Sunday, September 25, 2016

speed of light - Microsecond trading with neutrinos


The Spread Networks corporation recently laid down 825 miles of fiberoptic cable between New York and Chicago, stretching across Pennsylvania, for the sole purpose of reducing the latency of microsecond trades to less than 13.33 milliseconds (http://www.spreadnetworks.com/spread-networks/spread-solutions/dark-fiber-networks/overview). The lesson I would draw from this is that, in the near future, oil and natural gas extraction won't be the only lucrative use of ocean platforms.


So here's my question - since trades are occurring on the scale of tens to hundreds of microseconds, and considering the amount of money involved, can one use neutrino beams to beat the limitation due to having to travel the great-circle/orthodromic distance between two trading hubs? I'm imagining something similar to the MINOS detector (http://en.wikipedia.org/wiki/MINOS), where a neutron beam was generated at Fermilab in Batavia, Illinois, and detected ~735 km away, ~700 meters under the ground in a Northern Minnesota mine.


Is it possible to beat a signal traveling at the speed of light across the great-circle distance from, say, New York to Tokyo, using a neutron beam traveling the earth? Is it realistic to talk about generating these beams on a microsecond time-scale?


Addendum - Over what distances can you reasonably detect a neutrino beam?



Answer



Whether or not neutrinos would be suitable for rapid trading, people have seriously considered their utility for signalling in difficult environments. I read an article a while back about a paper (published in Phys. Lett. B, but I can't access that from here) by Patrick Huber which proposed using neutrinos for through-the-earth communication to submarines as an alternative to ELF, where bandwidths become competitive. The submarine would pick up the modulated cherenkov radiation produced by the generation of muons in seawater. This certainly allows faster-than-great-circle transmission times, but this is not the reason why the technique is attractive. The preprint indicates that calculated antipode to antipode bandwidth is only 10 b/s which doesn't seem ready for high-intensity trading.


Addendum:


If we consider a continuous lossless fibre optic link between antipodes around the equator, the transmission time will be about 99 ms, whilst the through-earth travel time (at $\approx{c}$) is 42 ms. Obviously this counts for nothing if you have high-latency equipment at either end.



Whilst the improvement in transmission time hardly seems worth it, it occurs to me that this would be a useful technique for communicating between either side of a huge, highly oblate structure such as a wide but thin disk-shaped megastructure, however that's veering in to sci-fi territory.


Saturday, September 24, 2016

scattering - Nature of Cooper pairs


Some people say it is bound state, some say it is not. Which is more accurate? Problem is that I read in some books, including Ziman, that Cooper pairs are bound states but my teacher says that it is not true and that Bardeen had to explain it many times even to his peers.Now, i know that it has something to do with the resonance in scattering cross section...but oversimplifications with hand-waving about phonons that mediate interaction creating an attractive force and a bound state, like some kind of electron-electron molecule just make me angry. I know it is phonon mediated but it is not that simple, right?




condensed matter - Anyons only in 2+1 spacetime dimensions - better explanation


Regrading why anyons exist only in 2+1 spacetime dimensions (which have an arbitrary phase on exchange), I read the reason that the paths for exchange in 3D are deformable into each other while in 2D, they may not be deformed into each other. So what is next? How to prove from this, that there can be arbitrary phase generated on exchange? I understand that in 2D, we have braids, but can I get a proof for an arbitrary phase on exchange which specifically takes into account the topological equivalence of paths in 3D and not in 2D.



Answer




Please, let me first refer you to the original paper by Leinaas and Myrheim where the existence of anyonic statistics was first predicted before its actual discovery. All the ingredients of the understanding of the special properties of the two dimensional case already exist in this old paper, however, I'll try to cast it in a more modern terminology:


By passing to a polar center of mass coordinates, the configuration space of two scalar identical particles in $\mathbb{R}^n$ can be represented as:


$$\frac{\mathbb{R}^n \times \mathbb{R}^n}{\sim } = \mathbb{R}^n \times \mathbb{R}^+\times \frac{S^{n-1}}{\mathbb{Z}_2}$$


