text:
Talk:Q-function

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Some solutions

I don't understand what you mean by discrete solution. --Zvika (talk) 09:20, 21 July 2009 (UTC)
The values for, e.g., Q(0) or Q(1) or even Q(0.5). Tomeasy T C 17:56, 21 July 2009 (UTC)
Ah, that's a good idea. Will do. --Zvika (talk) 04:33, 22 July 2009 (UTC)

Bounds

There is an improved bound for Q(x), given at [1]:

${\displaystyle Q(x)\approx \varphi (x){\bigg (}{\frac {1}{x}}-{\frac {1}{x^{3}}}+{\frac {1\cdot 3}{x^{5}}}-{\frac {1\cdot 3\cdot 5}{x^{7}}}+{\frac {1\cdot 3\cdot 5\cdot 7}{x^{9}}}+\cdots {\bigg )},}$

where each partial sum gives either upper or lower bound, depending on the sign of the last summand. The sum itself is divergent however, so it cannot be used to calculate the integral with arbitrary precision.

Another approximation due to Laplace:

${\displaystyle Q(x)={\frac {\varphi (x)}{x+{\frac {1}{x+{\frac {2}{x+{\frac {3}{x+{\frac {4}{x+\ldots }}}}}}}}}}}$

… stpasha »  19:09, 1 October 2009 (UTC)

Sounds interesting and worthy of inclusion, if you can find a RS for it. --Zvika (talk) 18:19, 3 October 2009 (UTC)
I just noticed that the bounds given in the article correspond to Laplace’s continuous fraction terminated at k=0 and k=1 respectively:
${\displaystyle Q(x)\approx {\frac {\varphi (x)}{x}}\quad {\text{and}}\quad Q(x)\approx {\frac {\varphi (x)}{x+{\frac {1}{x}}}}}$
But you're right, a reliable source which attributes second formula to Laplace would be nice to have.  … stpasha »  18:45, 3 October 2009 (UTC)

eta?!

In the second paragraph of the article, shouldn't the eta be a mu? -- UKoch (talk) 15:11, 24 February 2014 (UTC)

Yes, it should be a mu, and ${\displaystyle \varphi (x)}$ should be ${\displaystyle \phi (x)}$, to be consistent with the notation on the Normal Distribution page. I will fix this. David C Bailey (talk) 08:06, 21 November 2014 (UTC)
Thanks! -- UKoch (talk) 17:57, 23 November 2014 (UTC)

When you have finished reviewing my changes, please set the checked parameter below to true or failed to let others know (documentation at {{Sourcecheck}}).


text:
# Where Is My Friend?

A friend of mine, one of those world-traveler, rugged outdoorsman, live-out-of-a-tent-and-backpack sorts, has just sent me an interesting email:

From: whatchootalkinboutGrylls@hotmail.com
Subject: Check out this pic! Guess where I am?!
To: rubio@[redacted]
Date: 04/08/17 11:33 AM

Dude, check out this epic pic:

It's pretty awesome, isn't it? I think we captured the scene perfectly.
That's obviously just a thumbnail; for full resolution, open the attachment!
I'll see you in a couple months, man - keep it real!
— WG

[Attachment(s):fullrespic.doc]

Where the picture should have been I just saw a blank space, but I know sometimes embedded images don't work all that well; good thing he provided the attachment. I figured I was just a click away from seeing an amazing photo in all its full-res glory. But on opening the attachment, instead of finding an image I found ... well ... I'll just show you what it said. Since one of you apparently is partially to blame for this, it's only fair you help me figure out where my friend is!

________

Yo, I know you like puzzles. One of the guys I'm with out here is a puzzler himself — he's even on that puzzle website you were telling me about. Anyway, he helped me put together a bit of a puzzle for you. See if you can figure out how to get through my trail of clues to find my picture and figure out where I'm at!

The trail begins with some clues for you to figure out....

### 1: One thing leads to another

1. I heard you sing Flying High Again? (5)
2. Weird old camel’s place in Maryland. (2*,6)
3. 904
4. IIX857, a synthetic hormone, makes a heavenly body. (####,8)
5. Unusual array: song, nose, tongue, ear, eye, or skin, maybe. (1,7,5)
6. Perhaps a stout tankard made of hot and heavy metal with an edge. (3,5)
7. A measure of temperature just above zero? (3,6?)
8. (½)(12)=6 (4,5)
9. A vehicle full of blind drunk denial heads west on Rodeo Drive for date, maybe. (8,4)
10. Live! Exit around center aisle! (5)
11. Strike her head suddenly, then bug-eye; then, almost unendingly wrong, start sprinting thither. (6’1,10)

### 2: Looking for connections

Those weren't just simple standalone clues.
There are sets of answers that go together. Figure out how, or your trail ends here!

1 & 4 & 11
2 & 5 & 7 & 10
3 & 8
6 & 9

Do you have a hunch how these are connected?
Blaise de Vigenére says: C M P X V G V P V :)

I'm going to call the way these are connected, LINKINGS.
When you're sure you know how LINKINGS work, proceed to ...

Taking your final answers from Part 1, collect all the letters from all the answers together. Sort them alphabetically, discarding duplicates. You should end up with 13 letters. They are your key to unlock the next step:

$\small\texttt{LFCHA A̕DBJZVZB TP WRZLHRZ O}$

If you've got the right answers from Part 1 ...
If you've correctly figured out the LINKINGS for Part 2 ...
If you've deciphered the code above ...
If you're reading me Loud and Clear ...

Then you should be able to figure out what to do with the following leads:

• You're going to need 5 specific letters, and I've just given you what they are.
• You're going to need a grid to put them in, and I've just given you what it looks like.
• There's only one way the letters can go into the grid.
• Make sure ALL your letters make it into the grid!

When you've got everything up to here squared away, proceed to ...

### 4: How to unlock the solution, from A to Z

You've filled in a grid, more or less. Time for the next step.
Take the grid's outermost letters and sort them alphabetically, discarding duplicates.
They're now your key to unlock the next clues:

Ypw'pt becsvzlz yevf gc jjpt ujiaq nzgcybfvmkmyzp amr zqyz ajb jvliua. :)

• Q: Infhipqe ugbbnbcb vzovv e abvqv dq fjtwb-obok kyiov #255,153,051? (4,7,6)
• V: Xmeh fic cmwi aqise
• E: Boxp wwggv Dzkpikw ugfwt. (5)
• J: Nasv svq pzlz (5)
• A: Ao gbbuaqk Nmlojsnqa qipqtbeiiz qwjxmvftil ua Sdzpy, Ppxidvc. (5)
• B: Kyru pyunrf fic cmwi vgzpvm dovt
• X: Uaactpjas gcmevuyo dos cr Iyrfzxyn nkrs ecstdcs. (5)
• W: Wrsqzdk Nsnecc Aoucfg wovvl tqnrvmq fpnpwivbx hc. (5)
• Q: Nu Hmdry yvq a eqrsql’g mjgcf. (5)
• W: Tidg hyz disux, kxhs kcc njpxp
• U: Swinr pbtx, kxhs epkbft (3)
• D: Mjvgkn: gn fzmafrbtz. (2)
• N: Sjzxp oyiv jd pbtx wzr
• V: Umsnlgr bqrbrbcr’t didqeoxz. (5,3)
• B: Io Tsuq, n blhzes qj ugervm aatgw umawwzqtff mv hvcczlcf krqfnzct. (5)
• Z: Cbuwqgf Qcvw bpzia QFDE NnosvwKqahvm jebfize nbu Amrnwpi Aas ivaes Liiz bf Xdsljcrw. (5)
• M: Pcnvpdma gzk bik dl pbkr. (6)
• O: Oyiv ouo gtsu bnfk jle
• T: Rezf V qcpc VJKM
• N: evunvgtjpjw() ieogkcr dtiifrr joycl qzmdszfr. (2)
• C: Tig zmdl tzmqt dnym
• I: Swinr pbtx, bqahy xjuf
• N: Rmugvvm wet qv va, csicypt. (5)
• A: Tidg cez alvg itqise
• I: Kebpmvs vb "Rtc-azg, Wqd!" (3)
• K: Owocr QT gpmauvpn, cppjcevbx kjadg fmonav xmhftivf. (5)

Bv, fic mpti btvbx. Tmu xqr'b wacn rfau vs la jwkc rhju cmf, oik D'k psqzqpvbx dr npy.
mushf: k-C-L-Y-t (rro)

Whew! What comes next?

### 5: Things aren't always black and white

Four of your answers from Part 4 are, well, colorful.
It's time to add some color to those LINKINGS! Here's how:

Sort your colorful answers alphabetically by the index letters of their corresponding clues.
Then apply these cryptical directions* to the answers in their sorted order, to find which clue (and thus what color) goes with each LINKING. $$\def\T#1{\small\text{#1}}\begin{array}{rcl}\hline\T{first clue}&~~~~&\T{painlessly behead}\\\T{second clue}&&\T{no novice in speech}\\\T{third clue}&&\T{cut to the core}\\\T{fourth clue}&&\T{remove hind end; make bland; remove hind end}\\\hline\end{array}$$                     * They're cryptic-like more or less, but not necessarily entirely fair. Deal with it. :)

The fifth LINKING also has a color; its color is: SILVER.
When you've found the colors for all five LINKINGS, proceed to ...

### 6: Hex marks the spot

For an effective hex, one must put the right things together in the right way.
It's time to use something I gave you in Part 4.

There are numbers, 1 through 5:

• Each has a location.
• Each location identifies a LINKING.
• Each LINKING identifies a hex code.
• Each LINKING also has a color, which itself identifies a hex code.
• XOR the two hex codes together.

