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# Orthogonal Vectors in a 2D Lattice with minimum area

I came across an interesting problem in my research (not a mathematician). Here it goes:

Suppose, there is a 2D lattice $\Lambda$ in the X-Y plane with basis vectors $\vec{a}$ and $\vec{b}$, which are not orthogonal to each other. I want to find a rectangle in this lattice, whose area is the minimum of all possible rectangles. So far, this is what I have:

Consider two vectors in the lattice:

$\vec{v_1} = m_1 \vec{a} + n_1 \vec{b}$

$\vec{v_2} = m_2 \vec{a} + n_2 \vec{b}$

If they are orthogonal, then $\vec{v}_1 \cdot \vec{v}_2 = m_1 m_2 \left| \vec{a} \right|^2 + n_1 n_2 \left| \vec{b} \right|^2 + (m_1 n_2 + n_2 m_1) \vec{a} \cdot \vec{b} =0$

The area subtended by the two vectors is the norm of the cross product, i.e. $\left| \vec{v}_1 \times \vec{v}_2 \right| = \left| \vec{a}_1 \times \vec{b}_2 \right| \left|m_1n_2 - m_2n_1 \right|$

Therefore, the area is minimized if $\left|m_1n_2 - m_2n_1 \right|$ is minimized and greater than zero. And the constraint is:

$m_1 m_2 \left| \vec{a} \right|^2 + n_1 n_2 \left| \vec{b} \right|^2 + (m_1 n_2 + n_2 m_1) \vec{a} \cdot \vec{b} =0$

I am stuck after this! May be some kind of optimization with a constraint using Lagrange multipliers? But there are too many variables and $m_1, m_2, n_1, n_2 \in \mathbb{Z}$. Any suggestions will be greatly appreciated. Thank you for your time.

• I will work on it. Note that there is no guarantee of existence without some constraint, essentially that all lattice vectors have squared length rational, better integral...Hmmm; actually the standard condition is that all inner products be integers. – Will Jagy Oct 14 '14 at 2:14
• Thank you, Will. I agree that there is no guarantee that a solution exists. I might be able to think of the constraints for existence as I improve my intuition for this problem. Thanks! – Srikanth Oct 14 '14 at 2:30
• Sri.. put second answer with a necessary condition. It can be used to prove non-existence of orthogonal vectors in some lattices. – Will Jagy Oct 15 '14 at 3:20
• It turns out that the condition I give in my second answer is necessary and sufficient for existence of a pair of orthogonal vectors in your lattice. Calculations put into third answer, with some examples. None guaranteed to be optimal, though, just to exist. – Will Jagy Oct 15 '14 at 21:51
• Thank you for putting in so much effort on this, Will. I didn't even know what a gram matrix is and how it may be useful before this discussion. I think I understand the existence conditions now. I will try this on a few examples and will let you know! Thank you again. – Srikanth Oct 17 '14 at 16:28

There seems to be an answer if the lattice can be rotated and scaled so that all inner products of lattice vectors are integers (positive definite..) and there is no common integer factor of the entries in the gram matrix; $\gcd(A,B,C)=1.$ We also may demand that $A$ be the squared length of the shortest nonzero lattice vector.

http://en.wikipedia.org/wiki/Unimodular_lattice#Definitions

As far as i can tell, in this case, the best answer is a non-primitive lattice, given by identity $$\left( \begin{array}{rr} 1 & 0 \\ -B & A \end{array} \right) \left( \begin{array}{rr} A & B \\ B & C \end{array} \right) \left( \begin{array}{rr} 1 & -B \\ 0 & A \end{array} \right) = \left( \begin{array}{cc} A & 0 \\ 0 & A (AC - B^2) \end{array} \right)$$ This is a short version of Gauss-Legendre composition for integral binary quadratic forms.

Oh, almost forgot. You actually want the lattice reduced. Furthermore, if $B$ is a multiple of $A,$ then the lattice is already $SL_2 \mathbb Z$ equivalent to a diagonal one.

• I will need a little more time to completely understand this! I will accept this as an answer as soon as possible... Thanks, Will. – Srikanth Oct 14 '14 at 17:22
• @Srikanth, this is the setting, number theory, where I can give a substantial answer. As an example that does not fit this, take vector $v=(1,0)$ and $w=(-5,\pi).$ There is an evident orthogonal pair, $v, 5v+w,$ but I do not currently know much about this situation...for example, if there is an orthogonal pair, is the shortest vector one of them? Don't know yet. If so, that would give a quick test for existence. – Will Jagy Oct 14 '14 at 18:29
• @Srikanth so, look up "binary quadratic form reduction" the concepts are the same. Given a symmetric matrix $G$ and a column vector $X,$ we get a quadratic form from $X^T G X.$ – Will Jagy Oct 14 '14 at 18:59
• Thank you for the resources, Will. I get your solution now but I am still not sure about the existence condition. I will take a look at the binary quadratic form reduction. Thank you! – Srikanth Oct 15 '14 at 0:59

THIRD answer: It turns out the condition I gave in my second answer is necessary and sufficient for existence; I can also show a two parameter family, I'm afraid to minimize could be algorithmic but not formulaic. i will have time for that aspect later. Given Gram matrix with $A,B,C$ as before, we have the equation $$A \alpha \beta + B (\alpha \delta + \beta \gamma) + C \gamma \delta.$$ So need to have $iA + jB + kC = 0,$ with $j^2 - 4ik = w^2.$

So, take integer parameters $s,t,$ then $$\alpha = 2i s, \; \; \beta = 2 it, \; \; \gamma = (j-w)s, \; \; \delta = (j+w)t.$$ With these values, the vector $( \alpha \beta, \alpha \delta + \beta \gamma, \gamma \delta)$ is a scalar (rational) multiple of $(i,j,k),$ and we have constructed orthogonal vectors in the lattice. It is possible that $\gcd(\alpha, \beta, \gamma,\delta)> 1$ for some values of $(s,t)$ but not others.

