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Hi, I am a string theorist and a publicist.


Feb
28
awarded  Nice Answer
Feb
24
awarded  Enlightened
Feb
24
awarded  Nice Answer
Feb
21
awarded  Nice Answer
Feb
20
comment Is the wave-particle duality a real duality?
@StanShunpike - the wave-particle duality has never been a "problem", it's been a defining feature and virtue of quantum mechanics (or "quantum theory") since the beginning. Even non-relativistic quantum mechanics describes the objects so that they have both wave-like and particle-like properties. Quantum field theory does the same - while it's also easier to describe all the collections of particles as a "quantization of a classical field" (or "of classical waves"). But when it comes to basic conceptual properties like the co-existence of wave-like and par-like properties, QFT changes nothing
Feb
19
comment How to get Planck length
No, it doesn't sound like astrology (it sounds like fundamental and demonstrable science) and your question should have contained "doesn't it" rather than "isn't it".
Feb
17
awarded  Enlightened
Feb
17
awarded  Nice Answer
Feb
14
comment Why quantum electrodynamics?
Something is traveling in between the source and the other charge that reacts. It's the perturbations of the fields. If the charges are static (time-independent), then the fields inside are also static, so they appear not moving. But if the charged source starts to wiggle, there are literally electromagnetic waves propagating in between, by the speed of light. In QED, electromagnetic waves are reinterpreted as a collection of an integer number of photons, they also move by the speed of light.
Feb
14
comment Why quantum electrodynamics?
Dear @Primeczar, good if something has been englightening. The time delay - already existing in classical electrodynamics - is the time needed for the wave to get from one place to another. If you say something, the sound also spreads by a finite speed so there is a time delay. If the charged source of the field is static, you can't say at which moment the charge created the fields we feel now. But when things are moving, the answer is just one - it was after a delay $\Delta t = \Delta x/c$.
Feb
14
comment Why quantum electrodynamics?
E.g. you may define the field $H(x,y,z)$ which is the average of $|\vec E|^2$ in a ball of radius 1 meter around $x,y,z$. The equations describing the evolution of $H$ may look nonlocal and allow superluminal propagation of the signal but it's because $H$ is "complicated" in terms of the more elementary fields - and in terms of the elementary ones, the interactions may be seen to be local (the action is never faster than the speed of light).
Feb
14
comment Why quantum electrodynamics?
An unphysical degree of freedom, like the electrostatic potential, is one that is used in the equations we use to describe physics but that can't be measured. For example, the potential and the whole vector $A_\mu$ is unphysical because of gauge invariance, and only some derivatives like $F_{\mu\nu}$ are physical. A nonlocal variable is a functional of fields that depends on values of fields at many points in the space or spacetime, whose distance is strictly positive.
Feb
14
comment Locally every force admits a potential?
Yes, in actual Nature, friction (to the leading approximation) may be reduced to statistics of lots of interactions that are ultimately electrostatic i.e. conservative. But the more general claim isn't right. The expression for the total energy may be rather complicated and not just $p^2/2m+V$ where $V$ is a potential energy. There may be (and in general, there are, even in Nature) various more complicated functions and products of coordinates and momenta (or coordinates' derivatives, velocities etc.) and any sufficiently general form of the energy means that the forces are non-conservative.
Feb
14
comment Why quantum electrodynamics?
Electrodynamics (just like general relativity) is compatible with special relativity so no information or impulse or true matter may propagate faster than light or instantaneously, and quantum electrodynamics doesn't change anything about this fact. There are various descriptions in classical and quantum electrodynamics where it may "look" that something is instantaneous but this is always a result of using nonlocal or unphysical degrees of freedom. Elmg and gravitational waves/impulses are propagating exactly at the speed of light in the vacuum, as the equations show (both in class+quantum).
Feb
13
comment Why quantum electrodynamics?
Did you mean interaction between charged particles? Even in classical electrodynamics, CED, the interactions are due to the field in between. This force in may be described in QED using virtual particles, and different gauges, etc., this makes no sense to explain this aspect of QED in isolation from a comprehensive introduction to QED as a whole, sorry.
Feb
13
answered Why quantum electrodynamics?
Feb
8
comment Why are atoms of the same element exactly the same?
If the atoms are in the same excitation state, e.g. ground states, the wave function is exactly symmetric or antisymmetric (depending on whether the number of elementary fermion in each of them is even or odd). If there are two atoms that are in different excitation states, they effectively behave as distinguishable particles, so their wave function is neither symmetric nor antisymmetric in general. However, one may always describe the different levels as a multiplet of one particle, and then the full wave function is still a symmetric/antisymmetric wave function, now a matrix.
Feb
8
comment Why are atoms of the same element exactly the same?
Dear @BjornW, if you have two magnesium atoms, each of them has 12 electrons and the accurate description of the situation is in terms of a wave function of positions of 24 electrons (plus two nuclei) and this wave function is perfectly antisymmetric under the $S_{24}$ permutation group of the 24 electrons. Alternatively, you may overlook the exact internal structure of the atoms. Then you have a wave function that is a function of locations of 2 atoms.
Feb
7
comment Where does the $\gamma_5$ here come from?
Hi, I am confident that the equations you wrote down - if nothing is added - can't lead to any inconsistency. If you can derive an inconsistency from the equations in the book only, it's bad. But if you add any extra guesses about $\psi_L$ etc. that I mentioned, there may be (independent) signs everywhere so you must be careful what's the correct sign at each place. But even if there are real sign errors in the book, you should still be able to "do the same thing right", i.e. write a similar self-consistent set of equations without errors.
Feb
7
comment Why are atoms of the same element exactly the same?
You may use approximate Hamiltonians or treatments where the perfect antisymmetry or perfect symmetry is violated but they're always at most approximations or fudges. The exact treatment doesn't allow any deviation from the perfect symmetry or antisymmetry.