Recent content by Zack K

Starting with finding the probability of getting one of the states will make finding the other trivial, as the sum of their probabilities would be 1. Some confusion came because I never represented the states ##|\pm \textbf{z}\rangle## as a superposition of other states, but I guess you would...

It seems a semester of vector calculus had me fixated on the definition of a conservative field. Scalars are in place for gravitational and electric fields to distinct them from each other (or some better wording than what I said).

I seem to have forgotten that g was associated with a vector field, thank you. So it is written as ##\textbf{F}=m\textbf{g}= -m\nabla u##. I have never seen this expression. So are you saying ##u= \lim_{m \to 0} \frac{U}{m}##(U as potential energy)? Could you point me to where I can find this...

If you are asking how your coordinates transformed, then you do a matrix multiplication of that with your (x,y,z) column vector and see the result.

If we have a conservative vector field, then we can describe it as ##\textbf{F}=\nabla\phi## where ##\phi## is some potential. This here is the derivation of Newtons law of gravity: Where ##\nabla u## is the gravitational potential. If we were to ignore it as a gravitational field, why is it...

I should have clarified. I know ##dS= r^2sin\theta d\theta d\phi##. I want to be able to "interpret" it as ##d\textbf S=(dr e_r,rd\theta e_\theta,r\sin\theta d\phi e_\phi)##, or some equivalent form (I know that is not how you interpret it as it is not a Cartesian vector), so that I can evaluate...

Thank you, I've learned quite a bit from this thread. I think I can finish from here on.

That was a typo. Evaluating the right hand side: ##\int_0^aB_n\sqrt{\frac{2}{a}}^2sin^2(\frac{n\pi x}{a})dx = B_n\frac{(2\pi n - sin(2\pi n))}{2\pi n}##. If calculated right, will let me solve for ##B_n##

I see. So all the terms on the right except for what is ##\psi_n## will go to zero? So then $$\int_0^aAx(a-x)\psi_ndx=\int_0^aB^2_n\psi^2_ndx$$

I guess I could get the coefficients by using ##\int_0^a \psi(x,0)\psi_ndx=0## because the functions would be orthogonal. If this is true, still how would I relate the coefficients to the energy and energy probability?

I remember that two functions are orthogonal if ##\int_a^b f(x)g(x)dx=0##. So then do the eigenfunctions form a basis of funtions?

It was said that : ##\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x)## So expanding the first three n terms: ##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3##. Where I am assuming ##\psi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})##. This is the relation that I do not know. How is ##E_n=\frac{n^2...

I'm not sure how to find the probability of being in an energy state. Are we using the real valued eigenfunction for n=1 then finding its probability distribution?

##\langle E_1\rangle=\int_{0}^a A_1\psi^*_1\hat HA_1\psi_1 dx## (I think?).

"Calculate the probability that the measurement of the energy would yield the value ##E_n##". If it helps,the energy level question I was referring to is "in particular, calculate the numerical values of the probability to measure ##E_1,E_2## or ##E_3##"