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June 14[edit]

Long run equilibrium of infinite-dimensional open quantum systems[edit]

To start off, consider the simplest quantum system: a two-dimensional Hilbert space, or "qubit". The classical states are the 0 and the 1. If the system is isolated, then there exists a unitary time evolution operator which can be represented as a 2x2 matrix, e.g.

$U={\begin{bmatrix}a&b\\-b&a\\\end{bmatrix}}$

which acts on $|0\rangle ={\begin{bmatrix}1\\0\\\end{bmatrix}}$ and $|1\rangle ={\begin{bmatrix}0\\1\\\end{bmatrix}}$ and linear combinations thereof. The state of the system after n time steps is found by applying $U^{n}$ to the initial state.

For example, $U^{2}|0\rangle =(a^{2}-b^{2})|0\rangle -2ab|1\rangle$ , and subsequently the probability of observing the system in the 0 state is $|a^{2}-b^{2}|^{2}$ .

When the system is "open", the operator U "keeps track of the history". Denote the history of a state as a sequence of 0s and 1s, for example, $|*0\rangle$ is a sequence whose most recent state is the 0 state. Denote the new operator $U_{o}$ .

Now, for instance, $U_{o}|*0\rangle =a|*00\rangle -b|*01\rangle$ , and ${U_{o}}^{2}|*0\rangle =a^{2}|*000\rangle -ab|*001\rangle -b^{2}|*010\rangle -ab|*011\rangle$ .

The difference is that $|*000\rangle$ and $|*010\rangle$ , although both representing the 0 state, have different histories and therefore the probability of observing the system in the 0 state is now $|a^{2}|^{2}+|-b^{2}|^{2}$ .

It can be shown that as you take $\lim _{n\rightarrow \infty }{U_{o}}^{n}$ , the probability of observing the qubit in the 0 state (the sum of norm-squared probability amplitudes of all sequences whose most recent element is the 0 state) converges on ${\frac {1}{2}}$ .

In fact, for a quantum system described by an m-dimensional Hilbert space, it can be shown that the probability of observing the system in any given state approaches ${\frac {1}{m}}$ --all states are equiprobable in the long run.

What I want to know is what happens when the quantum system is described by an infinite-dimensional Hilbert space, such as a quantum field. Is the long run probability distribution of states still uniform in that case? My guess is that the answer is no, but that there is a long run equilibrium which is different from the uniform distribution over possible states.--49.183.144.209 (talk) 11:57, 14 June 2019 (UTC)[reply]

It depends on the system: you're asking about the long-range stability of a dynamical system, e.g. its Lyapunov stability; and this is a qualitative property that we can quantify only if we know the system equation - in other words, the time-evolution operator.

Whether the system is quantized, or not-quantized, you want to conduct stability analysis on the system equation; the quantization might introduce non-linearith into this analysis; but the methods are the same. It's a fascinating field, and there are a lot of open research questions. Wikipedia has articles on asymptotic analysis, a list of complex analysis topics; and so on that might point you in the right direction. I'm trying to think of a specific quantum mechanics book that has a section on stability-analysis - maybe I can recall a good one after checking the bookshelf. But I don't think you even need a quantum-physics book; even a book on linear systems - like Proakis & Manolakis - presents extensive mathematical analysis tools to help you prove whether an arbitrary system, representable as an operator with an infinite impulse response, converges or diverges asymptotically.

Unless the your operator implies that the individual elements interact with each other's values, it won't matter whether your state-space has one, two, or an infinite number of elements; their long-range stability depends on the operator only.

If the elements do interact, you've entered that branch of mathematics where the solutions depend strongly on tiny perturbations of the initial conditions.

Nimur (talk) 15:26, 14 June 2019 (UTC)[reply]

Thanks for your answer. I think the most important point is that the long range stability depends on the operator only, not the initial state. If I understand you correctly, you can form a unistochastic matrix from the 2x2 unitary matrix, and this unistochastic matrix can be interpreted as the transition matrix of a Markov chain, and it has an equilibrium distribution that doesn't depend on the initial state, and that remains true even if the state space is infinite.

What do you mean by "elements interacting with each other's values"?--49.183.144.209 (talk) 16:46, 14 June 2019 (UTC)[reply]

By "interactions" between elements, I mean to say that the system is coupled, e.g. a "coupled system of differential equations," or, in other words if the next transition of a state-variable depends on the value of a different state-variable. In mathematical jargon, this is true if and only if the matrix representation of the state-space transition equation contains non-diagonal elements. Here's Decoupling Systems, part of the Linear Systems class from MIT's math department; they describe a few methods of complex analysis that allow you to diagonalize any system of ODEs, or to determine that it is non-diagonalizable using only real (non-complex) solutions.

Then, to prove whether the system is stable or unstable, we can simply check if the magnitude of the real part of the diagonalized matrix is strictly less than unity.

This whole process has some kind of passing analogue in the mandelbrot fractal generator, which is itself an example of a test of convergence- or divergence- on one particular system's time-evolution operator. In that case, a system with two initial conditions (the real- and the imaginary- parts of c at time zero) can behave quite in a complicated manner, even though the time-evolution operator is almost trivial.