Furthermore, minimization of the determinant $|\alpha \delta - \beta \gamma|$ needs work, although it is a multiple of $w$ by construction. Here is a start: $$\alpha \delta - \beta \gamma = 4iwst.$$ Need to think about what that means, with varying GCD's and the possibility of zero values for $s,t.$ SIGH. Added: no, if one of $s,t$ is zero, one vector in the orthogonal pair is just the zero vector, so we may rule out that possibility. Good. So, it may not be smallest, but $s=t=1$ gives an orthogonal pair.

$$\left( \begin{array}{rr} 2i & j-w \\ 2i & j+w \end{array} \right) \left( \begin{array}{rr} A & B \\ B & C \end{array} \right) \left( \begin{array}{rr} 2i & 2i \\ j-w & j+w \end{array} \right) = \left( \begin{array}{cc} 4i^2 A + 4i(j-w)B + (j-w)^2C & 0 \\ 0 & 4i^2 A + 4i(j+w)B + (j+w)^2C \end{array} \right)$$ If $i=0,$ thus $jB + kC=0,$ we get

$$\left( \begin{array}{rr} j & k \\ 0 & 1 \end{array} \right) \left( \begin{array}{rr} A & B \\ B & C \end{array} \right) \left( \begin{array}{rr} j & 0 \\ k & 1 \end{array} \right) = \left( \begin{array}{cc} j^2 A + 2jkB + k^2C & 0 \\ 0 & C \end{array} \right)$$ When $j=0,$ the combination of $\gcd(i,j,k)=1$ and $j^2 - 4 i k = w^2$ allows us to demand, in integers, $$i = x^2, \; \; \; k = -y^2, \; \; \; w = 2 x y,$$ with $x^2 A - y^2 C = 0.$ Then $$\left( \begin{array}{rr} x & -y \\ x & y \end{array} \right) \left( \begin{array}{rr} A & B \\ B & C \end{array} \right) \left( \begin{array}{rr} x & x \\ -y & y \end{array} \right) = \left( \begin{array}{cc} A x^2 - 2 B x y + C y^2 & 0 \\ 0 & A x^2 + 2 B x y + C y^2 \end{array} \right)$$

Anyway, these show existence for any explicit $(i,j,k)$ triple of integers such that $iA + j B + k C = 0.$ These also give upper bounds on the determinants of the change of basis matrices, again $|\alpha \delta - \beta \gamma|.$ In the case that the lattice cannot be scaled to an integral lattice, I am not confident about giving an explicit recipe for the smallest determinant that works; I suggest using these as upper bounds for a computer search.

• Minimizing a determinant interestingly enough is the same as lattice reduction, at least for two by two. I believe that is the case anyway from my memories of past musings... Think in terms of orthogonalizing the two vectors (so basically swap two elements of one vector, determinant becomes dot product of the two vectors) – adam W Oct 15 '14 at 21:13
• @adamW, I know that; the real problem is that we have unpredictable common divisors. We are free to demand $\gcd(i,j,k) = 1,$ but we do not have complete control over the common divisors of the $\alpha, \beta, \gamma, \delta$ I suggest above. – Will Jagy Oct 15 '14 at 21:16

A second answer. This is getting pretty good. Given a lattice with Gram matrix $$\left( \begin{array}{rr} A & B \\ B & C \end{array} \right),$$ in order for there to be an orthogonal pair of lattice vectors, it is necessary that $A,B,C$ be linearly dependent over $\mathbb Q,$ that is, there are ordinary integers $i,j,k$ such that $$iA + j B + k C = 0.$$ This is still not quite enough. It is also necessary that there be another ordinary integer $w,$ which is allowed to be zero or nonzero, with $$j^2 - 4 i k = w^2.$$ This is a pretty strong restriction. However, as mentioned before, it is satisfied by $$\left( \begin{array}{rr} 1 & 0 \\ 0 & \pi \end{array} \right),$$ in which we are given orthogonal basis vectors but we cannot scale this to an integral lattice; here we can take $i=0, j=1,k=0.$

PROOF, the $i,j,k$ thing: $$\left( \begin{array}{rr} \alpha & \gamma \\ \beta & \delta \end{array} \right) \left( \begin{array}{rr} A & B \\ B & C \end{array} \right) \left( \begin{array}{rr} \alpha & \beta \\ \gamma & \delta \end{array} \right) = \left( \begin{array}{cc} D & 0 \\ 0 & E \end{array} \right)$$ The zero entries in the product give the equation $$A \alpha \beta + B (\alpha \delta + \beta \gamma) + C \gamma \delta.$$ So need to have $iA + jB + kC = 0$ with some multiplier $n$ and $$n i = \alpha \beta, \; \; nj = \alpha \delta + \beta \gamma, \; \; nk = \gamma \delta.$$ If you multiply it out, $$n^2 j^2 - 4nink = n^2(j^2 - 4ik)= (\alpha \delta - \beta \gamma)^2$$

General comment, given just two irrational numbers, we can find out whether their ratio is rational by getting giant decimal accuracy and writing the continued fraction for the ratio. To find integers $i,j,k$ that solve $iA + jB + kC = 0,$ we use PSLQ with which I have no experience.

• For this problem, one need not look only at the Gram matrix since the orthogonal vectors need not be in the same basis, only in the same lattice. I would guess that you have such a habit of using the Gram matrix from looking at binary quadratic forms. – adam W Oct 15 '14 at 2:19
• @adamW, I edited in an example I was thinking of before, which shows that orthogonal lattice vectors does not imply that one can scale to an integral lattice. Anyway, completely new question for me. – Will Jagy Oct 15 '14 at 2:29
• My thought is that only $A$ and $B$ need be linearly dependent (or $B$ and $C$) in order to find orthogonal vectors. – adam W Oct 15 '14 at 2:38
• @adamW, that would appear to be stronger than my restriction. My dinner is ready, I will type in the proof of necessity after; quite elementary. – Will Jagy Oct 15 '14 at 2:46
• Now I see though that $i$ or $k$ zero is the same thing which you did note. – adam W Oct 15 '14 at 2:47


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## Differential Geometry Seminar: Collapsing hyperkähler manifolds

Seminar | October 16 | 1-2 p.m. | 891 Evans Hall

Valentino Tosatti, Northwestern

Department of Mathematics

Consider a projective hyperkähler manifolds with a surjective holomorphic map with connected fibers onto a lower-dimensional manifold. In the case the base must be half-dimensional projective space, and the generic fibers are holomorphic Lagrangian tori. I will explain how hyperkähler metrics on the total space with volume of the torus fibers shrinking to zero, collapse smoothly away from the singular fibers to a special Kähler metric on the base, whose metric completion equals the global collapsed Gromov-Hausdorff limit, which has a singular set of real Hausdorff codimension at least 2. The resulting picture is compatible with the Strominger-Yau-Zaslow mirror symmetry, and can be used to prove a conjecture of Kontsevich-Soibelman and Gross-Wilson for large complex structure limits which arise via hyperkähler rotation from this construction. This is joint work with Yuguang Zhang.

