Recent content by jshtok

Many thanks for the reference, I have read the second cited article and was taken away by its clear and enriching exposition.

Thank you very much for the explanations. I think I have a clearer picture now.

Thank you for the detailed explanation. Indeed, if I adjust my earlier analogy, from ## \cos(z) \rightarrow \hat{x},\; \sin(z) \rightarrow \dfrac{1}{m\omega}\hat{p}##, to ## \cos(z) \rightarrow \sqrt{\dfrac{m\omega}{2}}\hat{x},\; \sin(z) \rightarrow \dfrac{1}{\sqrt{2m\omega}}\hat{p}##, and set...

Thank you for your input. Indeed, looking at the Heisenberg picture is a good hint, but the position and momentum operator are both combinations of sine and cosine functions of \omega*t there, unlike the analogy here. You are right, of course, that there are no non-zero commutators in cos(z)...

Hello everyone, I have noticed a striking similarity between expressions for creation/annihilation operators in terms position and momentum operators and trigonometric expressions in terms of exponentials. In the treatment by T. Lancaster and S. Blundell, "Quantum Field Theory for the Gifted...

Check out this material: http://applet-magic.com/relamomentum.htm The 'Relativistic Case' part discusses the Lagrangian of bodies with linear and angular motion, and the preservation of the angular momentum. Goldstein is also mentioned in end notes.Hope this helps, Joseph Shtok

I suggest to use the commutator identities: [AB,C] = A[B,C] + [A,C]B [A,BC]=B[A,C] + [A,B]C (one possible source: http://www.cchem.berkeley.edu/chem120a/extra/commutator.pdf) Thus, [Lx2,Ly2] = [LxLx, Ly2] = Lx[Lx,Ly2]+[Lx,Ly2]Lx Then you compute the [Lx,Ly2], which decomposes as a sum of...

Actually, there is an excellent book by Lee Smolin, named "The Trouble with Physics". On one hand, its prose, not a textbook; on other, it gives the reader an excellent exposition of the development of ideas in physics, starting from Copernicus and Kepler, analysing any significant milestone...

First glance: There are some conditions on A to allow diagonal of D=A'VA be composed of eigenvalues of V. If this is the case, V and D have the same determinant and trace (first is a product of eig.values and the second is the sum). Same determinant implies det(A'VA) = det(V), which means...

I am a bit rusty on vector fields, but I bielive the situation is as follows: (1) Notice that in every point r=(x,y,z) in space the field F(r) is a vector along the ray coming from the center of coordinate through this point. So you can picture the entire field like a blast from singe central...

You are right, I confused the order of inclusion and this also led me to think there is a problem with f=0. Anyway, here is a more rigor version of my proof: if g is not identically zero, then both kernels are of the same dimension and therefore they are the same space. So, the kernel...

You have confused the order of inclusion in your solution. You have K(g)\subset K(f), where K(g) is the kernel, or nullspace, of the linear functional g (that's how elements in Hom (V,F) are called in the daily life). Now, notice that the statement is wrong if f is identically zero; so we must...

brydustin, the question concern matrices over the finite field \mathbb{Z}_r, which has r elements. Hence the finite count.

Please explain this to me. Take any basis B in which the matrix X has certain determinant value d. Now multiply all vectors in B by 2. That will cause all the representations of X_i to become twice as small, which will cause a coefficient of 2^{-n} to the determinant.

I believe you do mean to count the matrices over the field \mathbb{Z}_r. The count is done as follows: group A of matrices are of form [a,b; c,d] with a\neq 0 and d=\dfrac{bc}{a}. This is (r-1)r^2=r^3-r^2 matrices. Groups B, C are of form [0,b;0,d] and [0,0;c,d] - each has r^2 members...