Wikipedia:Reference desk/Archives/Mathematics/2018 June 26

= June 26 =

Generalizing to the continuous case, take II
A unistochastic matrix can be constructed from a unitary matrix by taking the norm-squared of each element of the unitary matrix.

Can this process be generalized to unitary operators on general Hilbert spaces, to create "unistochastic operators"? For convenience, call the map from a unitary matrix $$U$$ to a unistochastic matrix $$D$$, $$m$$, i.e. $$D=m(U)$$.

The motivation is this: As a unistochastic matrix is a stochastic matrix, it can be interpreted as the transition matrix of a Markov chain. Markov chains with unistochastic transition matrices have unique properties. I would like to see how these properties are preserved or changed when generalizing to general Markov processes (as opposed to finite-state).

This is what I've tried:

A unitary matrix $$U$$ can be written in the form $$U = e^{iH}$$ where $$H$$ is a Hermitian matrix, or self-adjoint operator in general. An example of a self adjoint operator on the space of functions is $$-\alpha\frac{d^2}{dx^2}$$.

$$U=e^{-i\alpha \frac{d^2}{dx^2}} = \sum_{k=0}^\infty \frac{1}{k!}(-i)^k\alpha^k \frac{d^{2k}}{dx^{2k}}$$

If we approximate $$\frac{d^{2k}}{dx^{2k}}f(x)$$ with $$\frac{1}{h^{2k}} \sum_{j = 0}^{2k} (-1)^j \binom{2k}{j} f(x + (2k - j) h)$$ (see finite difference)

then

$$Uf = \sum_{k=0}^\infty \sum_{j = 0}^{2k} \frac{1}{h^{2k}} \frac{1}{k!}(-i)^k\alpha^k (-1)^j \binom{2k}{j} f(x + (2k - j) h)$$

This sort of looks like an infinite matrix multiplication. My guess is that

$$Df = m(U)f = \lim_{h\rightarrow 0}\sum_{k=0}^\infty \sum_{j = 0}^{2k} \left(\frac{1}{h^{2k}} \frac{1}{k!}\alpha^k \binom{2k}{j}\right)^2 f(x + (2k - j) h)$$

but I have no idea how to evaluate this, and I could be completely barking up the wrong tree anyway. PeterPresent (talk) 09:17, 26 June 2018 (UTC)
 * I'm having a bit of trouble with the example you're giving, and perhaps that stems from an error in the Self-adjoint operator section in our article. First, I think you need to qualify the 'space of functions' a bit to get a Hilbert Space; say C∞ functions on [0, 1] or square integrable C∞ on (−∞, ∞) for example. Otherwise the inner product, which I'm assuming to be
 * $$\langle f|g\rangle = \int f \bar{g} dx$$
 * isn't defined. Second, that the operator $$-\alpha\frac{d^2}{dx^2}$$ is self-adjoint relies on integration by parts and the assumption that the value of the function is 0 at the endpoints. But for the space to be closed under the action of the operator you'd need to restrict to a subspace where all derivatives, or all even derivatives at least, are 0 at the endpoints. (Later on in the article they talk about a multiple the first derivative being an operator as well.) You could take the subspace of odd periodic C∞ functions on [0, 1] for example. In any case, the article seems to gloss over this point so it's not clear what space the operator is supposed to be acting on. Again I'm way out of my comfort zone here so maybe someone who deals with Hilbert space operators on a regular basis can chime in on this point.