rbamler@berkeley.edu


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text:
Home My Page Projects Code Snippets Project Openings SML/NJ
 Summary Activity Forums Tracker Lists Tasks Docs Surveys News SCM Files

# SCM Repository

[smlnj] View of /sml/trunk/src/cm/Doc/09-files.tex
 [smlnj] / sml / trunk / src / cm / Doc / 09-files.tex

# View of /sml/trunk/src/cm/Doc/09-files.tex

Tue Jun 5 19:10:21 2001 UTC (19 years ago) by blume
File size: 4669 byte(s)
index-file generation in CM; small changes to c-calls API

% -*- latex -*-

\section{Files}
\label{sec:files}

CM uses three kinds of files to store derived information during and
between sessions:

\begin{enumerate}
\item {\it Skeleton files} are used to store a highly abbreviated
version of each ML source file's abstract syntax tree---just barely
sufficient to drive CM's dependency analysis.  Skeleton files are much
smaller and (for a program) easier to read than actual ML source code.
Therefore, the existence of valid skeleton files makes CM a lot faster
because usually most parsing operations can be avoided that way.
\item {\it Binfiles} are the SML/NJ equivalent of object files.  They
contain executable code and a symbol table for the associated ML
source file.
\item {\it Library files} (sometimes called: {\em stablefiles}) contain
dependency graph, executable code, and symbol tables for an entire CM
library including all of its components (groups).  Other libraries
used by a stable library are not included in full.  Instead,
references to those libraries are recorded using their (preferably
anchored) pathnames.
\end{enumerate}

Normally, all these files are stored in a subdirectory of directory
{\tt CM}. {\tt CM} itself is a subdirectory of the directory where the
original ML source file or---in the case of library files---the
original CM description file is located.

Skeleton files are machine- and operating system-independent.
Therefore, they are always placed into the same directory {\tt
CM/SKEL}. Parsing (for the purpose of dependency analysis) will be
done only once even if the same file system is accessible from
machines of different type.

Binfiles and library files contain executable code and other
information that is potentially system- and architecture-dependent.
Therefore, they are stored under {\tt CM/}{\it arch}{\tt -}{\it os}
where {\it arch} is a string indicating the type of the current
CPU architecture and {\it os} a string denoting the current operating
system type.

As explained in Section~\ref{sec:stable}, library files are a bit of
an exception in the sense that they do not require any source files or
any other derived files of the same library to exist.  As a
consequence, the location of such a library file should be described
as being relative to the location of the original CM description
file if that description file still existed''.  (Of course, nothing
precludes the CM description file from actually existing, but in the
presence of a corresponding library file CM will not take any notice
of that.)

{\em Note:} As discussed in section~\ref{sec:toolparam:cm}, CM sometimes
looks for library files in {\tt CM/}{\it version}{\tt /}{\it arch}{\tt
-}{\it os}.  However, library files are never {\em created} there by
CM.  If several versions of the same library are to be provided, an
administrator must arrange the directory hierarchy accordingly by
hand''.

\subsection{Time stamps}

For skeleton files and binfiles, CM uses file system time stamps
(i.e., modification time) to determine whether a file has become
outdated.  The rule is that in order to be considered up-to-date''
the time stamp on skeleton file and binfile has to be exactly the
same\footnote{CM explicitly sets the time stamp to be the same.} as
the one on the ML source file.  This guarantees that all changes to a
source will be noticed---even those that revert to an older version of
a source file.\footnote{except for the pathological case where two
different versions of the same source file have exactly the same time
stamp}

CM also uses time stamps to decide whether tools such as ML-Yacc or
ML-Lex need to be run (see Section~\ref{sec:classes}).  However, the
difference is that a file is considered outdated if it is older than
its source.  Some care on the programmers side is necessary since this
scheme does not allow CM to detect the situation where a source file
gets replaced by an older version of itself.

\subsection{Index files}
\label{sec:indexfiles}

CM can optionally generate a human-readable index file for each
description file.  An index files alphabetically lists all symbols
defined or imported within the given library or library component.
Index-file generation is normally disabled.  To enable it, {\tt
CM.Control.generate\_index} must be set to true (see
section~\ref{sec:api}).

With index-file generation enabled, index files will be written for
all description files involved every time CM performs a dependency
analysis.  (In other words, it is a side-effect to other CM operations
such as {\tt CM.make} etc.)  If the name of the description files is
{\tt $p$/$d$.cm}, then the corresponding index file will be in {\tt
$p$/CM/INDEX/$d$.cm}.


text:
## Functional form

Functional form

The following functional abstraction, termed an interatomic potential function or force field in chemistry, calculates the molecular system’s potential energy (E) in a given conformation as a sum of individual energy terms.

${\displaystyle \ E=E_{\text{covalent}}+E_{\text{noncovalent}}\,}$

where the components of the covalent and noncovalent contributions are given by the following summations:

${\displaystyle \ E_{\text{covalent}}=E_{\text{bond}}+E_{\text{angle}}+E_{\text{dihedral}}}$

${\displaystyle \ E_{\text{noncovalent}}=E_{\text{electrostatic}}+E_{\text{van der Waals}}}$

The exact functional form of the potential function, or force field, depends on the particular simulation program being used. Generally the bond and angle terms are modeled as harmonic potentials centered around equilibrium bond-length values derived from experiment or theoretical calculations of electronic structure performed with software which does ab-initio type calculations such as Gaussian. For accurate reproduction of vibrational spectra, the Morse potential can be used instead, at computational cost. The dihedral or torsional terms typically have multiple minima and thus cannot be modeled as harmonic oscillators, though their specific functional form varies with the implementation. This class of terms may include improper dihedral terms, which function as correction factors for out-of-plane deviations (for example, they can be used to keep benzene rings planar, or correct geometry and chirality of tetrahedral atoms in a united-atom representation).

The non-bonded terms are much more computationally costly to calculate in full, since a typical atom is bonded to only a few of its neighbors, but interacts with every other atom in the molecule. Fortunately the van der Waals term falls off rapidly. It is typically modeled using a 6–12 Lennard-Jones potential, which means that attractive forces fall off with distance as r−6 and repulsive forces as r−12, where r represents the distance between two atoms. The repulsive part r−12 is however unphysical, because repulsion increases exponentially. Description of van der Waals forces by the Lennard-Jones 6–12 potential introduces inaccuracies, which become significant at short distances. Generally a cutoff radius is used to speed up the calculation so that atom pairs which distances are greater than the cutoff have a van der Waals interaction energy of zero.

