So a photon on the Earth surface experiences an acceleration of 9.81 ##ms^{-2}## too? So its radius of curvature ##r=\frac{c^2}{9.81}\approx10^{16}## m?
But what if it travels straight down towards the Earth? It cannot move faster than ##c##, so its acceleration would have to be zero.
So given...
Gravity can be described not as a force but a curvature of spacetime. I assume this can’t be done to the other 3 fundamental forces. If so, then we cannot treat gravity in a way similar to the other forces. Why then does QFT postulate the existence of gravitons? Why does it attempt to treat...
My conclusion is because I use ##p^4=p^2p^2## to arrive at the error expression for ##p^4##, I cannot later use that expression to argue or conclude that ##p^4\neq p^2p^2##.
The two terms in the middle simplifies to ##r^2(2-\frac{r}{a})e^{-3r/2a}##, ignoring any constant factor.
I used...
This is weird because I get the error expression for ##p^4## in post #7 by assuming ##p^4=(p^2)^2##.
Ignoring the constant factor ##-4\pi\hbar##,
##\langle f|p^2(p^2g) \rangle=\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} - r^2\hat{p}^2g\dfrac{df}{dr} \right)\right|_0^\infty+\langle p^2f|p^2g...
For ##g=\psi_{100}## and ##f=\psi_{200}##,
##\left.\left( r^2 f\dfrac{dg}{dr} - r^2g\dfrac{df}{dr} \right)\right|_0^\infty=\left.\left(\frac{1}{8\sqrt{2}\pi a^5}r^3e^{-3r/2a}\right)\right|_0^\infty=0##, where ##a## is a constant.
All operators for observables must be hermitian. If ##\hat{p}^4## is not hermitian, then what would you obtain when you measure ##p^4## or ##E^2##? Would you get complex-valued measurements? What would it mean?
Everything involved is always differentiable.
Taking ##f=\psi_{200}=(2-\frac{r}{a})e^{-r/2a}## and ignoring all constant factors,
$$r^2\frac{df}{dr}=(4r^2-\frac{r^3}{a})e^{-r/2a}$$
$$\frac{1}{r^2}\frac{d}{dr}r^2\frac{df}{dr}=(\frac{r}{a}-\frac{2}{a}+\frac{16}{r})e^{-r/2a}$$
Yes the boundary term does not vanish, indeed. But how can this be? All the hydrogen radial wave functions are infinitely differentiable. Doesn't it contradict the theorem?
It is, because no matter how many times you differentiate ##e^{-r}##, multiplying and dividing by ##r^2##, in whatever order of these 3 operations, the result is still proportional to ##e^{-r}##.