# Tag Info

12

Another use of the SLE approach seems to be (I haven't read the papers below much beyond their abstracts) as a tool to probe for the presence of conformal invariance in various systems, when a direct (numerical or experimental) verification is difficult. In this approach, (i) one extracts suitable non self-crossing paths, (ii) one determines (empirically) ...

11

SLEs can be used in order to define a certain kind of QFT. You can check M. Douglas' talk, from page 28 forward: Foundations of Quantum Field Theory (PDF). There's also another great article, Conformal invariance and 2D statistical physics. You may also like SLE and the free field: Partition functions and couplings. Finally, there's an approach to try and ...

10

Probabilites are (squares of) probability amplitudes, which can be obtained as inner products of vectors on Hilbert space: $\langle X|Y \rangle$. Under a transformation U, the ket transforms as $$|Y\rangle \rightarrow |Y'\rangle = U |Y\rangle$$ and the bra as $$\langle X| \rightarrow \langle X'| = \langle X| U^{\dagger}$$ So if U is unitary, ...

10

Matty Hoban pointed me to a paper (PDF here) by Itamar Pitowsky from 1991 which looks the geometry of correlation polytopes and their symmetries. I haven't read the paper in full, but glancing through it, on page 400 (page 6 of the actual paper) under the statement of results the author seems to say that the cardinality of the symmetry group is $n! 2^n$ ...

10

This relationship is called the Born rule and it's one of the postulates of quantum mechanics. There have been various attempts to derive it, but none of them having been terribly convincing so at the moment we have to assume it is true. Fortunately experiment supports this assumption. The rule was originally suggested by Max Born (hence the name) in 1926 ...

8

With quantum mechanics, you have to ask your questions very, very carefully. Is it possible to have a Sun-sized star in your pocket? It depends on what you mean by "in". Do all of the atoms of the star need to be entirely in your pocket, or is it sufficient that some part of each atom's wave function be inside your pocket? It is possible to have any ...

8

If you want the high school answer, then no, the numbering does not matter. If all faces are equally-likely, the probability is the same regardless of how you number the die, and similarly all derived quantities (such as variance or probability to be greater than 8) are the same because the underlying distribution is the same. If you want the real answer, ...

8

There are two that I know of in the context of state estimation. The first is for estimating the mean of $P$ and is a Metropolis-Hasting MCMC algorithm here: Optimal, reliable estimation of quantum states. The second is also mainly for computing the mean (but can do other functions -- including the characteristic function of the region you are interested ...

7

You're quite correct that assuming space is continuous there is an infinite way of arranging the atoms that make up e.g. me. But there are two reasons why a copy of me isn't infinitely unlikely. Firstly on a practical level, if you moved around some of the atoms in me no-one would notice, so you don't have to match me exactly. That means somewhere in an ...

7

If $p=1$ and $d$ is non-empty, then clearly the random walk is not transient, because since you use the $\ell^1$ norm, there is a closed surface which always reflects the walker back into the enclosed finite region. If, however, $p = 0$ then the walk is transient, since simple random walks in 3 dimensions are transient. Now, if there is a $p_c<1$ its ...

7

I think a reasonable first approximation can be made like this: choose an arbitrary orientation for the die, and figure out, if the die were released in that orientation with its lowermost point resting on a surface, which side would it fall on? That can be easily calculated; you just draw a line going straight down from the center of mass, and whichever ...

7

You might like this 110-page paper by me and Alex Arkhipov, which is all about a quantum bosonic analogue of Galton's board (we even use the same graphic you did -- see Section 1.1!). In particular, we gave strong evidence that such a board (with an arbitrary configuration of "pegs," and with multiple entry points for the "balls") is exponentially hard even ...

7

Standard deviation adds uncertainties to the measured value: $23.3\pm 0.4\,{\rm m}$. One can quickly look at the error (which has units of ${\rm m}$ in my case) and think, The value could be as low as $22.9\,{\rm m}$ or as high as $23.7\,{\rm m}$ without much thinking. Modifying this to being a percentage of the value would be confusing. Plus it would be ...

6

This answer is really just an amplification of Yvan Velenik's answer, as I feel what the direction of study he briefly mentioned deserves a bit more space here. A remarkable 2006 Nature Physics paper by Bernard et al kicked off the little subject of finding SLE-like curves in turbulent systems -- here, the authors did a large numerical study of 2D ...

6

I am currently dealing with generalized probabilistic theories too, and I had the same problem. For example, I read papers like this one: http://arxiv.org/abs/1012.1215 I don't know a good online reference for this kind of math, but a book that I liked reading very much was Charalambos D. Aliprantis and Rabee Tourky: Cones and Duality This book helped me ...

6

All concepts one works with in theoretical physcis are idealizations. One needs idealizations in order to treat a system in a mathematically adequate way. In particular, probabilities, are like any measurement values idealized objects that can take arbitrary real numbers (in [0.1] for probabilities) as values. Restricting them to be rational (as relative ...

6

INCLUDING AN EXTENSION $\psi_o$ is, as mentioned previously, the normalisation constant which is calculated by doing the integral $\int_V|\psi|^2dV$ and setting its value equal to 1 (hence normalization). This will give you the equation for $\psi_o$. If your interest is to find the probablity amplitude for a particle in a volume V, for example, then you ...

6

Before trying to understand quantum mechanics proper, I think it's helpful to try to understand the general idea of its statistics and probability. There are basically two kinds of mathematical systems that can yield a nontrivial formalism for probability. One is the kind we're familiar with from everyday life: each outcome has a probability, and those ...

6

Part of you problem is "Probability amplitude is the square root of the probability [...]" The amplitude is a complex number whose amplitude is the probability. That is $\psi^* \psi = P$ where the asterisk superscript means the complex conjugate.1 It may seem a little pedantic to make this distinction because so far the "complex phase" of the ...

5

An application of calculation based on SLE to condensed matter physics is descibed in this talk entitled Conformal Restriction in Condensed Matter Physics by Eldad Batelheim

5

Are you looking for a proof? If so, this link (which has some sign errors as pointed out in the comments) proves it as follows (without the sign errors): We start by differentiating the definition of the probability with respect to time only:  \frac{\partial P(x,t)}{\partial t} = \frac{\partial}{\partial t}\left (\psi^*(x,t) \psi(x,t)\right) = \left[ ...

5

Why is it that the amplitude of a probability wave is the sign of "a single particle"? Take a spring or coil of any kind. Look at it from the side, or even better, project a shadow of it onto a piece of paper. In both cases you will see what looks like a sinusoidal wave complete with peaks, troughs, and zero points. But the spring has a smooth, constant ...

5

As you mention in your question, this is the definition of an independent event. If the coin is indeed fair, then it is irrelevant what has happened in the past in order to determine the probability of future events. This does not change in quantum mechanics. In fact, QM gives us the first concept of truly random events. As an explicit example of a quantum ...

5

The physics aspect of your question is answered by saying that quantum mechanics doesn't change anything in this situation. The conclusions you get from basic statistics continue to be valid. Now, as for why that is: I have a feeling that: $P(h | 100 h) < P(t | 100 h)$ because of the push towards equilibrium in the entropy of the system. Am I wrong ...

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