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Entanglement distance and quantum mechanics length scales

Why do we care about the quantum mechanics scale? Distance is no object for entangled particles. The connection between entangled particles is non-local and instantaneous, so distance does not matter....
KDP's user avatar
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Is Bell's Theorem Wrong?

People tend to overcomplicate this. When we have two entangled polarised photons and pass them through polarising analysers (A and B) the correlation probability is given by $$p = \cos^2(\theta_A - \...
KDP's user avatar
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Schroedinger equation applicable only in electron

Protons and neutrons are not elementary particles (they are made up of quarks), while (as far as we know) electrons are. So protons and neutrons are similar to any other object made up of multiple ...
Stefania's user avatar
1 vote

How do operators on kets and wavefunctions correspond?

Is it by showing that this holds for position and momentum operators, and then have the result follow from any observable having to be a function of these? Yes, naturally. Your text or instructor ...
Cosmas Zachos's user avatar
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Is it that electron of an atom can be found anywhere in the space?

While it is true that free atom doesn't have a boundary, very, very, very large majority of the atom is in a small area. The boundary is not complitely sharp just extremely sharp. The same goes for ...
Alien from future's user avatar
3 votes

Non-locality of the wavefunction in QM and Twistor theory

The discussion in the comments seems to be a classical example of miscommunication. People don't exactly define what they means by the terms that are used. Before, attempting to answer the OP question,...
flippiefanus's user avatar
3 votes
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Why doesn't Car-Parrinello molecular dynamics require an SCF calculation?

It is not entirely true that Car-Parrinello (CP) Molecular Dynamics (MD) doesn't require a self-consistent field (SCF) calculation at all. The method must start close to the SCF solution (actually a ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
4 votes
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Symbol denoting parity eigenvalue

In nuclear physics the symbol is usually $P$. In that context the parity is rarely abbreviated without also using the total angular momentum. For example, the deuteron has $J^P=1^+$. I am trying to ...
rob's user avatar
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Hund's rules in Helium level scheme

Hund's rules can only predict the energetic ordering of states arising from the same configuration. For example, $E(2 \, {^3S})<E(2 \, {^1S})$ can be guessed from the first rule, because the ...
dennismoore94's user avatar
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Are there wave functions that are neither symmetric nor antisymmetric?

I think Maxwell- Boltzmann particles are neither symmetric nor anti-symmetric. This particular discussion is given in the section-8.1 of the book, 'Fundamentals of Statistical Mechanics' by B.B.Laud.
Rajesh R's user avatar
2 votes

Wave packet as a field configuration acting like a particle's wave function?

There is a mathematical equivalence between two different physical situations. The first situation is a wavepacket in a classical field. This describes some localized packet of energy in an ordinary ...
Andrew's user avatar
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Wave packet as a field configuration acting like a particle's wave function?

Before the sentence you quoted, Caroll writes: Think of a state in which just one mode is involved, the mode with $k = 0$. Since $k = 2\pi/\lambda$, that’s a mode with “infinite wavelength”—basically ...
alanf's user avatar
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How do photon emitters and photon detectors work?

You right that we cannot observe a photon as it goes past a slit without absorbing it using a normal photon detector as is sometimes depicted in some YouTube videos. What is really done is that ...
KDP's user avatar
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On Griffith Quantum example 2.1: normalization of wave function in time.

If you are reading Griffith's Quantum Mechanics, there is an after-class exercise in the first chapter that says: for any two (normalizable) solutions to the Schrödinger equation (with the same $V(x)$)...
guoxu's user avatar
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1 vote

When is the temperature relevant for a quantum many body system?

If the system is at $T=0$, all particles should be at the ground state and no excited state is possible. However, this is clearly not true for an arbitrary quantum system, given that if we solve ...
Roger V.'s user avatar
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The usage of temperature in quantum mechanics

Statistical mechanics can be derived from suitable models in the decoherent limit of quantum theory. In many quantum experiments, such as single particle interference, the square amplitudes of states ...
alanf's user avatar
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1 vote

The usage of temperature in quantum mechanics

Many-body quantum mechanics is also often referred to as Quantum statistical physics, which is a more telling name in the sense that it points out that we are considering systems where one has to take ...
Roger V.'s user avatar
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8 votes
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Jellium Hamiltonian in the thermodynamic limit

To quote the standard reference on the Jellium:$^1$ The physical hamiltonian is recovered by letting $\kappa \to 0$ after going to the thermodynamic limit $L\to \infty$. The procedure is justified a ...
Tobias Fünke's user avatar
4 votes
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Some confusion about understanding the relativistic quantum mechanics

The group is still the Poincare group (Lorentz+ translations). The tricky thing is that we need to find a way for that group to act on the Rays in such a way that it preserves the probability. There ...
Josh Newey's user avatar
1 vote

When is the temperature relevant for a quantum many body system?