Nimur (talk) 22:30, 14 June 2019 (UTC)[reply]

I have another question:

A unitary matrix can be transformed into a unistochastic matrix by norm-squaring each element of the matrix. A unistochastic matrix is a stochastic matrix and can therefore be interpreted as the transition matrix of a Markov chain. This transition matrix has an equilibrium distribution which does not depend on the initial state (and is, in fact, always the uniform distribution over the states).

Can a similar analysis be performed beginning with a unitary operator on an infinite-dimensional Hilbert space? Is it the case that the equilibrium distribution is still uniform over the states (my guess is no)? Is it still the case that the equilibrium distribution does not depend on the initial state (my guess is yes)?--49.183.144.209 (talk) 04:19, 15 June 2019 (UTC)[reply]

This sounds like a hard question... I think the same analytical methods apply; I think the equilibrium distribution depends on the transition matrix, and I am not sure if the additional qualification that it is unistochastic puts constraints on the outcome (but I bet it can be proven one way or the other!); and I can't answer the last part of your question because I was not able to answer the earlier bit. Mostly, though, I'm looking for a good book to help you make progress on these topics...

After perusing MathWorld for a bit, I found Horn's Theorem, which provides a strict condition for the existence of an Hermitian matrix with specified eigenvalues (... and the physicist in me is leaping to the end-goal here, which is to say that these are the asymptotal or steady-state solutions - there's at least a passing relation to what you're seeking);... our encyclopedia also has an entry on the Schur–Horn theorem, but when the mathematicians start introducing symplectic geometries, that all becomes a bit unfamiliar to me...

Nimur (talk) 16:10, 15 June 2019 (UTC)[reply]

Weight lifting question.[edit]

Is it bad to weight lift 2 days in a row? A lot of professional weight lifters like Bruce Lee do a Mon/Wed/Fri schedule or Tu/Th/Sat schedule. The logic is that muscles need 48 hours to heal. But what is the argument against weight lifting 2 days in a row? Because I also hear, weight lifting just makes muscle cells bigger, bu if you want new muscle cells to be created, you need to lift when you're sore, so that could be M/T and Th/Fr schedule. Shrug. 67.175.224.138 (talk) 15:41, 14 June 2019 (UTC).[reply]

This depends on how close to the limit you're exercising. If you're using the shocking method to lift yourself out of a plateau as Schwarzenegger explains here, then you'll need a larger resting period because a single bout of such an exercise session is going to be extremely intense. The next day you should exercise other muscles to avoid injury as the muscles you used yesterday won't be able to handle anything close to what you are normally used to. Count Iblis (talk) 16:22, 14 June 2019 (UTC)[reply]

I don't think that this question qualifies as medical advice, but it's nevertheless the kind of thing that you should consult a qualified person on. Get ahold of a professional trainer (pretty much every gym has them) to help direct your progress. You can definitely hurt yourself lifting incorrectly, but even without that, it's very likely that you won't get the desired outcome without the proper advice. I know a few semi-pros and they all swear by the advice trainers gave them. Matt Deres (talk) 22:35, 14 June 2019 (UTC)[reply]

Incandescent bulbs shining brighter[edit]

I stayed in a dorm when I went to university and each room had two desk areas, each with a desk lamp attached to the wall by way of an articulated arm. The bulbs were incandescent (this was in the 90s). Someone had heard of a neat trick involving such set ups, which turned out to be true. If you smack the shade of the light hard enough so that the entire system is shocked, the bulb will burn brighter - quite a bit brighter actually. None of us had access to a light meter, but I'd bet they were easily twice as bright. Unsurprisingly, some of the bulbs died shortly afterwards (I'll say, over the course of the next week or two), while others kept working fine for the rest of the semester. For whatever reason, that "experiment" recently came to my mind. What the heck was actually happening? I described the set up as best I can recall, in case that had something to do with it, but I presume the effect was caused by shocking the bulb itself...? When it gets right down to it, none of it makes much sense, but we - ahem - conducted several iterations of the test, almost all of which worked. Matt Deres (talk) 22:31, 14 June 2019 (UTC)[reply]

An Incandescent light bulb often has a coiled coil tungsten filament. When hot a mechanical shock could cause turns to weld together resulting in lower resistance and higher power since supply voltage is constant. DroneB (talk) 23:05, 14 June 2019 (UTC)[reply]

This question reminded me that as a kid I would try to "repair" clear glass, incandescent bulbs with broken filaments by turning them around and around until the broken ends crossed over. Applying power at that moment would cause a weld to occur, and the globe would work again. Never noticed whether the repaired ones were brighter or not. These would normally be single, "straight" wire filaments, not coiled ones. HiLo48 (talk) 00:13, 15 June 2019 (UTC)[reply]

Neat - thanks! That actually makes sense. Matt Deres (talk) 15:15, 15 June 2019 (UTC)[reply]