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Seeing your last question, why would you even consider this scenario with entangled photons? Seems like the same question can be perfectly asked with one single photon. But anyway, ignoring the possible optical issues in the scenario, like the exact color of the film you use, or what Nanite has commented about, either way you will not see a superposition of ...

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Mathematically you have a superposition of two Eigenstates and you use it because you don't know all parameters of the source. And yes, you use a probabilistic source, otherwise you will not do any experiment with it. Using your special source you get always the next result: Measuring one of the photons you know immediately the Eigenstates of the twisted ...

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No, a non-physical conscious observer is not required to explain the collapse of a quantum wavefunction (a perspective called the "Von Neumann-Wigner interpretation"), because wavefunctions don't ever actually collapse. The quantum state of the combination of a quantum system and its environment always evolves purely unitarily, with no true collapse. The ...

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Stated in a simpler way, kets are the generalisation of vectors to complex and potentially continuous/infinite dimension space (Hilbert spaces). Yet you can keep in mind the image of a vector to begin with. When you multiply a vector by a nonzero real number, its direction does not change. The same is valid for vectors from a Hilbert space. If you multiply ...

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Obviously I cannot know what Dirac is thinking, but I think it is just that his direction does not correspond exactly to your direction. We "know", just as Dirac does, that quantum states are members of some Hilbert space $\mathcal{H}$. We also know that scalar multiplication should not change the state, so $\lvert \psi \rangle$ and $\lambda \lvert \psi ... 2 The general form, equivalent to eqn (5) but for outside the well, is$\Psi_o = A_o \dfrac{e^{-k_o r}}{\sqrt{r}} + B_o \dfrac{e^{k_o r}}{\sqrt{r}}$But just like we know$\Psi_i$is not infinite at the origin, we also know that$\Psi_o$doesn't go to infinity at large values of r. So we know$B_o$must be 0. 1 Evidently this is a homework problem. One way to approach the integral is to look at the symmetry of the integrand. To get started: the integrand has two factors. one of them,$y$, is odd under inversion of the axis. Once you have the symmetry of the integrand, you can make quick progress with the integral. Make a sketch of the integrand. That will ... 2 To "see another World" would require doing a measurement that involves (partially) reversing the effect that led to the split. In practice this is impossible to realize because the observer is a macroscopic object itself and it will decohere very fast. Decoherence means that the system becomes correlated with the environment and that poses a big problem ... 9 You're distracted by fancy language. Instead of "photon emitter" think "light bulb." (I suppose you could think "laser pointer" instead, since that's an easier way to make all the light exactly the same color than a light bulb plus a prism.) Instead of "photon detector" think "eyeball" or "camera." You've probably noticed that you can't detect a photon ... 2 The interpretation of the double slit experiment is very strange and i want to understand how they did it before I give up my concept of reality. The double slit experiment does not refute the idea that reality is objective. It does refute the idea that each photons goes through one slit or the other. Rather there are multiple versions of each photon ... 0 In QM the Schrödinger equation, is the equivalent of Newton's law in Classical Mechanics. The Schrödinger equation describes the state of a quantum system (i.e. atoms, subatomic particles etc.), and how the quantum system changes over time. I think you are getting confused because there are two main places where the term wave appears. (1) The Double Slit ... 3 This question is really classical. If you model two-slit interference in Maxwell's electrodynamics, the same thing happens: opening the second slit causes the intensity at some points on the screen to decrease. It happens with water surface waves too, and any other kind of wave. Destructive interference at one point is always matched by constructive ... 1 First of all, not every Hamiltonian admits a basis of eigenvectors. An operator has to be either compact or with compact resolvent to admit a basis of eigenvectors. With those type of operators, eigenvalues can accumulate (i.e. being distinct, but getting arbitrarily close) only at zero (compact) or infinity (compact resolvent). A discrete eigenvalue can ... 2 Yes,the photoelectric effect can be explained without photons! One can read it in L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995, a standard reference for quantum optics. Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting ... 0 To begin with, entropy is a classical thermodynamics concept. Different statistical frameworks , assuming some postulates, can define an entropy. The basic formulation of entropy defined by statistical mechanics where kB is the Boltzmann constant, equal to 1.38065×10^−23 J K−1. The summation is over all the possible microstates of the system, and p_i ... 0 In quantum statistical mechanics entropy is not defined via the probability density of a single state but through the density matrix which talks about the "non-quantum uncertainty" over the states. This is the "probability density" over which all the entropy theorems are proven in the quantum world. That is, if we know the system to be in a sharp quantum ... 2 Yes, the two are intimately related. One way, as in QMechanic's answer, is via Wick rotations, but in general there is a lot more freedom once you allow integration contours to go over into the complex plane. In my area, strong field physics, the use of complex time to understand tunnelling problems is everyday bread and butter for many people, and it is the ... 1 If you open up a new illuminated slit the interference pattern that results is different than it would be without the slit. If you open a new slit and keep the intensity of the light illuminating the screen the same energy that would formerly have been absorbed by the screen now arrives at the detectors. So the amount of energy that arrives at the detectors ... 1 A free quark wavefunction does not exist. Instead, inside nucleons quarks are relativistic and asymptotically free, which means that they only behave like individual particles for sufficiently large momentum/energy transfer. Imagine a classical solid state analog: if you apply a very small, slowly acting force to a single atom of a crystal, the force will ... 0 For particles to go faster than the speed of light requires that Lorentz invariance must be broken. It is possible to formulate theories that violate Lorentz invariance, and such theories do indeed predict that particles going faster than the speed of light will emit radiation and lose energy as a result. This effect was invoked by Andrew Cohen and Sheldon ... 1 Hints to the question (v2): First note that the operator norm$||A||=||UA||=||AU||$of an operator$A$is invariant if we compose with an unitary operator$U$from either left or right. Therefore$\dot{\rho}(t)$is not the zero-operator:$|| \dot{\rho}(t) || = || [H, \rho(0) || \neq 0. $0 You don't really get rid of the superposition. What happens is that the system gets entangled with the environment. This makes the system behave as if it were described by classical physics, but in reality it is still a quantum mechanical system and the superpositions are still there albeit hidden due to the entanglement with the environment. If the cat is ... 0 Quantum mechanics says there is a superposition. There is no reason to doubt QM on this point. The quantum processes of entanglement and decoherence make it impossible for us to be aware of the superposition. We will only be aware of one alternative, which fits well with our experience. 0 Schroedinger's cat is one of the worst examples of scientific visualization. It is deceptive in that it pretends that a cat (on the time and temperature scale of the observer) can be treated as a quantum system. Moreover, it pretends that state superposition is an objective quantity of the quantum system, when, in reality, it's merely a statement about the ... 1 Disproof of Joy Christian’s “Disproof of Bell’s theorem” Florin Moldoveanu Committee for Philosophy and the Sciences, University of Maryland, College Park, MD 20742 http://arxiv.org/pdf/1109.0535v3.pdf 6 Yes, quantum tunnelling in the double well potential can be solved in a Wick-rotated Euclidean formulation $$S_E[x]~=~\int \! dt_E \left[ \frac{1}{2}\left(\frac{dx}{dt_E}\right)^2 - (-V) \right],$$ see e.g. Ref 1. Here$t_E=it_M$denotes Euclidean time. The Euclidean action is in turn interpreted as the usual kinetic minus potential term with a potential ... 1 In quantum mechanics velocity is not an easy concept. Here particle motion is replaced by a wave. Momentum, is easier to define in quantum mechanics. A complex wave$\exp(ikx)$describes a particle with momentum$p=\hbar k$where$\hbar$is Plancks constant divided with$2\pi$. It is fundamental in quantum mechanics that momentum cannot strictly be defined ... 1 The superposition principle of quantum mechanics is not destroyed by quantum (hamiltonian) unitary evolution operator$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$as per @ACuriousMind's answer. Event if you dont know of the evolution operator in terms of the hamiltonian (which can be derived easily from the Schrodiger equation), still the fact that the ... 4 The time evolution operator of a quantum system is (in units with$\hbar = 1$) $$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$$ and the "stationary states" are the eigenstates of this operator, i.e. eigenstates of the Hamiltonian. If you are given a collection of stationary states (not a basis of the space, mind you)$\{\lvert \psi_E \rangle\}$with$H ...

