Mathematically what is the difference between pure separable state and entangled state ?  Can anyone explain with equations?
I am currently trying to establish a clear picture of pure/mixed/entangled/separable/superposed states. In the following I will always assume a basis of $|1\rangle$ and $|0\rangle$ for my quantum systems. This is what I have so far:   superposed: A superposition of two states which a system $A$ can occupy, so $\frac{1}{\sqrt{2}}(|0\rangle_A+|1\rangle_A)$ seperable: $|1\rangle_A|0\rangle_B$ A state is called separable, if its an element of the (tensor)product basis of system $A$ and $B$ (for all possible choices of bases) entangled: $\frac{1}{\sqrt{2}}(|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B)$ is not a state within the product basis (again for all possible bases). mixed state: Is a statistical mixture, so for instance $|1\rangle$ with probability $1/2$ and $|0\rangle$ with probability $1/2$ pure state: Not a mixed state, no statistical mixture   I hope that the above examples and classifications are correct. If not it would be great if you could correct me. Or add further cases, if this list is incomplete.   On wikipedia I read about      Another way to say this is that while the von Neumann entropy of the whole state is zero (as it is for any pure state), the entropy of the subsystems is greater than zero.   which is perfectly fine. However I also read on wikipedia a criterion for :     Another, equivalent, criterion is that the von Neumann entropy is 0 for a pure state, and strictly positive for a mixed state.   So does this imply that if I look at the subsystems of an entangled state, that they are in a mixed state? Sounds strange... What would be the statistical mixture in that case?  Moreover I also wanted to ask, whether you had further illustrative examples for the different states I tried to describe above. Or any dangerous cases, where one might think a state of one kind to be the other?
How to make wget faster or multithreading? I've just upgraded my computer hardware(cpu + motherboard + graphic card + memory + hard disk), so that install a new OS is needed. I tried to download debian-6.0.6-amd64-netinst.iso with wget command but the speed is so slow that I could not bear. 4Kb/s ~ 17 Kb/s, slow like a running turtle, or even more slower if I use Chrome.  I've read the help information of wget, it seems like there are no options could make it more faster.  Is there anyway to make wget faster? Or is it possible to make it multi-threading download?  PS: my bandwidth is 4M. I use this command:  wget -c url  http://hammurabi.acc.umu.se/debian-cd/6.0.6/amd64/iso-cd/debian-6.0.6-amd64-netinst.iso Each site got a &quot;hot&quot; tab, where its contents are calculated based on . However take a look on the screenshots below and see for yourself that &quot;hot&quot; got nothing to do with many of the questions listed there. :    :    As you can see, there are questions with negative score, no answers, and even closed. I won't be surprised to see spam in there as well. This can't be any worse. What I suggest, instead of fixing the formula or patching it, simply have the hot tab show the same questions as the Hot Network Questions sidebar. (from the specific site only, of course.) Since the hot questions &quot;pool&quot; (as seen ) is limited to 100 for the whole network, either expand it or have the per-site list show questions with less hotness points - it would still be better than what we have now. How can I accept the Microsoft EULA agreement for ttf-mscorefonts-installer? How to inject Spring Bean for factory method requiring MyClass.class parameter I had some NEF files in a folder that I didn't do anything with. When importing them in Lightroom, no previews and a "file unsupported or damaged message" So I tried with Nikons ViewNX, nothing. Even not in DxO or PhotoMechanic.  I had a look with EXIFTool and the corrupt file prints me only this:  ExifTool Version Number         : 8.39 File Name                       : _DSC5559.NEF Directory                       : /Users/tunafish/Desktop File Size                       : 12 MB File Modification Date/Time     : 2010:11:29 12:46:56+01:00 File Permissions                : rwxrwxrwx Error                           : File format error   I saw another question on photo.stack  and tried to copy over the makernotes from a good file (_DSC5556.NEF) with  exiftool -tagsfromfile _DSC5556.NEF -makernotes _DSC5559.NEF   but EXIFTool gives me this error  Error: Not a valid TIFF - _DSC5559.NEF 0 image files updated 1 files weren't updated due to errors   I have uploaded 2 NEF's, with _DSC5559.NEF being the corrupt one here: Spatial query to return multiple values I have a woodland scheme advice layer that has a polygon for each area of advice (brown layer in the picture). I also have a polygon layer of UK local authorities (red boundary layer in the picture). I would like to perform a spatial query which returns the local authority name for each woodland advice scheme. This would work fine if each woodland advice scheme was found in just one local authority boundary. However many are found across a boundary of two different authorities. For instance, one scheme sits across the boundaries of the authorities of Fife and Angus. The scheme is the yellow selected polygon in the picture and the red line is the authority boundary. How do I get the spatial join to return a value of Fife; Angus (or something similar to identify two values) instead of just Fife or Angus. Probability distribution in the coupon collector's problem I'm trying to solve the well known Coupon Collector's Problem by explicitly finding the probability distribution (so far all the methods I read involve using some sort of trick). However, I'm not having much luck getting anywhere as combinatorics is not something I'm particularly good at.   The Coupon Collector's Problem is stated as:  There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes one has to buy to collect all types of coupons?  