Gaussian wave packet (quantum mechanics)?
613346
The previous responses are right but ko is not a potential. It's the (average) momentum for a particle in free space. With psi defined by:\n\npsi(x) = A exp(ikx - x^2/(2sigma))\n==>\npsi*(x)psi(x) = |A|^2 exp(-x^2/sigma)\n\nYou need the integrals:\n\n&#92;int_(-oo)^(+oo) exp(-x^2/sigma) dx = sqrt(pi*sigma)\n&#92;int_(-oo)^(+oo) exp(-x^2/sigma) x^2 dx = (sqrt(pi)/2) sigma^(3/2)\n\n<x^2> is the second expression divided by the first:\n\n<x^2> = 1/(2 sigma)\n\nSince H = p^2/2m, \n\nH psi(x)\n= -(1/2m)(d^2/dx^2) psi(x)\n= -(1/2m) [(ik-x/sigma)^2 - 1/sigma] psi(x)\n= (1/2m) [k^2-x^2/sigma^2 + 2ikx/sigma + 1/sigma] psi(x)\n\n==>\n\n<H>\n=(1/2m)[k^2-<x^2>/sigma^2 + 2ik<x>/sigma + 1/sigma]\n=(1/2m)[(k^2+1/sigma-(1/2sigma)]\n=(1/2m)[(k^2+1/(2sigma)]
620361
yes, it is true:\n\nIdeally, the Gaussian kernel is K(s,t) = exp(-(s2 + t2)/(2C2))/(2*Pi*C2)\n\nIn polar coordinates s = r*cos(theta), t = r*sin(theta), this expression is K(r, theta) = exp(-r2/2C2)/(2*Pi*C2)\n\nThe value of C determines the effective range of K; points (s,t) beyond a distance of about 3*C barely contribute to K.  K() is essentially flat beyond 3*C.  The quartic approximation capitalizes on this.  The only symmetric quartic polynomial equal to 1 at r=0, to 0 at r=3, and with derivative 0 at r=3 is q(r) = (1-(r/3)2)2\n\nTherefore Spatial Analyst's "Gaussian" kernel is a multiple of K(r, theta) = (1-(r/3C)2)2 suitably normalized to have an area of 1.  These values are very quick to calculate and so the density is quick to compute.
558518
The problem is that there are not two wavefunctions for entangled particles. There is only one, but that one wave function describes *both* of the particles. So the observation collapses that one wave function as fixes the state of the system. This is one difficulty of the 'particle' view of things. There is too much of a tendency to think of a two particle state as two one particle states. This is wrong.
14608
Quantum mechanics is a fundamental physical theory that replaces Newtonian mechanics and classical electromagnetism at the atomic and subatomic levels and is the underlying framework of many fields of physics and chemistry, including condensed matter physics, quantum chemistry, and particle physics. Often, it is the answer to questions when general relativity fails.
880133
There isn't a definite boundary line.  A molecule of several atoms can exhibit quantum-mechanical behavior under appropriate experimental conditions.  Of course, the heavier the molecule, the more it obeys classical mechanics as opposed to quantum mechanics.  The "correspondence principle" tells us that at a large enough scale, all processes show classical behavior.
950482
OK, Drummanmatthew.  It's been a long time since I did this but I think I remember enough to help you out.\n\nFrom what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi)  (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.\n\nIn other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz.  To me, it is the y part of this that gives it away.  It clearly has to be something plus something.  You need the bit with the z to differentiate away when you diff wrt x.  It seems to me that -yz^2 will do the trick.  Diff wrt x and it goes, diff wrt y and you get -z^2.  Diff wrt z and you get -2yz, as required.  To get the rest of it right you also need a -xy^2 by similar logic.\n\nSo, I reckon that phi = -y(x^2 + z^2) + c\n\nThe constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).\n\nI'm not absolutely certain about the - but I think this correct apart from that.\n\nHope this helps.
406372
The following things are not random, but appear to be:\n\nFlipping coins, throwing dice, the weather, etc.\n\nAll of the above are ruled by the laws of physics. But in those cases, the results might as well be random because the equations governing them are so complex that in practice we can't find exact solutions for them.\n\nHowever, one might ask whether there exist any physical system that it truly random - or do the equations of physics predict exact results for everything? (The only reason we don't know the future, in that case, would be because the equations are too difficult to solve.)\n\nThe answer is actually quite surprizing. Subatomic particles - that is, things smaller than atoms - are truly random. It is impossible, even in principle, to ever calculate what they will do next. All we can do is calculate a probability. That's from an area of physics called Quantum Mechanics. So, in this area of physics, instead of asking a question like, "Where will the particle land?", we ask, "What's the chance of the particle landing over here? Or over there? Or anywhere?" Quantum mechanics produces some truly "spooky" effects - such as the fact that an electron, trapped at the bottom of a bowl, can get out of the bowl without having to get up over the sides! That's a phenomenon known as tunneling, and is due to the fact that there's a small chance of the electron being outside the bowl, even if the electron doesn't have enough energy to get over the walls. That means that it must, at some point, be seen outside the bowl. But it's energy means that it can't go over the walls! So, it basically "teleports" from inside the box to outside the box. There are other quantum mechanical effects due to probability as well - but I'm sure you can find those with a quick internet search...
