# Tag Info

1

I think it is important to emphasize that the notion of 'virtual particles' is a very dangerous one, which seems to lead to countless (unnecessary) misconceptions. It appears to have originated from the diagrammatic technique that can be used to carry out perturbative quantum field theoretic computations (i.e. Feynman diagrams), but it is crucial to keep in ...

-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 ...

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 ...

1

I'm essentially interpretting your question as "are there any cannonical commutation relationships (CCRs) where the Lie bracket is a different scaling constant of the identity matrix: $$[\hat{x},\,\hat{p}] = i\,\alpha\,\mathrm{id}\tag{1}$$ or, equivalently, are there any pairs of canonically commuting observables where the scaling constant $\hbar$?" ...

3

The uncertainty principle may be stated more generally for two observables $A$ and $B$ as: $$$\Delta A \Delta B \geq \dfrac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right|,$$$ where $\langle \hat{C}\rangle$ is the expected value of the observable $C$ and $[\cdot\,,\cdot]$ is the commutator (see here for details). From ...

0

I now know (as pointed out by 'Javier Badia') that as this inequality is like any other normal inequality, the units on both sides of the inequality must be equal.

2

This has to do with what we really mean when we ask if something is "at rest." Specifically, we usually mean something like this, "Is the particle's momentum in our frame of reference exactly zero?" As it turns out, we can't actually know a particle's momentum exactly in quantum mechanics because of the Uncertainty Principle. Supposing that we could would ...

0

As you already realized, you are messing up classical and quantum concepts. The uncertainty principle for two general observables for a particular quantum state is given by $$\Delta A\Delta B\geq\frac{1}{2}\left|\langle [A,B]\rangle\right|,$$ where $\Delta$ is the standard deviation and $\langle\cdot\rangle$ is the expected value of an observable for the ...

0

Time is not directly related to energy itself, but it is definitely related to many aspects of energy. For example, the direction of time (from past to future) can be determined by the flow of energy in the universe. This concept is known as entropy. Our universe is gradually moving from a state of energy concentration (where some regions of our universe ...

0

Here is a model for the relation between energy and momentum, based on classical physics: A particle at rest receives a linear acceleration. After this acceleration process it is moving, it has kinetic energy which corresponds to a momentum. For calculating the energy, you consider the length of the path the force of acceleration was applied: ekin = f x s ...

3

Energy is the name physicists give to the Noether Charge that is conserved when a physical system's description through its Lagrangian is unchanged by time shifts. Or, in more everyday language, most physics does not depend on where one puts the $t=0$ time co-ordinate origin. The laws are invariant when we shift our time origin back and forth. Noether's ...

-1

The units of energy is $\dfrac{ML^2}{T^2}$ so that way energy is inversely proportional to square of time. But in most equations of energy, time is never present. Energy is a simpler way to relate work done in complicated systems with variable force. That is found by work-energy theorem. Work is $$\text{force*distance}$$. There is also Noether's theorem ...

0

Assume you have quantum mechanical oscillator, e.g. a particle in a potential V(qx)∝q2x. Now the position of the particle shall be measured by having photons scattered from the particle and then detect k vector and phase of the photons (in fact an indirect quantum measurement). Note that in a quantum mechanical potential the particle is in a stable ...

0

The tentative answer to your question is "No" - not all quantum interactions produce a back action. There is an entire field of study under the heading of "Quantum Non Demolition Measurement". A simple example of this is illustrated by the Bomb Detector thought experiment.

1

If we have two arbitrary quantum-mechanical operators $\hat A$ and $\hat B$, then the corresponding observables have to satisfy the Robertson-Schrödinger uncertainty relation: $$\Delta A \Delta B \geq \frac12 \lvert\langle [\hat A,\hat B] \rangle\rvert$$ This equation implies that it is impossible to measure both $A$ and $B$ to arbitrary precision at the ...

1

While I agree with Noah Steinberg's answer, there are some other points. Usually when quantizing a system, we label our states using quantum numbers, which are as close to classical parameters (also called c-numbers or commuting numbers) as you can get. The calculation is usually easiest when you can find the largest set of commuting observables, and ...

3

If two observables are compatible it means the eigenstate of one observable is also an eigenstate of the other and that the commutator of the two operators is 0. This means that if two observables are compatible one can make a measurement of the first observable and then measure the second observable without changing the state of the particle.

1

I will answer this part In addition, we know that the Hamiltonian represents the sum of kinetic and potential energy in a system.However, I'm not quite sure why, intuitively, the time dependent version of the Schrodinger equation becomes Hψ=iℏ ∂/∂t ψ(r,t). Quantum mechanics was developed slowly, because experiments showed that light came in quanta ...

0

You ask a couple of different questions: 1) You say "by Heisenberg's uncertainty, we cannot measure the exact momentum or position of a particle/wave ever". No, Heisenberg's uncertainty principle doesn't say that. It says that IF you measure the position of a QUANTUM particle with precision Δx, i.e. you localize the particle within an interval Δx, then the ...

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