112,213 reputation
7165328
bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 4 months
seen 3 hours ago

Hi, I am a string theorist and a publicist.


May
11
answered $su(1,1) \cong su(2)$?
May
11
comment Quantum Entanglement - What's the big deal?
The only Bell whom you may find in the history of quantum computing - the page above - is the Bell in "Bell Labs", and please be aware that his name was Alexander Graham Bell, not John Bell or what was the name of your "hero".
May
11
comment Quantum Entanglement - What's the big deal?
Terry, good to hear you. Bell's contribution to quantum computing was zero - i.e. vastly smaller than the string theorists' contributions. Quantum computing's links with string theory is actually a hot subject. Quantum computers are based on the regular quantum mechanics known from 1925, especially Pauli's insights about spin, and it started as an applied physics or engineering discipline in 1970 when quantum codes began to be constructed. Feel free to search through the timeline of quantum computing en.wikipedia.org/wiki/Timeline_of_quantum_computing
May
10
comment How many fundamental fields / constraints are in Maxwell's Equations?
What do you exactly want to know about them? "What about" isn't a well-defined question. I said that the auxiliary components are needed for the Lorentz-invariant dynamic of the spin-one field. They are needed for the most general situations. If you pick some less clever, special, e.g. static, situations only, you may become unable to prove what I proved (or sketched a proof) above. But that doesn't mean that my proof has a problem.
May
8
awarded  Good Answer
May
5
comment Basic question about superspace, Grassmann numbers and world sheet supersymmetry
In that case, $\theta$ and $\bar\theta$ must be treated together and be equivalent to $\theta_A$ with different components $A$. It's still true that the derivative of $\theta$ bilinear is $\theta$ linear.
May
4
comment Do light and sound waves have mass
I assure you that phonons are exactly as real as photons, they are conceptually the very same thing, obey the same quantization rules, form a Hilbert space isomorphic to a Fock space, and their behavior is described by quantum field theories. Sound is a vibration of the underlying material, and in the same way, light is a vibration of the electromagnetic field. They only differ by the substance that vibrates and the elementary degrees of freedom. The case of light only differs by the substance's being simpler or more fundamental/elementary -fields that exist even in the vacuum.
May
4
awarded  Nice Answer
May
2
comment Basic question about superspace, Grassmann numbers and world sheet supersymmetry
Dear @leastaction, thanks for reading. Concerning $\bar\theta\theta$, I believe that there is a mistake in your formula. It seems like a chiral superfield that only depends on $\bar \theta$, look at the $\psi$ term, so that should be written as an argument on the left hand side and the last $B$ term should actually be $\bar\theta\bar\theta$, and that's meant to represent $\epsilon_{AB}\bar \theta^A\bar\theta^B$, perhaps with a factor of $1/2$ or $i/2$ or whatever is their convention. But these chiral fields should only contain $\theta$ or only $\bar\theta$.
May
2
comment entanglement status of late hawking radiation in AMPS thought experiment
Dear @ChrisHanney, the whole radiation you can see at +infinity results by unitary transformation from the initial state, so it's the same information "transformed". I wouldn't call it "entanglement". One isn't entangled with his own future. The early and late Hawking radiation also come after each other but for them, their entanglement makes sense because different directions generically make an early Hawking particle and a late Hawking particle spacelike-separated. So it makes sense to talk about them as existing at the same moment, and they are entangled.
May
2
answered Basic question about superspace, Grassmann numbers and world sheet supersymmetry
May
1
answered entanglement status of late hawking radiation in AMPS thought experiment
May
1
comment What does it mean for a particle to have spin of 2?
Dear @Al.Ka - to really understand it, you have to comprehend the basics of quantum mechanics. Observables (quantities) like the angular momentum are given by operators... In particular, the angular momentum generates rotations around an axis,and rotation by 4.pi has to be trivial - 2.pi is allowed to change the sign. That's why the spin components are always multiples of hbar/2 or 1/2 in the usual conventions. The spin - without a component - is the maximum value of a component in the multiplet. For massless particles, we must measure th angular momentum componen along the direction of motion
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.