Where: $\sim$ is the identical particles equivalence relation, the $\mathbb{R}^n$ on the right hand side is the center of mass coordinates, $\mathbb{R}^+$ is the radial coordinate, and $ \frac{S^n}{\mathbb{Z}_2}$ are the angular coordinates. The equivalence relation corresponding to the exchange in the angular coordinates is simply the identification of the antipodal points on the sphere's surface, which accounts for the factor ${\mathbb{Z}_2}$ in the denominator. The central of mass and the radial coordinates are transparent to the exchange, thus we may concentrate on the angular coordinates, which actually consist of the real projective spaces:


$$ RP^n = \frac{S^n}{\mathbb{Z}_2}$$


These spaces are not simply connected, their fundamental groups can be easily deduced from the contractibility of the closed loops on the sphere with the identification of the antipodal points as indicated in the question:


$$\pi_1(RP^n ) = \mathbb{Z}_2, n>1$$


$$\pi_1(RP^1 ) = \mathbb{Z}$$


The usual way to quantize on a non-simply connected manifold is to construct wave functions on the universal covering with appropriate transformation properties upon the exchange:


$$ \psi(x^a) = e^{i\phi} \psi(x) $$



Where $x^a$ is the antipodal point to $x$. Of course, the transformation can be only a phase multiplication, because the choice of the point or its antipode should not change the expectation values of the observables since both points correspond to the same physical point on the configuration space.


Moreover, since $S^n$ is simply connected then, the map $S^n \rightarrow RP^n$ is a covering map, therefore, $ RP^n = \frac{S^n}{\pi_1(RP^n ) }$. The phase transformation must be a representation of $\pi_1(RP^n )$. In our case, when $n>2$, $\mathbb{Z}_2$ has only two phase representation, the trivial representation and the alternating representation.


However, in the two dimensional case, we may choose a representation $\gamma$ in which the generator $\Gamma$ of $ \mathbb{Z}$ is represented by an arbitrary constant phase $e^{i\phi}$, then the representation of an arbitrary element in $ \mathbb{Z}$ will be:


$$\gamma(\Gamma^n) = e^{in\phi}$$


Now, please remember that in the geometric picture of the fundamental group, the generators are represented by closed loops, thus the representation $\gamma$ assigns a phase to every closed loop (such that the composition law of the loops is reflected in the multiplication of the phases). This assignment can be described as the existence inequivalent quantizations corresponding to the set of maps:


$$ \mathrm{Hom}(\pi_1(RP^n ), U(1))$$


When $n>2$, this set contains only two classes, (Bosons and fermions), while when $n=2$, this set is infinite.


In summary, there will be a set of possible quantizations, in each set the angular wave equation should be solved with a different condition upon the exchange of the antipodal points on the sphere. The solution will correspond to a different particle type.


Leinaas and Myrheim, solved the problem of the identical two dimensional isotropic harmonic oscillator with wave functions transforming with an arbitrary phase upon the antipodal identification and found that the spectrum depends on the transformation phase factor.


Now, according to the classification theorem of flat connections, every (projective) representation of the fundamental group can be associated to a flat connection $A$ such that:



$$\gamma(\Gamma) = e^{\int_{\Gamma} A_{\gamma}}$$


This connection is essential for the construction of the quantum operators corresponding to the classical functions on the phase space according the Koopman - Van Hove representation:


$$ \hat{O}_{\gamma} = O - i \hbar X_{O} -A_{\gamma} (X_{O} )$$


Where $O$ is a function on the phase space, $ \hat{O}_{\gamma}$ is the corresponding quantum operator, and $X_O$ is the corresponding Hamiltonian vector field.


By the way, when the configuration space is $S^1$, this flat connection is the famous Aharonov-Bohm connection.


For further reading on the classification of inequivalent quantizations, please see the following two articles by N.P. Landsman and by Doebner Šťovíček, and Tolar . Actually, the first author (Landsman) has reservations (footnote 13 in the article) over the customary explanation using parallel transport and prefers the induced representation reasoning that I tried to follow.


quantum field theory - Given expectation values for E and B, can you find an associated state?


When we quantize the electromagnetic field, we develop the concept of the field operator $A(\vec{r},t)$ and the simultaneous eigenstates of momentum and the free field Hamiltonian (i.e., each eigenstate is given by specifying the number of photons with momentum $k$ and polarization $\mu$). We can then construct the operators for the electric and magnetic fields, and we can calculate their expectation values for an arbitrary state.


