Recent content by gitano

I was trying to compute the time derivative of the following expression: \mathbf{p_k} = \sum_i e_{ki}\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)!} \mathbf{r_{ki}}(\mathbf{r_{ki}\cdot \nabla})^n \delta(\mathbf{R_k}-\mathbf{R}) I am following deGroot in his Foundations of Electrodynamics. He says...

After more consideration, I'm beginning to think that z|lm\rangle is not an eigenstate of L^{2}. Even though [L_{z},L^{2}] = 0 and share |lm\rangle as eigenstates, since they are not a complete set of commuting observables, all the eigenstates of one are not necessarily eigenstates of the...

I was just thinking, since z|lm\rangle is an eigenstate of L_{z} and has the same eigenvalue as |lm\rangle , does this degeneracy mean that z|lm\rangle is also an eigenstate of L^{2} with the same eigenvalue as |lm\rangle , namely \hbar^{2}l(l+1) ? In general, if an observable...

Sorry guys I changed the problem to correctly read: Is z|lm> an eigenstate of L^2, not of L_z.

By the way, how do you get the equations to be inline with the text?

Homework Statement Is z|lm\rangle an eigenstate of L^{2} ? If so, find the eigenvalue.Homework Equations L_{z}|lm\rangle = \hbar m|lm\rangle L^{2}|lm\rangle = \hbar^{2} l(l+1)|lm\rangleThe Attempt at a Solution So since L_{z} and L^{2} are commuting observables, they have are...

So the fact that you are cutting off the taylor expansion at the second term is ok because you are only dealing with an infinitesimal change in \langle x' | \alpha \rangle ? So it is still exact and not an approximation?

Hi, At one point in Chapter 1 of Sakurai he is deriving the momentum operator in the position basis - I just don't see how he makes some of the mathematical leaps (at least leaps for me) between the following expressions \int dx' |x' + \Delta x'> < x' | \alpha > = \int dx' | x' > < x' -...

Hi, I was reading through Jackson's Electrodynamics trying to reason through example 5.5 for the vector potential of a circular current loop of radius a centered at the origin. I pretty much understand everything except when he defines the current density as J_{\phi} = I\sin \theta'...

I am also curious that if in general you can express a generalized coordinate in terms of a constant of motion (the coordinate is cyclic) is it also wrong then to make such a substitution in the Lagrangian as is incorrect in the Kepler problem.

Hi, I have a question that has been bothering me for a while now. For Kepler's problem we know that angular momentum M_{z} is conserved and that the angular velocity \frac{d\phi}{dt} is equal to \frac{M_{z}}{mr^{2}}. When we substitute for \frac{d\phi}{dt} in the expression for energy, we...