The electrostatic terms are notoriously difficult to calculate well because they do not fall off rapidly with distance, and long-range electrostatic interactions are often important features of the system under study (especially for proteins). The basic functional form is the Coulomb potential, which only falls off as r−1. A variety of methods are used to address this problem, the simplest being a cutoff radius similar to that used for the van der Waals terms. However, this introduces a sharp discontinuity between atoms inside and atoms outside the radius. Switching or scaling functions that modulate the apparent electrostatic energy are somewhat more accurate methods that multiply the calculated energy by a smoothly varying scaling factor from 0 to 1 at the outer and inner cutoff radii. Other more sophisticated but computationally intensive methods are particle mesh Ewald (PME) and the multipole algorithm.

In addition to the functional form of each energy term, a useful energy function must be assigned parameters for force constants, van der Waals multipliers, and other constant terms. These terms, together with the equilibrium bond, angle, and dihedral values, partial charge values, atomic masses and radii, and energy function definitions, are collectively termed a force field. Parameterization is typically done through agreement with experimental values and theoretical calculations results. Norman L. Allinger’s force field in the last MM4 version calculate for hydrocarbons heats of formation with a rms error of 0.35 kcal/mol, vibrational spectra with a rms error of 24 cm−1, rotational barriers with a rms error of 2.2°, C-C bond lengths within 0.004 Å and C-C-C angles within 1°.

Later MM4 versions cover also compounds with heteroatoms such as aliphatic amines.

Each force field is parameterized to be internally consistent, but the parameters are generally not transferable from one force field to another.

Functional form
Scroll to top


text:
# Fully-Dynamic Coresets

@inproceedings{Henzinger2020FullyDynamicC,
title={Fully-Dynamic Coresets},
author={Monika Henzinger and Sagar Kale},
booktitle={ESA},
year={2020}
}
• Published in ESA 2020
• Computer Science, Mathematics
With input sizes becoming massive, coresets -- small yet representative summary of the input -- are relevant more than ever. A weighted set $C_w$ that is a subset of the input is an $\varepsilon$-coreset if the cost of any feasible solution $S$ with respect to $C_w$ is within $[1 {\pm} \varepsilon]$ of the cost of $S$ with respect to the original input. We give a very general technique to compute coresets in the fully-dynamic setting where input points can be added or deleted. Given a static… Expand
2 Citations

#### Figures and Topics from this paper

Coresets for Clustering with Missing Values
• Computer Science
• ArXiv
• 2021
The first coreset for clustering points in R that have multiple missing values (coordinates) is provided, which exhibits a flexible tradeoff between coreset size and accuracy, and generally outperforms the uniformsampling baseline. Expand
Robust Coreset for Continuous-and-Bounded Learning (with Outliers)
• Computer Science, Mathematics
• ArXiv
• 2021
This paper proposes a novel robust coreset method for the continuous-and-bounded learning problem (with outliers) which includes a broad range of popular optimization objectives in machine learning, like logistic regression and k-means clustering and can be efficiently maintained in fully-dynamic environment. Expand

#### References

SHOWING 1-10 OF 27 REFERENCES
New Frameworks for Offline and Streaming Coreset Constructions
• Computer Science, Mathematics
• ArXiv
• 2016
This work introduces a new technique for converting an offline coreset construction to the streaming setting, and provides the first generalizations of such coresets for handling outliers. Expand
Clustering High Dimensional Dynamic Data Streams
• Mathematics, Computer Science
• ICML
• 2017
Data streaming algorithms for the k-median problem in high-dimensional dynamic geometric data streams that guarantee only positive weights in the coreset with additional logarithmic factors in the space and time complexities are presented. Expand
On Coresets for k-Median and k-Means Clustering in Metric and Euclidean Spaces and Their Applications
• K. Chen
• Mathematics, Computer Science
• SIAM J. Comput.
• 2009
These are the first streaming algorithms, for those problems, that have space complexity with polynomial dependency on the dimension, using $O(d^2k^2\varepsilon^{-2}\log^8n)$ space. Expand
Approximating k-Median via Pseudo-Approximation
• Computer Science, Mathematics
• SIAM J. Comput.
• 2016
A novel approximation algorithm for $k-median is presented that achieves an approximation guarantee of$1+\sqrt{3}+\epsilon$, improving upon the decade-old ratio of$3+\ epsilon$by exploiting the power of pseudo-approximation. Expand Coresets in dynamic geometric data streams • Mathematics, Computer Science • STOC '05 • 2005 This work develops streaming (1 + ε)-approximation algorithms for k-median, k-means, MaxCut, maximum weighted matching (MaxWM), maximum travelling salesperson, maximum spanning tree, and average distance over dynamic geometric data streams. Expand On coresets for k-means and k-median clustering • Mathematics, Computer Science • STOC '04 • 2004 This paper shows the existence of small coresets for the problems of computing k-median/means clustering for points in low dimension, and improves the fastest known algorithms for (1+ε)-approximate k-means and k- median. Expand Improved Combinatorial Algorithms for Facility Location Problems • Mathematics, Computer Science • SIAM J. Comput. • 2005 Improved combinatorial approximation algorithms for the uncapacitated facility location problem and a variant of the capacitated facility locations problem is considered and improved approximation algorithms are presented for this. Expand Fully Dynamic Consistent Facility Location • Computer Science, Mathematics • NeurIPS • 2019 The cost of the solution maintained by the algorithm at any time is very close to the cost of a solution obtained by quickly recomputing a solution from scratch at time$t\$ while having a much better running time. Expand
An Improved Approximation for k-Median and Positive Correlation in Budgeted Optimization
• Mathematics, Computer Science
• ACM Trans. Algorithms
• 2017
This work improves upon Li-Svensson’s approximation ratio for k-median by developing an algorithm that improves upon various aspects of their work and develops algorithms that guarantee the known properties of dependent rounding but also have nearly bestpossible behavior—near-independence, which generalizes positive correlation—on “small” subsets of the variables. Expand
Unifying and Strengthening Hardness for Dynamic Problems via the Online Matrix-Vector Multiplication Conjecture
• Mathematics, Computer Science
• STOC
• 2015
It is shown that a conjecture that there is no truly subcubic (O(n3-ε) time algorithm for this problem can be used to exhibit the underlying polynomial time hardness shared by many dynamic problems. Expand


text:
Space with complex distance

I am interested in a mathematical space with specific properties, but I am not sure if such a space can be consistently defined. I would appreciate any guidance or ideas. If this space is known, what is its name? If this space cannot be defined, then why? If it can be defined, then what would its mathematical definition look like? Finally, if it can exist only in a certain number of dimensions, then what is this number? (Mostly interested in 2 to 4 dimensions).

