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Address Mobile Tech Unit, Macon, GA 31204 (478) 569-6575 http://www.lomaintech.com

confidence interval standard error estimate Bolingbroke, Georgia

As shown in Figure 2, the value is 1.96. Hyattsville, MD: U.S. These limits were computed by adding and subtracting 1.96 standard deviations to/from the mean of 90 as follows: 90 - (1.96)(12) = 66.48 90 + (1.96)(12) = 113.52 The value Therefore the confidence interval is computed as follows: Lower limit = 16.362 - (2.013)(1.090) = 14.17 Upper limit = 16.362 + (2.013)(1.090) = 18.56 Therefore, the interference effect (difference) for the

This probability is small, so the observation probably did not come from the same population as the 140 other children. However, with smaller sample sizes, the t distribution is leptokurtic, which means it has relatively more scores in its tails than does the normal distribution. However, the mean and standard deviation are descriptive statistics, whereas the standard error of the mean describes bounds on a random sampling process. Review of the use of statistics in Infection and Immunity.

The age data are in the data set run10 from the R package openintro that accompanies the textbook by Dietz [4] The graph shows the distribution of ages for the runners. Confidence intervals The means and their standard errors can be treated in a similar fashion. Dividing the difference by the standard deviation gives 2.62/0.87 = 3.01. The ages in one such sample are 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55.

Later in this section we will show how to compute a confidence interval for the mean when σ has to be estimated. Because the age of the runners have a larger standard deviation (9.27 years) than does the age at first marriage (4.72 years), the standard error of the mean is larger for n is the size (number of observations) of the sample. If σ is known, the standard error is calculated using the formula σ x ¯   = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the

The confidence interval is then computed just as it is when σM. We know that 95% of these intervals will include the population parameter. The graph shows the ages for the 16 runners in the sample, plotted on the distribution of ages for all 9,732 runners. The standard error of the mean of one sample is an estimate of the standard deviation that would be obtained from the means of a large number of samples drawn from

Table 2. A practical result: Decreasing the uncertainty in a mean value estimate by a factor of two requires acquiring four times as many observations in the sample. All such quantities have uncertainty due to sampling variation, and for all such estimates a standard error can be calculated to indicate the degree of uncertainty.In many publications a ± sign In general, you compute the 95% confidence interval for the mean with the following formula: Lower limit = M - Z.95σM Upper limit = M + Z.95σM where Z.95 is the

Figure 1. These measurements average $$\bar x$$ = 71492 kilometers with a standard deviation of s = 28 kilometers. The values of t to be used in a confidence interval can be looked up in a table of the t distribution. The first step is to obtain the Z value corresponding to the reported P value from a table of the standard normal distribution.

Generated Wed, 05 Oct 2016 07:22:13 GMT by s_hv977 (squid/3.5.20) When the true underlying distribution is known to be Gaussian, although with unknown σ, then the resulting estimated distribution follows the Student t-distribution. Thus in the 140 children we might choose to exclude the three highest and three lowest values. Assumptions and usage Further information: Confidence interval If its sampling distribution is normally distributed, the sample mean, its standard error, and the quantiles of the normal distribution can be used to

The shaded area represents the middle 95% of the distribution and stretches from 66.48 to 113.52. Normal Distribution Calculator The confidence interval can then be computed as follows: Lower limit = 5 - (1.96)(1.118)= 2.81 Upper limit = 5 + (1.96)(1.118)= 7.19 You should use the t Assume that the weights of 10-year-old children are normally distributed with a mean of 90 and a standard deviation of 36. However, to explain how confidence intervals are constructed, we are going to work backwards and begin by assuming characteristics of the population.

Faculty login (PSU Access Account) Lessons Lesson 2: Statistics: Benefits, Risks, and Measurements Lesson 3: Characteristics of Good Sample Surveys and Comparative Studies Lesson 4: Getting the Big Picture and Summaries Compare the true standard error of the mean to the standard error estimated using this sample. A natural way to describe the variation of these sample means around the true population mean is the standard deviation of the distribution of the sample means. There is much confusion over the interpretation of the probability attached to confidence intervals.

Assume that the weights of 10-year-old children are normally distributed with a mean of 90 and a standard deviation of 36. When the sample size is smaller (say n < 30), then s will be fairly different from $$\sigma$$ for some samples - and that means that we we need a bigger The sample mean plus or minus 1.96 times its standard error gives the following two figures: This is called the 95% confidence interval , and we can say that there is If you look closely at this formula for a confidence interval, you will notice that you need to know the standard deviation (σ) in order to estimate the mean.

Please try the request again. Two data sets will be helpful to illustrate the concept of a sampling distribution and its use to calculate the standard error. The notation for standard error can be any one of SE, SEM (for standard error of measurement or mean), or SE. Copyright © 2016 The Pennsylvania State University Privacy and Legal Statements Contact the Department of Statistics Online Programs ERROR The requested URL could not be retrieved The following error was encountered

Confidence intervals provide the key to a useful device for arguing from a sample back to the population from which it came. In this scenario, the 400 patients are a sample of all patients who may be treated with the drug. If we draw a series of samples and calculate the mean of the observations in each, we have a series of means. Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some

This is also the standard error of the percentage of female patients with appendicitis, since the formula remains the same if p is replaced by 100-p.


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# Physical meaning of elliptic-hyperbolic transitions

+ 4 like - 0 dislike
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I am interested in the interpretation of PDE that are not strictly elliptic, parabolic,or hyperbolic: ones that change type in sections of the domain, e.g. the generalised Tricomi equation

$A(x) u_{,xx} + u_{,yy} = 0$

will be elliptic in regions where $A(x) \geq 0$ and hyperbolic when $A(x) < 0$.

I am aware there is a great deal of literature surrounding the topic (e.g. transonic flows), but mostly regarding well-posedness and such, not so much about the actual physics of what is going on.

For a start, I usually think of hyperbolic equations as having wave-like components in their solutions -- this usually means there is some kind of 'time' involved. What does it mean for a PDE in spatial variables to be hyperbolic? It seems weird to me to think of 'travelling waves' in the $(x,y)$ plane. What about an elliptic PDE involving both spatial and temporal variables?

Can someone explain what is physically happening here? In the non-linear case [$A(x) \rightarrow A(x,u)$] does this kind of elliptic-hyperbolic transition imply something about shocks?

As a simple example, if we have $A(x) =x$ and the domain is something like $D = \{x,y: -a \leq x \leq a, -b \leq y \leq b\}$, what are the 'wave-like' portions doing in the $x < 0$ region? What physics are they conveying?

Do you know of a case of the generalised Tricomi equation that arises in physics? If so, please point to one. If not, why do you expect a physical interpretation exists?

A fair comment, Arnold. The actual example I am interested in has a somewhat complicated backstory, so I have just written the Tricomi equation as a simple example. Here is a GR example from Stewart, [J. M. Stewart, Classical Quant. Grav. 18 (2001) -- unfortunately paywalled]: Consider the line element:

$ds^2 = -dt^2 + d \rho^2 + \rho^2 \left( d \phi - \Omega d t \right)^{2} + dz^2$, for constant $\Omega$.

Then the Laplace equation for a static variable $u(\rho,\phi,z)$ reads

$\rho^{-1} \left( \rho u_{,\rho} \right)_{,\rho} + \left( \rho^{-2} - \Omega^2 \right) u_{,\phi \phi} + u_{,zz} = 0$,

which is hyperbolic for large $\rho$, and elliptic for small $\rho$. Again -- I don't understand what could be meant by 'wave-like' behaviour in spatial variables.

This particular (simple) example suffers in the respect that there exists a frame which makes the problem purely elliptic -- but you can easily imagine a situation where that is not possible.

Still looking for an answer to this; another example that may provide additional context. If one takes the time independent Euler and Continuity equations of fluid dynamics and linearises about a background which is rigidly rotating, it can be shown that the only remaining equation of motion reads

$u_{,xx} + u_{,yy} + \left( 1 - M^2 \right) u_{,zz} = 0$,

where the velocity field is irrotational and we have written $v^{\alpha} = \nabla^{\alpha} u$, and $M$ is the Mach number (see e.g. Bateman 1929). Such equations are elliptic when $M \leq 1$ and hyperbolic when $M > 1$. Still haven't a clue what it means for the variable $u$ to be 'hyerpbolic'.

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## College Physics (4th Edition)

We can write an equation for the frequency of a wave: $f = \frac{v}{\lambda}$ We can see that a longer wavelength results in a lower frequency. Since option (a) has the longest wavelength, option (a) has the lowest frequency. The correct answer is: (a)


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# Converting a table to “for loop”

I used “Table” to calculate and plot the following complicated function. First, we choose an initial value for $$G$$ and other parameters, but at later times $$G$$ changes with function $$I_a$$.

