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10

Quantum entanglement is the property of two objects $A,B$ – more precisely two subsystems – or a relationship between these two objects whose quantities or observables aren't independent of each other. It means that there exist some quantities $a_j$ and $b_k$ describing $A,B$, respectively, such that the probability distribution for these observations ...


7

Experimental determination of $c_i$ values starts with preparing multiple identical systems, then making measurements. From all the measurements, one determines the probabilities, which are the $|c_i|^2$. The square root of the probabilities will tell you the $c_i$ to within a phase factor of the form $e^{i\beta}$, where $\beta$ is real, and may or may not ...


6

Take your first relation for the 3 Pauli matrices individually: $$\sigma_1(t)=U^\dagger\sigma_1(0)U$$ $$\sigma_1(t)=U^\dagger\sigma_2(0)U$$ $$\sigma_3(t)=U^\dagger\sigma_3(0)U$$ Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 ...


4

I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a ...


4

Yes. For example, consider the harmonic oscillator potential $V(x)$, where the ground state has energy $\hbar \omega / 2$. Then the ground state of the potential $V(x) - \hbar \omega / 2$ has zero energy. This works because in quantum mechanics, like in classical mechanics, absolute energies don't matter. You can always add or subtract constants. ...


4

Once a measurement ( observation ) is made on a quantum system the system will be in an eigenstate of that property, so if the energy of an electron is measured the electron will afterwards remain in an energy eigenstate ( until some other measurement or interaction occurs ), but if the angular momentum is measured the electron system will afterwards remain ...


3

Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...


3

The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows: $$ \left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle $$ where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...


2

$|q\phi|\ll mc^2$ means that the rest mass of the particle is way larger than the electrostatic potential energy. This is the usual assumption of non-relativistic mechanics, which is equivalent to $v\ll c$. You can see this with the virial theorem: $$ \frac{1}{2}mv^2\sim q\phi $$ which means that $q\phi\ll mc^2$ iff $v\ll c$. ...


2

The key is in your words "to ... appear". I believe that it's a perceptual issue with how your brain processes the two kinds of images: a smooth rendering or a pixelated rendering. There is another possibility. In order to be sensitive to single photons, the detector is also going to be sensitive to very low-level noise. An image taken with a bright ...


2

As everyone points out commutation is not a shorthand for equivalence, as, given your relations, the Jacobi identity, [A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0 dictates that when the first two terms vanish, the third must too, so that B must commute with [C,A], non vanishing in general, as remarked repeatedly. Lie algebra commutators do, nevertheless, parameterize ...


2

The delta function is not really a function, it is a distribution, In the strict sense both $\delta (x)$ and $e^{ikx}$ are not normalizable when $n=m$ One way to prove your equations is to use fourier transforms Using Placherels theorem the fourier transform $F([f(x)]k)$ for the function $f(x)$ is given by ...


2

If $\mathbf{x}=\left(x_{1},x_{2},x_{3}\right)$ is a 3-vector rotated to $\mathbf{x}^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)$ then this rotation is expressed via special unitary matrices $U \in SU\left(2\right)$ as follows : \begin{equation} \mathbf{X}^{\prime}\equiv \begin{bmatrix} ...


2

Everything you write is correct and there is no inconsistency. When you write $\left<x\right|Hf(x)g(x′)\left|x′\right>$ you just, indeed, put the numbers on the right. But numbers commute so there is nothign wrong with it. Note that you have really no conclusion from it. You can't even transform back to operators on the form ...


2

Electrons and holes occupy their states according to the Fermi-Dirac distribution, which has a single parameter $E_f$, the Fermi level (assume a fixed temperature). Provided $E_f$ is in within the band gap and far from the band edges, the (energy integral of) Fermi-Dirac takes an exponential form $\propto e^{E_f}$ for electrons and $\propto e^{-E_f}$ for ...