Target properties:

1. A distance from any point to the chosen unique center of coordinates is a real number (radius-vector).
2. An infinitesimal distance in any direction perpendicular to the radius vector is an imaginary number. (Perhaps better stated this way: the distance between the ends of two radius-vectors of the same length separated by an infinitesimal angle is an imaginary number.)

Example: the circumference of a circle with the radius $R$ and the center in the center of coordinates is $2\pi\cdot R \cdot i$ where $i$ is the imaginary unit.

1. Only one imaginary unit regardless of the number of dimensions (e.g. no quaternions).

EDIT: Per the comments below, the following bullet is not well defined or even necesary. Therefore it is here only as a visual illustration of the expected symmetry.

1. Rotational symmetry in any direction, but only around the center of coordinates.

I need your help to understand if this space is mathematically possible or not. I would appreciate any answers, comments, suggestions, or requests to clarify the question.

EDIT: Clarifications/updates based on responses. The bullets #2 and #3 above probably should be combined this way (with no imaginary numbers):

1. The space must be locally asymptotic to a hyperbolic metric space with the following metric:

$ds^2 = ds_1^2 -ds_2^2$

where $ds_1$ is the radial coordinate and $ds_2$ is the rotational Euclidian distance (e.g. $ds_2^2 = dx^2 + dy^2 + dz^2$ where $x$, $y$, and $z$ are rotational coordinates). Local (infinitesimal) geodesics can be defined in the usual way by the vanishing partial second order derivatives along the geodesic.

Global geodesics should be segments of logarithmic spirals (including degenerate by either coordinate, such as radius segments or circular segments around the origin). Something like this (see the @pregunton comment below):

$r = ae^{b\theta}$

Geodesics may not be unique, but may depend on the number of revolutions around the origin. This is OK and expected.

• how would the triangular inequality work ? – mercio Sep 24 '17 at 7:18
• @mercio: Well, I guess the same way as on the complex plane (using absolute values), unless you mean something else. I don't think it would be a problem locally (over a relatively small area), would it? While I don't see it being a problem globally either, it is not really required to work globally (unless it is mandatory in general). Am I missing your point? Do you have a particular case in mind? This space must be metric locally, but not necessarily globally (such as over areas including the center of coordinates or large angles). If this makes any sense :) – safesphere Sep 24 '17 at 7:35
• One thing that I think fulfils your requirements is having the distance between two points in the plane be the complex number whose modulus is the Euclidean length of the unique segment of (possibly degenerate) logarithmic spiral centered at the origin and passing through the two points, and whose argument is the characteristic angle of that logarithmic spiral (i.e., $\arctan(1/b)$ if the spiral's equation is $r=ae^{b\theta}$ in polar coordinates). – pregunton Sep 24 '17 at 7:52
• In this way, the distance between two points lying on a circle centered at the origin is the length of the arc of circle (degenerate spiral) joining them multiplied by $i$, and the distance between two points lying on a ray from the origin is the length of the segment connecting them. The distance is evidently invariant under rotations around the origin. You can easily extend this definition to more than two dimensions by taking the unique plane formed by the two points and the origin (the plane is not unique if the points are aligned, but then choosing any plane passing through them works). – pregunton Sep 24 '17 at 7:52
• @pregunton: This is interesting, because I already had this spiral defined there for a different reason. The fact that this spiral appears again shows that it relates to the properties of this space. I need to sleep on this, but it looks promising. Meanwhile, would there be any mathematical inconsistency or concern regarding this space? – safesphere Sep 24 '17 at 8:24

The update makes it much clearer what is wanted for the metric.

Let $d\Theta^2$ denote the spherical metric (induced by the Euclidean) on $S^{n-1}=\{ x\in {\Bbb R}^n: \|x\|=1\}$ and write $r=\sqrt{x_1^2+\cdots + x_n^2}$. The standard Euclidean metric on ${\Bbb R}^n$ may then be written as follows: $\sum_i dx_i^2 = dr^2 + r^2 d\Theta^2$.

The central idea in the current post is to change the sign on the angular part, i.e. we consider the punctured Euclidean space $X={\Bbb R}^n\setminus\{0\}$ equiped with the following $(1,n-1)$ pseudo-Riemannian metric: $$g= dr^2 - r^2 d\Theta^2 = 2 dr^2 - \sum_{i=1}^n dx_i^2$$

We wish to describe geodesics in $(X,g)$. In relativistic terminology, a tangent vector with $g(v,v)>0$ is time-like, $g(v,v)<0$ is space-like, while $g(v,v)=0$ corresponds to a ligth-cone vector.

It is of interest to note that the metric is invariant under the orthogonal group which implies that there is angular momentum conservation: A geodesic starting at some given position in space and in a given direction will always stay in the span of those two directions, i.e. it suffices to restrict our attention to those two dimensions. So let us write: $$g= dr^2 - r^2 d\phi^2$$ with $(r,\phi)$ being standard polar coordinates in the plane.

Geodesics in normal Riemannian geometry are paths between points that are extremal for the length $\int \sqrt{g(\dot{x},\dot{x})} dt$. Normalizing to constant speed it is equivalent to be extremal for the action functional: $$S = \int g(\dot{x},\dot{x}) dt = \int {\cal L} (r,\phi,\dot{r},\dot{\phi}) dt,$$ with the Lagrangian ${\cal L} = \dot{r}^2 - r^2 \dot{\phi}^2$. So we take extremality of this action to define geodesics in the present context. An extremal path verifies Lagrange's equations: $$2 \ddot{r} = \frac{d}{dt} \frac{\partial L}{\partial \dot{r}} = \frac{\partial L}{\partial r} = - 2 r \dot{\phi}^2 \;\; \mbox{and} \; \; \frac{d}{dt} \left( r^2 \dot{\phi} \right) = \frac{d}{dt} \frac{\partial L}{\partial \dot{r}} = \frac{\partial L}{\partial r} = 0.$$ The last implies the above-mentioned angular conservation: $r^2 \dot{\phi}=A$ for some constant $A$. Similarly, there is conservation of energy (Hamiltonian): $$E = \dot{r}\frac{\partial L}{\partial \dot{r}} + \dot{\phi} \frac{\partial L}{\partial \dot{\phi}} - L = 2 L - L = L$$ So $\dot{r}^2 - r^2 \dot{\phi}^2 = E$ for some constant $E$.