``Y = 1; r1 = .3; r2 = .3; G = 0.35; points = Table[L = G - Y; w = Sqrt[-4 + G^2 + 2 G Y + Y^2];  Ia = 1/(    8 w^2) (Cosh[2 L t] + Sinh[2 L t]) (8 + 2 G^2 - 2 w^2 +       4 G Y + 2 Y^2 - 2 (G^2 - w^2 + 2 G Y + Y^2) Cosh[2 r1] -       8 Cosh[2 r2] - 8 Cosh[2 t w] - 2 G^2 Cosh[2 t w] -       2 w^2 Cosh[2 t w] - 4 G Y Cosh[2 t w] - 2 Y^2 Cosh[2 t w] +       G^2 Cosh[2 (r1 - t w)] + w^2 Cosh[2 (r1 - t w)] +       2 G Y Cosh[2 (r1 - t w)] + Y^2 Cosh[2 (r1 - t w)] +       4 Cosh[2 (r2 - t w)] + G^2 Cosh[2 (r1 + t w)] +       w^2 Cosh[2 (r1 + t w)] + 2 G Y Cosh[2 (r1 + t w)] +       Y^2 Cosh[2 (r1 + t w)] + 4 Cosh[2 (r2 + t w)] -       4 G w Sinh[2 t w] - 4 w Y Sinh[2 t w] -       2 G w Sinh[2 (r1 - t w)] - 2 w Y Sinh[2 (r1 - t w)] +       2 G w Sinh[2 (r1 + t w)] + 2 w Y Sinh[2 (r1 + t w)]) +    Integrate[(    E^(2 L (t - u) -       2 (t + u) w) (E^(2 u w) (-G + w - Y) +        E^(2 t w) (G + w + Y))^2)/(4 w^2), {u, 0, t}] // Chop;    G = 1/(1 + Ia/.05);   {{t, Ia}, {t, G}}, {t, 0, 10, .05}]; Iapoints = Map[First, points]; varpoints = Map[Last, points];   ListLinePlot [Iapoints,PlotRange -> All]  ListLinePlot [varpoints,PlotRange -> All] ``

The output seems good to this point. But if I change the initial values as

``Y = 2.2; r1 = .3; r2 = .3; G = 1.96; ``

and

``G = 2/(1 + Ia/5) ``

then the output does not seem good. I couldn’t upload the figure because I don’t know how imgur works!

Anyway, I think the irregularities in the plots are due to error accumulation. I want to try a for loop that does the same job and see if the errors are still there or I can get a good plot for any parameters. But I don’t know how to write a “for loop” for this problem.


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XeLaTeX: WordInitial, Inner and WordFinal seem to match everywhere

I am creating a blackletter font and would like to use it with XeLaTeX. I try to use initial/medial/final substitution for letters, like s/ſ, but the context seems to match even when it should not.

For example, I try to execute substitution of s by ſ using the init and medi tables, but with fontspec and Contextuals={Inner}, word-initial and -final “s”es are replaced by “ſ”, too. The corresponding applies for all substitutions of initial and final forms.

\documentclass{article}
\usepackage{fontspec}
\defaultfontfeatures{Script=Latin,Language=German,Contextuals=Inner}
\setmainfont{Jena1330.otf}
\begin{document}
smstsms
\end{document}

gives a document containing ſmſtſmſ in the correct font, instead of smſtſms, as I would expect.

Am I doing something wrong? How does XeLaTeX decide if a letter is initial, medial or final?

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I ignored this problem now by making Jena1330 fit for vanilla pdfLaTeX, by using a manually generated pl/tfm file. (., , and so on are still considered part of the word there, though.) –  Anaphory Aug 11 '12 at 0:33
Most OpenType layout engine handle isol/init/medi/fina properly only for specific scripts (Arabic, Syriac, Mongolian, etc.), for other scripts they will apply them unconditionally, which seems to be the case with the ICU layout engine used by XeTeX. AFAIK, only some Adobe applications handle them for Latin script.
So for compatibility with both LuaLaTeX and XeLaTeX I have to implement them as both calt and init/medi/fina, since LuaLaTeX has a problem with calt, but supports the in-word contexts for Latin script. –  Anaphory May 12 '12 at 9:16
I'm surprised that they work with LuaLaTeX, AFAIK as I know it doesn’t, the calt issue might be a bug, but without the actual font I can’t help much. –  Khaled Hosny May 12 '12 at 9:47


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How to write design and contrast in DESeq2 to find the up-regulated genes where the design have interactions
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0
Entering edit mode
Hanan • 0
@c51d74dc
Last seen 23 days ago
Israel

Hello I have the following design in DESeq2: ~GT+TP+TP:GT . GT has two levels (K,W) . TP has two levels (0,2). I am looking for the genes that are up-regulated in samples where GT=W and TP=2. How do I define the contrast that these genes will have the highest log2FC. Thank you

DESeq2 • 172 views
0
Entering edit mode
@mikelove
Last seen 1 day ago
United States

First, see the vignette section on interations, and the examples in ?results.

Beyond that, if you have further questions, setting up the statistical design is important. It's worth meeting with a local statistician or someone familiar with R to help you in this step. Unfortunately, I have to restrict my time on the support site to software related questions.


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Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# The molecular formula of lithium carbonate is Li2CO3. What is the molecular weight of this drug?

Its formula mass is 73.89 g*mol^-1.

As you know this salt is commonly used to treat mania. The formula mass is simply the sum of the molar masses of the contributing atoms: i.e. (2xx6.94+12.011+3xx15.999)*g*mol^-1 = ??


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# Maths competition

1. Jul 28, 2008

### Kushal

1. The problem statement, all variables and given/known data

1) If the repeating decimal 0.8451$$\bar{51}$$ is represented by the fraction a/b, where a and b are positive integers with no common factors greater than 1, find the sum a + b.

2) if log(log(log(log x))) = 0 and log represents the base ten logarith, what is the value of x.

3. The attempt at a solution

1) i don't understand how to proceed. some hints would be appreciated.

2)i kno that if, logba = x

then, a = bx

thanks

2. Jul 28, 2008

### nicksauce

For b why don't you show us how you can your answer, writing out every step in full.

For a, it is the same as finding a number 0.51515151... (and then adding 84 and dividing by 100). Do you know how you would find such a number?

3. Jul 28, 2008

### Kushal

2) log(log(log(log x))) = 0

log(log(log x)) = 100 = 1

log(log x) = 101 = 10

log x = 1010

x = 101010

oooh ok.... now i see my mistake!!! i tried doing the calculation in my head.... so foolish of my part

errm, for #1
d'you mean that i should break the 0.8451515151 into 0.84 + 0.005151515151 ?!

thnks

4. Jul 28, 2008

### nicksauce

Yes that's how I would approach the problem. Now can you find a fraction to represent 0.005151515.... ?
If you don't know how, you should read the section entitled "Fraction from a repeating decimal" here
http://en.wikipedia.org/wiki/Recurring_decimal


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# Schottky locus in genus 2

Let $\phi_g : \mathcal{M}_g \rightarrow \mathcal{A}_g$ be the period mapping from the open moduli space of genus $g$ Riemann surfaces to the moduli space of $g$-dimensional principally polarized abelian varieties over $\mathbb{C}$. Thus for a Riemann surface $S$ the image $\phi_g(S)$ is the Jacobian of $S$. The Schottky problem consists in determining the image of $\phi_g$.

It is classical that $\text{Im}(\phi_2)$ is exactly the set of abelian varieties that are not isomorphic to a product of elliptic curves. This is asserted in many places, but I have not been able to find a nice discussion of it in the literature. Does anyone know one? The more down-to-earth, the better.

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This will need expansion by a more knowledgable person, but as memory serves, it was proved by Mayer and Mumford that the closure in Ag of the locus of traditional Jacobians is the set of products of Jacobians. This is probably exposed first in a talk in the 1964 Woods Hole talks on James Milne's site. (I see Mumford credits it there, on page 4 of his talk, in part three of the Woods Hole notes, to Matsusaka and Hoyt. Apparently Mayer and Mumford computed the closure in the Satake compactification.) But let us try to explain this more in dim two.

A two diml ppav is a compact 2 torus A containing a curve C carrying the homology class a1xb1 + a2xb2, where the aj,bj are a basic symplectic homology basis of H1(A).

It follows from the topological Pontrjagin product that the induced map from the Albanese variety of C to A, has topological degree one, hence is an isomorphism. (I.e. the map from the Cartesian product of C with itself g times to A, has image whose class is the g fold Pontrjagin product of [C], which equals g! times the fundamental class of A. Hence the induced map from the g fold symmetric product of C, has image with exactly the fundamental class of A. Hence this map has degree one as does that induced from the Jacobian.)

Since it also induces the identity map on C, it also preserves the polarization.

Let me speculate on the special cases. If C is reducible it is known (Complex abelian varieties and theta functions, George Kempf, p. 89, Cor. 10.4) that A is a product of elliptic curves. If C is irreducible and singular then I guess the normalization map extends to a map of the Albanese of C to A. But that seems to imply the image of C in A does not span, a contradiction.

So it seems that any irreducible curve C contained in a two diml ppav A and carrying the class of a principal polarization, is smooth and induces an isomorphism from the Albanese (i.e. Jacobian) of the curve to the ppav.

I hope there is some useful information in this.

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This was very helpful. Thanks! –  G Fiori Jan 25 '12 at 20:21

By (a possible) definition, a principal polarization on an abelian surface is a curve with self-intersection 2. So, if smooth, it is a genus two curve and the abelian surface is a jacobian. You have to rule out the case of a singular irreducible curve and the remaining possibility is a union of two elliptic curves meeting at a point and then you show that in this case, the surface is the product of the two curves.

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According to MR0364265 (51 #520) Oort, Frans; Ueno, Kenji "Principally polarized abelian varieties of dimension two or three are Jacobian varieties" (J. Fac. Sci. Univ. Tokyo Sect. IA Math. 20 (1973), 377–381) covers this (I don't have access to the paper and am going solely by the review).

