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Jan
14
revised In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
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Jan
14
comment In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
There is absolutely nothing "weak" about the measurements of the energy of particles at the LHC. - BTW the term "particle" isn't just some modern deviation of particle physicists. Check Wikipedia - all elementary particles like photon, electron, even proton (despite compositeness) etc.- the particles that the OP is probably talking about - are defined as "particles" of some sort in the first sentence. Fermions were always called particles; bosons (photons in particular) were called particles as soon as the confusion about the existence of light quanta evaporated.
Jan
14
comment In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
@CuriousOne - the very claim that there exists something like "weak measurements" is a speculative and highly controversial subfield of "foundations of quantum physics" and virtually no particle physicist would actively say that they're doing "weak measurements". The LHC detectors are making standard measurements of the energy of the final particles. ... Jupiter is an extended object which may be approximated by a point mass. But particles in quantum field theory are exactly point-like, this is how quantum field theory differs e.g. from string theory, so they're nothing like Jupiter.
Jan
14
comment How should Christoffel symbols be written (in LaTeX)?
Right. Most authors probably do use $\Gamma^\mu_{\alpha\beta}$ and if they ever use any form with differently raised or lowered indices, they do follow particular conventions about the ordering of the indices. Whenever the indices are not "1 index above a pair of 2", then it must be explained what is the relationship with the normal form of the symbol.
Jan
14
comment In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
@CuriousOne - in the context of quantum mechanics, the words "particles" and "quanta" are basically synonymous and be sure that both are used - one of the reasons why the part of "quantum physics" studying events at the LHC etc.is known as "particle physics". It makes no sense to say that one exists and the other doesn't so I am afraid that your comment only conveys confusion, not any useful information that would have anything to do with the fact that we can't get the "which slit" information from the timing.
Jan
14
revised In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
added 1254 characters in body
Jan
14
answered In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected
Jan
14
revised Off-diagonal matrix elements of the Hamiltonian = transition rate?
edited title
Jan
14
comment Density matrices vs Pauli matrices
Dear @ACuriousMind, none of the Pauli matrices can ever occur as a density matrix of any system, see my answer. ;-)
Jan
14
answered Density matrices vs Pauli matrices
Jan
14
awarded  Enlightened
Jan
14
awarded  Nice Answer
Jan
13
awarded  Yearling
Jan
11
comment Why there is no Gibb's phenomenon in QM?
Using ideas related to your convolution and Fourier tricks, $\langle \delta,\delta\rangle=\infty$ basically because this inner product involves an extra integral over the whole axis, to convert a function into a scalar, and the integral of $1$ over the real axis diverges. The first answer on the math forum is just wrong,especially for a mathematician, because for them, a distribution is a linear form on the space of ordinary functions and $\delta$ isn't one, so one should say that $\int\delta^2$ is ill-defined.
Jan
11
comment Why there is no Gibb's phenomenon in QM?
Dear Rajesh, convolution isn't the same thing as a point-wise product. The simplest relationship is that the Fourier transform of convolution is the pointwise product of the Fourier transforms. So $\delta *\delta = \delta$ follows from $F(\delta * \delta) = F(\delta)$ which is equivalent to $1\cdot 1 = 1$ for three constant functions of $x$ or $p$. But the inner product is not convolution. In particular: The result of the inner product is a function, the result of convolution is a whole function (interpreted as state vector in QM).
Jan
11
comment Why there is no Gibb's phenomenon in QM?
Hi Rajesh, you can't "define" the dot product of $\delta(x)$ with itself to be one, it's just like trying to "define" $5+5$ to be $3$. But $5+5=10$ may be calculated and in the same way, the dot product of $\delta(x)$ with itself is the integral of $\delta(x)^2$ which is infinite because it's $\delta(0)$ if one treats one of the factors as the test function $f(x)$, and $\delta(0)$ is infinite which may be explained in tons of ways. If you "redefined" the inner products, you would destroy the distributive law on the Hilbert space and other things. You just cannot redefine calculable results.
Jan
10
awarded  Nice Answer
Jan
9
comment Limitations of 2D point-mass Dynamics of the solar system
If you have at least 3 massive bodies, Newton's theory - without any relativistic effects - also prohibits elliptical orbits. I honestly don't understand what may be hard to get about this simply proposition.
Jan
8
answered Is this a correct demonstration for why elements above untriseptium cannot exist?
Jan
8
comment Limitations of 2D point-mass Dynamics of the solar system
On one hand, you're saying that you "model" things - which sounds like simulations, a numerical calculation of the differential equations, including the planet-planet forces. But on the other hand, you say that the trajectories are ellipses if not circles. These two statements contradict each other. If you actually simulated the planetary motion, you would know that the trajectories are not ellipses or circles.