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Jan
26
comment Apparent paradox concerning Heisenberg's uncertainty principle
The whole story is full of many other wrong statements. For example, electrons aren't moving along straight lines in magnetic fields even in a classical theory. They move along circles or spirals, right? Also, all the claims that "the uncertainty is zero" are absolutely unjustified. To say that the uncertainty is zero means to be able to measure it, and know that the measurement is precise etc. In reality, it never is quite precise.
Jan
24
awarded  Good Answer
Jan
24
comment Entanglement up to a finite distance
If you mean experimental games with that, quantum teleportation has been verified at 143 kilometers, see e.g. phys.org/news/… - Maybe the record was beaten since that time. At any rate, no expert doubts that entanglement is able to fully sustain itself at least up to the radius of the visible Universe.
Jan
23
comment specific heat in solids and liquids
I must have overlooked this essential part of the question, sorry.
Jan
21
answered What is “mass” in particle physics?
Jan
21
answered specific heat in solids and liquids
Jan
20
awarded  Nice Answer
Jan
19
answered Local and global U(1) gauge symmetries of Hamiltonian
Jan
18
comment In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
Dear @annav - great but this is a question about particle (or similar) physics and that discipline is a subfield of quantum mechanics. Everywhere in quantum mechanics, all the information about physical systems is obtained from measurements and measurements always affect the measured system and one can't ever avoid this influence. All these things are absolutely universal postulates of quantum mechanics i.e. including particle physics. So what some people say about the "truth" of avoiding this influence is as off-topic here as astrology, isn't it?
Jan
18
comment In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
Dear @annav - the cloud chambers etc. measure some inaccurate information about position as well as momentum but they do so "strongly". There is no generalization of the Copenhagen notion of the measurement needed here (or anywhere in productive physics). When the position of a particle is measured only up to some poor accuracy, the motion (momentum) of the particle is not changed much. But it still follows all the rules of the normal measurement. At any rate, people talking about "weak measurement" and "active particle physicists" are almost disjoint groups. HEP isn't about flapdoodle.
Jan
16
comment How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
All physical quantities ("observables") are expressed as operators in quantum mechanics. But this is just a fact about the mathematical formalism. We may still talk about the measured values of the quantities, phenomenologically, and that's what the OP did, and we may talk about these things without knowing that the observables are represented by operators. The operator representation of observables like $x,p$ is needed for a clear proof of the uncertainty principle but it is not needed for the formulation of the uncertainty principle or the formulation of questions as as the OP's question.
Jan
16
comment How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
It is not ill-defined in the quantum theory. $\Delta x$ is simply equal to $\hat x(t+\Delta t) - \hat x(t)$. Your comment about "eigenstates" is pure confusion. Almost no states are eigenstates of a given operator and there's nothing wrong about this fact - that's why the superposition principle says that all linear superpositions are equally allowed. One only gets an eigenstate after a measurement. But if one does a measurement, he still needs to know what operator he is measuring, and $\Delta x$ is as good an operator to measure as any other operator.
Jan
16
revised How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
added 237 characters in body
Jan
16
comment How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
I don't think that the OP has any wrong notion of velocity. Velocity is the final position over the initial position, and this difference divided by the small duration. It's only you, not the OP, who is questioning that. The OP is totally right that one may keep on using this understanding of velocity in QM. Talking about Schr. picture doesn't change anything - one may still define the form of the operator of velocity at a given time by converting the expression from the Heisenberg picture by the Heis-Schr transformation.
Jan
16
comment How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
The equation $\dot x = v$ is the definition of the velocity - first written by Newton - and it is always true. In the Heisenberg picture of quantum mechanics, it holds as an operator equation. In non-relativistic physics, one has $p = mv$, again valid as the operator equation. In relativity, we would have $p = m_0 v / \sqrt{1-v^2/c^2}$ which would act as a nonlocal operators on the wave functions and it's not what we need in relativity - one of the reasons why we need to switch to quantum field theory to describe particles in a relativistic way.
Jan
16
answered How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
Jan
16
comment How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
Your formula isn't the formula for "the" momentum (and what appears inside isn't "the" velocity) but it is a formula for the average momentum and/or velocity along the trajectory $\Delta x$. The uncertainty principle constrains the uncertainties of $x,p$ at the same moment. But even if one ignores this thing, e.g. by taking $\Delta x$ infinitesimal, there is no contradiction.The uncertainty principle simply says that $x$ and $\Delta x$ cannot have well-defined values at the same moment. One must distinguish $\Delta x$ as the uncertainty of $x$, and as the change of $x$ between the moments.
Jan
16
comment How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position?
Apologies, I had to downvote it. The quantity "velocity" is exactly as well-defined as the position or the momentum, it's $\hat v = \hat p / m$ in non-relativistic physics, for example, and it is equal to the time derivative of the position. This is what the Heisenberg equation of motion says - and this equation holds as an operator equation. (One may describe physics in pictures not focusing on time dependence of operators but that only introduces confusion here.)
Jan
16
comment In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
Dear @annav - do you say that traces e.g. in cloud chamber represent a "weak measurement"? I don't see what's weak about it. The path we record represents an approximate but "strong" measurement of the position and the momentum. Even if you know the character of final states (channel),a particle is moving the interaction point e.g. as an s-wave. To see the particle in a particular direction involves a radical "collapse" ie modification of the initial (s-wave) wave fn, so it in no way fits the description of a weak measurement. Quite generally, a weak measurement doesn't appear in real physics.
Jan
16
comment Counting degrees of freedom for gravitational waves as a gauge field
OK, so the sentence would say that "groups encoded by at least one-dimensional Lie algebras cannot be discrete". ;-) I don't know whether any physicist would agree that a Lie algebra/group may be zero-dimensional (similarly for vector spaces) but be my guest.