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


Nov
26
answered Hidden observers in Double Slit experiments - Do they matter?
Nov
26
comment Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
Also, it's true that it would be "useless" to define the operator to be the same because then you would have no nontrivial dynamical equations that could predict the future. That's what you are probably saying. However, even if it would be useless, one may still ask whether it would be legitimate. I think the answer is No, see e.g. Qmechanic's answer. $[H,t]=0$ is different from the nonzero $\partial_t,t=1$ so the operators can't be the same. ($\partial_t$ isn't really an operator acting on the states like $\psi(x,y,z)$ but "extended" states $\psi(x,y,z,t)$ etc.: another "distortion".)
Nov
26
comment Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
Right, your point understood, Hans. When one says "Hamiltonian is by definition XY", there are different angles what the "definition" may mean. Of course, one may constructively define the Hamiltonian for particular systems, like $p^2/2m+V(x)$, in which case it's this expression by definition. More generally, we want to define it as whatever is needed for Schrödinger's or Heisenberg's equations to hold. The latter approach is more general. Still, when we say "the equation holds", it is not the same thing as saying "the operators are the same" because $\psi$ can't be canceled.
Nov
26
comment Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
+1, good point about $[H,t]=0$ etc.
Nov
26
comment Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
Dear Hans, $\partial /\partial t$ is surely not a "Hamiltonian by definition". It is by definition the limit of $[object(t+dt)-object(t)] / dt$ in the limit $dt\to 0$. Schrödinger's equation is only valid for state vectors satisfying the right dynamical equations - not for all objects and not even for time-dependent elements of the Hilbert space in physics - and it is a nontrivial law of physics (a constraint), not a "definition" of anything. You could perhaps say the opposite thing, the Hamiltonian is by definition the operator generating time translations - but "is" isn't symmetric here.
Nov
26
answered Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
Nov
24
comment Is this a simple Lie algebra?
I agree. The first commutator verifies that $\sigma_a$ are Pauli matrices. $\eta_1$ isn't a Pauli matrix, it is ${\rm diag}(+1,-1)$, as implicitly seen from the last two equations. So one deals with $4\times 4$ block diagonal matrices with a combination of Pauli matrices in each $2\times 2$ block, and that's clearly $SU(2) \times SU(2)$, with your right generalization. Just to answer the OP's question, it is a product so it is not simple.
Nov
24
comment Non-minimal coupling of electromagnetic field
For example, the covariant derivative $\nabla_\mu V^\mu$ of a vector may be written as a partial derivative $1/\sqrt{g} \partial_\mu (V^\mu \sqrt{g})$ - there should be $-g$ everywhere in the square roots, right? - which explains the first identity for the box of the scalar. A few more are needed to deal with the curvature etc. You will also have to learn why the Riemann curvature tensor is encoded in the "commutator" of two covariant derivatives as acting on a vector etc. Many identities but it's just maths.
Nov
24
comment Non-minimal coupling of electromagnetic field
This is straightforward maths - mathematical identities - and maybe it doesn't really belong to Physics SE. You should first try to understand why your first equation of motion for the scalar should have been written intelligently as $\nabla^\mu\nabla_\mu \phi$ in terms of covariant derivatives. This will require you to learn how to raise and lower indices by the metric tensor (you avoided this notation, suggesting you don't know this basic point), the Christoffel symbol, covariant derivatives, and various identities how to translate between those things.
Nov
18
awarded  Peer Pressure
Nov
18
comment What is the rationale behind representing a state function by a complex valued function in QM?
The commutator $[x,p]=i\hbar$ contains an $i$ - and it must contain an $i$ because the commutator of two Hermitian operators is anti-Hermitian - so it can't be represented by real operators on a real vector space. Equivalently, $\cos(kx)$ instead of $\exp(ikx)$ wouldn't remember whether one is moving left or right, or the sign of energy etc. One couldn't write first-order oscillating equations without $i$. See also Why complex numbers are fundamental in physics, motls.blogspot.com/2010/08/…
Nov
17
answered Inertial Mass of a scalar field
Nov
17
revised Is the spring constant k changed when you divide a spring into parts?
added 2 characters in body
Nov
16
answered Is the spring constant k changed when you divide a spring into parts?
Nov
16
answered Are there irreducible tensors of half integral degree in quantum mechanics?
Nov
16
comment Distinguishing two quantum states practically
Dear @Pratik, thanks for your interest. Quantum measurements end up being mundane things. For example $|0\rangle$ and $|1\rangle$ above refer to the electron with spin up or down. You send the electron into a magnetic field that deflects it up or down depending on the spin (the electron is a small magnet), and then you detect it BEEP up or down (after a second, or any amount of time). That's the measurement - a sort of destructive one. Most quantum measurements are "destructive"; to say the least, every measurement in quantum mechanics changes something about the state of the measured object.
Nov
16
comment Distinguishing two quantum states practically
Dear @lurscher, "yes, but then you have the problem..." - I personally don't have any problem related to these words. If you can't produce the exact same experimental situation arbitrarily many times, then you can't accurately measure any properties of the wave function describing the situation. The wave function simply isn't an observable, in the technical - and also informal - sense. This proposition isn't a "problem"; instead it is one of the "most fundamental postulates and deepest insights of [quantum] physics". At most, it may be someone's psychological problems - see Psychology SE.
Nov
16
comment Distinguishing two quantum states practically
Dear @Pratik, there's obviously no "general" prescription for how long time a measurement takes. Different experiments are different. Many of them can be fast but it is obviously nonsense to invent a general duration, like 2 nanoseconds, for "every" experiment.
Nov
16
comment What is the physical meaning of the terms in the multipole expansion?
Just to be sure, the dipole moment is called the dipole because the simplest way to realize a charge distribution with a nonzero dipole moment - but vanishing all other moments - is a pair (therefore "di") of nearby charges. Similarly for quadrupoles, octupoles etc. (powers of two). However, that doesn't mean that the number of charges always has to be a power of two (and they have to be pointlike): almost any charge distribution carries almost all the multipole moments. Just trust the formulae instead of words, and don't misinterpret the words.
Nov
16
comment What is the physical meaning of the terms in the multipole expansion?
In particular, it's not true that everything that isn't of the form of "the exact naive children's dipole" carries a vanishing dipole moment - which seems to be the (incorrect) assumption in Revo's question. @Revo, try this analogy: this assumption is analogous to saying that one kilogram is the mass of the platinum prototype in France, or whatever it is. Now, your reasoning is that the mass of a person has to be zero because the person isn't even made of platinum: she can't be the platinum stick. This sounds like a joke but your logic applied to the dipole moment is exactly the same.