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}##.
I would have thought all measurements, in one way or another, involve microscopic particles and are hence quantum in nature. If there are two kinds of measurements—one, quantum and the other, classical—then how do you tell them apart properly?
I guess you meant this part:
All these I understand, but it does not mention how the condition ##\psi(x+a)=e^{iKa}\psi(x)## is motivated. Be aware that it is only by first assuming this condition to be true that we could get ##\psi(\vec{x}+N a)=e^{iKNa}\psi(\vec{x})##.
If we start by not...
So is it just pure coincidence that the double-slit interference for light and electrons are described by the same equation as the classical water-wave interference?
Could you explain quantum coherence in simple terms?
A water wave of wavelength ##\lambda## should behave as a particle with momentum ##p=\frac{h}{\lambda}##. How can we observe this particle? And how can we collapse the wave function of this particle so that it passes through either the left...
Electrons passing through a double slit is in a superposition of passing through the left slit and the right slit, thereby producing an interference pattern on the screen. But when a detector is placed to detect which slit the electrons pass through, the interference pattern is destroyed.
How...
Since you understood the argument with the Wronskian, could you explain to PeterDonis that it doesn't impose the following condition on ##K##:
##e^{i K a} = e^{- i K a}##?
(Note that for the case of the free particle, K=k. K is defined by the Bloch's condition [5.49], while k is defined by the...
It is not a stationary state. But this is not a counter example because [5.48] is the time-independent Schrodinger solution. There exists an allowed value of k such that a solution to the time-independent Schrodinger equation is not a solution to the Bloch's condition.
Could you explain how the...
There exists an allowed value of k such that a solution to the time-independent Schrodinger equation [5.48] is not a solution to the Bloch's condition [5.49], but Griffiths said (according to @PeterDonis) this is impossible: all solutions to the Schrodinger equation is a solution to the Bloch's...
##\psi_{\lambda_1}## and ##\psi_{\lambda_2}## have the same E but different ##\lambda##, same eigenvalue for H but different ones for D! (Different ones in general: ##\lambda=\pm1## are the exceptions.) This is proven by B&J:
YES!!! YES!!! YES!!! This is what I have been saying several times...
Ok, let's hear the opinion of more people first.
[5.48] is Schrodinger equation. [5.49] is Bloch's theorem or Bloch's condition.
[5.62] is in post #65:
You are right that we need to take into account the Dirac-delta potential at ##x=0## and ##x=a##. But it turns out that [5.59] is still the...