Find the values for 1 to 5.
Put them together to find an image that should look very familiar
... but with some things missing, and some things new.

Useful URLs (among plenty of possibilities):
http://www.colorhexa.com/
http://xor.pw/

One more thing:
My friend wants me to tell you this message before you continue. He was going on about some Lord of the Rings quote — something about a phial, and light, and Eärendil’s star, and that his message was like that only the opposite? He's weird sometimes.
Anyway, he says, "King DEDEDE says OOOOOOh when he drinks his C0C0A0." I have no idea what he's talking about, but he made me promise to put that in here so there you go.

Moving on!

### 7: Being hexed may leave you feeling down

Part 4 gave you answers from A to Z.
Part 5 used four of them.
Time to use five more.

As you look down at your notes you should notice that some of your answers to Part 4 seem familiar from Part 3: some exactly, and some in part.
Now would be an excellent time to put what's missing into the image found in Part 6.
You'll find, as you write things down, that you're able to fill about 4½ columns using 5 answers.

That's another five answers down ...

### 8: Feeling down might lead to a cross word or two

Cross six more answers off the list.
(Do I really need to say more?)

By the way, remember the theme of this puzzle you found way back in Part 2?
The six entries you just crossed off? Yeah....

### 9: Be of good cheer — the end is near!

Eleven entries left. They spell out the key that will let you decipher the final text below.
If you're not sure how to find that key, it may help to remember where they came from and what I've told you about these entries previously.

(Tfh "cldc ld", en'h g tzqy!)
(Qsq "txcdra", irl dkxxn)
Zkzoky ^^ klmt lnljr.

Iy sfy'de kyrhunz nyme, yho'mi som nyi dizbk oqy — viekdamocefihhj!

Ef tacj taign psg sailpp htpv juleyu mz tay xvud B arzq yho zr Bakn 6 tsypeykixy, xrtibt yii sze xggxk sjorvq ig nyi pptx. Klmt xggxk sjorvq rxjiieegnj xte ictxgrx C gvambmvh kon... ueh zop, uk pmsm, C'd vqaws ks feef psg wayii ut bm.

Csak uuto mt Iuix 7. Defydfqr mbfwq fbpv ezspyiw? Uf mbv kdiw C xehe ril ams t nptuctf tvaslqfvp pntqpq gkcu, xtoly nsglw vv xte thjaqrl zfv fhx Xfaz ceovw zufvvvqd hhv xtrhoxl rioy. Z amnm sfy fo plzxq, ig iihqr, mbv E-L egnic xemnvve fhl veoh hz klasx zzzq agmnids, uok qmkx nyi adw hlqnekyu iztkcvw xopyi gmsx, ueh fhx ymiz nngsidew yexdixm ltbek wrwq - sh ck paodm cmwe qRoBj Tay iieuen zw, brxxzgfaufp, e bnz cdkgr mux. M fhbhb cau'ef rkdex nyi tizbvv delicyfihh zw m ghiu mypkimiyegn fzqr mbv xtufveeul B jlx un mbv iyabf sspy.

HHV PMSM NYMZG:
Bz psg'rx uj snsxmjmhe tm Z ey, tauk iypms jtat bh klq diuu qgsm vv hdiocek kon HLXE.
Fhlxiftbhx enonn klq ckijwiokx rrp chhjmpekcek vuln klmt wjrh, ihtn ciftxl xsqs bh klmt xggxk siuti, mnw qyc?

### Postscript

Hope you enjoyed the puzzle, and the picture.
Happy trails, dude!
— WG

________

So, I couldn't even get the first bits done!
I could really use a hand, and since my friend had *ahem* some help... it seems only fitting that I get some too.

Can you help me figure out:
What is this a picture of?
Where is my friend?

Part of a entry: I, for one, welcome our new ...
Metapuzzles answer: What word does the very last letter you solved for represent?

• (For other solvers: There's a spreadsheet here if you'd like to contribute.) – Deusovi Apr 15 '17 at 21:09

This answer is half a record of my solution path and half the "intended" solution path.

Here is the spreadsheet used to solve.

Thanks to Silenus and Sp3000 for the help solving.

### Part 1

The clues solve to USING, MD LOCALE, 904, 1714 ASTEROID, A SENSORY ORGAN, PUB ORDER, ONE DEGREE?, HALF DOZEN, CALENDAR WORD, EXIST, and HITHER'S ANTITHESIS.

These are all

clues to more words. For instance, 1714 ASTEROID is SY, EXIST could be BE, and HITHER'S ANTITHESIS could be YON. But what about the rest? There seem to be too many options. Hm... let's continue on for now.

### Part 4

Wait, part 4? Yes. A couple educated guesses (armed with the knowledge that the Vigenère key is alphabetical) reveal the key to be abceimnorvy. That reveals a ton of clues, but let's not worry about those just yet. We know that the key from 3 is also alphabetical, and a superset of part 4's key.

### Part 3

Time for more educated guesses! Just adding two more letters (L and S) gives the message LEADS P̕ROVIDED TO UNRAVEL X. Well, it seems like the "leads" will be helpful if we want to do any "unraveling"... and we need five letters... let's take LPTUX! (Er, LP̕TUX. Apparently P brought a diacritic along with it.)

The message mentions 5x5 grids... and we need 5 letters... Hey wait, those letters are pentominoes!

And this is where the puzzle finally starts to pick up. Suddenly, we can do a lot of things!

### Parts 2 and 3 (again)

We can decode the part 2 message with the keyword PENTOMINO(ES) to get the confirmation, NICE HUNCH.

We can assemble those pentominoes into a 5x5 grid into exactly 1 way (up to reflection/rotation).

We can solve the rest of the part 1 "subclues" and fit the answers into the pentominoes crossword-style.

(And we get the message "MAYBE CLEAR" for confirmation, which currently describes our thoughts on the puzzle!)
We can decode the part 3 and 4 messages the "right way". Speaking of part 4, it's time to solve some clues...

### Part 4 (for real this time)

We have 26 clues to solve. Luckily for us, 11 of them are from part 1. The rest are regular (non-contrived (usually)) cryptic clues. (Details on the sheet.)
We also have an image.

This'll be important later.

### Part 5

Four of our solutions from the part 4 clues have colors: YELLOW, GREEN TEA, BLUES, and DEEP SAFFRON PRIMER. The instructions tell us what to do with them: remove Y/OW, remove GREEN and take a homophone, take the centre letter, and remove DEE/R from the edges and SAFFRON from the middle.

This gives us letters: ELL, T, U, and P (prime). That's four of our pentominoes! We can now assign colors to them.

### Part 6

Look up the colors in each location based on the letter of the pentomino there. They look slightly different from the actual colors of the pentominoes. XORing them together gives us ASCII values kh30K, which lead to this image:

### Part 7 and 8

By this point, you may notice that CLEAR and MAYBE are solutions to two part 4 clues. And so are MCMIV and ERYON, which read downwards. We can cross off those words, then add letters in the edges of the X to cross off 7 more words: YES, MESSY, IS, SO, VISON, ALESI, BASSO. (All the added letters are S, incidentally.) This completes the grid (and gives confirmation for missing part 1 answers, if you're missing a clue or two).

### Part 9

We've used all clues from part 4 but the ones copied from part 1. Those clues' letters are EGIKOPRSUWY, which anagram to KEY IS GROW UP.

Decoding part 9 with GROW UP gives a helpful message that GROW UP is a cluephrase, not the literal key. So we try MATURE... and it works!
It gives us a confirming message:

### Part 9

So, we do what the message says. Take the "down" words, find their clue letters, make them alternate capital and lowercase, and we get this:

...Oh.
Well, I guess I should've expected that.

### Now what?

Let's see, what information have we not used? There are the three hex codes DEDEDE, 000000, and C0C0A0. There's the centre square of the d-pad.

And... there's the steganography tag. The image isn't all the same shade of white. Coloring the image in the colors provided gives this nice picture:

Of course! It's a polar bear blinking in a snowstorm, and that's why we couldn't see it. And polar bears live at the north pole, which is Rubio's friend's location.

### One Last Thing

So, what do we do with the D-pad? Ignore everything else, as the note says. We have four arrows, all of which have an S. This must be a sort of "compass rose", though of course every direction points south. Meaning the center would be the north pole itself, with the letter N. That means north is the answer we need for the upcoming metapuzzle.

