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bio website motls.blogspot.com
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Hi, I am a string theorist and a publicist.


May
1
comment Blackbody cavity relationship between energy of oscillators and EM radiation
The EM radiation is the excitation by these oscillators. The EM field may be divided to lots of harmonic oscillators and each of them may oscillate. The amplitude measures the strength of the EM wave at a given frequency and direction. In quantum physics, the oscillators have quantized spectrum. The energy above the minimum is an integer multiple of $E=hf$, the energy of the photon, and the integer may be interpreted as the number of photons in the state (photons with the given frequency and direction).
May
1
comment What does it mean for a particle to have spin of 2?
Shouldn't you have asked the same question - what it means - already when you were told that the electron has spin 1/2? Because if you followed what that meant, you must follow the same sentence about spin 2, too.
May
1
comment Tensor product of operators in QM
Dear Frank, even interacting systems composed of two subsystems have a Hilbert space that is a tensor product, and admits wave functions that are tensor products, too. However, the mutual interaction exactly means, as you suppose, that the Hamiltonian doesn't have the additive form $H_A\otimes 1+1\otimes H_B$. It contains terms that depend both on $A$ and $B$ degrees of freedom - which can therefore be hiding in neither of the two terms. It follows that such an interacting union of 2 subsystems will develop entanglement even if the initial state is not entangled (ie. if it is a tensor product)
May
1
comment Poisson brackets in curved spacetime
The variables $\phi$ entering this equation are not spacetime coordinates but phase space coordinates - which contains all the information about the state at one moment (both coordinates and, effectively, velocities). So you must know what the phase space is and what the Poisson bracket is. The equation above is totally universal for all systems where a Hamiltonian and a Poisson bracket may be defined. The Poisson bracket for positions and momenta on curved spaces is modified, indeed. But in AdS, we often consider fields and the phase space is parameterized by fields, not coordinates+momenta.
Apr
30
comment Tensor product of operators in QM
because e.g. $\psi'_A\otimes \psi_B = (1/i\hbar) H_A\psi_A \otimes \psi_B = (1/i\hbar) (H_A\otimes 1) (\psi_A\otimes \psi_B)$. So the Hamiltonian is an example of that "derivative" because it's a time derivative.Similarly, the momentum is the derivative with respect to space, angular momentum is the derivative with respect to the angles (rotations), like in Noether's theorem.
Apr
30
comment Tensor product of operators in QM
Dear Frank, thanks for the "non equal" fix! And concerning the question: absolutely. This is an example I should have written. The Hamiltonian is effectively the "derivative of the state vector with respect to time" over $i\hbar$, as Schrödinger's equation says. If the wave function is $\psi_A\otimes \psi_B$, a tensor product of two independent subsystems, then the time derivative acts on this product via the Leibniz rule, $(\psi_A\otimes \psi_B)' = \psi'_A\otimes \psi_B+\psi_A\otimes \psi'_B$ and this can also be written as $(1/i\hbar)(H_A\otimes 1+1\otimes H_B)$ acting on $\psi$
Apr
30
comment Why is parallel component of velocity along position vector considered rate of change of position?
Upvoted. One may also say that the motion of the point mass may be divided to the motion in the radial direction, and the motion in the transverse direction - which is locally the same thing as the angular direction "around" the center of coordinates. It's only the former that changes the distance - the distance isn't changing if we rotate around the circle - and it's only the former that contributes to the inner product.
Apr
30
comment How do particles “know” when to decay?
Just to be sure. The percentage of the particles (of the same kind) that have decayed is nothing else than the probability that one particular of them has decayed.
Apr
30
comment How do particles “know” when to decay?
Yes, the percentage of particles at a given time that will have decayed is computable. But the precise integer may fluctuate. So if you expect $N$ particles to decay in average by a certain moment, what you get in reality will be $N\pm \sqrt{N}$ or so – the error is comparable to the square root of $N$, quite a usual thing in random processes. And yes, also, we have the theories that allow us – at least in principle and often in practice – calculate the half-times for all particles or nuclei etc. It's understood why some isotopes decay more quickly, others less quickly, and others are stable.
Apr
30
revised Tensor product of operators in QM
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Apr
30
answered Tensor product of operators in QM
Apr
30
answered How do particles “know” when to decay?
Apr
29
comment What experiment would disprove string theory?
Right. In other words, if string theory were realistically proven wrong, it would almost certainly take the rest of physics to the bottom of the ocean with it. There's nothing wrong about this fact; it just means that string theory has become the state-of-the-art foundation of physics.
Apr
28
awarded  Nice Answer
Apr
28
revised What experiment(s) have or can refute the existence of an electron-particle “system” over the separate existence of a neutron within itself?
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Apr
28
comment What experiment(s) have or can refute the existence of an electron-particle “system” over the separate existence of a neutron within itself?
Dear @docscience, experiments can only refute a "position" if it is sufficiently well-defined to make at least some predictions that are not guaranteed a priori. With any understanding of the "composition" we have seen, and with some knowledge what the composition means according to quantum mechanics, your theory makes the prediction that the neutron will behave like the hydrogen atom or be indistinguishable, and easy experiments are surely enough to refute this prediction. You may refuse the assumptions "what QM implies about composite states" but then you have to write a whole new theory.
Apr
28
revised What experiment(s) have or can refute the existence of an electron-particle “system” over the separate existence of a neutron within itself?
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Apr
28
answered What experiment(s) have or can refute the existence of an electron-particle “system” over the separate existence of a neutron within itself?
Apr
28
comment Quantum Entanglement - What's the big deal?
Entanglement is just the most general correlation described in terms of this more accurate and more compatible-with-experience theory, quantum mechanics. The precise way to calculate the correlations and probability distributions are given by the laws of QM. The laws of classical physics - any classical theory - would be wrong. But it's not "weird" for some theories to be wrong. Most theories people can invent are wrong. This whole ritual of saying that QM or entanglement is "strange" is just an obsessive religious exercise of people who refuse to accept modern physics.
Apr
28
comment Quantum Entanglement - What's the big deal?
But my point is that none of these aspects of quantum mechanics is "weird" in the sense that it would contradict some experience we have really had in the previous centuries. The only thing that these features and predictions of quantum mechanics disagree with is classical physics - an approximate theory that was invented to describe people's observations up to 1925. But classical physics doesn't directly follow from our experience in any way, of course. Quantum mechanics is more, and not less, compatible with our everyday experience than classical physics.