Recent content by CalcYouLater

Are different electrical frequencies allocated to specific applications because of their properties or to regulate the spectrum so that each assigned frequency range is unique to one entity? I guess a related question is: Are we going higher in frequency ranges for things like automobile...

This is excellent advice. All of these points should be considered, especially numbers 1&2 in my opinion.

Thanks for the reply. I feel that I have a grandiose idea of what a first job would be like. I realize that it is silly to day-dream about my first job in that way, and it is helpful to read your post. While I would like to believe the same, I do not have any experience in the job market...

Thanks for the advice. I can certainly agree with your post.

I graduated with a BA in physics in December 2012 in the United States and am currently searching for a job. I have a connection at a company that manufactures bearings, seals, and rings among other things for a variety of industries. Some examples of these industries are: medical systems...

What follows assumes you are in America. I think you are much better off than you think. Searching for a job with degrees in math and physics can be a bit over-whelming at first. The more you search and read job descriptions, the more you learn how to tailor your resumes and cover letters to...

Thanks a lot for all the help.

Ok, and the reason it is not an eigenstate is because the first term (m=0) vanishes and the act of the operator does not result in a multiple of the original state.

Thanks very much vela. In a related issue, I just came across an expression that I do not understand involving the angular momentum operator Lz. I hope it is a typo in the book. Here it is along with the sentence directly below it...

Homework Statement I have some questions about the eigenstates of an operator. A state is an eigenstate of an operator if the application of the operator on the state results in a constant complex multiple of the state. The constant complex multiple will be the eigenvalue. For instance...

Wow, thanks again for all the help. I feel as though I have a better grasp on the idea now.

Thank you both again for the help. I think I get it now. I calculated the probability per unit volume for the wave function:\psi_{2,1,-1}(r,\theta,\phi)=\frac{1}{\sqrt{24{a_0}^{5}}}r{e^{\frac{-r}{2{a_0}}}}Y_{1,-1}(\theta,\phi) at (a_0,\frac{\pi}{4},\frac{\pi}{3}) and I got a very large...

Ok, that makes sense. I should have thought about what normalization does. So in that case, (assuming the properly normalized wave function above) the probability per unit volume would simply be: a_{0}^{2}e^{-2}sin^2(\frac{\pi}{6}) That leads me to another question. If I then wanted the...

Thank you for the response fzero. So, if I understand correctly, evaluating the square of the wavefunction at a particular point will give the probability of finding the electron at that point. For instance, if the following were a normalized wavefunction...

Homework Statement I have a particular normalized energy eigenstate equation of a hydrogen atom. I am asked to find the probability per unit volume of finding the electron at a specific point. The point is r=a_0, theta=pi/6, and phi=pi/3. Homework Equations The Attempt at a...