 * Not sure about the overall question about whether the the operator U could be turned into some kind of stochastic matrix. I suppose the result would be a stochastic process of some kind, and it seems to me that it would be similar to Brownian Motion since the $$\frac{d^2}{dx^2}$$ appears in the diffusion equation. --RDBury (talk) 02:54, 28 June 2018 (UTC)
 * Thanks for your response. There is a dearth of information online about this. I think one of the reasons this is difficult to define is that $$m$$ is not invariant to a change of basis, i.e. $$m(P^{-1}UP) \ne P^{-1}m(U)P$$ in general. PeterPresent (talk) 15:57, 29 June 2018 (UTC)

Confusion about prime-generating Diophantine equation
The formula for primes article gives a Diophantine equation for generating primes, describing it as "a polynomial inequality in 26 variables, and the set of prime numbers is identical to the set of positive values taken on by the left-hand side as the variables a, b, …, z range over the nonnegative integers", the equation in question being:



\begin{align} & (k+2) (1 - {} \\[6pt] & [wz + h + j - q]^2 - {} \\[6pt] & [(gk + 2g + k + 1)(h + j) + h - z]^2 - {} \\[6pt] & [16(k + 1)^3(k + 2)(n + 1)^2 + 1 - f^2]^2 - {} \\[6pt] & [2n + p + q + z - e]^2 - {} \\[6pt] & [e^3(e + 2)(a + 1)^2 + 1 - o^2]^2 - {} \\[6pt] & [(a^2 - 1)y^2 + 1 - x^2]^2 - {} \\[6pt] & [16r^2y^4(a^2 - 1) + 1 - u^2]^2 - {} \\[6pt] & [n + \ell + v - y]^2 - {} \\[6pt] & [(a^2 - 1)\ell^2 + 1 - m^2]^2 - {} \\[6pt] & [ai + k + 1 - \ell - i]^2 - {} \\[6pt] & [((a + u^2(u^2 - a))^2 - 1)(n + 4dy)^2 + 1 - (x + cu)^2]^2 - {} \\[6pt] & [p + \ell(a - n - 1) + b(2an + 2a - n^2 - 2n - 2) - m]^2 - {} \\[6pt] & [q + y(a - p - 1) + s(2ap + 2a - p^2 - 2p - 2) - x]^2 - {} \\[6pt] & [z + p\ell(a - p) + t(2ap - p^2 - 1) - pm]^2) \\[6pt] \end{align} $$

But how can the left hand side ever be positive considering that (k+2) is being multiplied by one minus all of those terms? It just doesn't make sense! Earl of Arundel (talk) 17:52, 26 June 2018 (UTC)


 * It could if the terms being squared are all 0, which looking at the article right above this statement is what's really being sought. Writing it like this is apparently just a way to make it be expressed as an inequality.  –Deacon Vorbis (carbon &bull; videos) 18:06, 26 June 2018 (UTC)


 * Yes, that is just another way of saying that all of the square terms are 0. For example, the first equation in the set is:
 * $$\alpha_0= wz + h + j - q = 0$$

So when that is zero, the first square is 0. Bubba73 You talkin' to me? 18:12, 26 June 2018 (UTC)


 * I still don't see how this formula could be used to generate primes. The equation $$P(N) = N^2 + N + 41$$ produces primes by plugging in values for N on the interval [0, 39]. How could the Diophantine equation be so rearranged to work in that sort of manner? Earl of Arundel (talk) 19:23, 26 June 2018 (UTC)


 * It is not a practical way of generating primes because it is so hard to find a set of the 26 variables that will satisfy the equation. Bubba73 You talkin' to me? 19:59, 26 June 2018 (UTC)


 * I was afraid you were going to say that. Well, at least I learned something new today. :p Thanks for the prompt responses everyone. Cheers! Earl of Arundel (talk) 20:20, 26 June 2018 (UTC)


 * You might like the generating primes article. Bubba73 You talkin' to me? 20:27, 26 June 2018 (UTC)


 * Yes, sieving might be the answer although for extremely large primes it seems like it might not be very practical. I'm inclined to think randomly selecting primes of a certain form and then subjecting them to the Miller-Rabin test or what have you would be more efficient. Earl of Arundel (talk) 20:44, 26 June 2018 (UTC)


 * It depends on how many primes you need and how large they are. There is PrimeGrid which looks for large primes of certain forms.  Bubba73 You talkin' to me? 02:24, 27 June 2018 (UTC)