space? I have no doubt one can concoct measures that make it possible, but I'm asking if it's done in practice. I'm particularly thinking of the notion that quantum states correspond to vectors on the unit sphere in Hilbert space. If n>7 or so, the volume of the unit n-ball decreases as n increases, with a limit of 0 as n approaches infinity. So would we say the unit sphere in Hilbert space has zero volume? What about the unit cube?

I thought about the unit sphere because if it had positive volume, one could say from the curse of dimensionality that the volume was entirely concentrated at the surface, so in some sense quantum states as unit vectors couldn't be distinguished from random points in the unit ball, and I wondered if that could mean anything physically. I don't actually know any QM so am trying to make some sense of it by understanding little aspects like this. Thanks.

173.228.123.121 (talk) 11:17, 5 March 2018 (UTC)
 * There is no extension of the Lebesgue measure to an infinite-dimensional Hilbert space other than the uninteresting assignment of either zero or infinity. --Jasper Deng (talk) 11:49, 5 March 2018 (UTC)


 * Jasper, thanks, that makes sense. (Edited): is that a theorem, about not being able to put a non-trivial measure on Hilbert space?  What about on the unit sphere or unit ball in Hilbert space?  Those are not themselves Hilbert spaces.  The thought is that picking a random point on the unit sphere (i.e. making a physical measurement under the Born rule) implies there's a probability measure on the unit sphere to pick from, and I guess the Born rule itself gives a measure. 173.228.123.121 (talk) 22:49, 5 March 2018 (UTC)
 * You can probably come up with a measure, but it won't satisfy the properties of a Lebesgue measure. That's all I'm saying. See . Basically, compact sets are hard to come by in an infinite-dimensional topological vector space.--Jasper Deng (talk) 03:32, 6 March 2018 (UTC)
 * Not that I understand this stuff, but we have an article about Infinite-dimensional Lebesgue measure's; it basically states the theorem above and gives an outline of a proof. It also points to some alternative measures which don't have all the properties of Lebesgue measures. (The Stack Exchange post points to the same article but it seems worth adding it here too.) --RDBury (talk) 06:53, 6 March 2018 (UTC)

Roots of cyclotomic polynomial for z^7 = 1
Our article Root of unity says, concerning the sixth-degree cyclotomic polynomial giving the primitive 7th roots of unity,


 * As 7 is not a Fermat prime, the seventh roots of unity are the first that require cube roots. There are 6 primitive seventh roots of unity; thus their computation involves solving a cubic polynomial, and therefore computing a cube root. The three real parts of these primitive roots are the roots of a cubic polynomial....


 * Why does this involve a cubic polynomial, rather than a 6th degree one?
 * How is this cubic polynomial found?

Thanks in advance, Loraof (talk) 16:41, 5 March 2018 (UTC)


 * After factoring out the obvious root of $$z = 1$$, the remaining roots form 3 conjugate pairs, with 3 distinct real parts - the roots of a cubic:
 * $$8 (\mathrm{Re}(z))^{3} + 4 (\mathrm{Re}(z))^{2} - 4 \mathrm{Re}(z) - 1 = 0$$


 * Use $$z = x + i y$$ and then $$y^{2} = 1 - x^{2}$$


 * $$\begin{align}0 & = z^{7} - 1 \\

& = (z - 1) (z^{6} + z^{5} + z^{4} + z^{3} + z^{2} + z + 1) \\ & = (x + i y - 1) ((x + i y)^{6} + (x + i y)^{5} + (x + i y)^{4} + (x + i y)^{3} + (x + i y)^{2} + x + i y + 1) \\ & = (x + i y - 1) (y (6 y^{4} x + y^{4} - 20 y^{2} x^{3} - 10 y^{2} x^{2} - 4 y^{2} x - y^{2} + 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1) i - y^{6} + x^{6} + 15 y^{4} x^{2} + 5 y^{4} x + y^{4} - 15 y^{2} x^{4} - 10 y^{2} x^{3} - 6 y^{2} x^{2} - 3 y^{2} x - y^{2} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1) \\ & = (x + i \sqrt{1 - x^{2}} - 1) (\sqrt{1 - x^{2}} (- 8 x^{3} - 4 x^{2} + 4 x + 1) (2 x + 1) (- 2 x + 1) i + x (- 4 x^{2} + 3) (- 8 x^{3} - 4 x^{2} + 4 x + 1)) \\ & = (- 8 x^{3} - 4 x^{2} + 4 x + 1) (x + i \sqrt{1 - x^{2}} - 1) (\sqrt{1 - x^{2}} (2 x + 1) (- 2 x + 1) i + x (- 4 x^{2} + 3)) \end{align}$$
 * --catslash (talk) 17:23, 5 March 2018 (UTC)


 * Wow, catslash, thanks for the detailed answer! Loraof (talk) 18:04, 5 March 2018 (UTC)