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This was very helpful to me, and I had trouble deciding whether to accept it or roy smith's answer. Thanks! –  G Fiori Jan 25 '12 at 20:22

Geoffrey Mess' paper "The Torelli group of genus 2 or 3 surfaces" provides two proofs of this fact---one cohomological and the other topological. But I understand neither. Maybe somebody could provide some commentary on his demonstrations?

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I just looked at his paper, and I also understand neither of his proofs. But they look like exactly what I want! Hopefully an expert will come along and unpack them. –  G Fiori Jan 24 '12 at 21:50
Yes, this argument shows that the image of $\phi_2$ is dense in $\mathcal{A}_2$. But why is its complement as I described above (ie the Jacobians of curves of compact type)? –  G Fiori Jan 24 '12 at 17:39


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# Indium Antimonide Has a Band Gap of 0.23 Ev Between the Valence and the Conduction Band. Find the Temperature at Which Kt Equals the Band Gap. - Physics

Short Note

Indium antimonide has a band gap of 0.23 eV between the valence and the conduction band. Find the temperature at which kT equals the band gap.

#### Solution

Given:
Band gap between the conduction band and the valence band, E = 0.23 eV
Boltzmann's constant, k = 1.38 × 10−23 J/K
We need to find the temperature at which thermal energy kT becomes equal to the band gap of indium antimonide.
∴ kT = E

$\Rightarrow 1 . 38 \times {10}^{- 23} \times T = 0 . 23 \times 1 . 6 \times {10}^{- 19}$

$\Rightarrow T = \frac{0 . 23 \times 1 . 6 \times {10}^{- 19}}{1 . 38 \times {10}^{- 23}}$

$\Rightarrow T = \frac{0 . 23 \times 1 . 6 \times {10}^4}{1 . 38}$

$\Rightarrow T = 0 . 2666 \times {10}^4 \approx 2670$ K

Concept: Energy Bands in Conductors, Semiconductors and Insulators
Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 23 Semiconductors and Semiconductor Devices
Q 3 | Page 419


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# Gradient of a vector field?

What does it mean to take the gradient of a vector field? $\nabla \vec{v}(x,y,z)$? I only understand what it means to take the grad of a scalar field... thank you.

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–  draks ... Jun 11 '12 at 8:42
The short answer is: the gradient of the vector field $\sum v_i(x, y, z)e_i$, where $e_i$ is an orthonormal basis of $\mathbb{R}^3$, is the matrix $(\partial_i v_j)_{i, j=1, 2, 3}$. –  Giuseppe Negro Jun 11 '12 at 8:48
The long answer involves tensor analysis and you can read about it on books such as Itskov, Tensor algebra and tensor analysis for engineers. –  Giuseppe Negro Jun 11 '12 at 8:49
Thanks, @GiuseppeNegro! –  fred Jun 11 '12 at 11:49
Another possible explanation is that the dot is missing between $\nabla$ and $\vec v$, and the "gradient" is actually divergence. –  user31373 Jun 11 '12 at 14:27

Gradient of a vector field (or a multi-valued function $f: R^m\to R^n$) is jacobian of the multi-valued function $f$, where each row $r_i$ of the $\text{Jacobian}(f)$ represents the gradient of $f_i$ (remember, each component $f_i$ of the multi-valued function $f$ is a scalar).
The gradient of a vector is a tensor which tells us how the vector field changes in any direction. We can represent the gradient of a vector by a matrix of its components with respect to a basis. The $(\nabla V)_{\text{ij}}$ component tells us the change of the $V_j$ component in the $\pmb{e}_i$ direction (maybe I have that backwards). You can check out the Wikipedia article for the details of calculating the components.


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# martingale proof

• May 5th 2010, 06:38 AM
willowtree
martingale proof
I need to do the proof for the second condition but I'm not getting the right answer.

$\{S_n , n\geq 1\}$} is a martingale if

1) E[| $S_n$|]< $\infty \forall n$
2) E[ $S_{n+1}|S_1,...,S_n$] = $S_n \forall n$

$S_n$= $X_1X_2...X_n$

Suppose E[| $X_n$|] < $\infty \forall n$ and E[ $X_n$]=1

Prove that $\{S_n , n\geq 1\}$} is a martingale

I can prove the first condition but for the second one I keep getting E[ $S_{n+1}|S_1,...,S_n$] = $S_n+1$ which is wrong. Can anyone shine a light on this problem? Thanks

(ps sorry for any bad use of LaTex, this is the first time I have used it!)
• May 5th 2010, 08:38 AM
matheagle
Quote:

Originally Posted by willowtree
I need to do the proof for the second condition but I'm not getting the right answer.

$\{S_n , n\geq 1\}$} is a martingale if

1) E[| $S_n$|]< $\infty \forall n$
2) E[ $S_{n+1}|S_1,...,S_n$] = $S_n \forall n$

$S_n$= $X_1X_2...X_n$

Suppose E[| $X_n$|] < $\infty \forall n$ and E[ $X_n$]=1

Prove that $\{S_n , n\geq 1\}$} is a martingale

I can prove the first condition but for the second one I keep getting E[ $S_{n+1}|S_1,...,S_n$] = $S_n+1$ which is wrong. Can anyone shine a light on this problem? Thanks

(ps sorry for any bad use of LaTex, this is the first time I have used it!)

If this is a product and not a sum, then

E[ $S_{n+1}|S_1,...,S_n] = E( X_1X_2\cdots X_{n+1}|S_1,...,S_n)=X_1X_2\cdots X_nE(X_{n+1}|S_n)$

and I'm guessing you need $E(X_1)=1$

YUP, I see it there.
I found your $E(X_1)=1$

$=X_1X_2\cdots X_nE(X_{n+1}) =S_n (1)=S_n$
• May 5th 2010, 12:40 PM
willowtree
ahh i seee, thank you, i was doing something a bit backwards with the stuff inside the brackets!


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## PT-Symmetric Talbot Effects

Hamidreza Ramezani, D. N. Christodoulides, V. Kovanis, I. Vitebskiy, Tsampikos Kottos

We show that complex PT-symmetric photonic lattices can lead to a new class of self-imaging Talbot effects. For this to occur, we find that the input field pattern, has to respect specific periodicities which are dictated by the symmetries of the system. While at the spontaneous PT-symmetry breaking point, the image revivals occur at Talbot lengths governed by the characteristics of the passive lattice, at the exact phase it depends on the gain and loss parameter thus allowing one to control the imaging process.

http://arxiv.org/abs/1209.2349

Optics (physics.optics); Other Condensed Matter (cond-mat.other); Quantum Physics (quant-ph)

## PT-Symmetric Electronics

J. Schindler, Z. Lin, J. M. Lee, Hamidreza Ramezani, F. M. Ellis, Tsampikos Kottos

We show both theoretically and experimentally that a pair of inductively coupled active LRC circuits (dimer), one with amplification and another with an equivalent amount of attenuation, display all the features which characterize a wide class of non-Hermitian systems which commute with the joint parity-time PT operator: typical normal modes, temporal evolution, and scattering processes. Utilizing a Liouvilian formulation, we can define an underlying PT-symmetric Hamiltonian, which provides important insight for understanding the behavior of the system. When the PT-dimer is coupled to transmission lines, the resulting scattering signal reveals novel features which reflect the PT-symmetry of the scattering target. Specifically we show that the device can show two different behaviors simultaneously, an amplifier or an absorber, depending on the direction and phase relation of the interrogating waves. Having an exact theory, and due to its relative experimental simplicity, PT-symmetric electronics offers new insights into the properties of PT-symmetric systems which are at the forefront of the research in mathematical physics and related fields.

http://arxiv.org/abs/1209.2347
Other Condensed Matter (cond-mat.other); Optics (physics.optics); Quantum Physics (quant-ph)

## Bragg solitons in nonlinear PT-symmetric periodic potentials

Mohammad-Ali Miri, Alejandro B. Aceves, Tsampikos Kottos, Vassilios Kovanis, Demetrios N. Christodoulides

It is shown that slow Bragg soliton solutions are possible in nonlinear complex parity-time (PT) symmetric periodic structures. Analysis indicates that the PT-symmetric component of the periodic optical refractive index can modify the grating band structure and hence the effective coupling between the forward and backward waves. Starting from a classical modified massive Thirring model, solitary wave solutions are obtained in closed form. The basic properties of these slow solitary waves and their dependence on their respective PT-symmetric gain/loss profile are then explored via numerical simulations.

http://arxiv.org/abs/1209.0787
Optics (physics.optics); Mathematical Physics (math-ph); Exactly Solvable and Integrable Systems (nlin.SI); Quantum Physics (quant-ph)

## Light Transport in Random Media with PT-Symmetry

Samuel Kalish, Zin Lin, Tsampikos Kottos

The scattering properties of randomly layered optical media with $${\cal PT}$$-symmetric index of refraction are studied using the transfer-matrix method. We find that the transmitance decays exponentially as a function of the system size, with an enhanced rate $$\xi_{\gamma}(W)^{-1}=\xi_0(W)^{-1}+\xi_{\gamma} (0)^{-1}$$, where $$\xi_0(W)$$ is the localization length of the equivalent passive random medium and $$\xi_{\gamma}(0)$$ is the attenuation/amplification length of the corresponding perfect system with a $${\cal PT}$$-symmetric refraction index profile. While transmitance processes are reciprocal to left and right incident waves, the reflectance is enhanced from one side and is inversely suppressed from the other, thus allowing such $${\cal PT}$$-symmetric random media to act as unidirectional coherent absorbers.

