# Recent content by Happiness

1. ### B Is gravity a force?

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...
2. ### B Is gravity a force?

Could you explain further? If light is massless, then its motion should be unaffected by Newtonian gravity.
3. ### B Is gravity a force?

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...
4. ### I Why is p^4 not Hermitian?

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...
5. ### I Why is p^4 not Hermitian?

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...
6. ### I Why is p^4 not Hermitian?

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.
7. ### I Why is p^4 not Hermitian?

Does this mean that a measurement of ##p^2## is only approximately real valued? What does that even mean?
8. ### I Why is p^4 not Hermitian?

##p^2## is hermitian then. The contradiction still persists.
9. ### I Why is p^4 not Hermitian?

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?
10. ### I Why is p^4 not Hermitian?

c(r) has the factor ##r^2##, so it is exactly 0 at r=0.
11. ### I Why is p^4 not Hermitian?

Everything involved is always differentiable. Taking ##f=\psi_{200}=(2-\frac{r}{a})e^{-r/2a}## and ignoring all constant factors, $$r^2\frac{df}{dr}=(4r^2-\frac{r^3}{a})e^{-r/2a}$$ $$\frac{1}{r^2}\frac{d}{dr}r^2\frac{df}{dr}=(\frac{r}{a}-\frac{2}{a}+\frac{16}{r})e^{-r/2a}$$
12. ### I Why is p^4 not Hermitian?

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?
13. ### I Why is p^4 not Hermitian?

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}##.
14. ### I Why is p^4 not Hermitian?

Yes they are since hydrogen radial wave functions all have the ##e^{-r}## term.
15. ### I Why is p^4 not Hermitian?

##p^2## is hermitian does indeed hold for all states, including ##l=0##, but not ##p^4##.