• Huzzah! Nice teamwork, and thank you for patiently slogging through the rougher patches. I can now tell my friend I know exactly where he is, thank you so much for your help! – Rubio Apr 16 '17 at 5:35

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###### Question:

(a) Briefly explain three (3) methods of material selection. (3 marks) (b) With suitable example, explain three (3) types of design (6 marks)

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##### Utilice el ciclo de Born-Haber los datos de las tablas de los apendices para calcular laenergia reticular de CaO. (AH;635.5 k] /mol) (7 pts:) (2 pts por Ia contestacion,pts por el diagrama Born-Haber los clculos matematicos)"2587 kJmol2631.8 kJlmol2809 kJmol-3445.3 KJlmol
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##### How do you solve and write the following in interval notation: x ≤ -4?
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##### A report states that the mean household income last year for acertain rural county was $46,200 and the medianwas$54,500. If a histogram were constructed for the incomesof all households in the county, would you expect it to be skewedto the right, to the left, or approximately symmetric?Select one:A. skewed leftB. skewed rightC.approximately symmetric
A report states that the mean household income last year for a certain rural county was $46,200 and the median was$54,500. If a histogram were constructed for the incomes of all households in the county, would you expect it to be skewed to the right, to the left, or approximately symmetric? Select ...
##### Show that the function satisfies the wave equation:W = In(2x + Zct)
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##### Use Green's thcorem in order to compute thc line integral1" dy-y" drwhero C is the boundary of the square [0, 1] * [0, 1] traversed in the counterclockwise way.
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##### Part II: Dana Boar, controller of Digital Electronics Canada, developed the figures requested by...
Part II: Dana Boar, controller of Digital Electronics Canada, developed the figures requested by her boss and president of Digital Electronics Canada, Hans Fritz. The numbers allowed her to see how the projected sales volume for 2017 related to breakeven, and examine the relative profitability of t...
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##### It is given that for the function $f(x)=x^{3}-6 x^{2}+a x+b$ on $[1,3]$, Rolle's theorem holds with $c=2+frac{1}{sqrt{3}}$. Find the values of $a$ and $b$ if $f(1)=f(3)=0$.
It is given that for the function $f(x)=x^{3}-6 x^{2}+a x+b$ on $[1,3]$, Rolle's theorem holds with $c=2+frac{1}{sqrt{3}}$. Find the values of $a$ and $b$ if $f(1)=f(3)=0$....


text:
# How to suppress certain .bib fields (e.g. year) only for certain entry types (e.g. inproceedings)

How can I suppress the year from appearing in the bibliography for entries of type @inproceedings? In my specific case the year is implied in the booktitle (e.g. FOCS '09) so it is redundant when I need to save space. I found http://www.latex-community.org/forum/viewtopic.php?f=50&t=11447 and the "second solution" looks promising but I am not sure how to translate it to only suppress the year for one type of entry. I am guessing \ifentrytype might be helpful. I am using biblatex+biber.

-

\documentclass{article}

\usepackage{biblatex}

\AtEveryBibitem{%
\ifentrytype{inproceedings}{%
\clearfield{year}%
}{%
}%
}

\usepackage{filecontents}

\begin{filecontents}{\jobname.bib}
@misc{A01,
author = {Author, A.},
year = {2001},
title = {Alpha},
}
@inproceedings{B09,
author = {Buthor, B.},
editor = {Cuthor, C.},
year = {2009},
title = {Bravo},
booktitle = {FOCS'09},
}
\end{filecontents}

\nocite{*}

\begin{document}

\printbibliography

\end{document}

-
Thanks. I didn't realise it would be that straightforward. –  Raphael Jul 15 '11 at 5:17

Or, you can do this before the data even gets to biblatex. Assuming biber 0.9.8+; put this in your biber:conf:

<?xml version="1.0" encoding="UTF-8"?>
<config>
<sourcemap>
<maps datatype="bibtex" map_overwrite="1">
<map>
<per_type>INPROCEEDINGS</per_type>
<map_step map_field_set="YEAR" map_null="1"/>
</map>
</maps>
</sourcemap>
</config>

-


text:
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text:
Krein space

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Let be a complex linear space on a which a Hermitian sesquilinear form is defined (i.e. a mapping such that and for all , ). Then (or, more exactly, ) is called a Krein space if in there are two linear manifolds such that

 (a1)

and are Hilbert spaces (cf. Hilbert space) and . It is always assumed that (otherwise or is a Hilbert space); is called the indefinite inner product of the Krein space . If, in particular, , then is a -space or Pontryagin space of index (cf. also Pontryagin space); in the sequel, for a -space it is always assumed that .

Using the decomposition (a1), on the Krein space a Hilbert inner product can be defined as follows:

 (a2)

Although the decomposition (a1) is not unique, the decompositions of the components are uniquely determined and the Hilbert norms, generated by different decompositions (a1) according to (a2), are equivalent. All topological notions in a Krein space, if not stated explicitly otherwise, refer to this topology. In the Hilbert space , the orthogonal projections onto and are denoted by and , respectively. Then for the operator , called a fundamental symmetry, one has

 (a3)

and has the properties: , . Conversely, given a Hilbert space and in it an operator with these properties (or, more generally, an operator with , ), then an indefinite inner product is defined on by (a3) (or, respectively, by the relation

 (a4)

and is a Krein space. Because of this construction, Krein spaces are sometimes called -spaces.

If, more generally, a Hilbert space and a bounded self-adjoint, not semi-definite, operator in are given, the relation (a4) with defines a Hermitian sesquilinear form on . This form can be extended by continuity to the completion of the quotient space with respect to the norm (). This completion, equipped with , is a Krein space containing as a dense subset.

If is a real and locally summable function on which assumes positive and negative values on sets of positive Lebesgue measure, then the space of all (classes of) measurable functions (cf. Measurable function) on such that and equipped with the indefinite inner product () is a Krein space. More generally, if is a real function which is locally of bounded variation and not isotone on and denotes its total variation, then the space , of all measurable functions such that and equipped with the indefinite inner product () is a Krein space.

Further, a complex linear space with a Hermitian sesquilinear form , which has negative squares (that is, each linear manifold with for , , is of dimension and at least one such manifold is of dimension ), can be canonically imbedded into a -space by taking a quotient space and completing it (see [a4], [a2], [a9], [a11]).

The indefinite inner product on the Krein space gives rise to a classification of the elements of : is called positive, non-negative, neutral, etc. if , , , etc. A linear manifold or a subspace in is called positive, non-negative, neutral, etc. if all its non-zero elements are positive, non-negative, neutral, etc. The set of all, e.g., non-negative elements is not linear, but it contains subspaces, and among them maximal ones, called maximal non-negative subspaces. All maximal non-negative subspaces of the Krein space have the same dimension (as ). A subspace of (with the decomposition (a1)) is maximal non-negative if and only if it can be written as , where , the angular operator of , is a contraction from into . A dual pair of subspaces of is defined as follows: is a non-negative subspace, is a non-positive subspace and . Any dual pairs is contained in a maximal dual pair (maximality of dual pairs is defined in a natural way by inclusion); in a maximal dual pair the subspace (respectively, ) is maximal non-negative (respectively, non-positive) (R.S. Phillips).

Using the indefinite inner product, orthogonality can be defined in : are called orthogonal if ; if , then . Some properties of orthogonality in a Hilbert space are preserved; however, there are also essential differences; e.g., can contain non-zero vectors; coincides with if is neutral, and is equivalent to .

For a densely-defined linear operator in the Krein space an adjoint (sometimes called -adjoint) is defined by (, ). If denotes the adjoint of in the Hilbert space (see (a2)), then evidently . Now in the Krein space classes of operators are defined more or less similarly to the case of a Hilbert space: is symmetric if , self-adjoint if , dissipative if (), contractive if (), unitary if is bounded, and , etc. Also, new classes of operators arise: E.g., is a plus-operator if implies , and a doubly plus-operator if and are plus-operators. In a Krein space a densely-defined isometric operator (i.e. for all ) need not be continuous. As in a Hilbert space, self-adjoint and unitary, symmetric and isometric, dissipative and contractive operators are related by the Cayley transform. E.g., if , and , then is unitary.

The spectrum of a self-adjoint operator in a Krein space is not necessarily real (it can even cover the whole plane), but it is symmetric with respect to the real axis. Similarly, the spectrum of a unitary operator is symmetric with respect to the unit circle.

The indefinite inner product sometimes gives a classification of the points of the spectrum of an operator: An eigen value is said to be of positive type (negative type, etc.) if the corresponding eigen space is positive (negative, etc.).

If , are isolated eigen values of the self-adjoint operator in a Krein space, then for the corresponding Riesz projections , one has , and if, e.g., , then the restrictions and have the same Jordan structure. If in a -space the symmetric operator has a real non-semi-simple eigen value , then the corresponding algebraic eigen space can be decomposed into a direct orthogonal sum: , where is a positive subspace contained in the geometric eigen space of at , and is invariant under with ; if are the lengths of the Jordan chains of , one puts ; if is a non-real eigen value of , one defines as the dimension of the corresponding algebraic eigen space. Then , where the sum extends over all eigen values of in the closed upper half-plane. In particular, the length of any Jordan chain of is , and the number of eigen values of in the open upper half-plane, and also the number of non-semi-simple eigen values of , does not exceed .

Specific results for Krein spaces are statements about the existence of maximal non-negative (or maximal non-positive) subspaces, which are invariant under a given operator. The first general result of this type was proved by L.S. Pontryagin in 1944, stating that a self-adjoint operator in a -space has a -dimensional non-positive (that is, a maximal non-positive) invariant subspace. Subsequently, similar results were proved for various classes of operators in Krein spaces. E.g., a bounded linear operator in a Krein space has a maximal non-negative invariant subspace if is compact and, additionally, is self-adjoint or dissipative or unitary or a plus-operator, etc. (see [a2], [a4]). One possibility for proving these results, e.g. for a unitary operator , is to establish the existence of a fixed point of the fractional-linear transformation

where is a contraction from into (an angular operator) and is the matrix representation of with respect to (a1). By different methods also in other cases the existence of a maximal non-negative invariant subspace has been proved, e.g.: 1) is unitary and is uniformly bounded for all ; 2) for all , , and ; and 3) is bounded, self-adjoint and there exists a polynomial such that (). In many cases these maximal non-positive invariant subspaces can be specified by properties of the spectrum of . E.g., if is bounded, self-adjoint and is compact, then can be chosen such that . There are also results about the existence of a common invariant maximal non-positive subspace for a commuting family of operators, e.g.: A commuting family of bounded self-adjoint operators in a -space has a common maximal non-negative invariant subspace (M.A. Naimark; for applications in the representation theory of groups in -spaces see [a19]). Phillips asked ([a16]) if a dual pair of subspaces of which are invariant under a commutative algebra of bounded self-adjoint operators in the Krein space can always be extended to a maximal dual pair whose subspaces are still invariant under (which would imply that each bounded self-adjoint operator in has a maximal non-negative invariant subspace). Only partial solutions to this problem are known (cf. [a4], [a2], [a14]).