First case: If $\dot{\phi}=0$ at some instant of time then $A=0$ and $\phi$ is a constant of motion. We may solve to get: $\dot{r}=\pm\sqrt{E}$ which is just a linear motion in time, $r(t) = \pm\sqrt{E}(t-t_0)$ (the geodesic ceases to exist when $r(t_0)=0$).

Second case: When $\dot{\phi}\neq 0$, it has a constant sign (same as the sign of $A$). We may then by the implicit function theorem write $r=r(\phi)$ so that $\dot{r} = r'(\phi) \dot{\phi}$. Then $(r'^2-r^2) \dot{\phi}^2 = E$ and inserting the angular momentum conservation we deduce the following equation for the trajectories: $$r'^2 - r^2 = \frac{E}{A^2} r^4$$

Subcases:

a) $E=0$: We get $r'=\pm r$ or $r(\phi) = \exp (\pm (\phi-\phi_0))$.

b) $E<0$ (space-like trajectories): Write $r=1/u$ and solve the resulting ode for $u$. You end up with (modulo mistakes in my calculations): $$r(\phi) = \frac{A/\sqrt{-E}}{\cosh(\phi-\phi_0)}$$

c) $E>0$ (time-like trajectories): $$r(\phi) = \frac{A/\sqrt{E}}{\sinh(\phi-\phi_0)}$$

Symmetries: A part from the rotational symmetry I don't think that there are any other. The fact that $(r,\phi)$ is identified with $(r,\phi+ 2\pi)$ gives a topological constraint which prevents us from doing Lorentz-like transformations.

My answer to the original post:

A suggestion: In ${\Bbb R}^n$, write $r\cdot r'$ and $|r|$ for the Euclidean scalar product and length, respectively.

Define two infinitesimal (Riemannian) pseudo-metrics between infinitesimal close vectors $r$ and $r+dr$ (with $r\neq 0$):

$$ds_1 = \left| \frac{r}{|r|} \cdot dr \right| \; \; {\rm and} \; \; ds_2 = \left|dr - \frac{r}{|r|}\left( \frac{r}{|r|} \cdot dr\right) \right|$$ $ds_1$ measures the radial distance, $ds_2$ the rotational. If you want to represent them as complex numbers you may set $dz=ds_1+i ds_2$. Then, $|dz|$ (modulus of complex number) corresponds to the infinitesimal Euclidean distance.

You may then measure "complex" path lengths: If $r(t)$, $t\in [0,1]$ is a $C^1$ curve then $$L={\rm len}_{\Bbb C} (r,[0,1]) = \int_0^1 \left| \frac{r}{|r|} \cdot \dot{r} \right| dt + i \int_0^1 \left|\dot{r} - \frac{r}{|r|}\left( \frac{r}{|r|} \cdot \dot{r}\right) \right|dt$$ separates the usual length of the curve into the radial part (real) and the rotational part (imaginary). The modulus of $L$ is equivalent (though not necessarily equal) to the usual Euclidean length of the path.

Depending on your purpose with defining a complex distance, there might be an ambiguity as to the definition of the distance between two finite vectors $r_1$ and $r_2$ as you would have to specify what a geodesic is in this picture. A perhaps natural choice is to define a geodesic as a path that minimizes Euclidean distances. Then geodesics are straight line segments and the complex distance may be (partially) calculated as follows:

Let $r'$ be the point on the line segment $[r_1;r_2]$ closest to the origin (could be one of the end-points) and let $a\geq 0$ be the distance from the line through $r_1$, $r_2$ and the origin. For $0< u\leq v$ write $\Theta(u,v) = u\ln \frac{v+\sqrt{v^2-u^2}}{u}$. Then if $r'$ is not one of the end-points: $$d_{\Bbb C}(r_1,r_2) = \left(|r_1|+|r_2|-2|r'| \right) + i \left(\Theta(a,|r_1|+ \Theta(a,|r_2|)-2\Theta(a,|r'|)\right)$$ while if $r'$ is one of the end-points you get the simpler: $$d_{\Bbb C}(r_1,r_2) = \left||r_2|-|r_1|\right| + i \; a \left| \ln \frac{|r_2|+\sqrt{|r_2|^2-a^2}}{|r_1|+\sqrt{|r_1|^2-a^2}}\right|$$ separating into how much you move radially and rotationally along the geodesic. In the limit $a\rightarrow 0$ the imaginary part vanishes as wanted since $r_1$ and $r_2$ are proportional in that case.

• Hi Hans Henrik, thank you so much for working on my problem! Please don't be discouraged by my lack of response. I understand most of the math involved, but I'm not a mathematician, so understanding your solution will take me a bit of time :) To answer your question on the purpose of the complex distance, this space must be locally asymptotic with the metric space of the hyperbolic metric $ds_1^2 - ds_2^2$ where $ds_1$ is asymptotic to the radial coordinate and $ds_2$ is asymptotic to the angular linear coordinate. I'm a bit over my head here, please let me know if this makes no sense :) – safesphere Sep 25 '17 at 18:52
• Accordingly, local (asymptotic or infinitely small, whichever is the appropriate term) geodesics perhaps should be defined by the vanishing partial second derivatives along the geodesic. Globally geodesic are probably logarithmic spirals (including degenerate by either coordinate). This is what I thought and also mentioned by pregunton in the comments, but please do not hesitate to correct, if this seems wrong. This way a geodesic may not be unique, but may depend on the number of revolutions around the origin. This is OK and expected, unless it creates a logical inconsistency. – safesphere Sep 25 '17 at 19:10
• Clarification: By "angular linear" I meant "rotational". Please excuse my lack of familiarity with the conventional terminology :) – safesphere Sep 25 '17 at 19:16
• Ok, at least that gives more sense to the question. I don't have the time right now but will try to get back to this later on. You should perhaps write some more details regarding this in your question. It looks like seeking geodesics (or rather stationary paths) for the integral of ${\rm Re}\; dz^2$. In fact the representation as complex numbers is irrelevant. – H. H. Rugh Sep 25 '17 at 19:39
• Perfect. Thanks so much! I have updated the question as requested. – safesphere Sep 26 '17 at 8:17


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# Is TeX as word and logo a trade mark?

A few days ago I talked with someone about writing a book about TeX and we came to the question if TeX as a word and its Logo (and the related things like LaTeX, XeTeX, …, package names etc.) are (registered) trademarks or can I use it freely as any other word like dog, house, moon, …?

