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1

There are no such things as particles in the physical world. The correct description of "small things" in classical mechanics is that the dynamics of the motion of the center of mass of an extended object is the only relevant physical quantity while internal degrees of freedom like rotation, vibration, magnetization, temperature etc.. can be ignored. That ...


0

If you have the photoelectron momentum-space wavefunction $\psi(\mathbf p)$ and you have projected out the contribution from the bound states, then the momentum-space distribution $|\psi(\mathbf p)|^2$ gives you the three-dimensional distribution of velocities as measured e.g. by a velocity-map imager. To get beyond that, it depends on what it is you want do ...


3

In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics.


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I guess there is not that much to grasp, unless you can really understand dark spots on an electron diffraction pattern. Very roughly explanation would be to interpret wave functions of a particle in a potential well as "standing waves", or as two interfering waves reflected from the walls of the well. Increasing the energy leads to higher harmonics, which ...


0

The simplest answer to your question is probably the single-slit experiment. If you shoot two electrons consecutively in the same direction towards a plate pierced with a slit, then two dots will appear at different positions on the screen placed behind the plate. The difference in the positions is almost entirely due to the quantum mechanical uncertainty. ...


3

So, Quantum Mechanics implies that a particle has no trajectory whatsoever It depends what "whatsoever" means and what "particle" means and what "trajectory" means. All these words in physics depend on the framework. For distances larger than nanometers and energies larger than some kilo electron volts or so, the classical framework is what defines ...


2

Quantum systems do not have a position. This is intuitively hard to grasp, but it is fundamental to a proper understanding of quantum mechanics. QM has a position operator that you can apply to the wavefunction to return a number, but the number you get back is randomly distributed with a probability density given by $|\Psi |^2$. I can't emphasise this ...


0

Quantum physics has momentum. It's just not something fundamental. Saying that quantum physics doesn't have momentum because it's just the waveform evolving is like saying that an airplane doesn't have wings because it's really just a bunch of atoms.


0

If you have a conserved probability (such as in nonrelativstic quantum mechanics), then you get a conserved current, the probability current. In almost all situations it will do for you what you want a velocity to do. Just don't try to get it to do more than you want by expecting it to be too classical. For instance the expectation value of momentum could ...


2

It comes from the mathematical framework of Quantum Mechanics. A general expectation value is the result of computing a state $\omega$ over some observable $A$. Mathematically speaking $A$ is a self-adjoint operator from the C*-algebra $\mathfrak A$ of observables and $\omega$ is a state over $\mathfrak A$, i.e. a normalised positive linear functional on ...


4

$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\bra}[1]{\langle #1 \rvert}$As a special instance of the Born rule stating that, given a state $\ket{\chi}$, the probability to find it in a state $\ket{\psi}$ is given by (for normalized states) $\lvert \bra{\chi} \psi \rangle \rvert^2$, it is an axiom in the standard formulations of quantum mechanics that ...


1

The operator you want will be given in position space by $$ \mathcal{O} \psi(x) = \begin{cases} \psi(x) & a < x < b \\ 0 & \text{otherwise} \end{cases} $$ This is a linear operator on $\psi$, and the only ways to get $\mathcal{O} \psi = \lambda \psi$ is for the support of $\psi$ to be either contained entirely within $(a,b)$ (in which case ...


0

$\newcommand{\bra}[1]{\langle #1 \rvert}\newcommand{\ket}[1]{\lvert #1 \rangle}$What you seek is, in general terms, a projector onto a subspace. Given a set of basis kets $\ket{\psi_i}$ for a subspace, the projector onto that space is simply $$ \sum_i \ket{\psi_i}\bra{\psi_i}$$ It is easy to see that the $\ket{\psi_i}$ are eigenvectors with eigenvalue $1$ ...


2

This is a mathematical result. The Schmidt decomposition tells you that there are bases for two parties $A$ and $B$ such that $$ \sum_{ij} \lambda_{ij} |i_{A}\rangle |j_{B}\rangle = \sum_k \nu_k |\tilde{k}_A\rangle |\tilde{k}_B\rangle $$ with some orthonormal bases $|i_A\rangle,|\tilde{i}_A\rangle, |i_B\rangle, |\tilde{i}_B\rangle$. If you compare the two ...


2

Let us make clear that the problem If proton spin emergence from quarks and gluons is mysterious, why is silver atom spin not? is a modelling problem. The spin both of the proton and the silver atom is measured and known to identify them. John's answer covers it, the energy carried by the virtual quarks and gluons within the proton are much larger ...


0

This is a very interesting question. I don't know if there is a general and definitive answer, but I'll try to make some comments. I apologize if this ends up rambling; I'm finding this out as I write this answer. Operators have dimensions, since their eigenvalues are physical quantities. For bras and kets it gets more complicated. First, you cannot in ...