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Okay, I think I understand now where the confusion is coming from. If we use $j=1$ objects, then we no longer are in the fundamental representation. A $j=1$ object can be a composite object of two spin-$1/2$ objects, and as such, isn't a 3-dimensional representation of $SU(2)$.

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I try to measure the energy and the position of the system simultaneously The states with definite energy are not states with definite position so there is no particle state of both definite energy and definite position.

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At the first look the question seemed very interesting, but later I found the mistake. You said you are measuring the position of the particle precisely. But how? You can tell that the particle is inside the well but you can not know the exact position of the particle. For more info read ...

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If you are considering a system of a single particle in a potential well with infinitely high walls and with finite width, the energy operator is $H = \frac{p^2}{2m} + V(x)$ where $V(x)$ is the potential energy operator, vanishing inside the well and infinite outside it. Being that $\frac{p^2}{2m}$ does not commute with $x$, how are you saying that ...

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For massive particles the intuition of thinking of spin as a rotation is correct. In the rest frame, a massive ($M^2>0$) particle has momentum $$p_0^\mu=(M,0,0,0).$$ Remember that the quantity $p^2=p_\mu p^\mu$, for arbitrary $p_\mu$ is invariant under Lorentz transformations. In the case above the subgroup of the Lorentz group leaving $p_0^\mu$ ...