Examples

• using in book titles: Start with TeX or Become a TeXnician
• using in corparate names: LaTeX production company or TeX Help Line

Are there any restrictions in using “TeX” commercially (assuming the products/services are TeX related)?
How about the derived words like “LaTeX”, “XeTeX” etc.

Edit: As Patrick said, the country could be important … I’m living in Germany and want to sell from there too but I could imagine that there are readers from Swiss and Austria for German books too.

• See also en.wikipedia.org/wiki/TeX#License – it claims that "The American Mathematical Society has once tried to claim a trademark for TeX". Oct 2 '11 at 14:08
• there is no restriction, as long as it is rendered either as TeX or with the logo, and as long as the reference is to the actual TeX(-related) software. that said, it is not likely that any lawyers will be sent after you; however, beware of peer pressure, which can be considerable. @Jukka -- i actually handled the letter from the Honeywell lawyers, asking that DEK and AMS cease and desist from use of the name. this is a case where peer pressure did have a strong effect; a compromise was worked out by which the TeX form of the name was agreed to be acceptable, if used consistently. Oct 2 '11 at 18:18
• @barbarabeeton Now that I've migrated this to the main site, I think your comment would make an excellent answer. Oct 3 '11 at 10:15
• Trademarks are local, you should state where (which country) you want to use LaTeX, TeX etc. Even if it isn't registered by the AMS, it could be registered by other people. Oct 3 '11 at 11:44
• Simple answer: Ask the lawyer of your publisher! Note that 'TeX' and 'LaTeX' might be legally handled quite differently even when the first is part of the second. It normally about the possibility of confusion. There are already a lot of (La)TeX books around which didn't got sued (AFAIK), so I don't think it is an issue. Oct 3 '11 at 22:05

There is no restriction, as long as it is rendered either as TeX or with the logo, and as long as the reference is to the actual TeX(-related) software.

That said, it is not likely that any lawyers will be sent after you; however, beware of peer pressure, which can be considerable.

At the time of the attempted registration, I actually handled the letter from the Honeywell lawyers, asking that DEK and AMS cease and desist from use of the name. This is a case where peer pressure did have a strong effect, since Honeywell was then a computer manufacturer, and DEK was already a leading light of computer science. A compromise was worked out by which the TeX form of the name was agreed to be acceptable, if used consistently.

There has not been a later attempt to register the trademark; however, Knuth's standing internationally is sufficient that if the name were "misused", there would probably be at least a few reasonably local interested persons to object. In fact, this is what we count on when complaining about PDF copies of The TeXbook found on the web; in most parts of the world, this is sufficient to get them removed.

• Would you care to explain why a PDF copy of The TeXbook is considered "misuse" of the TeX name/logo? Oct 3 '11 at 17:39
• @Werner: I think it is just an example of relying on peer pressure to handle such situation. Oct 3 '11 at 18:09
• @Werner, Khaled's interpretation is exactly what i meant. Oct 3 '11 at 19:38
• Please extend this history of TeX the trademark in a future issue of TUGboat so we have a reliable source for Wikipedia (I doubt that an answer on SX meets the standards of Wikipedia). Oct 3 '11 at 22:39
• I always thought that the X was actually a \chi and that why it was written as a capital x becomes they look more alike Jun 24 '12 at 11:38

Totally unrelated to TeX, but it might be of interest to note that the supermarket group Carrefour holds various trademark registries for the name "TeX" and multiple logos, some of which closely resemble the TeX logo:

In most countries, this trademark is registered for textile (not texlive as I was about to write) products such as shirts, pants, sheets, underwear, baby clothes, etc. But in France at least, the list is quite longer, including for example scientific devices, games, medical devices, lighting/warming/cooling devices, security devices (for cars, bikes, etc.), jewelry, printing devices, leather, furniture, cleaning and cooking devices, toys, etc.

This should not prevent from publishing books about TeX and LaTeX, but it might be a problem if we wanted to sell a TeX (or TeX.SX) t-shirt for example.


text:
Skip to main content

8: Confidence Intervals

In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.

• 8.1: Prelude to Confidence Intervals
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
• 8.2: A Single Population Mean using the Normal Distribution
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution.
• 8.3: A Single Population Mean using the Student t-Distribution
We rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation ss as an estimate for σσ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.
• 8.4: A Population Proportion
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different.
• 8.5: Confidence Interval - Home Costs (Worksheet)
A statistics Worksheet: The student will calculate the 90% confidence interval for the mean cost of a home in the area in which this school is located. The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval.
• 8.6: Confidence Interval -Place of Birth (Worksheet)
A statistics Worksheet: The student will calculate the 90% confidence interval the proportion of students in this school who were born in this state. The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval.
• 8.7: Confidence Interval -Women's Heights (Worksheet)
A statistics Worksheet: The student will calculate a 90% confidence interval using the given data. The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean.
• 8.8: Confidence Intervals (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
• 8.9: Confidence Intervals (Summary)
In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known.

Contributors and Attributions

• Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/30189442-699...b91b9de@18.114.

This page titled 8: Confidence Intervals is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

• Was this article helpful?


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There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum umber of triangles with vertices at these points is

# There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum umber of triangles with vertices at these points is

1. A

2. B

${\mathrm{p}}^{2}\left(\mathrm{p}-1\right)$

3. C

${\mathrm{p}}^{2}\left(4\mathrm{p}-1\right)$

4. D

${\mathrm{p}}^{2}\left(4\mathrm{p}-3\right)$

Register to Get Free Mock Test and Study Material

+91

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${\mathrm{p}}_{{\mathrm{C}}_{1}}×{\mathrm{p}}_{{\mathrm{C}}_{1}}×{\mathrm{p}}_{{\mathrm{C}}_{1}}+{\mathrm{p}}_{{\mathrm{C}}_{2}}×2{\mathrm{p}}_{{\mathrm{C}}_{1}}×{3}_{{\mathrm{C}}_{1}}={\mathrm{p}}^{2}\left(4\mathrm{p}-3\right)$


text:
# Binomial Probability & Tables

• May 26th 2008, 07:19 AM
16sammy
Binomial Probability & Tables
I hate math assignments and it's due tomorrow and the teacher is stupid and won;t help people. I was away when it was explained so i have no idea what to do!!

