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1

Starting from scratch I would propose an order of topics to study as follows: Kinematics (motion) Dynamics (forces) Rotational kinematics and dynamics Collisions (momentum and impulse) Vibrations and waves Thermodynamics Electricity (DC) Electricity (AC) Magnetic fields and forces Electromagnetic waves Light (optics, photons) Quantum mechanics ...


4

Even in special relativity, the claim that light always travels at a constant speed of c is only true in inertial reference frames, in non-inertial coordinate systems like Rindler coordinates, the coordinate velocity (change in position coordinate with respect to coordinate time) may vary. In general relativity, all coordinate systems covering large regions ...


3

If they are much far away will they have relative velocity of separation greater than speed of light and if so how can we even detect such galaxies. We can't detect such galaxies. Redshift goes to infinity at the cosmic horizon, and we cannot see beyond. Note that the cosmic horizon is different from the Hubble sphere: At the former, relative velocities ...


2

These galaxies are not out of reach. The Hubble Sphere is the volume of space surrounding an observer where everything inside the sphere moves away from the observer with a speed less than c. Following this logic it immediately follows that the Hubble Sphere is equivalent to the cosmological event horizon that @Hritik is discussing in his answer. ...


3

Such galaxies cannot be detected. and Quantum entanglement is just a correlation between two bits of information and it no way does, and no way can be used to transfer information faster than the speed of light. Any light sent from such galaxies towards ours would show a wavelength of infinity after considering Doppler shift. In short these galaxies can ...


1

It was not a wave. The wavefunction is not a wave. It fulfills the Schrödinger equation in the position representation, and although that looks similar to what one usually writes as "wave equation", and produces similar interference phenomena, it is not a wave in any physical sense. The wave function is not a physical object, it is merely a way of writing ...


2

We don't know we only assume that it was a wave, and our assumption works well in some cases, and works badly in other cases. For instance if we put on the way of the particle a beam-splitter, we believe that we get a splitting of the wave, into a reflected wave and a transmitted wave. I.e. although we speak of one particle, we believe that we get two ...


2

The way you wrote it, they are distinguishable (unless $a=b$ of course). For the particles to be indistinguishable their wavefunction must be of the form $$ \psi(r_1,r_2) = \frac{1}{\sqrt{2}} [ \psi_a(r_1)\psi_b(r_2) \pm \psi_a(r_2)\psi_b(r_1) ]$$ where the sign depends on the fermionic/bosonic nature of the particles. If the particles are described by a ...


0

Planck derived his postulate particularly for blackbody radiation because that is what he was studying. If energy is quantized as $E=n\hbar\omega=nh\nu$ (depending on your preference), then a bit of work leads to his famous Planck function, \begin{align} B_\lambda&=\frac{2hc^2}{\lambda^5}\,\frac{1}{e^{hc/k_BT\lambda}-1}\\ ...


0

You're right, but the effect you're interested in is not manifest as a change in the Coulomb potential. Basically you define an interaction potential which is a function of time and space during the reaction and that changes accordingly. The Coulomb potential is a purely radial potential defined over all distances from the centre of the atomic system, ...


1

Even in the case of a single electron (as opposed to large number approximating a continuous wave), the probability distribution on the measurement shows double slit interference, rather than what would be expected in the case of a single slit. So even though there is but a single electron, the physics of its journey is affected by both slits. We have, ...


0

You will break an entanglement once you change the spin of one component independently.


5

I will give you some rigor, since it seems what you are looking for. The rigor takes the form of the very famous spectral theorem for self-adjoint operators. First of all, a simple example to elucidate: a purely discrete spectrum, that forms an orthonormal eigenbasis of the hilbert space (i.e. the operator $H$ is either compact or with compact resolvent). ...


0

There is never a double-slit interference pattern in the total pattern of signal photons at D0, such a pattern at D0 can only be found by taking a coincidence count with some detector that detected the idler photons. See my answer here for details--I think your question is basically a duplicate of that one.


11

Entanglement is simply a particular kind of quantum multiparticle state: it happens to be the "most common" kind of state in the sense that if you choose a random quantum superposition from a multiparticle state space, it will almost surelybe (in the measure-theoretic sense) entangled, so it's a little curious why entanglement takes some effort to observe in ...


1

The underlying framework of physics is quantum mechanical, this is the current level of physics development. Classical theories emerge from this underlying framework in a consistent and understood manner. Quantum mechanics has differential equations that have to be solved for specific situations . The square of the solutions of these equations give the ...


1

Consider a two-state system. Particle A can either be in state $|\!\uparrow\rangle$ or state $|\!\downarrow\rangle$. If there are two independent particles in the system, then each particle can either be $|\!\uparrow\rangle$ or $|\!\downarrow\rangle$ so we label the overall system states $|\!\uparrow\uparrow\rangle$, $|\!\uparrow\downarrow\rangle$, ...


