A single photon, passing through a beamsplitter (BS), ends up in the state $\frac{1}{\sqrt2}(a_1^\dagger + a_2^\dagger)\lvert\operatorname{vac}\rangle$, which is a coherent superposition of the two possible spatial modes. Let us consider its momentum.  When the photon goes straight its momentum doesn't change. However, when the photon gets deflected, some amount of momentum must have been transferred to the BS, which therefore acquires some (small) amount of momentum in the opposite direction of the photon (see image below). This new tiny momentum will correspond to some phononic excitation of the BS crystalline structure, or something like that.     Because the photon ends up in a superposition of the two output modes, the complete output state must be something of the form: $$ \frac{1}{\sqrt2} \Big(\lvert\text{photon up}\rangle\lvert\text{BS steady}\rangle+\lvert\text{photon right}\rangle\lvert\text{BS momentum changed}\rangle \Big).$$ However, we do not care about the final state of the BS, so that the final state of the photon will be obtained by partial tracing over the BS degree of freedom.  And here comes the catch: a naive calculation would now lead to the conclusion that, after partial tracing, the final state of the photon is $$ \frac{1}{2} \Big(   \lvert\text{photon up}\rangle\langle\text{photon up}\rvert +   \lvert\text{photon right}\rangle\langle\text{photon right}\rvert \Big). $$ This is a completely mixed state, which means that the photon is not in a coherent superposition of the output modes.  Now, we know very well that this is not actually the case, as the fact that the photon does end up in a coherent superposition is easily tested and well enstablished. So what did we get wrong? The only logical conclusion (that I can see) is that $\lvert\text{BS steady}\rangle$ and $\lvert\text{BS momentum changed}\rangle$ are not orthogonal states, so that partial tracing does not lead to a totally mixed state for the photon.  So here comes the question: why are the two final states of the BS not orthogonal? Is there a clean way to see this?  Moreover, given that these two states are also necessarily not the same, this means that partial tracing will always lead to some small amount of mixedness in the final photon state. Does this mean that there is a fundamental limit in the purity of the final photon state?
In many experiments in quantum mechanics, a single photon is sent to a mirror which it passes through or bounces off with 50% probability, then the same for some more similar mirrors, and at the end we get interference between the various paths. This is fairly easy to observe in the laboratory.  The interference means there is no which-path information stored anywhere in the mirrors. The mirrors are made of 10^20-something atoms, they aren't necessarily ultra-pure crystals, and they're at room temperature. Nonetheless, they act on the photons as very simple unitary operators. Why is it that the mirrors retain no or very little trace of the photon's path, so that very little decoherence occurs?  In general, how do I look at a physical situation and predict when there will be enough noisy interaction with the environment for a quantum state to decohere?
What is the difference between integration and unit tests? Why do physicists believe that particles are pointlike? String theory gives physicists reason to believe that particles are 1-dimensional strings because the theory has a purpose - unifying gravity with the gauge theories.   So why is it that it's popular belief that particles are 0-dimensional points? Was there ever a proposed theory of them being like this? And why?   What reason do physicists have to believe that particles are 0-dimensional points as opposed to 1-dimensional strings? Why is 4 not an instance of Number? Just curious:    4 instanceof Number => false new Number(4) instanceof Number => true?   Why is this? Same with strings:    'some string' instanceof String returns false new String('some string') instanceof String => true String('some string') instanceof String also returns false ('some string').toString instanceof String also returns false   For object, array or function types the instanceof operator works as expected. I just don't know how to understand this.  [new insights]  Object.prototype.is = function() {         var test = arguments.length ? [].slice.call(arguments) : null            ,self = this.constructor;         return test ? !!