Suppose an NBA player makes 90% of his free throws. This player takes 20 free throws in the course of a game.
Find the probabilits of his making:

A) Atleast 19 of the 20 free throws.
B) Less than 15 of the 20 free throws
C) Exactly 90% od the 20 free throws.
• May 26th 2008, 07:35 AM
janvdl
Quote:

Originally Posted by 16sammy
I hate math assignments and it's due tomorrow and the teacher is stupid and won;t help people. I was away when it was explained so i have no idea what to do!!

Suppose an NBA player makes 90% of his free throws. This player takes 20 free throws in the course of a game.
Find the probabilits of his making:

A) Atleast 19 of the 20 free throws.
B) Less than 15 of the 20 free throws
C) Exactly 90% od the 20 free throws.

Lecturers don't explain things twice. Get used to it. Also, most students have very poor excuses for not coming to class. I'm not saying you're one of them though.

The probability of a success is 0,9

$
P(A) = {20 \choose 19} \left( 0,9 \right) ^{19} \left( 0,1 \right)^{1} + {20 \choose 20} \left( 0,9 \right) ^{20} \left( 0,1 \right)^{0}$

$
P(B) = 1 - {20 \choose 15} \left( 0,9 \right) ^{15} \left( 0,1 \right)^{5}$
$- {20 \choose 16} \left( 0,9 \right) ^{16} \left( 0,1 \right)^{4}$ $- {20 \choose 17} \left( 0,9 \right) ^{17} \left( 0,1 \right)^{3}$ $- {20 \choose 18} \left( 0,9 \right) ^{18} \left( 0,1 \right)^{2}$ $- {20 \choose 19} \left( 0,9 \right) ^{19} \left( 0,1 \right)^{1}$ $- {20 \choose 20} \left( 0,9 \right) ^{20} \left( 0,1 \right)^{0}$

90% of 20 throws equal 18 throws.

$P(C) = {20 \choose 18} \left( 0,9 \right) ^{18} \left( 0,1 \right)^{2}$
• May 26th 2008, 04:51 PM
mr fantastic
Quote:

Originally Posted by 16sammy
I hate math assignments and it's due tomorrow and the teacher is stupid and won;t help people. I was away when it was explained so i have no idea what to do!!

Suppose an NBA player makes 90% of his free throws. This player takes 20 free throws in the course of a game.
Find the probabilits of his making:

A) Atleast 19 of the 20 free throws.
B) Less than 15 of the 20 free throws
C) Exactly 90% od the 20 free throws.

1. So how long have you had to do this assignment?
2. What steps did you take to catch up with the work you missed?


text:
# 6.4: Normalization and Orthogonality

Although we aren’t yet going to learn rules for doing general inner products between state vectors, there are two cases where the inner product of two state vectors produces a simple answer. The first is not intrinsic to the mathematical representation, but rather something we will insist for state vectors that properly represent real physical states. For a complete state vector $$|\psi\rangle$$ to be a proper quantum mechanical state, it must satisfy the condition

$$\langle\psi \mid \psi\rangle=1$$

We say that this means that the state vector is normalized. It is possible to have non-normalized state vectors. For instance, in the equation

$$|+x\rangle=\frac{1}{\sqrt{2}}|+z\rangle+\frac{1}{\sqrt{2}}|-z\rangle$$

the two parts of the sum on the right side are themselves ket vectors. However, because they are valid state vectors multiplied by a constant, they are not normalized themselves. We will show later that this definition of $$|+x\rangle$$ is, however, normalized.

The second rule is that state vectors that represent different possible states corresponding to different possible measurements of a given observable must be orthogonal. Mathematically, this is expressed as:

$$\left\langle\phi_{1} \mid \phi_{2}\right\rangle=0$$

if $$\left|\phi_{1}\right\rangle$$ and $$\left|\phi_{2}\right\rangle$$ are two different states corresponding to definite states for a given observable. For example, the states $$|+z\rangle$$ and $$|-z\rangle$$ correspond to two states of the same observable, specifically, the $$z$$ component of angular momentum. The first corresponds to that component being measured along $$+z$$, the second to it being measured along $$−z$$. The orthogonality condition is then:

$$\langle+z \mid-z\rangle=0$$

As an example of doing these calculations with a more complicated state, consider the state $$|+x\rangle$$. If this state is properly normalized, then we should have $$\langle+x \mid+x\rangle=1$$. Do we? Well, first, we have to construct the bra vector that goes along with the ket vector:

\begin{aligned} &|+x\rangle=\frac{1}{\sqrt{2}}|+z\rangle+\frac{1}{\sqrt{2}}|-z\rangle \\ &\langle+x|=\frac{1}{\sqrt{2}}|+z\rangle+\frac{1}{\sqrt{2}}|-z\rangle \end{aligned}

Notice that in the case of a compound ket vector, to get the bra vector we just turn all ket vectors on the right side into bra vectors, and replace all the numbers with their complex conjugates (which is trivial here, since all the numbers are real). Now we have what we need to figure out the inner product. Just substitute in our expressions for $$|+x\rangle$$ and $$\langle+x|$$:

\begin{aligned} \langle+x \mid+x\rangle &=\left(\frac { 1 } { \sqrt { 2 } } \left\langle+z\left|+\frac{1}{\sqrt{2}}\langle-z|\right)\left(\frac{1}{\sqrt{2}}|+z\rangle+\frac{1}{\sqrt{2}}|-z\rangle\right)\right.\right.\\ &=\frac{1}{2}\langle+z \mid+z\rangle+\frac{1}{2}\langle+z \mid-z\rangle+\frac{1}{2}\langle-z \mid+z\rangle+\frac{1}{2}\langle-z \mid-z\rangle \end{aligned}

That looks very complicated, but now we can use the orthogonality condition we know is true for the $$z$$ states, as we’ve defined them as good states corresponding to the $$z$$ component of $$z$$ angular momentum. We know that $$\langle+z \mid+z\rangle=1,\langle-z \mid-z\rangle=1$$, $$\langle-z \mid+z\rangle=0$$, and $$\langle+z \mid-z\rangle=0$$ from normalization and orthogonality. Substitute these in:

\begin{aligned} \langle+x \mid+x\rangle &=\frac{1}{2}(1)+\frac{1}{2}(0)+\frac{1}{2}(0)+\frac{1}{2}(1) \\ \langle+x \mid+x\rangle &=1 \end{aligned}

So the state is properly normalized! I leave it as an exercise for the alert reader to show that $$|+x\rangle$$ and $$|-x\rangle$$ are orthogonal.

This page titled 6.4: Normalization and Orthogonality is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.