http://arxiv.org/abs/1205.1849
Optics (physics.optics)

## Bypassing the bandwidth theorem with PT symmetry

Hamidreza Ramezani, J. Schindler, F. M. Ellis, Uwe Guenther, Tsampikos Kottos

The beat time $${\tau}_{fpt}$$ associated with the energy transfer between two coupled oscillators is dictated by the bandwidth theorem which sets a lower bound $${\tau}_{fpt}\sim 1/{\delta}{\omega}$$. We show, both experimentally and theoretically, that two coupled active LRC electrical oscillators with parity-time (PT) symmetry, bypass the lower bound imposed by the bandwidth theorem, reducing the beat time to zero while retaining a real valued spectrum and fixed eigenfrequency difference $$\delta\omega$$. Our results foster new design strategies which lead to (stable) pseudo-unitary wave evolution, and may allow for ultrafast computation, telecommunication, and signal processing.

http://arxiv.org/abs/1205.1847
Classical Physics (physics.class-ph)

## Experimental observation of the dual behavior of PT-symmetric scattering

Zin Lin, Joseph Schindler, Fred M. Ellis, Tsampikos Kottos

We investigate experimentally parity-time $${\cal PT}$$ symmetric scattering using $$LRC$$ circuits in an inductively coupled $${\cal PT}$$- symmetric pair connected to transmission line leads. In the single-lead case, the $${\cal PT}$$-symmetric circuit acts as a simple dual device – an amplifier or an absorber depending on the orientation of the lead. When a second lead is attached, the system exhibits unidirectional transparency for some characteristic frequencies. This non-reciprocal behavior is a consequence of generalized (non-unitary) conservation relations satisfied by the scattering matrix.

http://arxiv.org/abs/1205.2176
Mesoscale and Nanoscale Physics (cond-mat.mes-hall)

## Exceptional Point Dynamics in Photonic Honeycomb Lattices with PT Symmetry

Hamidreza Ramezani, Tsampikos Kottos, Vassilios Kovanis, Demetrios N. Christodoulides

We theoretically investigate the flow of electromagnetic waves in complex honeycomb photonic lattices with local PT symmetries. Such PT structure is introduced via a judicious arrangement of gain or loss across the honeycomb lattice, characterized by a gain/loss parameter \{\gamma\}. We found a new class of conical diffraction phenomena where the formed cone is brighter and travels along the lattice with a transverse speed proportional to \{\sqrt{\gamma}\}.

http://arxiv.org/abs/1112.4734
Optics (physics.optics); Quantum Physics

## Experimental Study of Active LRC Circuits with PT-Symmetries

Joseph Schindler, Ang Li, Mei C. Zheng, F. M. Ellis, Tsampikos Kottos

Mutually coupled modes of a pair of active LRC circuits, one with amplification and another with an equivalent amount of attenuation, provide an experimental realization of a wide class of systems where gain/loss mechanisms break the Hermiticity while preserving parity-time PT symmetry. For a value PT of the gain/loss strength parameter the eigen-frequencies undergo a spontaneous phase transition from real to complex values, while the normal modes coalesce acquiring a definite chirality. The consequences of the phase-transition in the spatiotemporal energy evolution are also presented.

http://arxiv.org/abs/1109.2913
Other Condensed Matter (cond-mat.other); Quantum Physics (quant-ph)

## Unidirectional Invisibility induced by PT-Symmetric Periodic Structures

Zin Lin, Hamidreza Ramezani, Toni Eichelkraut, Tsampikos Kottos, Hui Cao, Demetrios N. Christodoulides

We show that parity-time (PT) symmetric Bragg periodic structures, near the spontaneous PT – symmetry breaking point, can act as unidirectional invisible media. In this regime, the re flection from one end is diminished while it is enhanced from the other. At the same time the transmission coefficient and phase, are indistinguishable from those expected in the absence of a grating. The phenomenon is robust even in the presence of Kerr non-linearities, and it can also eff?ectively suppress optical bistabilities.

http://arxiv.org/abs/1108.2493
Optics (physics.optics)


text:
# 1 Purpose

The purpose of this vignette is to assess the correctness of the functions of the OwenQ package.

As said in the main vignette, the fourth Owen cumulative function $$O_4$$, implemented under the name powen4 allows to evaluate the power of equivalence tests. This is perhaps the main application of the OwenQ package and we will particularly focus on this situation.

Our strategy runs as follows:

• To test the OwenT function, we will compare some values it returns to those returned by other implementations. Note that the other functions of the OwenQ package rely on OwenT when the number of degrees of freedom is odd. Therefore, by testing these other functions for an odd number of degrees of freedom, we implicitely test the OwenT function.

• We will compare a couple of values returned by OwenQ1 and OwenQ2 to the values obtained by numerical integration in Wolfram|Alpha.

• We will test powen4 by performing some power calculations relying on this function and comparing the results with the ones provided by SAS. We will also perform these power calculations with the help of the ipowen4 function, which evaluates the $$O_4$$ function by a powerful numerical integration, and we will compare the results.

• These power calculations will be performed for $$100$$ different scenarios, and we will conclude that powen4 is reliable for these scenarios. Then, to test the other functions of the OwenQ package, we will check that some mathematical relations hold for these $$100$$ scenarios.

The folder tests/testthat of the OwenQ package contains many comparisons with Wolfram|Alpha. The results we obtain with OwenQ are always the same as the ones obtained with Wolfram|Alpha up to a tolerance of at least $$10$$ decimal digits.

# 2 Owen $$T$$-function (OwenT)

The Owen $$T$$-function is implemented under the name OwenT. This is a port of the function owens_t of the C++ special_functions library included in the boost set of libraries. Some details about the C++ implementation are available here.

The libraries provided by boost are peer-reviewed, and this is enough to trust the reliability of OwenT. Nevertheless, below we compare the results of OwenT to the six values given in Patefield’s article, up to $$14$$ decimal digits. We have also checked that Wolfram|Alpha gives these values.

