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Hi, I am a string theorist and a publicist.


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reviewed Approve suggested edit on Free Body Diagram of Fluid Statics Problem
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Dec
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comment Heisenberg picture saves locality in QM?
Heisenberg picture is physically equivalent to Schrödinger's picture and in relativistic theories, namely quantum field theories and string theory, locality exactly holds in both and can be easily proved (and is a direct consequence of the Lorentz invariance). It's more manifest in the Heisenberg picture, indeed, because the equations are the same as the classical field equations which preserve locality. All comments about "nonlocality" are due to some people's fundamental misunderstanding of quantum mechanics. Locality was proved long before Deutsch learned physics. QFTs are called local QFTs
Nov
27
comment Hidden observers in Double Slit experiments - Do they matter?
Dear Colin K, I am writing the answer not only for the OP but for everyone who asks the same question which may be asked, believe me, even if one has been exposed to double-slit experiment. Fraggle, nothing "causes" probabilities to shrink. Probabilities by definition always describe "shrunk" outcomes. For example, if you throw a dice, the probability distribution is spread over the numbers 1,2,3,4,5,6. It is completely spread. But it doesn't prevent "3" from being the result. In fact, it is guaranteed that one sharp number will be the result if you throw dice.
Nov
26
revised Hidden observers in Double Slit experiments - Do they matter?
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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
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