Some new progress:
I still didn't find way to normalise this.. Perhaps problem doesn't require it idk, and I am analyze this equation:
$$
\alpha \triangleq\sqrt{E} \ \
\gamma \triangleq\sqrt{\frac{2m}{\hbar^2}}L \ \
tan(\gamma\alpha)=-\sqrt{\frac{\alpha^2}{V-\alpha^2}}\ \ (\gamma,V\ \ are\ \...
So I have come up with my solution(attempt) which is:
where
(
$$\psi_ 1 \triangleq Asin(kx),0<x<L$$
$$\psi_ 2 \triangleq Be^{-sx},x>L$$
$$k \triangleq \sqrt{\frac{2mE}{\hbar^2}} $$
$$s \triangleq \sqrt{\frac{2m(V-E)}{\hbar^2}} $$)
But this has a serious problem about boundary: I think...
Also, may I ask that, if potential V0 > 0 and it's a constant, so when the particle with E > V0 pass this potential barrier, the wave amplitude doesn't change but the wavenumber decrease right(Some energy convert to potential energy?) and where the energy transfer to?
So, it is trying to solve this eqs right, there are two variables , alpha and k, all related to E , so when this stands, there will be a valid E. the question is , the valid E for both of them are not stand together, how can there be "combined solutions"?
So I think I use the right approach and I get uncertainty like this:
And it's interval irrelevant(ofc),
So what kind of wave function gives us \h_bar / 2 ? I guess a normal curve? if so, why is normal curve could be? if not then what's kind of wave function can reach the lower bound
So I don't understand "due north from a position at colatitude ##\theta## " , whether how I translate it...
I keep getting that direction should be radial...(toward Earth CM)
This is my work:
Thank you so much!