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It seems that the dimension is wrong. $[\hbar] = J\cdot s$, $[e] = A\cdot s$, where $J=$joule, $s=$second and $A=$ampere. Then, $[\hbar/e]=Wb=T\cdot m^2$, with $T=$tesla and $m=$meter. The light speed $c$ is erroneous [see this wikipedia article http://en.wikipedia.org/wiki/Magnetic_flux_quantum]. Now you plug $\hbar\approx 10^{-34}J\cdot s$ and $e\approx ... 1 The best way to understand spin is actually to consider the Dirac Equation $$i\hbar \frac{\partial }{\partial t}\Psi=\left[c\sum_i{\alpha_i p_i}+mc^2\beta\right]\Psi$$ or more compactly: $$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$ The solutions to the Dirac equation are collections of complex valued fields called spinors. The spinor solution actually ... 1 Not sure about this but if you use that unitary operator to generate another ground state, then (if theres no infinite potential between the ground states) you can find some amplitude for tunneling between the two states and hence the ground state is some linear combination of the two with one of the combinations being lower than the original (and the other ... 2 In qutip we use use the scipy ode solvers and in particular the 12th order Adams-Bashforth method via zvode. Solving for the evolution of the density matrix is just a standard ode equation provided that you pick a basis representation to express the Liouvillian in matrix form, and the density matrix as a column vector by stacking the matrix columns, for ... 0 Looking at other similar questions... My understanding is that we can't use change events as a measure without knowing "when" measurements are made on one end we can't differentiate those changes from random changes (noise) on the other end. Why can't quantum teleportation be used to transport information? Improvements to my understanding are ... 0 As John Rennie, says, what-does-a-de-broglie-wave-look-like has helpful answers which you should read first, but I don't think they are complete. Do they behave like transverse waves? No - the wave function for a single particle with no spin from the Schrodinger equation is just a scalar so there is no direction connected with it. For example: can ... 1 You do not understand what does Von Neumann’s state vector (wave function) reduction mean. It is not about other eigenvalues which are not needed. It is about something else: after a quantum measurement the coherent superposition is destroyed. If the observer measured one of eigenvalues, then (the same) observer henceforth lives with only this possibility ... 4 I'd like to expand a little bit on the interpretation of commutators as a measure of disturbance (related to incompatibility, as touched on in the other answers). My interpretation of the commutator is that$[A,B]$quantifies the extent to which the action of$B$changes the value of the dynamical variable$A$, and vice versa. Let's assume that$A$is a ... 0 You use the letter S ambiguously to refer to both a system and the result of a measurement made on that system. It's important to distinguish them. So the situation is that two observers O1 and O2 independently make a measurement M of a system S and agree that the result of that measurement is X. It is tempting to conclude that the reason they agree is ... 1 Let's start with the Schrödinger equation: $$\mathrm i\hbar\frac{\partial}{\partial t}\left|\psi\right> = H\left|\psi\right>$$ Since$H$is self-adjunct, this also implies $$\mathrm -i\hbar\frac{\partial}{\partial t}\left<\psi\right| = \left<\psi\right|H$$ Now consider the most general quantum state, expressed by a density matrix$$\rho = \sum_k ... 0 My answer is more of comment on other correct answers: you cannot build a delta-function for the photon in 3D becase the longitudinal component of a massless vector field is missing. But that does not mean there is no useful and meaningful concept of a wave function in the single-photon sector. This is just a peculiar fact about free electromagnetic field, ... 0 There are several different waves associated with a photon. In QED the photon is associated with a classical solution of the (4-)vector potential. The vector potential contains features that are not physical, as a change of gauge is not reflected in any change of physical properties. Thus its role as a wave function might be somewhat questionable.But still, ... 1 It may be helpful to assign the students following HW problem : Suppose$A$and$B$be two observables i) What is the necessary condition that$A$and$B$can be simultaneously measured in an experiment without any uncertainty ? ii) Write down all the second degree polynomials in$A$and$B$which are again observables. iii) Suppose ... 1 You make many questions in one, all of them have their own answer. Just to clarify, nuclear decay and nuclear reaction are two totally separated and different things. Radioactivity occurs naturally, spontaneously. You have to sit and wait for the nucleus to decay. A nuclear reaction is forced, is something you obtain by, for example, shooting a particle ... 0 There is no such thing as “conduction of electric wave in conductor” (and I am unsure about where “electric waves” can be observed). There is a conduction of electric current in a conductor. One can say that electric potential in a piece of conductor is always the same (so the electric field is zero inside it), although it is not always so due to resistance, ... 4 At a basic level : 1) if$[A,B]=0$, and if$A$and$B$are infinitesimal generators of a symmetry (so also conserved quantities), this means that both$A$is invariant by$B$, and$B$is invariant by$A$. For instance,$[H,J_z]=0\$, means that the angular momentum is conserved during time evolution, and that the hamiltonian is invariant by rotation. As ...

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