A self-adjoint operator in the Krein space is called definitizable (positizable in [a4]) if and if there exists a polynomial such that (). Each self-adjoint operator in a -space has this property (where can be chosen to be with the minimal polynomial of , being a -dimensional non-positive invariant subspace of ); also, each self-adjoint operator in a Krein space for which and for which the Hermitian sesquilinear form () has a finite number of negative squares, is definitizable.

The non-real spectrum of the definitizable operator consists of at most finitely many eigen values, and has a spectral function, with possibly certain critical points [a13], [a2]. This means that there is a finite set (of critical points) such that on the semi-ring , consisting of all bounded intervals of with end points not in and their complements, a homomorphism with values in the set of all self-adjoint projections in the Krein space is defined, such that for : a) is a positive (negative) subspace if (respectively, ) on for some definitizing polynomial of ; b) is in the double commutant of the resolvent of ; and c) if is bounded, then and , . If, in particular, is bounded and (), then , and one has

for some bounded operator such that , , ().

If the spectrum of a definitizable operator is discrete, then the linear span of its algebraic eigen spaces is dense in ; if is compact and self-adjoint in a -space and , then there is a Riesz basis of consisting of eigen and associated vectors of (I.S. Iokhvidov).

There is a theory of extensions of symmetric operators to self-adjoint operators and of generalized resolvents in -spaces, and also in Krein spaces, which is similar to the Hilbert space situation. The same is true for dilation theory: Each bounded linear operator in a Krein space has a unitary dilation in some Krein space ([a2]). In this context one has the following result: Let , be Krein spaces, a simply-connected open domain with smooth boundary such that , , and let be a function which is holomorphic in whose values are bounded linear operators from to . Then there exists a Krein space and a unitary operator

such that

(T.Ya. Azizov, see [a2], [a6]; here unitary means that maps the Krein space continuously onto the Krein space , preserving the indefinite inner product).

Some of the first papers about Krein spaces or, more generally, spaces with indefinite inner product, were stimulated by problems of (quantum) mechanics ([a4], [a2]; see also [a18], [a17]). Operators in Krein spaces arise also in a natural way in problems in mathematical analysis. Some examples of these are: I) Consider the canonical system of differential equations on , where , are matrices, , , and let be the corresponding matrizant (cf. Cauchy operator): , . Then is -unitary (that is, unitary with respect to the inner product defined in by the matrix , see (a3)), and, e.g., in the stability theory for periodic equations the classification of the eigen values of into those of positive or negative type plays an essential role ([a5], [a8]). II) The integral operator , real and of bounded variation on the interval , (), is self-adjoint in the Krein space . III) The theory of dual pairs of subspaces of a Krein space and their extensions to maximal dual pairs is related to certain questions in the theory of extensions of dissipative operators in a Hilbert space to maximal dissipative ones. Phillips started these investigations in connection with the Cauchy problem for dissipative hyperbolic and parabolic systems (see [a2], [a4] for references). IV) With the monic operator polynomial , bounded self-adjoint operators in some Hilbert space , one can associate the so-called companion operator

which is self-adjoint in the Krein space , (), where is the inner product in and

If, e.g., and is compact and , the results about the existence of maximal non-negative invariant subspaces mentioned above imply that there exists a bounded linear operator in satisfying , and [a12]. In a similar way, if , and are matrices such that has positive and negative eigen values, the solutions of the matrix Riccati equation

with are in bijective correspondence with all maximal non-positive subspaces which are invariant under the self-adjoint operator in the -dimensional Krein space , equipped with the indefinite inner product (a4) (see [a8]). V) If is a formally-symmetric regular ordinary differential operator on the interval with symmetric boundary conditions at and , and is a summable function on which is not of constant sign (a.e.) on , then the differential equation leads to a self-adjoint operator in the Krein space with inner product . If is semi-bounded from below, the operator is definitizable. VI) Krein spaces can be associated with certain eigen value problems for ordinary differential operators containing the eigen value parameters in the boundary conditions. E.g., consider in the problem

which is supposed to have a limit point at and with a boundary condition at (, are functions which are holomorphic on some set and satisfying a symmetry condition). The solution of this problem can be represented as (), where, in general, is a self-adjoint operator in some Krein space and is the orthogonal projection from onto [a17]. VII) Certain classes of analytic functions are closely related to the theory of operators in -spaces. This concerns, e.g., functions which are defined and meromorphic in the upper half-plane (or the unit disc) and which are such that the kernel

(or

has negative squares (that is, for arbitrary and , the matrix has at most negative eigen values and for at least one choice of it has negative eigen values). Corresponding extrapolation or moment problems can be treated by making use of results of the theory of symmetric or isometric operators in -spaces (see [a12], [a2]).

References

 [a1] T.Ya Azizov, I.S. Iokhvidov, "Linear operators in spaces with indefinite metric and their applications" Russian Math. Surveys , 15 (1981) pp. 438–490 Itogi Nauk. i Tekhn. Mat. Anal. , 17 (1979) pp. 113–205 [a2] T.Ya Azizov, I.S. Iokhvidov, "Foundations of the theory of linear operators in spaces with indefinite metric" , Moscow (1986) (In Russian) [a3] T. Ando, "Linear operators in Krein spaces" , Hokkaido Univ. (1979) [a4] J. Bognár, "Indefinite inner product spaces" , Springer (1974) [a5] Yu.L. Daletskii, M.G. Krein, "Stability of solutions of differential equations in Banach space" , Amer. Math. Soc. (1974) (Translated from Russian) [a6] A. Dijksma, H. Langer, H.S.V. de Snoo, "Unitary colligations in Krein spaces and their role in the extension theory of isometries and symmetric linear relations in Hilbert spaces" S. Kurepa (ed.) et al. (ed.) , Foundational analysis II , Lect. notes in math. , 1247 , Springer (1987) pp. 1–42 [a7] A. Dijksma, H. Langer, H.S.V. de Snoo, "Symmetric Sturm–Liouville operators with eigenvalues depending boundary conditions" , Oscillation, Bifurcations and Chaos , CMS Conf. Proc. , 8 , Amer. Math. Soc. (1987) pp. 87–116 [a8] I. [I. Gokhberg] Gohberg, P. Lancaster, L. Rodman, "Matrices and indefinite scalar products" , Birkhäuser (1983) [a9] I.S. [I.S. Iokhvidov] Iohidov, M.G. Krein, H. Langer, "Introduction to the spectral theory of operators in spaces with an indefinite metric" , Akademie Verlag (1982) [a10] V.I. Istraţescu, "Inner product spaces. Theory and applications" , Reidel (1987) [a11] M.G. Krein, "Introduction to the geometry of indefinite -spaces and the theory of operators in these spaces" , Second Math. Summer School , 1 , Kiev (1965) pp. 15–92 (In Russian) [a12] M.G. Krein, H. Langer, "Ueber einige Fortsetzungsprobleme, die eng mit der Theorie hermitescher Operatoren im Raume zusammenhängen, I: Einige Funktionenklassen und ihre Darstellungen" Math. Nachr. , 77 (1977) pp. 187–236 [a13] H. Langer, "Spectral functions of definitizable operators in Krein spaces" D. Butković (ed.) et al. (ed.) , Functional analysis , Lect. notes in math. , 948 , Springer (1982) pp. 1–46 [a14] H. Langer, "Invariante Teilräume definisierbarer -selbstadjungierter Operatoren" Ann. Acad. Sci. Fenn A. I , 475 (1971) [a15] J. Milnor, D. Husemoller, "Symmetric bilinear forms" , Springer (1973) [a16] R.S. Phillips, "The extensions of dual subspaces invariant under an algebra" , Proc. Internat. Symp. Linear Spaces (Jerusalem, 1960) , Pergamon (1961) pp. 366–398 [a17] L. Bracci, G. Morchio, F. Strocchi, "Wigner's theorem on symmetries in indefinite metric spaces" Comm. Math. Phys. , 41 (1975) pp. 289–299 [a18] K.L. Nagy, "State vector spaces with indefinite metric in quantum field theory" , Noordhoff (1966) [a19] M.A. Naimark, R.S. Ismagilov, "Representations of groups and algebras in a space with indefinite metric" Itogi Nauk. i Tekhn. Mat. Anal. (1969) pp. 73–105 (In Russian) [a20] M.G. Krein, H. Langer, "On some mathematical principles in the linear theory of damped oscillations of continua" Integral Equations, Operator Theory , 1 (1978) pp. 364–399; 539–566 Proc. Internat. Symp. Appl. Theory of Functions in Continuum Mechanics, Tbilizi , 2 (1963) pp. 283–322
How to Cite This Entry:
Krein space. H. Langer (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Krein_space&oldid=18988
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098


text:
# align with long equations

I have a very long align* with lots of \intertext. Some of the equations are really long, like this:

What is the most sensible way to typeset this? So far I have always used the following:

&= \bra{\pi^-} \bra{\pi^-} \delta^{ab} \delta^{cd} \ket{\pi^+} \ket{\pi^+} A_s
+ \bra{\pi^-} \bra{\pi^-} \delta^{ac} \delta^{bd} \ket{\pi^+} \ket{\pi^+} A_t
+ \bra{\pi^-} \bra{\pi^-} \delta^{ad} \delta^{bc} \ket{\pi^+} \ket{\pi^+} A_u \,.