testData <- data.frame(
h = c(0.0625, 6.5, 7, 4.78125, 2, 1),
a = c(0.25, 0.4375, 0.96875, 0.0625, 0.5, 0.9999975),
Patefield = c(3.8911930234701e-02, 2.0005773048508e-11, 6.3990627193899e-13, 1.0632974804687e-07, 8.6250779855215e-03, 6.6741808978229e-02),
OwenT = numeric(6)
)
for(i in 1:nrow(testData)){
testData$OwenT[i] <- OwenT(testData$h[i], testData$a[i]) } print(testData, digits=14) ## h a Patefield OwenT ## 1 0.06250 0.2500000 3.8911930234701e-02 3.8911930234701e-02 ## 2 6.50000 0.4375000 2.0005773048508e-11 2.0005773048508e-11 ## 3 7.00000 0.9687500 6.3990627193899e-13 6.3990627193899e-13 ## 4 4.78125 0.0625000 1.0632974804687e-07 1.0632974804687e-07 ## 5 2.00000 0.5000000 8.6250779855215e-03 8.6250779855215e-03 ## 6 1.00000 0.9999975 6.6741808978229e-02 6.6741808978229e-02 We get the same results with OwenT. # 3 Owen $$Q$$-functions: comparing a couple of values The two Owen $$Q$$-functions $$Q_1$$ and $$Q_2$$ are defined by: $Q_1(\nu, t, \delta, R) = \frac{1}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac12(\nu-2)}} \int_0^R \Phi\left(\frac{tx}{\sqrt{\nu}}-\delta\right) x^{\nu-1} e^{-\frac{x^2}{2}} \mathrm{d}x$ and $Q_2(\nu, t, \delta, R) = \frac{1}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac12(\nu-2)}} \int_R^\infty \Phi\left(\frac{tx}{\sqrt{\nu}}-\delta\right) x^{\nu-1} e^{-\frac{x^2}{2}} \mathrm{d}x.$ They are implemented in the OwenQ package under the respective names OwenQ1 and OwenQ2, for integer values of $$\nu$$, following Owen’s algorithm (1965). In Wolfram|Alpha, these functions are not available, but we can evaluate them by numerical integration. Below we compare two values of OwenQ1 to the ones returned by Wolfram|Alpha. # wolfram: Integrate[(1+Erf[(3*x/Sqrt[3]-2)/Sqrt[2]])*x^(3-1)*Exp[-x^2/2],{x,0,5}]/2/Gamma[3/2]/2^((3-2)/2) OwenQ1(3, 3, 2, 5) ## [1] 0.6800117 Our value rounded to $$6$$ digits is the same as the one given by Wolfram. # wolfram: Integrate[(1+Erf[(3*x/Sqrt[1000]-2)/Sqrt[2]])*x^(1000-1)*Exp[-x^2/2],{x,0,30}]/2/Gamma[1000/2]/2^((1000-2)/2) print(OwenQ1(1000, 3, 2, 30), digits=16) ## [1] 0.008518809463589428 The two values are the same up to $$13$$ digits. Now we compare two values of OwenQ2 to the ones returned by Wolfram|Alpha. # wolfram: Integrate[(1+Erf[(3*x/Sqrt[3]-2)/Sqrt[2]])*x^(3-1)*Exp[-x^2/2],{x,5,Infinity}]/2/Gamma[3/2]/2^((3-2)/2) OwenQ2(3, 3, 2, 5) ## [1] 1.54405e-05 The two values are identical. # wolfram: Integrate[(1+Erf[(3*x/Sqrt[1000]-2)/Sqrt[2]])*x^(1000-1)*Exp[-x^2/2],{x,5,Infinity}]/2/Gamma[1000/2]/2^((1000-2)/2) print(OwenQ2(1000, 3, 2, 5), digits=16) ## [1] 0.8406201459600969 The two values are identical up to $$13$$ digits. # 4 Fourth Owen cumulative function $$O_4$$ (powen4) As seen in the other vignette, the fourth Owen cumulative function $$O_4$$, implemented under the name powen4, can be used to evaluate the power of equivalence tests. The powerTOST function below returns the power of the equivalence test for a so-called parallel design with equal variances, when considering the alternative hypothesis $$H_1\colon\{-\Delta < \delta_0 < \Delta \}$$, where $$\delta_0$$ denotes the difference between the two means. This function takes as arguments the significance level $$\alpha$$, the difference $$\delta_0$$ between the two means, the threshold $$\Delta$$, the common standard deviation $$\sigma$$ of the two samples, and the two sample sizes $$n_1$$ and $$n_2$$. powerTOST <- function(alpha, delta0, Delta, sigma, n1, n2, algo=2) { se <- sqrt(1/n1 + 1/n2) * sigma delta1 <- (delta0 + Delta) / se delta2 <- (delta0 - Delta) / se dof <- n1 + n2 - 2 q <- qt(1 - alpha, dof) powen4(dof, q, -q, delta1, delta2, algo=algo) } As you can see, the powen4 function has an argument algo. This is also the case for the other Owen functions, except for ptOwen. The algo argument can take two values, 1 or 2. Th default value is algo=2, and as we will see later, the evaluation is more reliable with this option. The value of algo corresponds to a small difference in the algorithm. With algo=1, the evaluation is a bit faster. And we will see that powerTOST is reliable for algo=1 when the value of n1+n2 is not too large. ## 4.1 Comparisons with SAS The table shown below contains $$100$$ results of power calculations with SAS version 9.4, for a total sample size $$n_1+n_2$$ going from $$20$$ to $$1200$$. The results are rounded to $$5$$ decimal digits. alpha delta0 Delta sigma n1 n2 powerSAS 1 0.05 0.0 1 1 10 10 0.39094 2 0.05 0.0 1 1 15 15 0.69541 3 0.05 0.0 1 1 20 20 0.85580 4 0.05 0.0 1 1 25 25 0.93426 5 0.05 0.0 1 1 30 30 0.97092 6 0.05 0.1 1 1 10 10 0.38272 7 0.05 0.1 1 1 15 15 0.67836 8 0.05 0.1 1 1 20 20 0.83661 9 0.05 0.1 1 1 25 25 0.91781 10 0.05 0.1 1 1 30 30 0.95889 11 0.05 0.2 1 1 10 10 0.35908 12 0.05 0.2 1 1 15 15 0.62954 13 0.05 0.2 1 1 20 20 0.78069 14 0.05 0.2 1 1 25 25 0.86796 15 0.05 0.2 1 1 30 30 0.92017 16 0.05 0.3 1 1 10 10 0.32287 17 0.05 0.3 1 1 15 15 0.55540 18 0.05 0.3 1 1 20 20 0.69322 19 0.05 0.3 1 1 25 25 0.78471 20 0.05 0.3 1 1 30 30 0.84909 21 0.05 0.4 1 1 10 10 0.27820 22 0.05 0.4 1 1 15 15 0.46531 23 0.05 0.4 1 1 20 20 0.58306 24 0.05 0.4 1 1 25 25 0.67174 25 0.05 0.4 1 1 30 30 0.74260 26 0.05 0.5 1 1 10 10 0.22970 27 0.05 0.5 1 1 15 15 0.36967 28 0.05 0.5 1 1 20 20 0.46208 29 0.05 0.5 1 1 25 25 0.53883 30 0.05 0.5 1 1 30 30 0.60600 31 0.05 0.0 1 1 10 30 0.70373 32 0.05 0.0 1 1 15 30 0.85764 33 0.05 0.0 1 1 20 30 0.92324 34 0.05 0.0 1 1 25 30 0.95452 35 0.05 0.0 1 1 30 30 0.97092 36 0.05 0.1 1 1 10 30 0.68648 37 0.05 0.1 1 1 15 30 0.83846 38 0.05 0.1 1 1 20 30 0.90604 39 0.05 0.1 1 1 25 30 0.94003 40 0.05 0.1 1 1 30 30 0.95889 41 0.05 0.2 1 1 10 30 0.63703 42 0.05 0.2 1 1 15 30 0.78255 43 0.05 0.2 1 1 20 30 0.85439 44 0.05 0.2 1 1 25 30 0.89501 45 0.05 0.2 1 1 30 30 0.92017 46 0.05 0.3 1 1 10 30 0.56192 47 0.05 0.3 1 1 15 30 0.69503 48 0.05 0.3 1 1 20 30 0.76944 49 0.05 0.3 1 1 25 30 0.81676 50 0.05 0.3 1 1 30 30 0.84909 51 0.05 0.4 1 1 10 30 0.47061 52 0.05 0.4 1 1 15 30 0.58470 53 0.05 0.4 1 1 20 30 0.65610 54 0.05 0.4 1 1 25 30 0.70594 55 0.05 0.4 1 1 30 30 0.74260 56 0.05 0.5 1 1 10 30 0.37365 57 0.05 0.5 1 1 15 30 0.46343 58 0.05 0.5 1 1 20 30 0.52475 59 0.05 0.5 1 1 25 30 0.57052 60 0.05 0.5 1 1 30 30 0.60600 61 0.05 0.0 3 4 50 50 0.96239 62 0.05 1.0 3 4 50 50 0.79849 63 0.05 2.0 3 4 50 50 0.34329 64 0.05 2.5 3 4 50 50 0.15288 65 0.05 0.0 4 4 10 90 0.81791 66 0.05 1.0 4 4 10 90 0.70346 67 0.05 2.0 4 4 10 90 0.43595 68 0.05 2.5 4 4 10 90 0.29819 69 0.05 0.0 4 7 100 100 0.98278 70 0.05 1.0 4 7 100 100 0.91512 71 0.05 2.0 4 7 100 100 0.64376 72 0.05 3.0 4 7 100 100 0.26169 73 0.01 0.0 4 7 100 100 0.90831 74 0.01 1.0 4 7 100 100 0.74924 75 0.01 2.0 4 7 100 100 0.37443 76 0.01 3.0 4 7 100 100 0.09290 77 0.05 0.0 4 4 185 10 0.84568 78 0.05 1.0 4 4 185 10 0.73025 79 0.05 2.0 4 4 185 10 0.45459 80 0.05 3.0 4 4 185 10 0.19001 81 0.05 0.0 4 8 185 100 0.98240 82 0.05 1.0 4 8 185 100 0.91417 83 0.05 2.0 4 8 185 100 0.64226 84 0.05 3.0 4 8 185 100 0.26103 85 0.01 0.0 4 10 250 250 0.96713 86 0.01 1.0 4 10 250 250 0.84523 87 0.01 2.0 4 10 250 250 0.46161 88 0.01 3.0 4 10 250 250 0.11288 89 0.01 0.0 4 14 500 500 0.97112 90 0.01 1.0 4 14 500 500 0.85433 91 0.01 2.0 4 14 500 500 0.47184 92 0.01 3.0 4 14 500 500 0.11536 93 0.01 0.0 5 35 600 600 0.11547 94 0.01 4.0 5 35 600 600 0.01658 95 0.05 0.0 5 30 600 600 0.78512 96 0.05 4.0 5 50 600 600 0.02642 97 0.01 0.0 5 6 1190 10 0.23194 98 0.01 4.0 5 6 1190 10 0.02739 99 0.05 0.0 5 9 1190 10 0.08256 100 0.05 4.0 5 9 1190 10 0.03115 This table is stored in a dataframe named SAS. We compare the SAS values to the ones obtained by our powerTOST function. power <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { power[i] <- powerTOST( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i] ) } Rounding our values to $$5$$ decimal digits, we find that our results are identical to the SAS results: identical(round(power,5), SAS$powerSAS)
## [1] TRUE