What I'm attempting:  Let $N$ be the random variable associated with the number of boxes one has to buy to find all coupons. Then $P(N=n)=\frac{|A_n|}{|\Omega _n|}$, where $A_n$ is the set of all outcomes such that all types of coupons are observed in $n$ buys, and $\Omega _n$ is the set of all the possible outcomes in $n$ buys. I think $|\Omega _n| = m^n$, but I'm not even sure about that anymore, as all my attempts so far led to garbage probabilities that either diverged or didn't sum up to 1. What influences findability on a post?   Upvotes Views Asked date Last active time   How is this done in the Stack Exchange internal search engine? And how does Google do it?  Why is it that when searching on Google for 'iPhone development stackoverflow' the fourth result is ? Mathematica creates a huge eps file Adverb between conjugated verb and infinitive? Is there a practical gain from elliptical/oval chainrings? I have seen competitive (amateur) cyclists with oval chainrings (front). What is the deal? Are they just for showing off, or is there a practical gain from it (different force transfer). Is there a way of finding a formula for $\sum\limits_{k=1}^n k^k$? Maybe I'm missing something really obvious, but I've looked around a bit on the Internet and I haven't been able to find anything.  So, what I'm looking for is a formula in closed form to generate the sequence $1,5,32,288,3413,\dots$ Electromagnetic waves can be shielded by a perfect conductor. What about gravitational fields or waves? Prove that an open interval and a closed interval are not homeomorphic Prove that an open interval $(a,b)$ and a closed interval $[c,d]$ are not homeomorphic.  I'm trying to prove this statement but have only vague ideas on how to start. How may I use the property of connectedness to show this? What is a word for someone who is experienced and wise, but who deliberately acts naïve?    I don’t intend it for sarcastic use; I’m trying to describe someone succeeding in making people think that that person truly is naïve.   Is there a word for this? Reverse the ordering of words in a string Testing for coefficients significance in Lasso logistic regression [A similar question was asked  with no answers] I have fit a logistic regression model with L1 regularization (Lasso logistic regression) and I would like to test the fitted coefficients for significance and get their p-values. I know Wald's tests (for instance) are an option to test the significance of individual coefficients in full regression without regularization, but with Lasso I think further problems arise which do not allow to apply the usual Wald formulas. For instance, the variance estimates neded for the test do not follow the usual expressions. The original Lasso paper:  suggests a bootstrap-based procedure to estimate the coefficients variance, which (again, I think) may be needed for the tests (section 2.5, last paragraph of page 272 and beginning of 273):  One approach is via the bootstrap: either $t$ can be fixed or we may optimize over $t$ for each bootstrap sample. Fixing $t$ is analogous to selecting the best subset (of features) and then using the least squares standard error for that subset  What I understand is: fit a Lasso regression repeatedly to the whole dataset until we find the optimal value for the regularization parameter (this is not part of the bootstrap), and then use only the features selected by the Lasso to fit OLS regressions to subsamples of the data and apply the usual formulas to compute the variances from each of those regressions. (And then what should I do with all those variances of each coefficient to get the final variance estimate of each coefficient?) Furthermore, is it correct to use the usual significance tests (for instance Wald's test which makes use of the estimated betas and variances) with the Lasso estimates of the coefficients and the bootstrap-estimated variances? I am fairly sure it is not, but any help (use a different test, use a more straightforward approach, whaterever...) is more than welcome. According to the answers  I suspect inference and p-values just cannot be obtained. In my case, p-values are an external requirement (although the use of L1 regularization was my choice). Thanks a lot EDIT What if I fit an OLS logistic regression using only the variables selected by a previous run of the Lasso logistic regression? Apparently (see ),  There's no need to run the model again after doing cross-validation (you just get the coefficients from the output of cv.glmnet), and in fact if you fit the new logistic regression model without penalisation then you're defeating the purpose of using lasso  But what if I do this with the sole purpose of being able to compute p-values while keeping the number of variables low? Is it a very dirty approach? :-) Why haven't I received the "Curious" badge? Moving from SQL 2005 [SQL_Latin1_General_CP1_CI_AS] to 2008 - will I lose any features by using 'backwards compatibility' We're moving from SQL 2005 [Instance and DB have collation of SQL_Latin1_General_CP1_CI_AS]  to SQL 2008 [which defaults to Latin1_General_CI_AS].  I completed a SQL 2008 R2 installation, and used default Latin1_General_CI_AS collation, with the restoration of the database still on SQL_Latin1_General_CP1_CI_AS. The excepted problems occurred - the #temp tables where in  Latin1_General_CI_AS whilst the db was in  SQL_Latin1_General_CP1_CI_AS and this is where I am now - I need advice on the pitfalls now please.  On installation of SQL 2008 R2, I have the option on installation to use 'SQL Collation, used for backwards compatibility' where I have the option to select the same collation as the 2005 database : SQL_Latin1_General_CP1_CI_AS.   This will allow me to not have problems with #temp tables, but are there pitfalls? Would I lose any functionality or features of any kind by not using a "current" collation of SQL 2008? What about when we move (e.g. in 2 years ) from 2008 to SQL 2012? Will I have problems then? Would I at some point be forced to go to Latin1_General_CI_AS? I read that some DBA's script complete the rows of complete databases, and then run the insert script into the database with the new collation - I'm very scared and wary of this - would you recommend doing this? Sometimes I wonder if he is actually/really  as lavish as he   pretends to be.   I know that this sentence can work without either actually or really, but if we were to use one of them to make an impact on the listener, which one should it be?