(test.filter(function(a){return a === self}).length)                : (this.constructor.name ||                   (String(self).match ( /^function\s*([^\s(]+)/im )                     || [0,'ANONYMOUS_CONSTRUCTOR']) [1] ); } // usage var Newclass = function(){};  // anonymous Constructor function var Some = function Some(){}; // named Constructor function (5).is();                     //=&gt; Number 'hello world'.is();           //=&gt; String (new Newclass()).is();        //=&gt; ANONYMOUS_CONSTRUCTOR (new Some()).is();            //=&gt; Some /[a-z]/.is();                 //=&gt; RegExp '5'.is(Number);               //=&gt; false '5'.is(String);               //=&gt; true I want to see all URLs that for packages that would be downloaded for apt install &lt;pkg&gt; without actually installing the package.  Normally, the URLs can be extracted from the screen output after installing the package (as above).  An example of a URL I want to list:    Is it only possible to get such list of URLs beforehand by looking through all the apt's package source lists? How to understand the combination formula? I'm reading an algorithm book. In which it mentioned:     Imagine (once again) you have n people lined up to see a movie, but   there are only k places left in  the theater. How many possible   subsets of size k could possibly get in? That’s exactly $C(n,k)$, of   course,  and the metaphor may do some work for us here. We already   know that we have $n!$ possible orderings of  the entire line. What if   we just count all these possibilities and let in the first $k$? The only   problem then is  that we’ve counted the subsets too many times. A   certain group of k friends could stand at the head of  the line in a   lot of the permutations; in fact, we could allow these friends to   stand in any of their $k!$  possible permutations, and the remainder of   the line could stand in any of their $(n–k)!$ possible  permutations   without affecting who’s getting in. Aaaand this gives us the answer!       $$C(n,k)= \frac{n!}{k!(n −k)!}$$      This formula just counts all possible permutations of the line (n!),   and divides by the number of  times we count each “winning subset,” as   explained.   I'm not quite understand why the reminder can in any $(n-k)!$ permutation, if he stands at the end of the queue, won't he let all these people in?  And why I should use $n!$ to divide by $k!(n-k)!$ ? Didn't see that consequence from the metaphor.  If you have better ideas to explain this formula, you can use them of course. These are my current disk partitions. /yedek is for some personal backup files. As you can see there is only Ubuntu on my PC. I will install Windows, too. Because of this I want to shrink /dev/sda3 . It currently 546GB and I want to split it: 300GB for / and 246GB for Windows (NTFS) is OK for me.  I'm using GParted for the first time. How can I shrink /?  I don't want to lose any datas from Ubuntu. In chrome, most of the time, when i middle click on a link (text or image), it opens a new tab and the current tab does not change. This is the expected behaviour.  However, some websites have links (i'd say 5%) that also switches the current tab to the new one. For example: on this , the name of the players in the main list have the problem.  By looking at the source, i didn't find anything anormal (the link is a standard html tag). But if i disable javascript, i have not the problem. I looked at the source, and didn't find the problematic script... Note also that if i open the tab with right click > open link in new tab, there is no problem.  I'd like to know if you find the reason of the problem and how i can fix it. I am running chrome 10.0.648.204 on windows 7 x64.  EDIT: the website above does not exhibit the behavior anymore, but I sill have the problem sometimes on other sites (which I don't remember) How does Cooper get out of the black hole? In the movie Interstellar,  Cooper (played by ) enters a black hole and sends a message to Murph (his daughter). Later, he is found by rangers, but I don't understand how he gets out of the black hole?  As far as I know, nothing can get out of a black hole, as it has extremely high gravity. Is the tabu package obsolete? Subfields of finite fields We know that if a finite field $F$ has characteristic $p$ (prime), then $F$ has cardinality $p^r$ where $r = [F:\mathbb{F}_p]$.   I'm now trying to say something about the possible cardinalities of subfields of $F$. I can see that there is a subfield of cardinality $p^s$ for each $s$ that divides $r$, given by the fixed field of the group generated by $\phi^s$, where $\phi$ is the Frobenius automorphism.  