730142
Possibly so. Possibly no. I'd say the Heisenberg uncertainty principle comes foremost into play here. Maybe also Schrödinger's cat. Is it dead or still alive? That would make a significant difference, you know. These questions can't just be taken so lightly. And how could you even consider Y!A as a forum to resolve such complex issues as this? More to the point, even assuming Planck scale dynamics can be fully understood with the mathematical tools we have at hand today, where exactly does that get us? We still don't know what topping to put on our pizza this weekend. Or whether a third eye would make a good chick magnet. Or even whether that darn cat is still alive or dead.... 'tis an enduring mystery. And we wouldn't want to take all the mysterious challenge out of life, now would we?....Meow.
8585
Hi,\n\nMagnetic Quantum Number "m" in relation to Azimuthal Quantum Number "l" can be depicted by- \n\nm = 2l + 1\n\nSo, for any subshell (defined by the value of L), there are (2L + 1) values of M possible given by-\n\nM = -L, -(L-1), ......, 0, ....., (L-1), L\n\nSo, from your given options, I think the second one- -L, 0, L  is the closest.
605311
chemistry, quantum mechanics...
791590
E = m x c^2,\nso \n\nmass = Energy / c^2 (speed of light squared)
145418
The word quamtum is derived from quanta which means packets.This is the theory that radiation or conduction or radiation of light or energy takes place in the form of quanta.
741723
De Broglie wavelength L = (h/mv) sqrt( 1-v^2/c^2).\n\nSquare, solve for v = hc /sqrt(mc^2 L^2 + h^2) \nH = Planck's constant\nc = the speed of light\nm = mass
1050831
Mathematical method of quantum phase space is very useful in physical applications like quantum optics and non-relativistic quantum mechanics. However, attempts to generalize it for the relativistic case lead to some difficulties. One of problems is band structure of energy spectrum for a relativistic particle. This corresponds to an internal degree of freedom, so-called charge variable. In physical problems we often deal with such of dynamical variables that do not depend on this degree of freedom. These are position, momentum, and any combination of them. Restricting our consideration to this kind of observables we propose the relativistic Weyl--Wigner--Moyal formalism that contains some surprising differences from its non-relativistic counterpart
31466
Mit has open courseware on their website.  They've been videotaping the classes, but they don't have this for the quantum physics classes yet.  If you just want basic information, then you can check out some of the other physics.
241834
I beleive this is the one:\n\nhttp://www.phy.ohiou.edu/~braslavs/Biophysics/Articles/Wei%20Zhang/Rosensweig2002.pdf
427866
>we know that physically one photon can not pass through the two slits at a time\n\nThat's kind of the whole point of considering the single photon case of the double slit experiment: it give results that don't match classical predictions, and so it becomes a good argument for quantum mechanics.  \n\nIf light were a classical particle, then yes, it can't go through both slits, and you'd end up with two overlapping Gaussian bands of light, and not an interference pattern if you summed up the results of many single photon experiments \n\nIf light were a classical wave, you'd get an interference pattern, but you'd expect the energy from it to be spread out all over the detector or photographic plate, just very faint.  \n\nWhat actually happens when this experiment is done is that the energy gets delivered to the detector all in one place, as if light is a classical particle.  But, the locations where the energy gets delivered, if the experiment is done one photon at a time and the results are stacked up, conforms to the interference pattern you'd expect from a classical wave.\n\nSo, we have something that matches neither a classical particle nor a classical wave in  behaviour.  \n\nBut, in quantum mechanics we have a description of something that does fit this behaviour.  That is, we have a wave which, when measured, has to provide energy in quantized steps.  To get from the wavelike portion to the particlelike portion, we have to recognize that there isn't a completely deterministic answer but rather a statistical one.  So, while this model descibes what a real photon does in the double slit experiment, it can't predict the exact behaviour of a specific given photon.  As best we can tell so far, this isn't just a shortcoming of the theory, it's actually how physical reality operates.\n\n\nEdit:  Ramblingman, it can be done with either an electron beam or a photon beam.  Both will show the same sort of effect, but traveling electrons will have a much shorter wavelength than light.  Quantum mechanics works for both things that were classically considered as waves, like photons, and things that were classically considered as particles, like electrons.