0

Really good question. Measurements have unit but in quantum mechanics, a measurement is the "evaluation" of an observable on a state (or a state on an observable) something like $$ \left\langle \psi | A | \psi \right\rangle ,\quad \psi\in\mathcal{H},\ A\in\mathcal{B}(\mathcal{H}) \ \text{self-adjoint}$$ A priori, there seem to be an arbitrariness in the ...


0

Gravity can change frequency. A light beam going towards a massive body is blue-shifted by the gravitational field. If it is escaping a gravitational body, then it is red-shifted.


2

The photon is an elementary particle. There are two ways do measure the frequency and therefore the energy of the photon since its energy E=h*nu . 1) using a diffraction grating which analyses the wavelengths in a beam of light , as below: This is the spectrum of iron. Each line is composed of zillions of photons with that frequency. If one sent one ...


2

Sure! In fact, this is one of the great strengths of atomic physics--if you and I are doing two experiments on rubidium atoms then (assuming we use the same isotopes) then our atoms will be exactly the same. Your rubidium atom and my rubidium atom are simply identical. So plenty of atomic physics is very repeatable, and such. As for the probabilistic ...


0

In regards to your question: Can the frequency of light change in propagation from one media to another, the answer is no. I found a previous response to a similar question that might help you: Think of it like this: At the boundary/interface of the medium, the number of waves you send is the number of waves you receive, at the other side, almost ...


0

There are several ways how the canonical ensemble can be derived (or maybe better say justified) in classical and quantum mechanics. Possible ways include: From the micro-canonical ensemble: This is essentially the approach outlined the answer by Sebastian Riese and addresses most directly your question. The short answer is yes, this is possible. However, ...


2

You understanding is correct. The Heisenberg Uncertainty Principle does not stem from the precision of our measuring apparatus; it is instead a fundamental property of the physical states which can be prepared. In essence it is the bandwidth theorem ($\Delta k\Delta x\geq 1/2$) translated into physical form by de Broglie's relation ($p=\hbar k$).


3

The word "relativistic" means "compatible with principles of Special Relativity". This usually implies that we can no longer use the "classical" picture of universal stationary space and time. Instead we talk about 4-dimensional space-time. The word "quantum" means compatible with principles of quantum mechanics. You can look them up on wikipedia etc. But ...


1

According to a list of levels from NIST, the ground state for mercury has quantum numbers $^1S_0$ (with the electron in the 6th shell). I usually have to look up how to read those symbols: the $^1$ tells you it's a spin singlet, the $S$ tells you that the orbital angular momentum $L=0$, and the $_0$ tells you the value of the total angular momentum quantum ...


0

The Hilbert space of quantum mechanics arises from considering irreducible representations of the C*-algebra of observables. When the C*-algebra of a physical system is commutative all its irreducible representations are one dimensional and therefore the corresponding Hilbert space is $\mathbb C$. Hence any classical physical system is associated to the ...


0

It's not about disturbing the photon. If you observe it, your detector gets entangled with it. It's possible for the photon going through the left slit and going through the right slit to end up in the same state and interfere, but if you get a detector involved then one of those will set off the detector and the other won't, so they won't be in the same ...


0

To explain why, you have to go back to something known as the ultraviolet catastrophe. In classical physics, one uses the Rayleigh-Jeans law, which states that "intensity is proportional to temperature divided by wavelength to the fourth power". The expression looks like this: where T is the temperature in Kelvins, lambda is the wavelength, and k is ...


0

completing the good Frank answer : Bell designed an experiment that cuts between forecasts of Einstein and Bohr. In a particular situation, Bohr's theory provides a result that other theories could not foresee. Technically, Bohr theory predicts $cosĀ²(angle1-angle2)$ correlations where other theories could not expect better than a line sloping at 0.5. ...


2

It all goes back to EPR experiment. In a paper published by Einstein, Podolsky and Rosen the authors argued that quantum mechanics is "incomplete". They argued that principles (such as the principle of locality) needed to be restored in order for it be a complete theory. The problem is, ultimately quantum theory is a "nonlocal" theory. What this means is ...


0

There is a method I like. It appears in the book H. Schulz: "Statistische Mechanik" (Harry Deutsch, 2005) (to my knowledge only available in German). It runs as follows: Consider a bath weakly coupled to a system, together they are prepared micro-canonically at energy $E$. The Hamiltonian of the system have only non-degenerate eigenenergies (otherwise we ...


2

I'm not an expert (mathematical physicist here), but as far as I know, the answer is no over really long distances and yes over shorter distances. There are inherently two problems that your question calls attention to: entangling particles over long distance maintaining this entanglement over longer time You only address the second one, but the first ...


0

The only book that I know can help build you intuition is "Lectures on Quantum Mechanics" by Weinberg. Weinberg's book is used to give a course in Advanced QM. It should not be your first read on the subject. The book is quite enlightening because it sheds lights on details that most QM books jump over without explaining the intuition or at least the ...