This seems to work, although I am not sure whether \quad is really the right measure to shift it.

Now I have tried to use some nested thing:

&=
\begin{aligned}[t]
&\bra{\pi^-} \bra{\pi^-} \delta^{ab} \delta^{cd} \ket{\pi^+} \ket{\pi^+} A_s
\\
&+ \bra{\pi^-} \bra{\pi^-} \delta^{ac} \delta^{bd} \ket{\pi^+} \ket{\pi^+} A_t
\\
&+ \bra{\pi^-} \bra{\pi^-} \delta^{ad} \delta^{bc} \ket{\pi^+} \ket{\pi^+} A_u \,.
\end{aligned}

That looks like this now:

Only with {aligned}[t] I got it to look correct. Without the [t] the equal-sign would be in the middle. With a simple gathered the three lines would be centered, that looked really weird:

The aligned seems to be a bit more typing. However, the semantics in the LaTeX code seem more clear and the output seems to be a bit more consistent, the + pretty accurately under the \langle. The spacing between = and \langle in the first line got a bit larger, that is probably a bad thing.

Is there a good way to do this?

• It may depend on what's to the left of the = sign. There are various other possibilities. Could we know the l.h.s.,so we can test?. Jul 9, 2016 at 18:37
• Please, make a full example. Jul 9, 2016 at 19:50

just using the align or align* environment should give you what you want:

\documentclass[preview]{standalone}
\usepackage{amsmath}
\usepackage{braket}
\begin{document}
\begin{align}
=&\bra{\pi^-} \bra{\pi^-} \delta^{ab} \delta^{cd} \ket{\pi^+} \ket{\pi^+} A_s\\
&+ \bra{\pi^-} \bra{\pi^-} \delta^{ac} \delta^{bd} \ket{\pi^+} \ket{\pi^+} A_t\\
&+ \bra{\pi^-} \bra{\pi^-} \delta^{ad} \delta^{bc} \ket{\pi^+} \ket{\pi^+} A_u \,.
\end{align}
\end{document}

gives you

Is that the thing you are looking for?

• That's pretty much what I have been doing so far. The slight problem are the three numbers on the side, but I use align* on the outer level anyway. Jul 9, 2016 at 18:35
• you could also suppress line numbering for specific line by adding \notag or give one number to multiple lines by putting them in an \begin{split} ... \end{split} environment. Jul 9, 2016 at 18:39
• Might want to use = {} &  or the spacing might be wrong. Jul 9, 2016 at 23:21

Not an answer; too long for a comment.

My quick reading seems to show that the only thing that changes from line to line in these formulas is the partition of abcd and the subscript on A . Is there a way to rewrite the discussion to separate out the constant parts from the variable parts. That might both make the alignment problem go away and make the argument easier for your readers.

If you want to keep the equations whole I think you should consider making some of the outer parenthesis and bracket pairs larger.

I don't think the intertext is a problem - but you could solve it by naming the left hand side of these equations and repeating the short name. That wouldn't make the block of lines much wider.


text:
# Two stationary positive point charges, charge 1 of magnitude 3.10 nC and charge 2 of magnitude...

###### Question:

Two stationary positive point charges, charge 1 of magnitude 3.10 nC and charge 2 of magnitude 1.95 nC , are separated by a distance of 51.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Part A

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

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##### A determinant is given as follows: I a b c I d e f = 5...
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# Communication Systems/What is Modulation?

Modulation is a term that is going to be used very frequently in this book. So much in fact, that we could almost have renamed this book "Principals of Modulation", without having to delete too many chapters. So, the logical question arises: What exactly is modulation?

## Definition

Modulation is a process of mixing a signal with a sinusoid to produce a new signal. This new signal, conceivably, will have certain benefits over an un-modulated signal.

${\displaystyle f(t)=A\sin(\omega t+\phi )}$

we can see that this sinusoid has 3 parameters that can be altered, to affect the shape of the graph. The first term, A, is called the magnitude, or amplitude of the sinusoid. The next term, ${\displaystyle \omega }$ is known as the frequency, and the last term, ${\displaystyle \phi }$ is known as the phase angle. All 3 parameters can be altered to transmit data.

The sinusoidal signal that is used in the modulation is known as the carrier signal, or simply "the carrier". The signal that is used in modulating the carrier signal(or sinusoidal signal) is known as the "data signal" or the "message signal". It is important to notice that a simple sinusoidal carrier contains no information of its own.

In other words we can say that modulation is used because some data signals are not always suitable for direct transmission, but the modulated signal may be more suitable.

## Types of Modulation

There are 3 basic types of modulation: Amplitude modulation, Frequency modulation, and Phase modulation.

amplitude modulation
a type of modulation where the amplitude of the carrier signal is modulated (changed) in proportion to the message signal while the frequency and phase are kept constant.
frequency modulation
a type of modulation where the frequency of the carrier signal is modulated (changed) in proportion to the message signal while the amplitude and phase are kept constant.
phase modulation
a type of modulation where the phase of the carrier signal is varied accordance to the low frequency of the message signal is known as phase modulation.

## Why Use Modulation?

Clearly the concept of modulation can be a little tricky, especially for the people who don't like trigonometry. Why then do we bother to use modulation at all? To answer this question, let's consider a channel that essentially acts like a bandpass filter: both the lowest frequency components and the highest frequency components are attenuated or unusable in some way, with transmission only being practical over some intermediate frequency range. If we can't send low-frequency signals, then we need to shift our signal up the frequency ladder. Modulation allows us to send a signal over a bandpass frequency range. If every signal gets its own frequency range, then we can transmit multiple signals simultaneously over a single channel, all using different frequency ranges.

Another reason to modulate a signal is to allow the use of a smaller antenna. A baseband (low frequency) signal would need a huge antenna because in order to be efficient, the antenna needs to be about 1/10th the length of the wavelength. Modulation shifts the baseband signal up to a much higher frequency, which has much smaller wavelengths and allows the use of a much smaller antenna.

## Examples

Think about your car radio. There are more than a dozen (or so) channels on the radio at any time, each with a given frequency: 100.1MHz, 102.5MHz etc... Each channel gets a certain range (usually about 0.22MHz), and the entire station gets transmitted over that range. Modulation makes it all possible, because it allows us to send voice and music (which are essential baseband signals) over a bandpass (or "Broadband") channel.

## non-sinusoidal modulation

A sine wave at one frequency can be separated from a sine wave at another frequency (or a cosine wave at the same frequency) because the two signals are "orthogonal".

There are other sets of signals, such that every signal in the set is orthogonal to every other signal in the set.

A simple orthogonal set is time multiplexed division (TDM) -- only one transmitter is active at any one time.

Other more complicated sets of orthogonal waveforms -- Walsh codes and various pseudo-noise codes such as Gold codes and maximum length sequences -- are also used in some communication systems.

The process of combining these waveforms with data signals is sometimes called "modulation", because it is so very similar to the way modulation combines sine waves are with data signals.


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Differential Equation with Integration

Function $f(x)$ is differentiable. Given further that

\begin{eqnarray*} f’(x)+xf’(x-1) &=& 4 \\ \int_0^1 f(xt) dt + \int_0^x f(t-1)dt &=& x^3 +x^2 +2x \end{eqnarray*}

Determine the function $f(x)$.

$\textbf{Attempt}$ I know $\frac{d}{dx} \int_0^x f(t-1)dt = f(x-1)$, but I am confused about what’s $\frac{d}{dx} \int_0^1 f(xt) dt$ . Maybe I can do $- \frac{d}{dx} \int_1^x f(xt) dt - \frac{d}{dx} \int_0^x f(xt) dt$ ?

You can set $xt=u$.

Setting $xt=u$ gives $x=\frac{du}{dt}$, so $$\int_0^1 f(xt) dt =\int_{0}^{x}f(u)\cdot\frac{du}{x}$$ So, differentiating the both sides of $$\frac 1x\int_{0}^{x}f(u)du+ \int_0^x f(t-1)dt =x^3 +x^2 +2x$$ gives $$-\frac{1}{x^2}\int_{0}^{x}f(u)du+\frac{f(x)}{x}+f(x-1)=3x^2+2x+2,$$ i.e. $$\int_{0}^{x}f(u)du=xf(x)+x^2f(x-1)-3x^4-2x^3-2x^2$$ Now, differentiate the both sides and use $f'(x)+xf'(x-1)=4$.

For the proposer's work:

Using the Leibniz differential rule one has: $$\frac{d}{dx} \, \int_{a}^{b} f(x, t) \, dt = f(x, b) \frac{d b}{dx} - f(x,a) \frac{d a}{dx} + \int_{a}^{b} \frac{d f}{dx} \, dt,$$ where $a = a(x)$, and $b = b(x)$. In this example $$\frac{d}{dx} \, \int_{0}^{1} f(x t) \, dt = \int_{0}^{1} t \, f'(x t) \, dt.$$

Since $$\int_0^1 f(xt) dt + \int_0^x f(t-1)dt = x^3 +x^2 +2x$$ then differentiation would lead to $$\int_{0}^{1} t \, f'(x t) \, dt + f(x-1) = 3 x^2 + 2 x + 2.$$ Seemingly this leads to just as much "confusion".

Alternatively one can take the path of the following.