Now suppose $K$ is a subfield of $F$. Then (since both are additive groups), Lagrange gives us that $|K|$ divides $|F|$, so $|K| = p^t$ for some $1 \leq t \leq r $ (alternatively, $K$ contains $\mathbb{F}_p$ and so is a vector space over $\mathbb{F}_p$ and is thus isomorphic to $\mathbb{F}_p^t$, where $t = [K:\mathbb{F}_p]$). By considering the multiplicative group of units of $K$ and $F$ respectively, we get that $ p^t - 1$ divides $p^r -1$. I want to make the leap to $t|r$, but I'm failing to see why this needs to be true. Any help would be appreciated. Applications of complex numbers to solve non-complex problems Recently I asked a question regarding the diophantine equation $x^2+y^2=z^n$ for $x, y, z, n \in \mathbb{N}$, which to my surprise was answered with the help complex numbers. I find it fascinating that for a question which only concerns integers, and whose answers can only be integers, such an elegant solution comes from the seemingly unrelated complex numbers - looking only at the question and solution one would never suspect that complex numbers were lurking behind the curtain!  Can anyone give some more examples where a problem which seems to deal entirely with real numbers can be solved using complex numbers behind the scenes? One other example which springs to mind for me is solving a homogeneous second order differential equation whose coefficients form a quadratic with complex roots, which in some cases gives real solutions for real coefficients but requires complex arithmetic to calculate.  (If anyone is interested, the original question I asked can be found here: )  EDIT:  I just wanted to thank everyone for all the great answers! I'm working my way through all of them, although some are beyond me for now! When to use static classes in C# Here's what :   static class CompanyInfo {     public static string GetCompanyName() { return "CompanyName"; }     public static string GetCompanyAddress() { return "CompanyAddress"; }     //... }       Use a static class as a unit of   organization for methods not   associated with particular objects.   Also, a static class can make your   implementation simpler and faster   because you do not have to create an   object in order to call its methods.   It is useful to organize the methods   inside the class in a meaningful way,   such as the methods of the Math class   in the System namespace.   To me, that example doesn't seem to cover very many possible usage scenarios for static classes. In the past I've used static classes for stateless suites of related functions, but that's about it. So, under what circumstances should (and shouldn't) a class be declared static? In case of shell-builtins (eg type itself):  $ type type type is a shell builtin  $ which type &lt;Doesn't return anything since it's a shell builtin, silently exits&gt;   In case of commands (normally) (eg python):  $ type python python is /usr/bin/python  $ which python /usr/bin/python   In case of which (which is a command located at /usr/bin/which)  $ type which which is hashed (/usr/bin/which) $ which which /usr/bin/which   Why does type which say that which is hashed? What is the significance of which being hashed and what does it actually mean? Consider the examples:    "My name is Ricky."   He said his name is Ricky.    The tense is not backshifted in #2 because his name is and will always be Ricky. Also consider:    Ricky says: "I live in Paris."   Ricky said that he lives in Paris.    The tense is not backshifted in #4 because he still lives in Paris.  For examples #2 and #4, it is possible to not backshift the tense because the reporting words are still true at the time of reporting, but what will you say about the following examples:        I asked what his name is.   I asked where he lives.      Here, is it correct for #5 and #6 to not backshift the tense?  If #5 and #6 are correct, then under what conditions is it correct to not backshift the verb like that? Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even ) How to evenly space vertices? I have a edge loop that has vertices on it. Is there a way I can separate the vertices on the the edge evenly without moving the end points? I would also like for the vertices to maintain the current shape. Can't read code blocks from the Stack Exchange android app How did Voldemort get his body back? In the first book it seems that Voldemort has been reduced to nothing but a ghostly wisp. But IIRC by the fourth book he somehow has a body that is carried around by Wormtail. In the first book he needs someone to inhabit, by the fourth, he can't really move around, but he does have a body back which can hold a wand.  How did this transition happen? Install Numpy for Python3.4.0 on Ubuntu 14.04.01 LTS How to suppress Terminated message after killing in bash? How can you suppress the Terminated message that comes up after you kill a process in a bash script?  I tried set +bm, but that doesn't work.  I know another solution involves calling exec 2&gt; /dev/null, but is that reliable? How do I reset it back so that I can continue to see stderr?