2

As an experimental particle physicist I stick to observables . The complicated mathematical functions which have been established as necessary to describe the quantum mechanical state function of the particles under consideration are not observable. By the postulates of quantum mechanics the complex functions of space time or energy momentum are not ...


1

You can try easily what is the good answer by calculating the difference in energy between the two participating levels. $E_{upper} - E_{lower} = -Rhc [\frac {1}{(n+1)^2} - \frac {1} {n^2}]$ $= Rhc [\frac {2n + 1}{n^4 + 2n^3 +n^2}]$ So, its obvious that the smaller n, the bigger $E_{upper} - E_{lower}$.


5

This is called "resolution of the identity", in case you wanted more information. The identity depends on the eigenvectors of $H$ forming a complete orthonormal basis. Thus, $$\langle k | k'\rangle = \delta_{k,k'}$$ Now consider the matrix $$O = \sum_q |q\rangle \langle q |,$$ where the ket $\left|q\right\rangle$ represents the wavefunction in the following ...


6

Quantum mechanical postulates So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms? The mathematical expression for each individual operator is sort of a postulate, but it should not be listed. The postulates define a (more or less) complete theory in that I can ...


0

Yes. Given any Hamiltonian $H'$, let $C_1=1$ be the identity operator and $h_{11}:=H'$ be a $1\times 1$-matrix. (If you want to have $h$ take values in a field, the fact that you propose a discrete sum over states will be insufficient if the Hamiltonian involves operators with a continuous spectrum like the standard momentum operator. I see you also leave ...


2

It's actually easier to see if you write them as Dirac braket form. Equation (1) is $$\langle k\lvert k'\rangle = \delta(k,k') = \int \mathrm{d}r\, \langle k\lvert r\rangle\langle r\lvert k'\rangle = \int \mathrm{d}r\, \psi^*(r;k) \psi (r;k')$$ Equation (2) is $$\langle r\lvert r'\rangle = \delta(r,r') = \int \mathrm{d}k\,\langle r\lvert k\rangle\langle ...


0

I believe that the important thing to remember is the basic difference between quantum and classical description of a system. In classical mechanics you describe it by defining some sort of equation of motion which gives you a position of all the constituents of the system described by function of time but for each of these particles you have one function. ...


0

According to the cited article Zitterbewegung is "a theoretical rapid motion of elementary particles". It never was observed. But let us say, that Zitterbewegung is real. The emission of photons is, or the reaction of an higher energy level of this particle to his environment, or takes place during acceleration processes. How ever, the particle get a higher ...


1

The polarization states can be effectively defined in terms of the Stokes operators $\hat{S}_i$ for $i=[0,3]$ $\hat{S}_0 = \hat{a}^{\dagger}_{x}\hat{a}_{x}+\hat{a}^{\dagger}_{y}\hat{a}_{y}$ $\hat{S}_1 = \hat{a}^{\dagger}_{x}\hat{a}_{x}-\hat{a}^{\dagger}_{y}\hat{a}_{y}$ $\hat{S}_2 = \hat{a}^{\dagger}_{x}\hat{a}_{y}+\hat{a}^{\dagger}_{y}\hat{a}_{x}$ ...


1

The wavelength of a photon with 1 eV of energy is 1.24 µm . Atomic spacing is about 4 orders of magnitude smaller. So I would say "yes".


0

Quantum mechanics has not been shown to be non-local. Rather, hidden variable theories that make the same predictions as quantum mechanics are non-local. Quantum systems can be described in terms of observables that evolve locally. Those observables are represented by Hermitian operators, not by hidden variables, i.e. - single numbers. In entanglement of the ...


1

Of course linear polarization is an observable, we measure it. Linear polarization along an axis x, and perpendicular to x, are the eigenstates of the observable "polarization along x". In fact, by measuring polarization along x, we measure the projection of the electric field along the axis x.


1

The difference between the whole system properties and its constituents can be explained on two-particle system. Consider positronium (electron+positron) in the state $l=1$. The quantum number $l$ describes the relative motion of constituents. This state has a magnetic moment, which belongs to the whole system. But when you consider the angular momentum ...


5

I don't intend to repeat the points already made in the other answers, consider this as a small addition to those, with the intention to give a more practical description (reminding some of the basic ideas) without getting into observer-philosophies. Quite clearly the act of observing, i.e. measuring a quantum system can be done via many different ...


4

Perhaps some additional information is in order to shed additional light... The whole discussion begs the question: If $\hbar$ is so convenient, why do we have $h$ around? As usual, "historical reasons". Planck originally invented $h$ as a proportionality constant. The problem he was solving was blackbody radiation, for which the experimental data came ...


0

Well, the initial problem was not so glorious... The problem existed with the so-called "blackbody radiation". There was for one, energy distribution per frequency as derived by classical mechanics, which yielded the Rayleigh-Jeans formula for blackbody radiation: $\rho_T(\nu) = \frac{8 \pi \nu^2 kT}{c^2} d\nu$ ($\nu$ here is the frequency. For some ...