Using $$\int_{0}^{1} f(x t) \, dt = \text{ with } u = xt \text{ then } \int_{0}^{x} \frac{f(u) \, du}{x}$$ and \begin{align} \int_{0}^{1} f(x t) \, dt + \int_{0}^{x} f(t-1) \, dt &= g(x) = x^3 + x^2 + 2 x \\ \int_{0}^{x} ( f(t) + x \, f(t-1) ) \, dt &= x \, g \\ \frac{d}{dx} \int_{0}^{x} ( f(t) + x \, f(t-1) ) \, dt &= \frac{d}{dx}(x \, g) \\ f(x) + x \, f(x-1) + \int_{0}^{x} f(t-1) \, dt &= \frac{d(x g)}{dx} \end{align} Differentiation again leads to \begin{align} f'(x) + x \, f'(x-1) + f(x-1) + f(x-1) &= \frac{d^2(x g)}{dx^2} \\ 4 + 2 \, f(x-1) &= 2(6 x^2 + 3x + 2) \\ f(x-1) &= 6 x^2 + 3 x = 3x \, (2x +1) \\ f(x) &= 6 x^2 + 15 x + 9. \end{align}


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# QMA definition of difference between probabilities intuition

I'm reading about the complexity classes related to quantum computation, currently I'm studying QMA class. A language is in QMA(c,s) if there exists a polynomial time verifier and polynomial $$p(n)$$ such that

if $$x\in L \implies$$ there exists quantum state $$\psi$$ such that the probability V accepts is greater than $$c$$

if $$x \notin L \implies$$ For all quantum states $$\psi$$ the probability that V accepts is less than $$s$$.

Usually it is remarked that $$c$$ and $$s$$ have the following property $$c-s = \frac{1}{poly(n)}$$. This last condition is what confuses me. What is the intuition behind imposing this condition on the difference of $$c$$ and $$s$$? Not understanding this makes me feel I haven't fully understood QMA class definition.

Something similar happens with the problem called Local Hamiltonian problem. Here two constants are defined such that $$a-b = \frac{1}{poly(n)}$$. If someone could explain the intuition behind this and if it is related to the previous question that would be great. Thanks!

While this question deals with the difference between $$\mathsf{BPP}$$ and $$\mathsf{PP}$$, it should still be useful for you to read through. The "short answer" is that the "gap" between $$c$$ and $$s$$ allows for standard amplification techniques to be used, which end up showing that $$\mathsf{QMA}(c,s) = \mathsf{QMA}(2^{-n},1-2^{-n})$$ (provided $$c - s$$ meets the condition you wrote down).
The "case study" of $$\mathsf{BPP}$$ vs $$\mathsf{PP}$$ should be interesting overall for your question, as the only difference between them is the presence / lack of the $$c - s \geq 1 / \mathsf{poly}(n)$$ condition. It's known that: $$\mathsf{BQP}\subseteq\mathsf{QMA}\subseteq \mathsf{PP}\subseteq\mathsf{PSPACE}$$ But: $$\mathsf{BPP}\subseteq\mathsf{BQP}$$ This is quite a large difference. I'm unsure of what is known about $$\mathsf{QMA}(c,s)$$ where $$c - s$$ is negligible (meaning $$n^{-\omega(1)}$$), where amplification techniques can't be used.


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## Tuesday, April 29, 2008

### [AD 14] Test of MSD and PSD in s*d model

Within the sd-model, it can be shown that the VAC depends only on the mean value d_mean and on the variance d_var=sigD2 of the step width distribution, not on the detailed distribution function. The PL-exponent is not at all affected by the step widths.

As a test, the correlation function Css(m) of the sign factors was chosen as

$C_{ss}(m)=\frac{\sigma_s^2}{1+|m|^{2-\beta}}$

This form assures a smooth transition to the required value of Css(0)=sigS2=1/4. The PL regime starts at m>>1. Then, the VAC of the correlated part of the random process yields

$C_{vv}(m)=\frac{\sigma_s^2}{1+|m|^{2-\beta}}\cdot\frac{\overline{d}^2\!+\!\delta_{m,0}\;\sigma_d^2}{\Delta&space;t^2}$

For simplicity, it was assumed for the evaluation that

$\overline{d}^2&space;=&space;\sigma_d^2$
Using the discrete MSD formula from [MT 6], one finds the expected behaviour:

A fast Fourier transform of Cvv yields the corresponding PSD (the frequencies are in Hz):

The simulation parameters were:

dT=0.1; // simulation time intervall
NMX=16*1024; // nr. of sim. time steps

sigN2=0.5e-4; // variance of white noise
dMean=0.5e-2; // mean step width
sigD2=dMean*dMean; // variance of step width
beta=1.25, 1.5, 1.75; // PL exponent of MSD

Note that a mean step width of about 5 nm seems to be a reasonable value, in order to get the sub- to superdiffusive transition time correct.


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## LaTeX Tips

### Resources

Resources I've found useful are:

### Tables that fill the page

Using tabularx you can get tables that fill the width of the page and automatically assign space for the columns without having to specify a parbox of such and such a width.

Just use X for the dynamic columns and specify the width of the table as \textwidth

\begin{tabularx}{\textwidth}{lX}
foo & lots of text here that wraps and goes over the line length \\
bar & Look another one! \\
\end{tabularx}

TheEarthWiki: LaTeXCompTips (last edited 2008-02-19 15:39:22 by localhost)


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## The Quantum-Classical Debate:

Today I will reply to the arguments Andrei made last week in his guest post. But let me first start by giving out the answer to the questions from two posts ago. First, on why there is no uncertainty relationship for spin, from any two non-commuting observables A and B one obtains:

$$\sigma_A \sigma_B \geq \frac{1}{2} |\langle [A,B] \rangle|$$

The key point of Heisenberg uncertainty relationship for position and momenta is to be pedantic and observe that the commutator is proportional with the identity operator $$I$$:

$$[x,p] = i \hbar I$$

which is not true in the case of spin. Because of this, in the spin case there exists states for which the right hand side value $$|\langle [A,B] \rangle|$$ is zero. The position-momenta uncertainty remains bounded from below for any state.

For the other question on the apparent violation of the uncertainty principle. here is what Heisenberg stated:

"If the velocity of the electron is at first known, and the position then exactly measured, the position of the electron for times previous to the position measurement may be calculated. For these past times, δpδq is smaller than the usual bound." and "the uncertainty relation does not hold for the past." I think this is not a well known or appreciated fact by the majority of physics community.

Then Heisenberg pointed out that these values can never be used as initial conditions in a prediction about the future behavior of the electron.

Now back to answering Andrei's challenge to quantum mechanics, Andrei discussed 3 points:

Objection 1: Classical, local theories have been ruled out by Bell’s theorem.
Objection 2: Classical theories cannot explain single-particle interference (double slit experiment), quantum tunneling, the stability of atoms or energy quantification in atoms or molecules.
Objection 3: Even if one could elude the previous points, there is no reason to pursue classical theories because quantum mechanics perfectly predicts all observed phenomena.

Let's analyze them in turn.

On Objection 1, I agree that classical, local theories have been ruled out by Bell’s theorem with only one loophole left: super-deterministic theories pursued by 't Hooft. Any statistical theory obeying Kolmogorov's axioms respects Bell's inequalities. It is interesting to see how different realistic quantum mechanics interpretations escape Kolmogorov's axioms: Bohmian interpretation is contextual while quantum mechanics in phase space uses negative probabilities.

On superdeterminism, one needs to deny free will and this is a very tall order. While I (and anyone else) cannot give a rigorous definition of free will, I know that I have it. Andrei contends that classical theory is deterministic. While true, this is both an insufficient and an irrelevant argument. Superdeterminism is only a pre-requisite step: you need to obtain from it quantum correlations, and so far I am not aware of any successful model. Second, determinism does not imply superdeterminism because the existence of chaotic evolution equations. Predicting weather is a classical example. I do not think superdeterminism has any chance of success.

On Objection 2, I again agree with its statement. Quantum mechanics arose out of the inability of classical mechanics to explain atomic phenomena. But instead of expanding on this let's reply to the concrete arguments Andrei raised. Let's start with:

" This is all nice, but classical physics is not the same thing as Newtonian physics of the rigid body. Let’s consider a better classical approximation of the electron, a charged bullet. The slits are made of some material that will necessarily contain a large number of charged “bullets”. As the test bullet travels, its trajectory will be determined by the field generated by the slitted barrier. The field will be a function of position/momenta of the “bullets” in the barrier. But the field produced by a barrier with two slits will be different than the field produced by a barrier with only one slit, so the effect with both holes open is NOT the sum of the effects with each hole open alone."

This argument is wrong on two counts. First, one can make an interference experiment with neutrons where the neutrons not passing through the slits will be simply absorbed. Using electrically neutral particles renders irrelevant Andrei's objection. Second, "the field produced by a barrier with two slits will be different than the field produced by a barrier with only one slit" is incorrect as shown by a simple order of magnitude analysis. The electric fields near the slit are relevant on an atomic distance scale, while the distance between slits is macroscopic. You are looking at about seven order of magnitude difference in the ratio of relevant distances which translates in terms of force into a ten to minus fourteen order of magnitude effect. But the interference pattern is macroscopic and the difference between two Gaussian distributions vs. interference pattern cannot be explained away by a force fourteen of orders of magnitude smaller than what it is needed.

The next objection is appealing to Yves Couder's results. Those are interesting experiments, but they are not a confirmation of quantum mechanics emergence from classical physics and I know no one who claims it so. As such the argument is irrelevant to the current debate.

On the free fall atomic model, I did not read those papers so do not know if the claims are correct or not. The author may simply have proposed a model and analyzed the consequences and never claimed that his model describes nature.  Based on the general information available it is immediately clear that that model has nothing to do with reality. Also peer review is no magic bullet for avoiding publishing incorrect results. In my area of expertise in the last 12 months I read 2 published papers which were pure unadulterated garbage: the errors were not subtle, but blatant and packaged in a dishonest way to bamboozle the reader. Moreover I have concrete proof that the authors were aware of their mistakes when they published it. Physics is not immune to charlatans, crooks, and incompetent reviewers.