0

The author is working by analogy with the 3D box case. The 3D box worked out easily because $\epsilon\propto k^2$, so that the number of states up to each energy could use the volume of an ellipsoid, which is well known. He/she makes the volume of an ellipsoid work in a "rough way" for the harmonic oscillator. He/she puts a state where each 3D box state is ...


0

WSC - Not all matter is made of fermions. Things such as helium-4 and carbon-12 are bosons, and there are are many other composite particles and molecules that are actually composite bosons. Any composite particle with an even number of fermions, and thus with an integer value of spin, is boronic, which to me seems a little moronic. I imagine that you are ...


1

My answer to the question is first a question. How do you photograph a photon? You can measure a photon by detecting it's presence on photographic film, or by using some sort of photomultiplier and digital detector, or by a handful of other ways. My point is that these all require the photon to be absorbed by the detector, thus they must be localized, and ...


0

Theoretically we can, I we had enough. But making and maintaining antimatter is hard. Also, if you aren't in a vacuum, you will just see bright light and energy.


0

From the way the question is worded, I would assume you can treat this system as a BEC wavepacket in a potential barrier, where the potential barrier is given by your harmonical trap. Since the trap is harmonic, this is in analogy to the well-known vibrational wavepackets. I would also recommend reading this paper for some more insight: ...


3

There are two things to consider: What does the potential look like? Is the wave function of the qubit narrow in the flux or charge basis? Potential shape The Hamiltonian of the transmon (a junction in parallel with a capacitor) is $$H_{\text{charge qubit}} = - E_J \cos(\phi) + \frac{(-2en)^2}{2C}$$ where $E_J\equiv I_c \Phi_0 / 2\pi$, $I_c$ is the ...


0

Can matter waves interfere with light? Light - matter interactions are possible, namely, light - atom interactions when you use laser light for atom diffraction: Atom Interferometry For this, I'm pretty sure not, in the same way light does not diffract with sound waves as they operate with different mechanisms. Light does diffract with sound ...


1

The point of this thought experiment has been widely taken out of context and misused by new age science supporters. Schrodinger initially considered this experiment to show the RIDICULOUSNESS of the situation, not because it was physically what is happening. Additionally, anyone who says that there is a line between the quantum and classical regime is ...


3

Vectors (which kets are) don't have adjoints, they have duals. Whether the dual of $\lvert n_1,\dots,n_n\rangle$ is denoted by $\langle n_1,\dots,n_n\rvert$ or $\langle n_n,\dots,n_1 \rvert$ is entirely conventional.


0

Sort of a silly answer, but a bunch of protons, electrons, and neutrons bound together (i.e. an atom or molecule) can certainly diffract through a double slit. This has been taken to ridiculous extremes in recent years: witness the double-slit diffraction of C284H190F320N4S12, a molecule with 810 atoms and over 5000 electrons. But if you're thinking ...


0

The transmon and Cooper pair box share the same design, but operate in opposite limits: Cooper pair box is operated under the condition $E_C\gg E_J$, so the charging energy dominates. While for transmon, it is $E_C\gg E_J$, and it is less sensitive to charge noise because of this parameter choice.


2

Energy is not a Lorentz invariant quantity, it is the zero-th component of the four-vector. Only proper orthochronous Lorentz transformations preserve the sign of the zeroth component, so if the energy is positive in one frame, a non-orthochronous Lorentz transformation would yield a frame in which the energy is negative. But we usually only allow the ...


1

Protons and electrons both obey the de Broglie hypothesis: wavelength = Planck constant / momentum. But protons do not act within the atom the same as electrons - they move in a tighter radius at higher speed. They have to be accelerated to reveal their wave nature, and as the momentum of a proton would be much greater than that of an electron at typical ...


-3

we can see negative energy solution as anti particle travelling in the -p direction in momentum space.creation operator have coefficents e^-ipx so it will create anti particle in the -p direction. here p is a four vector.four vector can be negative or positive.so we have solution in the positive p direction and solution in the - ve p direction.


0

Maybe a chemists perspective could be useful to understand why we take both energies in quantum mechanics? It's not an elegant answer but is easy to rationalise! In computational chemistry we use the variational principle to compute the orbital mixing coefficients of molecular orbitals from an atomic orbital basis set. In doing so along the way we end up ...


4

All $\mathrm{SU}(2)_k$ with $k>2, k\neq 4$ are universal. For a proof see http://arxiv.org/abs/math/0103200.


0

It is perhaps illuminating to think in terms of polarisation-entangled photons, where this result is sometimes recognised as the quantum Malus' law. A reference can be found here. As you will see from Eq. (3), this simply derives from transmission probability $\left|\mathcal{A}\right|^{2}=\left|\langle\Omega|\vec{a}'\rangle\right|^{2}$ (consistent with your ...



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