Now, suppose the expectation value of the electric field is $E(\vec{r},t)$ and the magnetic field is $B(\vec{r},t)$. Assuming $E$ and $B$ obey Maxwell's Equations, can we construct a state that has these expectation values? Is it unique, or could there be multiple states with the same expectation value for $E(\vec{r},t)$ and $B(\vec{r},t)$?


What if the expectation values are time independent (i.e., static fields $E(\vec{r},t)=E(\vec{r},0)$ and $B(\vec{r},t)=B(\vec{r},0)$ for all $t$)?



Answer



Of course one can construct states with any desired expectation values. This is no different from constructing a state of a simple harmonic oscillator with the desired expectation value of position and momentum, repeated for each field mode. Just make a wavepacket centered on the desired position and with the right phases. Note however that you cannot make a simultaneous eigenstate of the electric and magnetic fields since they don't commute with each other, but you can fix the expectation values.


I can prove the non-uniquess of such states just by giving an example: all states with a definite number of photons have zero expectation values $\langle \vec{E} \rangle = \langle \vec{B} \rangle = 0$. This follows because $\vec{E}$ and $\vec{B}$ are operators which change the photon number.


homework and exercises - The equivalent resistance betweeen A and B. I need the answer with proper explaination



enter image description here


how are we supposed to do this question if resistance is given as 1 ohm



Answer



You have two basic options:





  1. Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or




  2. Use Kirchoff's laws to derive the equivalent resistance.




If you choose option 1, I'll help you out by revealing the parallel resistors in this weirdly drawn network. I've labeled the resistors 1-5 from left to right to make the transformation easier to follow:


Starting arrangement:


Starting arrangement


Step 1 - rotate R2 counterclockwise by 90°:



Step 1 - rotate R2 counterclockwise by 90°


Step 2 - rotate R3 clockwise by 90°:


Step 2 - rotate R3 clockwise by 90°


Step 3 - rotate R4 counterclockwise by 90°:


Step 3 - rotate R4 counterclockwise by 90°


Voila - a familiar-looking network.


quantum mechanics - Hilbert space of harmonic oscillator: Countable vs uncountable?


Hm, this just occurred to me while answering another question:


If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.


On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.


Clearly they can't both be right, so where is the flaw in my logic?



Answer



This question was first posed to me by a friend of mine; for the subtleties involved, I love this question. :-)


The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as a limiting case. This is essentially done by considering a UV regulator for your wavefunctions in space. Let's solve the simpler "particle in a box" problem, on a lattice. The answer for the harmonic oscillator will conceptually be the same. Also note that solving the problem on a lattice of size $a$ is akin to considering rectangular functions of width $a$ and unit area, as regulated versions of $\delta$-functions.


The UV-cutoff (smallest position resolution) becomes the maximum momentum possible for the particle's wavefunction and the IR-cutoff (roughly max width of wavefunction which will correspond to the size of the box) gives the minimum momentum quantum and hence the difference between levels. Now you can see that the number of states (finite) is the same in position basis and momentum basis. The subtlety is when you take the limit of small lattice spacing. Then the max momentum goes to "infinity" while the position resolution goes to zero -- but the position basis states are still countable!



In the harmonic oscillator case, the spread of the ground state (maximum spread) should correspond to the momentum quantum i.e. the lattice size in momentum space.


The physical intuition


When we consider the set of possible wavefunctions, we need them to be reasonably behaved i.e. only a countable number of discontinuities. In effect, such functions have only a countable number of degrees of freedom (unlike functions which can be very badly behaved). IIRC, this is one of the necessary conditions for a function to be fourier transformable.


ADDENDUM: See @tparker's answer for a nice explanation with a slightly more rigorous treatment justifying why wavefunctions have only countable degrees of freedom.


Collapse in Quantum Field Theory?



I do not want answers telling me that wave-function collapse is not real and decoherence is the answer (I know the situation with that). I am asking a question purely on the basis if wave-function collapse is the correct method. My question is: in normal quantum mechanics superposition of the state (position, momentum) exists until the wave-function collapses (how or why it collapses is not important in this question), now in quantum field theory we can also have superposition as in the superposition of Fock space states with different particle number. Can the superposition also collapse here under collapse interpretations of quantum mechanics/quantum field theory?


Laymans answers would be mainly appreciated...




classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...