2

Schmidt decomposition is in general a singular value decomposition (SVD) and it is applied on wave vectors and not on density matrices. While dealing with bi-partite wave vectors we use SVD because there is no restriction that the size of the two systems in question are the same. So the matrix of the wave vector coefficients can be rectangular and SVD can ...


2

$kT$ is related to the kinetic translation energy by the equipartition theorem. You are saying that the mean kinetic energy, is much greater than the rest energy. The particle has a large or relativistic velocity. The limit $kT>> mc^2$ is called ultrarelativistic limit. It means you can approximate the energy momentum relation $E^2=(pc)^2+(mc^2)^2$ by ...


2

One possible application of AMO physics would be inertial guidance systems based on atomic interferometers--similar systems are currently being investigated for missile guidance and have shown much higher accuracy than other methods. Inertial navigation systems don't rely on a network of GPS satellites, which obviously wouldn't be present on Mars (yet!)


2

It's because always even number of Fermionic fields appear in the Hamiltonian. for example the Dirac Lagrangian for free electron: $\mathcal{L}=i\hbar\bar \psi(\gamma^\mu\partial_\mu-m)\psi$ has two $\psi$s. Invariance of the theory upon a global gauge transformation requires that each term in the Lagrangian have an even number of Fermionic fields. Since an ...


2

Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ...


2

This is not a simple problem with a single answer, but I'll try to give a few practical examples to demonstrate the thought process involved. The most basic way to determine "quantumness" is by comparing the length scales in the problem to the de Broglie wavelength: $$ \lambda = \frac{\hbar}{p} $$ where $p$ is the momentum of a particle. As an example, a ...


2

A crucial hypothesis is missed in your construction. Each $\phi_i$ must also satisfy $\phi_i \not \perp \psi_i$, otherwise $\langle \psi_i |E_i \psi_i\rangle >0$ is false. This point provides an answer to your last question as well. If $\psi$ is an added further vector, linearly dependent on the vectors $\psi_i$, the construction you made cannot be ...


2

The most effective way to think about such problems is by means of operator algebras. The best way to treat systematically quantum observables (and their associated unitary operators) is to collect them in a C$^*$-algebra, roughly speaking a Banach algebra where observables have a norm, can be summed and multiplied (with some additional technical ...


2

Yes, their outer product is defined as you said. Further, the product of operators is given by $$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$ Therefore, $$ [A \otimes B, F \otimes 1_B] = [A,F] \otimes [B,I_B] = [A,F]\otimes {\bf 0} = 0 $$


1

For a neutron of that speed, the uncertainty in the momentum is expected to be less than the momentum magnitude. Using the actual momentum will be an upper bound on the momentum uncertainty. That correlates to a lower bound on the position uncertainty. So, $\Delta x$ is lower bounded by $\hbar/(2p)$: $$\Delta x \ge \frac{\hbar}{2m_nv}.$$ $\Delta x$ could ...


1

Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...


1

First, let's clarify the expression of $|\phi\rangle$. The kets $|+\rangle_z$ and $|-\rangle_z$ are eigenvectors of $\hat{S}_z$ such that $$ \hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z $$ This means that in the $\{|+\rangle_z = ...


1

I think that the best answer is probably just "no". It's easy to give specific cases where you need/don't need quantum mechanics, and plenty of people attempt to give such a rule of thumb (e.g. relating to de Broglie wavelengths or Plank's constant). However, all of the affirmative answers to this question I've ever heard are either (a) too specific to be ...


1

This is a profound question at the heart of science. Indeed no little effort has been devoted to trying to understand whether the mathematics of quantum mechanics are simply a contrivance or reflect reality. Surprisingly, Bell discovered an experimental method to answer the question and the result is that quantum weirdness is an intrinsic aspect of nature. ...


1

How about $\Delta p\Delta x$? From The Physics of Stargates: Parallel Universes, Time Travel and the Enigma of Wormhole Physics, by Enrico Rodrigo:



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