On the ionic crystals argument, without quantum mechanics the collection of electrons and nuclei will behave like a plasma and not like a crystal. This is moderately easy to test: create a computer simulation of say 1000 atoms and use Maxwell's and Newtonian equations of motion only to model the interaction. Then try to find an initial configuration which will be stable. I think there is none. Prove me wrong with such a model and I'll concede this point.

On quantum tunneling, the argument is pure handwaiving. Let me make an analogy. I know how my microwave works. But an alternative explanation might be that little Oompa-Loompas inside it are heating the food and I cannot see them because they move really fast.  The point is that the argument needs to have more predictive power than a fuzzy non-committal: "A new theory could predict a much stronger force." Show me the money. Propose such a theory and then we can discuss its merits. I am not asking something impossible. Regarding tunneling, quantum mechanics provides testable predictions which were confirmed experimentally. I only hold any alternative theory to the same level of experimental confirmation.

On Objection 3, I somewhat disagree with it. It is worthwhile to pursue non-quantum toy theories to better understand quantum mechanics, but not to search for an alternative to quantum mechanics.

Let me answer the four sub-points raised by Andrei:

a.       If nature is not probabilistic after all, there is much to be discovered. Detailed mechanism behind quantum phenomena should be revealed, bringing out a deeper understanding of our universe, and maybe new physical effects.

There is no "sub-quantum" or "hidden variable" explanation to quantum effects. Quantum mechanics is at the core of Nature, and my work is about proving this rigorously and not as a result of personal beliefs.

b.      Quantum theories are not well equipped to describe the universe as a whole. There is no observer outside the universe, no measurement can be performed on it, not even in principle.

This is a sterile objection to quantum mechanics and this cannot be used to justified a realistic alternative theory. First, even in quantum mechanics there are realistic interpretations. Second, epistemic interpretation like Qbism avoids this because they only talk about Bayesian probabilities. This objection only applies to naively using quantum mechanics in cosmology. Loop quantum gravity is unaffected by this objection.

c.       Due to its inability to provide an objective description of reality, quantum mechanics may not be able to solve the cosmological constant problem. A theory that states clearly “what’s there” could provide a much better estimate of the vacuum energy. After all we are not interested in what energy someone could find by performing a measurement on the vacuum, but what the vacuum consists of, when no one is there to pump energy into it.

The statement underscores a deep misunderstanding of the vacuum. Vacuum is actually a very violent place filled with virtual particles due to interplay of relativity and quantum mechanics. See this poor quality but brilliant video of a Sidney Coleman lecture to understand why merging them is not a trivial thing as one may naively expect that adding symmetries to quantum mechanics always results in simplifications. See also the QCD "Lava Lamp" which was shown at the 2004 Nobel Physics lecture. The cosmological constant problem is not a problem of quantum mechanics. I am not an expert in string theory but I know it has at least a solution to this problem (I don't know if it got rid of Susskind's "Rube Goldberg" label).

d.      Quantum mechanics requires an infinitely large instrument to measure a variable with infinite precision. When gravity is taken into account, it follows that local, perfectly defined properties cannot exist, because, beyond a certain mass, the instrument would collapse into a black hole.

The same argument can be used in the case of classical deterministic physics.

As you can see I disagree with the points Andrei was making, but nevertheless I want to thank him for participating in this debate and I look forward to discuss his replies in the commenting section of this post. I think such debates are useful, and I feel that the professional community of physicists is not doing a good job in engaging the public or explaining what it is doing. Physicists are very busy people trying to get ahead in a very competitive field. However the outside world usually experiences an arrogant wall of silence.

1. Rebuttal - Part 1

Dear Florin,

I would like first to clear two important issues:

1. My opening post is about classical, deterministic, field theories in general. It is about a class of possible theories, not about a specific theory. So, when I am speaking about a field I do not refer to the electric field, or magnetic field, or a pressure field, or gravitational field. I am just speaking about theories which define a function, let’s call it Phi (x, t), ascribing to each point in space and time a set of vectors/scalars. Let’s not go into more detail and just see how this class of theories can deal with the arguments I presented.

2. My claim is not that classical field theories are true. I may believe that, but I am not prepared to defend such a belief. As I have clearly specified in the title I only argue for compatibility between them and QM’s predictions and the way I did this was to show that the non-existence arguments fail. So, by asking me to explain how some specific quantitative results are recovered from classical theories, you are trying to shift the burden of proof.

In light of the two clarifications above, I will answer to your rebuttal.
“On Objection 1, I agree that classical, local theories have been ruled out by Bell’s theorem with only one loophole left: super-deterministic theories pursued by 't Hooft.”

Classical field theories are also superdeterministic. 't Hooft’s model (the cellular automaton) is just a special case of a discrete field theory. As you let the lattice unit go to zero, you regain the continuous case. Therefore Bell’s theorem is irrelevant for them as well.

“Any statistical theory obeying Kolmogorov's axioms respects Bell's inequalities.”

Are classical field theories obeying those axioms?

“Bohmian interpretation is contextual”

So are classical field theories.

“On superdeterminism, one needs to deny free will and this is a very tall order. While I (and anyone else) cannot give a rigorous definition of free will, I know that I have it.”

Free-will is an incoherent concept, and it implies non-determinism, rendering your argument logically fallacious (circular reasoning).

“Superdeterminism is only a pre-requisite step: you need to obtain from it quantum correlations, and so far I am not aware of any successful model.”

Here is one point where you try shifting the burden of proof. You have admitted earlier that superdeterministic theories are not excluded by Bell’s theorem. This is all I want to prove here. My claim is that classical field theories are possible, not that they are true (even though they could be).

“Second, determinism does not imply superdeterminism because the existence of chaotic evolution equations.”

I have made clear in my opening statement that classical field theories imply correlations between distant systems (Alice and Bob and the source of particles). This is what superdeterminism means (as far as I know). If you have some other idea of superdeterminism in mind, please make it clear so I can adjust my arguments accordingly.

Chaos does not preclude correlations, just makes them more difficult to notice. In a deterministic field theory there is always a perfect correlation between all particles, which is another way of saying that the trajectory of each particle is a function of the state of all other particles.

Objection 2

“This argument is wrong on two counts. First, one can make an interference experiment with neutrons where the neutrons not passing through the slits will be simply absorbed.”

There is no such thing as “simple absorption”; all interactions are mediated by fields. Neutrons, which are composed of 3 electrically charged quarks couple with all known fields (gravitational, electromagnetic, week and strong). Hopefully you wouldn’t want to claim that they are like drops of glue, sticking as they encounter a nucleus.

1. Dear Andrei,

Indeed, I think we have a disagreement on the burden of proof. If you propose a new theory the burden of proof is on you. But if you want to justify researching an alternative, you are free to do whatever you want, physics is not a dictatorship. However, if you request money for grants, the burden of proof is again on you.
“Classical field theories are also superdeterministic.” Determinism is not superdeterminism. Superdeterminism is like in the Matrix: “You've already made the choice, now you need to understand it.”
“Are classical field theories obeying those axioms?” Yes.
“So are classical field theories.” No. See http://www.mth.kcl.ac.uk/~streater/lostcauses.html#I for definition of contextual.
“You have admitted earlier that superdeterministic theories are not excluded by Bell’s theorem. This is all I want to prove here.” It is well known that Bell’s theorem has this superdeterministc loophole.

2. Rebuttal - Part 3

Andrei: “Quantum theories are not well equipped to describe the universe as a whole. There is no observer outside the universe, no measurement can be performed on it, not even in principle.”

Florin: “This is a sterile objection to quantum mechanics and this cannot be used to justify a realistic alternative theory. First, even in quantum mechanics there are realistic interpretations.”

Those are either non-local (Bohm) or inconsistent (many worlds).

“Second, epistemic interpretation like Qbism avoids this because they only talk about Bayesian probabilities.”

QBism is worse than Copenhagen.

1. It speaks about the subjective experiences of observers. There has to be plenty of them outside the universe.

2. It is based on decision theory, which presupposes a brain that works in the way our brain works. Explaining the properties of an electron based on some hardly understood properties of a brain is as fallacious as creationist’s appeal to an all-powerful god. A proper scientific approach explains complex systems (like a brain) in terms of simpler constituents (particles), not the other way around. This objection would apply as well to the “free-will” assumption in Bell’s theorem.

3. Its “local” character is a joke. The theory is supposed to be local because it all boils down to the experience of one observer. So, if you put a QBist on a Klingon war-bird and take him to Alpha-Centauri and back in 5 minutes he will still claim that he lives in a local universe because it is after all just another experience in his consciousness! There is also no evidence that this “experience” is a local process.

“Loop quantum gravity is unaffected by this objection.”

As far as I can tell LQG has been falsified for some time. So now, that it is dead, it is really unaffected by anything. Anyway, I am not arguing that classical field theories are compulsory, just worthy of more research.

“Vacuum is actually a very violent place filled with virtual particles due to interplay of relativity and quantum mechanics.”

If this is really the case, then you need to explain the “small discrepancy” between the prediction of QM and observed value, of about 120 orders of magnitude. Or you can go anthropic. The problem has not been solved by string theory.

Andrei: “Quantum mechanics requires an infinitely large instrument to measure a variable with infinite precision. When gravity is taken into account, it follows that local, perfectly defined properties cannot exist, because, beyond a certain mass, the instrument would collapse into a black hole.”

