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Nov
18
comment How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?
that clears up where my confusion came from - what I was aiming for was $\langle p|\hat x|p'\rangle = -i\frac{\partial}{\partial p'}\delta(p'-p)$, which has a negative sign that gets re-absorbed by partial integration; I also don't know if it's fine as-is, ie under what assumptions $\lim_{\epsilon\to0}\int\frac{\partial}{\partial p}\delta_\epsilon(p-p') \psi(p') dp' = \frac{\partial}{\partial p}\lim_{\epsilon\to0}\int\delta_\epsilon(p-p') \psi(p') dp'$ holds...
Nov
18
comment How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?
conceptionally, the derivative of the $\delta$ function gets applied by partial integration (see here), so there's a minus sign missing in that expression; I tried to figure out why we still get the correct result, but apparently I botched that; I'll go through the equations again and see if I can't figure it out...
Nov
18
comment How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?
I think there are two sign errors which cancel each other: if I'm not mistaken, there's a minus sign missing when introducing $\frac{\partial}{\partial p}$ in the expression for $\langle p|\hat x|p'\rangle$, which gets 'fixed' by the missing minus sign when applying the derivative of the $\delta$ function
Nov
16
revised Why do Maxwell's equations contain each of a scalar, vector, pseudovector and pseudoscalar equation?
spelling
Nov
15
revised Why do Maxwell's equations contain each of a scalar, vector, pseudovector and pseudoscalar equation?
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Nov
15
comment Maxwell's Equations using Differential Forms
see also this answer, which would be appropriate here as well
Nov
15
answered Why do Maxwell's equations contain each of a scalar, vector, pseudovector and pseudoscalar equation?
Nov
14
answered Maxwell's Equations using Differential Forms
Nov
14
comment Maxwell's Equations using Differential Forms
@Danu: only from a 3-space perspective
Nov
14
comment Maxwell's Equations using Differential Forms
probably the most economic way to see what's going on is to write the field tensor as $F=E_idx^i\wedge dt+\star(B_idx^i\wedge dt)$ (sign?) and figure out how $d$ and $d\star$ act on 'fake 3-vectors' $f_idx^i\wedge dt$; $d$ should turn out to be the curl, $d\star$ should end up with the time derivatives in the spatial components and the divergence in the time component
Nov
14
comment Maxwell's Equations using Differential Forms
you're missing that we're also changing dimensions from 3 to 4; you can of course show that it all works out by computing $dF$ and $d\star F$ in a local basis of the exterior algebra, but there's probably a nicer way to show this...
Nov
14
revised Electromagnetic Field Tensor via Tensor Products?
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Nov
14
answered Electromagnetic Field Tensor via Tensor Products?
Nov
13
comment Electromagnetic Field Tensor via Tensor Products?
note that $F_{\mu\nu}$ can still be considered a linear transformation: ${\rm T}^*M\otimes{\rm T}^*M\cong{\rm Hom}({\rm T}M,{\rm T}^*M)$; if you accept the general premise (velocity as vector, momentum as covector), mass would be promoted from a simple number to a map ${\rm T}M\to{\rm T}^*M$ - in this case it's just a multiple of the metric, in the case of Lagrangian mechanics you'd use the fiber derivative; to fully geometrize Newtonian mechanics, you need additional structure - eq $(1)$ as-is is not well-defined from a differential-geometric perspective
Nov
13
comment Electromagnetic Field Tensor via Tensor Products?
the lower indices of $F$ are natural because that's what you get from gauge theory - in general, it's a Lie-algebra-valued two-form (and becomes a regular one after contraction with a generalized charge, a co-adjoint orbit); to see that the lower index for $p$ is natural, look at the Legendre-transformation from tangent (ie velocity phase space) to cotangent (ie momentum phase space) when going from the Lagrangian to the Hamiltonian formulation; also, minimal coupling and contraction with velocities to get an energy suggest that momentum should be a covector
Nov
13
comment Electromagnetic Field Tensor via Tensor Products?
note that the most 'natural' placement of indices is $p_\mu$ and $F_{\mu\nu}$
Nov
11
answered Motivation for form of Lagrangian
Nov
6
comment Wave function interaction
an accurate treatment needs QED, but the approximate Breit equation - in particular the decomposition of the full Hamiltonian - might be more in line with what you're looking for
Nov
6
comment Wave function interaction
wave functions describe systems as a whole, not individual particles
Oct
31
comment Photon “stuck” on the event horizon of a black hole
@JerrySchirmer: for what it's worth, the fact that massless particles cannot decay makes sense from a classical (relativistic) point of view: quantum-mechanical decay is random and has a finite probability to happen in a certain time interval; now, think about proper time for massless particles...