With the option algo=1, the results are identical as well:

power <- numeric(nrow(SAS))
for (i in 1:nrow(SAS)) {
power[i] <-
powerTOST(
alpha = SAS$alpha[i], delta0 = SAS$delta0[i],
Delta = SAS$Delta[i], sigma = SAS$sigma[i],
n1 = SAS$n1[i], n2 = SAS$n2[i],
algo = 1
)
}
identical(round(power,5), SAS$powerSAS) ## [1] TRUE ## 4.2 Comparison with numerical integration The ipowen4 internal function of the OwenQ package evaluates the fourth Owen cumulative function $$O_4$$ by a powerful numerical integration implemented in C++ (with the package RcppNumerical). Thus we have two completely different implementations of $$O_4$$. The ipowerTOST function below is obtained by replacing powen4 with ipowen4 in powerTOST. We will compare the results of powerTOST with the ones of ipowerTOST. ipowerTOST <- function(alpha, delta0, Delta, sigma, n1, n2) { se <- sqrt(1/n1 + 1/n2) * sigma delta1 <- (delta0 + Delta) / se delta2 <- (delta0 - Delta) / se dof <- n1 + n2 - 2 q <- qt(1 - alpha, dof) OwenQ:::ipowen4(dof, q, -q, delta1, delta2) } ### 4.2.1 Failures and successes of powen4 • For $$n_1=n_2=1000$$, $$\alpha=0.05$$, $$\delta=0$$, $$\Delta=5$$, the powen4 function with the option algo=1 abnormally takes some negative values: Note that the discrepancy between powen4 and ipowen4 occurs only for $$\sigma > 65$$. • For $$n_1=n_2=720$$, $$\alpha=0.05$$, $$\delta=0$$, $$\Delta=5$$, we observe a discrepancy between powen4 with the option algo=1 and ipowen4: There is no discrepancy between powen4 with the option algo=2 and ipowen4. The irregularity of powen4 with the option algo=1 suggests that it returns wrong values. • For $$n_1=n_2=700$$, $$\alpha=0.01$$, $$\delta=0$$, $$\Delta=5$$, we still observe a small discrepancy between powen4 with the option algo=1 and ipowen4, and we still observe an irregularity of powen4 (on the second figure below): With the option algo=2, the results of powen4 coincide with the ones of ipowen4. • For $$n_1 = n_2 = 600$$, $$\alpha=0.005$$, $$\delta=0$$, $$\Delta=5$$, we do not observe any discrepancy between powen4 and ipowen4: We conclude that powen4 with the option algo=1 is not reliable when the number of degrees of freedom is too large. As said before, the interest of the option algo=1 is that the evaluation is faster. For $$n_1=n_2=2500$$, the results of powerTOST with algo=2 still coincide with the result of ipowerTOST: alpha <- 0.05; delta0 <- 0; Delta <- 5 sigma <- 110 n1 <- n2 <- 2500 powerTOST(alpha, delta0, Delta, sigma, n1, n2) ## [1] 4.523596e-05 ipowerTOST(alpha, delta0, Delta, sigma, n1, n2) ## [1] 4.523596e-05 ## attr(,"err_est") ## [1] 5.022201e-19 ## attr(,"err_code") ## [1] 0 And even for $$n_1=n_2=5000$$: sigma <- 152 n1 <- n2 <- 5000 powerTOST(alpha, delta0, Delta, sigma, n1, n2) ## [1] 0.003612374 ipowerTOST(alpha, delta0, Delta, sigma, n1, n2) ## [1] 0.003612374 ## attr(,"err_est") ## [1] 4.010541e-17 ## attr(,"err_code") ## [1] 0 ### 4.2.2 Comparisons for $$n_1+n_2 \leq 1200$$ Below we compare the results returned by powerTOST to the ones returned by ipowerTOST for the parameters given in the SAS dataset. power <- ipower <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { power[i] <- powerTOST( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i] ) ipower[i] <- ipowerTOST( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i] ) } identical(round(power, 10), round(ipower, 10)) ## [1] TRUE Results are the same up to $$10$$ digits. And also with the option algo=1: power <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { power[i] <- powerTOST( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i], algo = 1 ) } identical(round(power, 10), round(ipower, 10)) ## [1] TRUE ## 4.3 Conclusion • The powen4 function with option algo=1 seems to return correct values for $$\nu \leq 1200$$. • The powen4 function with option algo=2 allows higher values of $$\nu$$. • In any case, we recommend to compare the result of powen4 with the one of ipowen4. # 5 Other Owen cumulative functions We previously validated the powen4 function for the $$100$$ scenarios of the dataset SAS. Now we test the functions powen1, powen2 and powen3 by checking the equality $O_1(\nu, t_1, t_2, \delta_1, \delta_2) + O_2(\nu, t_1, t_2, \delta_1, \delta_2) + O_3(\nu, t_1, t_2, \delta_1, \delta_2) + O_4(\nu, t_1, t_2, \delta_1, \delta_2) = 1.$ We restrict our attention to default setting algo=2. We check the above equality for the $$100$$ scenarios of the dataset SAS. f <- function(alpha, delta0, Delta, sigma, n1, n2) { se <- sqrt(1/n1 + 1/n2) * sigma delta1 <- (delta0 + Delta) / se delta2 <- (delta0 - Delta) / se dof <- n1 + n2 - 2 q <- qt(1 - alpha, dof) powen1(dof, q,-q, delta1, delta2) + powen2(dof, q,-q, delta1, delta2) + powen3(dof, q,-q, delta1, delta2) + powen4(dof, q,-q, delta1, delta2) } test <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { test[i] <- f( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i] ) } all(abs(test-1) < 1e-14) ## [1] TRUE We find that each of the $$100$$ sums is equal to $$1$$ up to a tolerance of $$14$$ digits. # 6 First Owen $$Q$$-function $$Q_1$$ (OwenQ1) We previously validated the powen4 function (implementation of the fourth Owen cumulative function $$O_4$$) for the $$100$$ scenarios of the dataset SAS. Now, to test OwenQ1, we will use the following equality: $O_4(\nu, t_1, t_2, \delta_1, \delta_2) = Q_1\left(\nu, t_2, \delta_2, \frac{\sqrt{\nu}(\delta_1-\delta_2)}{t_1-t_2}\right) - Q_1\left(\nu, t_1, \delta_1, \frac{\sqrt{\nu}(\delta_1-\delta_2)}{t_1-t_2}\right).$ We use this formula to perform the power calculations with OwenQ1 instead of powen4. powerTOST2 <- function(alpha, delta0, Delta, sigma, n1, n2) { se <- sqrt(1/n1 + 1/n2) * sigma delta1 <- (delta0 + Delta) / se delta2 <- (delta0 - Delta) / se dof <- n1 + n2 - 2 q <- qt(1 - alpha, dof) R <- sqrt(dof)*(delta1 - delta2)/q/2 OwenQ1(dof, -q, delta2, R) - OwenQ1(dof, q, delta1, R) } power2 <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { power2[i] <- powerTOST2( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i] ) } identical(round(power, 14), round(power2, 14)) ## [1] FALSE The values rounded to $$14$$ digits are the same as the ones we previously obtained with powen4. # 7 Second Owen $$Q$$-function $$Q_2$$ (OwenQ2) We previously validated powen2 for the $$100$$ scenarios of the dataset SAS. Now we test the OwenQ2 function by checking the equality $O_2(\nu, t_1, t_2, \delta_1, \delta_2) = Q_2\left(\nu, t_1, \delta_1, \frac{\sqrt{\nu}(\delta_1-\delta_2)}{t_1-t_2}\right) - Q_2\left(\nu, t_2, \delta_2, \frac{\sqrt{\nu}(\delta_1-\delta_2)}{t_1-t_2}\right).$ We check this equality for the $$100$$ scenarios of the dataset SAS. g <- function(alpha, delta0, Delta, sigma, n1, n2) { se <- sqrt(1/n1 + 1/n2) * sigma delta1 <- (delta0 + Delta) / se delta2 <- (delta0 - Delta) / se dof <- n1 + n2 - 2 q <- qt(1 - alpha, dof) R <- sqrt(dof)*(delta1 - delta2)/q/2 x <- OwenQ2(dof, q, delta1, R) - OwenQ2(dof, -q, delta2, R) y <- powen2(dof, q, -q, delta1, delta2) x - y } test <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { test[i] <- g( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS$n2[i] ) } all(abs(test) < 1e-15) ## [1] TRUE We find that each of the $$100$$ equalities hold true up to a tolerance of $$15$$ digits. # 8 Student cumulative distribution function (ptOwen) We finally test ptOwen, the implementation of the cumulative distribution function of the noncentral Student distribution. If $$T_1$$ follows the noncentral Student distribution with number of degrees of freedom $$\nu$$ and noncentrality parameter $$\delta_1$$, then for any $$t_1$$ the equality $\Pr(T_1 \leq t_1) = O_1(\nu, t_1, t_2, \delta_1, \delta_2) + O_2(\nu, t_1, t_2, \delta_1, \delta_2)$ holds for any $$t_2$$ and $$\delta_2$$. We check this equality for the $$100$$ scenarios of the dataset SAS. h <- function(alpha, delta0, Delta, sigma, n1, n2) { se <- sqrt(1/n1 + 1/n2) * sigma delta1 <- (delta0 + Delta) / se delta2 <- (delta0 - Delta) / se dof <- n1 + n2 - 2 q <- qt(1 - alpha, dof) x <- ptOwen(q, dof, delta1) y <- powen1(dof, q, -q, delta1, delta2) + powen2(dof, q, -q, delta1, delta2) x - y } test <- numeric(nrow(SAS)) for (i in 1:nrow(SAS)) { test[i] <- h( alpha = SAS$alpha[i],
delta0 = SAS$delta0[i], Delta = SAS$Delta[i],
sigma = SAS$sigma[i], n1 = SAS$n1[i],
n2 = SAS\$n2[i]
)
}
all(abs(test) < 1e-15)
## [1] TRUE

For each scenario, the equality holds up to a tolerance of $$15$$ digits.

# 9 References

• Patefield, M. (2000). Fast and Accurate Calculation of Owen’s T Function. Journal of Statistical Software 5 (5), 1-25.

