Recent content by McLaren Rulez

Of course. Let's start with the ##[l_2, l_3]## commutator ## \begin{align*} \langle m, l | [l_2^2, l_3^2]| l,m \rangle &= \langle m, l | l_2^2l_3^2 - l_3^2l_2^2| l,m \rangle \\ &= \langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle \\ \end{align*} ## Utilize the...

I feel silly for not trying the obvious. Thank you Dr. Claude.

That's okay, thank you for helping. I guess the idea is to somehow write ##l_1## and ##l_2## in terms of the ladder operators but some basic manipulations haven't really led me anywhere.

That is the full question. I believe it means that the commutator acting on the state ##|l,m\rangle## i.e. ##\langle m, l| [l_i^2, l_j^2] |l, m \rangle = 0##. Of course ##l## and ##m## are the usual quantum numbers corresponding to ##L^2## and ##l_3##

Homework Statement Show that ##|l, m\rangle## for ##l=1## vanishes for the commutator ##[l_i^2, l_j^2]##. Homework Equations ##L^2 = l_1^2 + l_2^2 + l_3^2## and ##[l_i^2,L^2]=0## The Attempt at a Solution I managed to so far prove that ##[l_1^2, l_2^2] = [l_2^2, l_3^2] = [l_3^2, l_1^2]##. I...

I think I see what you mean. You're saying that the entire oscillator system moves with some energy (and this is translational kinetic energy) and the oscillations themselves carry a different and unrelated energy. This makes sense - earlier, I assumed that the kinetic energy you were talking...

How is this different to allocating all 12J to the oscillator at rest and then releasing it? Am I then correct in transforming that to state that the amplitude and the phase are the two degrees of freedom here? I can understand how more energy leads to a larger amplitude but I'm struggling to...

Homework Statement A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of oscillator is A. ½kT B. kT C. ³⁄₂kT D. 3kT E. 6kT Homework Equations Equipartition theorem The Attempt at a Solution So I know the...

Ah I see! Thank you a million DrClaude. You've really helped me finally get it!

Thank you for the detailed document. I think I mostly get it except for how the ##L## value is known from a given ##M_L##. If I look at the microstates, for example, the fifth one in your table ##M_L = 0## but the ##L## value of this can be either ##0, 1## or ##2##. How is it that it is labelled...

Thank you. Sorry to ask more questions but how did you know that ##^3P, ^1D##, and ##^1S## were the available options? In other words, why is it that the way spin adds up also determine ##L##? For example, why can I not have ##S=1## and ##L=0## which is the state ##^3S## in spectroscopic notation?

Thank you for your reply. Can I check what exactly 3P, 1D, and 1S notation means? Sorry, if this is an obvious question but I'm guessing the 3 and the 1 refer to the triplet and the singlet but what are the P, D, and S?

Can anyone explain the second rule, because the Wikipedia page is not very clear? Hund's zeroth rule - Ignore all inner shells and focus on the outermost shell. Hund's first rule - Put the electrons such that they maximize spin, ##s##. So far so good. Hund's second rule appears to be simply...

I'm nearly at the end of this derivation but totally stuck so I'd appreciate a nudge in the right direction Consider a set of N identical but distinguishable particles in a system of energy E. These particles are to be placed in energy levels ##E_i## for ##i = 1, 2 .. r##. Assume that we have...

I meant the stationary states so only energy is allowed