Florin: “The same argument can be used in the case of classical deterministic physics.”

No, it cannot. According to a classical theory the particle is there, with well-defined properties, it does not matter if you measure them or not. There are no unpredictable quantum fluctuations either, so you need not use an infinitely large instrument.

In the end, I would also like to point out that, because "the uncertainty relation does not hold for the past." one cannot argue for the fundamental probabilistic character of QM based on uncertainty. Just another argument against a classical universe that bites the dust!

Andrei

1. “QBism is worse than Copenhagen.” No, it is a variant of Copenhagen.

“It speaks about the subjective experiences of observers.” In probability theory you have the frequentist approach and the Bayesian approach. Their predictions are identical. Qbism takes the Bayesian point of view and attempts to make sense of the world from within the world.

“It is based on decision theory” No, it is based on Bayes update of information theorem https://en.wikipedia.org/wiki/Bayes%27_theorem

“Its “local” character is a joke.” – I agree.

“As far as I can tell LQG has been falsified for some time.” – This is irrelevant for the discussion purpose. Compare LQG with Wheeler DeWitt.

“If this is really the case, then you need to explain the “small discrepancy” between the prediction of QM and observed value, of about 120 orders of magnitude.” The most natural value is actually zero. String theory does have a contrived solution to it. Vacuum polarization is another effect of the vacuum.
“No, it cannot. According to a classical theory the particle is there, with well-defined properties, it does not matter if you measure them or not. There are no unpredictable quantum fluctuations either, so you need not use an infinitely large instrument.” GR is a classical theory. The resolution of a microscope is given by the wavelength of light. Ultra short distances require ultra short wavelengths which demand very high energy, to the point that you will create a black hole.

“I would also like to point out that, because "the uncertainty relation does not hold for the past." one cannot argue for the fundamental probabilistic character of QM based on uncertainty.”
The probabilistic character of QM arises out of non-commutativity. It is this non-commutativity which demands the matrices in Heisenberg’s formalism, and those matrices (after diagonalization) contain the predictions for all experiments, not a single one. Hence the probabilistic character.

3. Andrei, thank you for your rebuttal. There is quite a lot of material and it will take me some time to reply. I'll try to do it this weekend but I cannot make hard promises due to other commitments, but I will reply to all your points.

4. Dear Florin,

For some reason the part 2 of my rebuttal did not get published, so it may appear that I have not responded to some points. I will upload it again on Monday. I do not have access to the text until then.

Andrei

1. Your part 2 is not on the server and must have gotten lost during posting. Looking forward to it.

5. Rebuttal - Part 2-1

“Using electrically neutral particles renders irrelevant Andrei's objection. Second, "the field produced by a barrier with two slits will be different than the field produced by a barrier with only one slit" is incorrect as shown by a simple order of magnitude analysis. The electric fields near the slit are relevant on an atomic distance scale, while the distance between slits is macroscopic. You are looking at about seven order of magnitude difference in the ratio of relevant distances which translates in terms of force into a ten to minus fourteen order of magnitude effect.”

As I have specified in the beginning of my rebuttal, I am not speaking of the electric field, but about some field Phi (x, t). Let’s discuss the general issues leaving for now the details regarding the strength of the forces involved, etc.

“But the interference pattern is macroscopic and the difference between two Gaussian distributions vs. interference pattern cannot be explained away by a force fourteen of orders of magnitude smaller than what it is needed.”

The force has nothing to do with this, but let’s just agree that for a generic field, Phi (with whatever strength you want), Feynman’s analysis is incorrect.

“The next objection is appealing to Yves Couder's results. Those are interesting experiments, but they are not a confirmation of quantum mechanics emergence from classical physics and I know no one who claims it so.”

Again, I did not make the claim that you imply. Those experiments only prove that Feynman’s analysis about how probabilities should add in “ANY” classical theory is incorrect.

“On the free fall atomic model, I did not read those papers so do not know if the claims are correct or not. The author may simply have proposed a model and analyzed the consequences and never claimed that his model describes nature. Based on the general information available it is immediately clear that that model has nothing to do with reality.”

What do you mean by “general information”? Can you point me to some paper discussing the classical atom with spin? If you cannot, you should drop this argument.

1. Dear Andrei, Sorry for the long delay. On the free fall atomic model all I had access to was the Wikipedia information on it. From there it is clear it does not describe reality.

On Yves Couder's example, the position is completely classical obeying Bell's inequalities. As such Feynman's argument is still valid.

6. Rebuttal - Part 2-2

“On the ionic crystals argument, without quantum mechanics the collection of electrons and nuclei will behave like a plasma and not like a crystal. This is moderately easy to test: create a computer simulation of say 1000 atoms and use Maxwell's and Newtonian equations of motion only to model the interaction. Then try to find an initial configuration which will be stable. I think there is none. Prove me wrong with such a model and I'll concede this point.”

A classical simulation of NaCl crystal growth:

http://tresen.vscht.cz/fch/tul/jz302065w.pdf

does not seem to produce such a plasma.

Anyway, you can take a system of charged metal spheres covered with some dielectric. Would such a system produce a “plasma” too?

“On quantum tunneling, the argument is pure handwaiving. Let me make an analogy. I know how my microwave works. But an alternative explanation might be that little Oompa-Loompas inside it are heating the food and I cannot see them because they move really fast. The point is that the argument needs to have more predictive power than a fuzzy non-committal: "A new theory could predict a much stronger force." Show me the money.”

Again, shifting of the burden of proof. If you think that tunneling rules out only some specific classical theories, then be careful and make that clear. Do not say “no classical theory can explain tunneling” and then argue from some details that are not general to any classical theory, like the strength of the electrostatic or weak force.

“There is no "sub-quantum" or "hidden variable" explanation to quantum effects. Quantum mechanics is at the core of Nature, and my work is about proving this rigorously and not as a result of personal beliefs.”

This is a declaration of personal believes, nothing to argue against.

Andrei

1. Thank you for the simulation paper. I am not an expert in those kinds of modeling so I cannot asses what they use inside them. It is very likely they use some sort of effective effects which arise out of QM. What I asked was different. use only Maxwell and Newton. If they had used only those, then I concede the point.

Again we disagree on the burden of proof. It the job of the challenger of an existing paradigm to prove he has something worthwhile.

On the no sub-quantum, see my next post which is in preparation right now.

7. I was proposing last June or July that the PBR result was correct in that D > 0, but that the strict result that D = 1 was not. As such while the ψ-epistemic interpretations are wrong, it is uncertain in that 0 < D <= 1. My argument centered around the uncertainty principle for spin.

For noncommuting operators X and Y we have that

ΔXΔY = ħ/2|< |[X, Y]| >|.

We may set X = L_y and Y = L_y and we have that [L_x, L_y] = ħL_z and = l(l + 1) – m^2 and so

ΔXΔY = ½ħ^2(l(l + 1) – m^2).

This means the uncertainty principle is within a discrete set of possible values, rather than a continuous distribution. My argument centered around this.

This idea was a bit of a sideline to what I have been working on, which is to find quotient group realizations of fields on a horizon that correlated with locality in spacetime with one dimension larger. The approach I took to this was a quotient group realization, and the G/H = K quotient would give the exact PBR result for H = id.

I came here and see that you are presenting something that is similar to what I was proposing last summer.

8. Hi Lawrence, see http://fmoldove.blogspot.com/2015/09/heisenbergs-matrix-mechanics-and.html

Originally I was reading Beltrametti and Cassineli to prepare for the post. Shortly after that we had a discussion where you challenged me on the discrete spectra case. I have to admit I had to think a bit to find the correct answer. Then I thought the problem would be of interest to a larger audience.

I still maintain there is no universal uncertainty principle for spin, only uncertainties for given states.

1. That would imply a different sort of relationship between the state and the operator.

I may return to that little project, which is something I thought of with respect to working on quotient groups for unitary qubits with black holes.

LC

2. Lawrence, I don't quite follow what you are saying. Do you say there is an uncertainty relationship for spin for any states? If so I can find a counterexample.

9. What I am saying is that this seems to imply some subtle difference in the relationship operators and states. I am not sure what that is.

LC

1. Same comment as in the next post: another novel idea, pursue it and see where it leads.

2. The idea is in relationship to quotient groups and SLOCC for qubit-black hole correspondence. I also think this has something to do with BMS supertranslations. The elementary case is the conformal theory with SO(2,1)/U(1), which is a special case with SL(2n,C)/SU(n). Another case is the case of the quotient group between the 9 rays in the F4 Kochen-Specker vs the 8 rays in the set of possible rays, 8 = 2^3, that can construct a classical ray trace path. The additional dimension is ~ R, that reduces the 9 KS rays to the 8 classical rays.

This is a bit of a side calculation, and it has some bearing upon the PBR result. The PBR result has been accused of having some ontology built into it. If 0 < D[ψ, φ] < 1 then all quantum interpretations have holes in them.

LC

3. The idea is in relationship to quotient groups and SLOCC for qubit-black hole correspondence. I also think this has something to do with BMS supertranslations. The elementary case is the conformal theory with SO(2,1)/U(1), which is a special case with SL(2n,C)/SU(n). Another case is the case of the quotient group between the 9 rays in the F4 Kochen-Specker vs the 8 rays in the set of possible rays, 8 = 2^3, that can construct a classical ray trace path. The additional dimension is ~ R, that reduces the 9 KS rays to the 8 classical rays.

This is a bit of a side calculation, and it has some bearing upon the PBR result. The PBR result has been accused of having some ontology built into it. If 0 < D[ψ, φ] < 1 then all quantum interpretations have holes in them.

LC