• Owen, D. B. (1965). A special case of a bivariate noncentral t-distribution. Biometrika 52, 437-446.

• Wolfram Alpha LLC. 2017. Wolfram|Alpha. (access July 17, 2017).


text:
Agda-2.5.1.2: A dependently typed functional programming language and proof assistant

Documentation

class (Functor m, Applicative m, Monad m) => HasBuiltins m where Source #

Minimal complete definition

getBuiltinThing

Methods

Instances

 MonadIO m => HasBuiltins (TCMT m) Source # Methods

Rewrite a literal to constructor form if possible.

The names of built-in things

Builtins that come without a definition in Agda syntax. These are giving names to Agda internal concepts which cannot be assigned an Agda type.

An example would be a user-defined name for Set.

{--}

The type of Type would be Type : Level → Setω which is not valid Agda.

The coinductive primitives.

Constructors

 CoinductionKit FieldsnameOfInf :: QName nameOfSharp :: QName nameOfFlat :: QName

Tries to build a CoinductionKit.

Builtin equality

Get the name of the equality type.

Check whether the type is actually an equality (lhs ≡ rhs) and extract lhs, rhs, and their type.

Precondition: type is reduced.

Revert the EqualityView.

Postcondition: type is reduced.


text:
# Collagen mutations result in Osteogenesis imperfecta

### Problem

Collagen is the most abundant protein in mammals and forms the main component of connective tissue. Over 28 types of collagen have been identified and described in humans. However, type I collagen constitutes over 90, percent of collagen. Type I collagen is found in bone, tendon, ligament, and skin. This protein is composed of three polypeptide chains, which wind around one another into a triple-helix stabilized by covalent cross-linkages. Collagen is distinctive in that its subunits are composed of a regular arrangement of Gly-Pro-X, where X can be any amino acid. Each amino acid has a precise role, as the glycine side chain is the only one that is small enough to fit into the center of the triple helix. The angle of the proline residue enables the polypeptide chains to fold into the triple helical structure.
Osteogenesis imperfecta, which affects 6 to 7 per 100, comma, 000 people worldwide, typically results from genetic mutations in the collagen genes. The most common form of osteogenesis imperfecta, type I, arises from a dominant mutation in the COL1A1 gene, which results in a deficiency in collagen production. Other types of osteogenesis imperfecta, including types II, III, and IV are caused by dominant mutations in the COL1A1 or COL1A2 genes that alter the structure of type I collagen. In all cases, people with osteogenesis imperfecta suffer from bones that fracture easily.
Where does the mutation that causes osteogenesis imperfecta first occur?
Please choose from one of the following options.
Get 5 answers correct in a row


text:
# Forum:No seriously though, we are going to merge with Illogicopedia

Forums: Index > Village Dump > No seriously though, we are going to merge with Illogicopedia
Note: This topic has been unedited for 1458 days. It is considered archived - the discussion is over. Do not add to unless it really needs a response.
Sorry guys but decision has been made. I've already called them. -
# 17:05, 14 November 2008 RAHB (Talk | contribs | block) blocked Kingkitty (Talk | contribs) with an expiry time of a eternity  (Fucking with the natural order of things.) (unblock) - 00:06, 15 November 2008 (UTC)
Merging? That sounds sexy. What are you wearing? I'm wearing a snowsuit, earmuffs, two pairs of mittens, winter boots, a tuque, a scarf, another scarf, and nothing underneath. Except for the long underwear. And the wool socks. And the regular-sized underwear. 00:18, 15 November 2008 (UTC)
Thus spoke Sarah Palin. Spake? Nah, probably spoke. 03:25, 15 November 2008 (UTC)

But they HATE us! WAAAAH! WAAAAH!-- 17:10, 15 November 2008 (UTC)

Why do certain people here want to merge with Illogicopedia? I don't get it. Also, Illogic peoples don't hate Uncyc, they just have gripes with, er, let's say 'certain things'. -- Hindleyite Converse 18:52, 15 November 2008 (UTC)
Well, we don't want to merge with you guys, your pages are full of adverts. oh. MrN  Fork you! 18:57, Nov 15
No. They're full of adverbs. Lovely adverbs. 19:01, 15 November 2008 (UTC)
Hah, the only reason Uncyclopedia is ad-free is because it's Wikia's 'pet project'. I reckon, with the domain shift, they are trying to use Uncyclopedia as some sort of saviour to their ailing financial issues. PS. Anyone who goes to the Wikia Illogicopedia is seriously behind the times, man. -- Hindleyite Converse 19:26, 15 November 2008 (UTC)
Yeah. The Linkifier!!!!!? 00:37, 16 November 2008 (UTC)
I'm pleased to see that Nerd42 found a niche there. I always thought we were too harsh to him. Anyway, it's good to know other humor exists. Out of curiosity, was it the deletionism that forced you chaps to escape our hallowed halls? It was the deletionism, wasn't it?-- 01:28, 16 November 2008 (UTC)
/me deletes Bradaphraser MrN  Fork you! 01:29, Nov 16
Not again...-- 01:37, 16 November 2008 (UTC)
/me protects Bradaphraser, redirects MrN9000 to CRP 06:07, 18 November 2008 (UTC)
/me replaces Modusoperandi with "penis penis lol lol lol" --UU - natter 11:50, Nov 18
/me replaces Under User with Goa Tse 09:17, 20 November 2008 (UTC)
Depending on how severe the Google pagerank penalty we incur from losing the uncyclopedia.org domain becomes, this might almost make sense. As of today, dnsscoop.com reports desciclopedia.org's and illogicopedia.org's page ranks as PR3, but uncyclopedia.org is now only PR2 - it also shows the Alexa rank drop to nada after Wikia redirected the domain. Maybe this isn't something we should be joking about until we have some idea of how much damage has been done. --Carlb 05:37, 18 November 2008 (UTC)

## MNM5150's Brilliant New Proposal

Why don't we all just get Illogicopedia accounts and vote for a merger there.-- 23:25, 17 November 2008 (UTC)

Because that would require ingenuity. Personally, I'm too lazy to be ingenuish. Or whatever. 02:12, 18 November 2008 (UTC)
Which Illogicopedia is everyone talking about now? The new one or the old one? I couldn't give a rat's ass what happens to the old one, as long as it's closed down :P
However, a merge still ain't happening or my name's not Chicka Chicka Slim Shady. Stop talking about it now or we'll bite the heads off your Jelly Babies. -- Hindleyite Converse 11:47, 18 November 2008 (UTC)

## Ilogicopedia is my second home

When my articles don't fit Uncyc, I move them to Illogicopedia, because it could be just plain nonsense I can post there, and I have to make funny here. 09:17, 20 November 2008 (UTC)

What an excellent summary of why we should steer well clear. Asahatter (annoy) 09:25, 20 November 2008 (UTC)
Disagree wholeheartedly, but then I would. Illogicopedia isn't JUST plain nonsense, I hate that stupid, childish image that we've created - it isn't like that anymore. Illogicopedia is funny too, you just have a slightly different taste in humour. And I would say we're moving well, well away from just plain stupid nonsense these days. We aren't Uncyclopedia's trash bin, the vast majority of articles deleted at Uncyclo would be deleted at Illogicopedia, only the community are more accepting and willing to improve articles at Illogicopedia. So yeah, it's not a juvenile playground any more. -- Hindleyite Converse 13:02, 20 November 2008 (UTC)
/me tries so hard not to laugh... 16:35, Nov 20
Wait, so are you saying Illogicopedia is LESS juvenile than Uncyclopedia?--Nytrospawn 16:58, 20 November 2008 (UTC)
I neither confirm nor deny the truthfulness of my false unstatement with double negatives about the Illogic of merging Uncyclo and Illogico...with SAVINGS!!! 17:19, Nov 20
This is becoming all too illogical for me to comprehend. Would you like a Rolo? -- Hindleyite Converse 22:22, 20 November 2008 (UTC)

Can't we stop these foolish accusations and just admit that both sites are the worst? 18:13, 20 November 2008 (UTC)

I masturbate to ED at night. --Nachlader 20:34, 20 November 2008 (UTC)
Ed Wood or Edward Norton? No! Keep that to yourself. Either way, I don't want to know. 22:09, 20 November 2008 (UTC)
Uncyclologic is the baddest wurst. -- Hindleyite Converse 22:22, 20 November 2008 (UTC)
Graham Norton? --Nachlader 22:48, 20 November 2008 (UTC)
No thanks, I just ate. Thanks for the offer though. -- Hindleyite Converse 22:53, 20 November 2008 (UTC)
Great! Then I can have the Irish homosexual all to myself. --Nachlader 23:33, 20 November 2008 (UTC)
Too late, MetalFlower called him first. -- Hindleyite Converse 18:29, 21 November 2008 (UTC)
What can I say? I love the Irish :P talk 18:40, 21 November 2008 (UTC)

## Kippers has an idea

Let's join forces and go to war with Wikia. Pretty please? -- Kip > Talk Works 20:47, Nov. 20, 2008