I will write my answer by addressing some of your statements in your question, and elaborate a bit around them. You start by saying: The Hamiltonian of the system is typically of the form (expression ...
Marius Ladegård Meyer's user avatar
1 vote

When is the temperature relevant for a quantum many body system?

The question in the title, the "when" question, is basically impossible to answer. I mean, it is only when your level of precision required for whatever experiment it is you are doing, ...
naturallyInconsistent's user avatar
0 votes

Defining the geometry of Bell inequalities

Behaviours are contained in a real (Euclidean) vector space. The set of behaviours is not a vector space, but rather a convex polytope contained in an ambient vector space. Consider a standard ...
glS's user avatar
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What would be the outcome of an experiment wherein the spin of a qubit is measured in two or more orthogonal directions simultaneously?

You can't "measure a qubit in two bases at the same time". A measurement necessarily involves a collapse of the wavefunction, so "measuring in the Z basis" means forcing the state ...
glS's user avatar
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2 votes

Time evolution of operators in the path integral formalism

If we limit ourselves for a moment to one-particle path integrals, then a path integral is essentially a matrix element of the evolution operator (I might be skipping some nuaces here): $$ G(x,t|x_0,...
Roger V.'s user avatar
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Good book on quantum photonics and applications

I suggest reading the paper "The concept of the photon" by M. Scully and M. Sargent. This is a good start. Then, I recommend reviewing your classical EM stuff and being very comfortable with ...
0 votes

Energy eigenvalue of hydrogen-like atoms using Laplace-Runge-Lenz vector

Your first task is to absorb all superfluous constants into your nondimensionalized variables, and do the same for the nice review by Valent which is required reading, if you cannot follow WP or Pauli....
Cosmas Zachos's user avatar
1 vote
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Bohmian mechanics, Leggett inequality, realism and nonlocality

There are several things to consider in the debate between realistic vs. indeterministic and Bohmian mechanics vs. no-go theorems (which includes the Leggett inequality). First of all, Bohmian ...
Caesar.tcl's user avatar
1 vote

Intuitive reason why bound states correspond to poles

I think this link will help you: https://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Scattering_III.htm ... if the scattering matrix $S_0(k)$ becomes infinite at some complex value of $k$, ...
JMadar's user avatar
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3 votes

Angular momentum quantum number $l$ either integer or half integer

The book is correct in using “either - or”. But the reason relies upon more fundamental instances of QM than the structure of angular momentum operator. More precisely a certain superselection rule. ...
Valter Moretti's user avatar
0 votes

Angular momentum quantum number $l$ either integer or half integer

The other answers and comments are all great; I will solely focus on the word either as emphasized in the question. The book is answering the question "what are the possible values of $l$?" ...
Quantum Mechanic's user avatar
1 vote

Why Consider Only Triplet States for Spin in $2$-Electron Systems?

This is slightly strange. In any case what matters is that the fermionic states should be fully antisymmetric w/r to permutations of the particles, and that means spin and spatial degrees of freedom. ...
ZeroTheHero's user avatar
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2 votes

Time evolution of state in first and second quantization

For a system with energy eigenvalues $\{\varepsilon_i\}$, the second quantized Hamiltonian is $\hat{H}=\sum_i \varepsilon_i \hat{c}^\dagger_i \hat{c}_i$. In Heisenberg picture the creation operator at ...
hft's user avatar
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1 vote
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Calculating the expectation value of the angular momentum operator

How do come up with $$ \frac{1}{\pi}\frac{\hbar}{i} \frac{1}{2\phi}\sin^2{2\phi \pi}? $$ The answer for the integral must be a number, not a function of $\phi$. The integral $$ \int_0^{2 \pi} \cos\...
mike stone's user avatar
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1 vote

How much does quantum uncertainty contribute to the uncertainty of earthquakes?