1

It is not true that these theories cannot coexist. To put things in context: Ever since Newton's time we have been thinking of things taking place in locations in space and time. Special relativity (SR) showed us that these are connected, and we really should be thinking of spacetime as the theater in which we live. General relativity (GR) simply gave some ...


7

To quote Stephen Gasciorowicz, Before evaluating these quantities to obtain an idea of their magnitude, we will introduce some notations that will be very useful. First, it is $h/2\pi$ rather than $h$ that appears in most formulas in quantum mechanics. We therefore define $$\hbar=\frac{h}{2\pi}=1.0546\times10^{-34}\,{\rm J\cdot s}$$ So basically it's ...


1

1) There exists the classical electromagnetic wave as described by Maxwell's equations. 2) The photoelectric effect showed that these electromagnetic waves are composed of photons, with energy E=h*nu , where nu is the frequency of the classical wave. Single photon experiments have been performed by limiting the intensity of the beam to one photon at the ...


2

Given that the $\mathcal E_i$ are linear operators, i.e., $\mathcal E_i(\alpha A + B)=\alpha \mathcal E_i(A) + \mathcal E_i(B)$ for all $A$ and $B$, it is true that $\mathcal E_1(\rho)=\mathcal E_2(\rho)$ for all $\rho\ge0$ implies that $\mathcal E_1=\mathcal E_2$. The argument goes as follows: Every operator $A$ can be written as $A=X+iY$, with ...


0

The step of evaluating your second expression for $Z$ and arriving at the path integral form, which is the first expression you mention, is a nontrivial but standard derivation in the path integral formulation of QFT. It is quite lengthy, but can be found in most books that cover the path integral formulation. The general idea it to split the time interval ...


2

The interaction between two charged partices occurs through a change of momentum. All very well, but the next question is how we calculate the momentum exchange, and this is where quantum field theory comes in. The interaction between two electrons is described as a disturbance in the quantum fields involved. Quantum field theory gives us an expression for ...


0

Thinking of the photon itself as a single particle, and imagining the building up diffraction patterns with one photon in the apparatus at a time will give you an idea of the correspondence between classical wavelike behaviour and the probability waves of the wavefunction, at least for bosons. The following, I believe, is what Akrasia's answer means when ...


-2

Yes, QM violates locality. The fact that QM is non-local was fully proved by the experiments carried by the group of Aspect in the years after 1980, in base of the Clauser, Horne, Shimony and Holt inequality. Until today we don't know how does the nature work in order to achieve this non-locality. What yes we know, is that we cannot set hand on that ...


0

Indeed, to the 1st order, the sum is 1. Note that $|c_b(t)|^2$ is on the 2nd order of the perturbation.


-2

The Schrodinger equation is non relativistic and yes, it will give different solutions in different frameworks. As the wavefunctions will be different, their square, which will give the probabilities of finding the state at a given (x,y,z) in time t, will be different. The relevant equations for relativistic situations are the Klein Gordon for bosons and ...


2

The Schroedinger equation is non-relativistic and it propagates effects at an infinite velocity to begin with. It is thereof nonsensical to even talk about "locality". Schroedinger's equation doesn't describe local physics any more than a first order diffusion equation describes the speed of sound. There is no technical issue here, at all, you are simply ...


0

The no-communication theorem is a no-go theorem from quantum information theory which states that, during measurement of an entangled quantum state, it is not possible for one observer, by making a measurement of a subsystem of the total state, to communicate information to another observer. The theorem is important because, in quantum mechanics, quantum ...


2

Then, there is the case that such an operator is defined on the full interval I assume that by "full interval" you mean the whole real line. First question: Do we then need any boundary conditions? Yes, as noted by Sam Bader, boundary conditions are part of the Hamiltonian. In my physics lecture we used so-called Born von Karmann boundary ...


4

What are phonons? Phonons aren't particles like electrons or protons are, phonons are quasi particles, these type of particles are just used to describe excitations of a field: in phonons case, phonons are used to describe elementary lattice vibrations which have certain frequency. Electron-Phonon Interaction: Basically Cooper pairs are just pairs of ...


3

You are quite correct. If you have a wavefunction $\Psi$ for the water molecule then swapping the two protons will multiply the wavefunction by -1. However the wavefunction is not observable, by which I mean that there is no experiment we could do that will measure the value of the wavefunction. Typically to measure some property $Q$ there will be an ...


1

I think you may have confused two separate things. By orbital we normally mean the distinct atomic orbitals like $1s$, $2s$, $2p$, etc. These actually only exist for hydrogen atoms (and single electron ions) so let's just consider hydrogen for now. In practice if we do an experiment with a sample of hydrogen its energy is well characterised and under normal ...


2

Quantum mechanics tells us the probability for an electron in a certain orbital to be in a certain position or momentum state. Inverting this to find the probability for the electron to be in a certain orbital given its position or momentum state requires some a priori idea of what orbitals are the most probable. This is stated succinctly by Bayes' theorem: ...



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