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This issue is a bit confused on textbooks, however the statement of the professor is physically wrong (mathematically all the procedure can be rigorously justified using the theory of distributions). The point is that the claimed position operator is not the position operator because it is not even self-adjoint (nor Hermitian) in the relevant Hilbert space ...


2

Here is the answer (I will not consider the constants on the denominator of your Fourier transform for simplicity, however they are there ;-) ). When you write the operator $\hat{\phi}$ you have to be careful. I will drop the hats, because it will be clearer I think (maybe here the hat stands for an operator and not for the fourier transform). Your operator ...


1

It seems that that you're converting from a volume ($R_0^3$) (is the minus sign a typo?) to a mass($M_0$). The constant of proportionality is a density, presumably the $\rho_0$ that you're given. In more general terms, the constants that are hidden by the $\propto$ symbol can have units too.


1

It would be highly unreasonable to obtain the uncertainty of the average of the measurements to be greater than the uncertainty of any one measurement (√5/2 > 1). After all, what's the point in taking averages if it just makes your readings more uncertain? I believe the ETS people used the argument that the harmonic sum of the individual variances should ...



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