Yes, but only if I can get more than 31 beans on my toast. -- Hindleyite Converse 22:24, 20 November 2008 (UTC)
You'll get 29 beans and you'll like it, dammit! - 00:19, 21 November 2008 (UTC)
Guys I'm eating cereal -
Aaah! You're a cereal killer! 00:48, 21 November 2008 (UTC)
...oh wait. I get it. You had to rip open the box to get to the insides. That's the joke, right?-- 01:45, 21 November 2008 (UTC)
I'm against going to war with Wikia. After all, I like Wikipedia just as much as I like Uncyclopedia. In fact, it wouldn't put too much of a fair point that I like all encyclopedias, whether serious or not, young or small. I like to call myself a pediaphile. --Nachlader 09:50, 21 November 2008 (UTC)
Wikia =/= Wikipedia.-- 11:49, 21 November 2008 (UTC)
Oh yeah?! Well, $Benson > Bradaphraser$ 11:59, 21 November 2008 (UTC)
...and?-- 12:25, 21 November 2008 (UTC)
My point exactly. 12:31, 21 November 2008 (UTC)
I know Wikia isn't Wikipedia, seesh there's no such thing as artistic licence these days... --Nachlader 14:29, 21 November 2008 (UTC)
Are they sanitary? I LOST my Username! (:() So I got a NEW one. (:)) Are the Admins still sexy as hell? Secret Agent Man 00:33, 22 November 2008 (UTC)
Oh, my. Here we go again! <cast smiles, freeze frame, roll credits> 02:56, 22 November 2008 (UTC)
<petition for a second series is thankfully rejected> --Nachlader 11:13, 22 November 2008 (UTC)

## No.

Merge us with Illogicopedia? Is this a joke? We have completely different styles - it just wouldn't work.

Wait, you aren't being serious, are you? Sorry.-- (CUN) 09:21, September 19, 2010 (UTC)

Check the date on the discussion. --Mn-z 12:56, September 19, 2010 (UTC)

### What's

Illogicopedia?  Avast Matey!!! Happytimes are here!*  ~  ~  19 Sep 2010 ~ 17:17 (UTC)

A very illogical place with bananas and waffles and eggspoons. And I feel as out-of-place there as I do here. ~ *shifty eyes* (talk) (stalk) -- 20100919 - 17:40 (UTC)

Went there recently. They don't seem to hate us; they're only grumbling, for example, resisting someone's offer to import UN:VFD cast-offs, because they wish they were good enough to be here. Whereas we already are.  18:53 19-Sep-10


text:
# XXVII International Workshop on Deep Inelastic Scattering and Related Subjects

Apr 8 – 12, 2019
Turin
Europe/Rome timezone

## Parton distribution functions from Lattice QCD

Apr 9, 2019, 9:50 AM
20m
Cavallerizza Reale - Aula Magna

### Cavallerizza Reale - Aula Magna

Via Verdi 9, Turin
Parallel Session Talk WG7: Future of DIS

Krzysztof Cichy

### Description

In this talk, I will discuss the computation of collinear PDFs in the framework of Lattice QCD, using the so-called quasi-PDF approach introduced by Xiangdong Ji in 2013 and intensively developed thereafter. We employed twisted mass fermions with light quark mass set to its physical value. The relevant matrix elements obtained on the lattice were non-perturbatively renormalized and converted to the $\overline{MS}$ scheme at the scale of 2 GeV. A matching process was applied together with nucleon mass corrections, leading to the reconstruction of light-cone PDFs. We observe a similar behavior between the lattice and phenomenological data, and for both types of polarized PDFs a nice overlap for a range of Bjorken-x values. This presents a major success for the emerging field of direct calculations of quark distributions using Lattice QCD.

### Primary authors

Prof. Martha Constantinou (Temple University) Karl Jansen (DESY) Ms Aurora Scapellato (University of Cyprus)


text:
# Solution: A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.

###### Problem

A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.

###### Solution

Recall: Graham's Law of Effusion allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.

This means that when comparing two gases:

Let's designate CO2 as gas 1 and the unknown as gas 2. We're given:
rateCO2 = 102 mL/s                  MCO2 = 12.01 g/mol C + 2(16.00 g/mol O) = 44.01 g/mol
rateunk = 83.3 mL/s                  Munk = ?

View Complete Written Solution


text:
March 29, 2017 10:10 am

Contents:

## Semi-classical approach

For a combination of a rotating radio-frequency (RF), with a frequency , and a static magnetic field, in a configuration known from nuclear magnetic resonance (NMR), the magnetic field is denoted as . The PauliSchrödinger equation for the neutron’s wavefunction is given by ,  which is solved with a separation ansatz denoted as , where denotes the neutron’s spatial wave function and the (time dependent) spinor wave function in matrix notation. This yields . Both sides must be equal to a constant given by , since the left side is only time depending and the right side only space dependent. The right side (only depending on the coordinate ), given by , and its solution is a plane wave given by , with .

However the spinor part remains much more complicated 1, where the equation , can be transformed with the substitution to . Now and are defined as and , which leads to . Next a unitary transformation is introduced, which transforms the equation in a system rotating around the -axis with a frequency of the magnetic field , yielding

When the equation is multiplied with from the left side and the exponential function is expanded in a power series (with ) as , and using (with ) we get

and in the same manner . This leads to the following simpler equation:

where the time dependency between the curly brackets has vanished. Now the following quantities are defined: and , which gives , where is the frequency of the rotating magnetic field, is related to the guide field and to the amplitude of the rotating field , as introduced above. Integration leads to the following equation:
. Since we get and in the non rotating system a solution is calculated as . Using the definition yields the final solution denoted as , with for instance accounting for an initial polarization pointing in -direction.

So far so good – the problem has been solved theoretically ! However what remains is to discuss the result and to understand its physical meaning. For this purpose, the second exponential has to be examined more precisely , discussing the linear expressions sequentially, we obtain and since we get . All together we finally obtain , with the time-independent unit-vector calculated as , with . In the rotating system, the neutron perceives the effective magnetic field .

In summary, the total wave function of the neutron can be expressed as . Using , and assuming the incident beam to be polarized in the -direction, i.e., and applying the Pauli matrices , and the identity matrix we finally get

which for yields , that is a free particle of energy   polarized in   direction ( ).

If the rotating magnetic field has a resonance frequency , the action of the static field is completely compensated and  has only a -component. This condition is referred to as frequency resonance. At this point we would like to consider the  -component of the polarization given by  which for frequency resonance reduces to . So if in addition the amplitude resonance for , namely , with being is the transmission time through the dynamical spin flipper with a flipper length , depending on the velocity of the neutron, is fulfilled we get  ; a spin flip occurred ! If we now calculate the wave function after the spin flip we get

with  , that is polarization in -direction. One can immediately see that compared to the total energy before the spin flipp, given by the neutron hast lost an amount of energy , by emission of a photon of frequency . In case of a flip from to the neutron absorbs an photon of the same energy from the rf-field.

However, in practice an oscillating field is used instead of a rotating one (though it is in principle feasible to implement a rotating filed). An oscillating field of frequency plus static magnetic guide field, denoted as , can be decomposed as the sum of two counter-rotating fields with frequencies :

The interaction representation used above describes the physics in a rotating frame with the field component ω with the Larmor precession of the spin in the static field. The other component is seen in this frame as a fast rotating field (), whose effect can be neglected, which is referred to as rotating wave approximation (RWA). A consequence of the rotating wave approximation is the so called Bloch Siegert shift, which originates from the second term 2. In 1940 Bloch and Siegert proved that the dropped part, oscillating rapidly, can give rise to a shift in the true resonance frequency such that . Now the frequency resonance is given by  and the amplitude becomes

The action of a resonance-frequency spin-rotator on the neutrons spin can be written using the unitary transformations , with the oscillation frequency of the resonance-frequency spin-rotator , and the polar angle of rotation . This is a rotation around -axes which is transformed into a frame rotating about the -axes at a frequency of . The phase of the resonance-frequency spin-rotator can be taken into account additionally, by an other unitary transformation .

## Coupling to quantized oscillating field

The Hamiltonian of a the neutrons energy in an oscillating magnetic field, due to the neutrons magnetic moment, is denoted as , with . Next operators and are defined as and , having the following action on the spin eigenstates and . The quantized magnetic field for a single mode is given by , using the use of the creation and annihilation operators and . Using these definitions the Hamiltonian is given by

with the coupling strength . Hence, the total Hamiltonian consisting of the magnetic energy and the energy of the radiation field, is given by

where the interaction Hamiltonian accounts for the coupling of the neutron with the radiation field, i.e. the energy exchange between the neutron and the quantized magnetic field. This Hamiltonian is referred to as Jaynes-Cummings (J-C) Hamiltonian 2, named after Edwin Thompson Jaynes and Fred Cummings.

The state vector of the oscillating field of frequency are represented by coherent states , which are eigenstates of and . The eigenvalues of coherent states are complex numbers, so one can write with . One can define a total dressed state 3 vector including not only the neutron system , but also the quantized oscillating magnetic field: , where for simplicity a single-plane-wave representation for the neutron’s wavefunction is chosen: At this point one can take a look at the action of the interaction Hamiltonian on a spin-up state 5:

Here represents the usual coupling strength of the neutron field interaction which yields after the interaction with RF-field : a spin flip due to emission of a photon of energy and a phase factor from the coherent state of the oscillating field. Now the time evolution of a dressed state, as defined above, is taken into account .

As a last step we can rewrite this as , where is associated with a corresponding state vector , just like in an atomic two-level system, where a certain energy level is associated with a ground state . Hence can be identified with (in analogy to the exited state ). Thus our equation yields .