When information is copied out of a quantum system undergoing interference that interference is suppressed: this effect is called decoherence https://arxiv.org/abs/quant-ph/0306072 Since macroscopic ...
alanf's user avatar
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1 vote

What would be the outcome of an experiment wherein the spin of a qubit is measured in two or more orthogonal directions simultaneously?

A measurement is a physical process that creates a record of some property of a physical system. Since a measurement is a physical process it is constrained by the laws of physics, which include the ...
alanf's user avatar
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1 vote

How much does quantum uncertainty contribute to the uncertainty of earthquakes?

Taken literally, earthquakes are macroscopic phenomena, where quantum effects are negligible. Note that the macroscopic size itself does not preclude quantum effects, but it usually implies ...
Roger V.'s user avatar
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3 votes
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Dipole term in light-matter interaction

These are not distinct things, but rather different side of the same notion. More systematically dipole, quadrupole and other moments arise when performing Multipole expansion of a charge distribution....
Roger V.'s user avatar
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7 votes
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In hetero- homodyne detection, what does it mean to operate at the Quantum Shot Noise limit?

There are good questions here. I don't have time to thoroughly answer all of them. What does it mean for a hetero- or homodyne receiver to be operating in the Quantum Shot Noise limit? In general ...
Jagerber48's user avatar
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2 votes
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The eigenvectors associated to the continuous spectrum in Dirac formalism

Dirac is dead, we cannot ask him what he really meant by this. However, the pretension that these eigenstates exist in some sense is prevalent in many texts on quantum mechanics. There are several ...
ACuriousMind's user avatar
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1 vote

Operator's definition in Dirac picture

I have a question about the definition of quantum operators in the Dirac picture. The definition is: $$A=\sum_i \sum_j \vert i \rangle A_{ij} \langle j \vert.\tag{1}$$ By deplacing the ket vector I ...
hft's user avatar
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4 votes

Operator's definition in Dirac picture

How did you get from the 1st equation to the second equation? The Second equation will be a number, not an operator. You can't just move the ket to the right. The top equation is an outer product but ...
Josh Newey's user avatar
2 votes

The eigenvectors associated to the continuous spectrum in Dirac formalism

As long as you allow elements of the associated rigged Hilbert space the answer is "yes."
mike stone's user avatar
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0 votes

What make the energy transition in NMR?

You are talking about MRI not NMR spectroscopy. NMR requires only 1 field. MRI uses more complicated magnetic field to produce 3-d image. You only need magnetic field for Zeeman effect, with electric ...
Alien from future's user avatar
4 votes

Does photocurrent depend on intensity or on number of photons?

The current depends on the number of photons with high enough frequency(energy). 1 radio photon and 1 gamma photon will give you 1 electron, because gamma photon has enough energy to expell an ...
Alien from future's user avatar
5 votes
Accepted

Does photocurrent depend on intensity or on number of photons?

You have it right. You can idealize it as each photon kicks off one electron. (It might be really that each photon kicks off one avalanche.) So current is proportional to number of photons. If the ...
mmesser314's user avatar
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1 vote

Quantum Mechanical Current Normalisation

You just need to normalise your solution to unity, after correcting your expressions. The meaning of this is known as the Born rule.
my2cts's user avatar
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1 vote

Dipole term in light-matter interaction

The coupling to the light field is related to your second dipole: an eletromagnetic field of a given frequency $\omega$ will couple two states of the electron that roughly of the order of $\omega$ as ...
Franklin Luis's user avatar
1 vote

Reference on partial wave expansion in the context of QFT

Apart from Weinberg QFT vol1, section 3.7, one can look at the paper: "On the general theory of collisions for particles with spin", Jacob and Wick 1959. A recent review that I found useful ...
-1 votes

How can quantum entanglement not be non-local?

The equations of motion of quantum theory, such as the Schrödinger equation or equations of motion for relativistic quantum theory don't predict collapse and aren't compatible with it. For this ...
alanf's user avatar
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How can quantum entanglement not be non-local?

There is another explanation. Hidden variables. If at their creation entangled electrons are already in pure state due to unknown physics this reproduces all experiments. Of course due to Bell's ...
Alien from future's user avatar

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