Recent content by ygolo

This is perhaps a philosophical question, but I am trying to make the transition from engineer to scientist, and I am trying to relearn how to think and ask questions. As an engineer, a lot of times, we can get away with making something that consistently worked without understanding it...

These are generalized coordinates, so p and P don't necessarily have to be linear momentum.

You pretty much have it. Except, I don't believe you calculated the possible l values properly. |1-0|=1, so only l=1 works. Now, all you have to do is use Hund's rules to find the order. This will lead to the answer in the book.

This ought to be simple, I think. But I haven't found a consistent way to think about things yet. Is it as simple as adding up all the spins of the elementary particles in the particle and checking whether the total has inter or half-integer spin? Homework Statement State whether the...

Ah, OK. Thanks for the correction. I was going outside the domain of \theta and \phi and the \phi transformation was just wrong.

Be careful that you are meeting the boundary conditions. I assume you are talking about an infinite square well. If this is the case, the wave functions that you use have to be 0 at x=a, and x=-a. I believe this gives the solutions: u _{n}(x)=\sqrt{\frac{2}{a}}sin ( \frac{n \pi}{2a} (x+a))...

I don't think this is true. Take: |1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\theta)e^{i\phi} P|1 1\rangle = (3/(8\pi))^{(1/2)}sin(\theta)e^{-i\phi} The \phi component screws up the relation.

This is a pretty brute force method. It simply stores all possibilities in the open squares, loops over all the solved squares, eliminating the number in the solved squares from the appropriate open squares, then loops over the open squares to see if they can then be moved the solved square...

First, to understand things better, what, more concretely, is the contents of [a]? Second, don't you still have to worry about what the weightometer reading gives you? In other words, shouldn't the equation be [a]*[tr]+[b]=[cr]. After which you solve for tr with [tr]=[a]^-1([cr]-[b]). Or...

Are you sure you mean the null matrix? Because the null matrix is the matrix with all zero entries. So if A is the null matrix, then I+AB=I=I+BA. I is invertible. So for your theorem, both the hypothesis and conclusion are true (in both directions). Making the theorem vacuously true.

General complex linear operators. --- Silly me. If |A-aI|=0 then A-aI is sigular, and therefore not of full rank, and therefore has a (non trivial) null-space, which means A has an eigenvector. Forgive me, it has been 13 years since I took linear algebra, and 12 years since Complex Analysis.

Well then the proof by the fundamental theorem of algebra falls short of proving the exisence of an eigenvalue then doesn't it? Good point! I was just curious, but I appreciate yours and the others' help. I'll try and decipher this after work.

I wonder why the science advisers and pf mentors have so far not replied this thread. Too basic? I am actually not sure I answered my first question fully, since I did not rigorously prove that the characteristic polynomial is non-constant. Is it? I think its slightly more subtle. Ax=ax...

For every Eigenvalue must there be a non-trivial eigenvector? If so, why? So then the next question becomes: For every Eigenvalue must there be a non-trivial eigenvector? If so, why?