Recent content by Oxvillian

I stumbled on this thread when struggling with and then googling the same issue - sorry to be 4 years late! I think we can resolve things by treating the solid angle subtended as a variable that is only defined modulo 4\pi. The decision to "reset" it from 2\pi to -2\pi as we cross through...

Ha well it's all just words of course :smile: But in an effort to convince you, I put it to you that the number of degrees of freedom of a system should be a kinematical characterization, independent of dynamics, and therefore independent of how many quadratic terms there happen to be in the...

I wouldn't call those degrees of freedom though. The correct statement of the equipartition theorem would be something along the lines that you get a kT/2 contribution to the heat capacity for every term that appears quadratically in the Hamiltonian, rather than for every degree of freedom.

Seems like you guys are making this one way too hard. Why not just work in the \sigma_x basis?

\mathbb{C}^2, of course! That's where electron spins live. I totally plead guilty to being a sloppy hand waving physicist. :oops:

Ha it's really not that bad! See for instance the chapter on addition of angular momentum from Shankar's book, where the addition of 2 spins is done like this. It's also kind of necessary to think this way when we block diagonalize the 4x4 matrix and extract the irreducible representations...

Hi fresh_42! I'm not sure which part you're objecting to - can you explain? The tensor product in question is just the product of the two original vector spaces - objects in that space are most certainly vectors! Now you're certainly free to arrange the components of those vectors in the rows...

Hi gnieddu! I think maybe the confusion here evaporates once you realize that the tensor product of two vectors is still a vector, and not a matrix. I would suggest writing \begin{bmatrix}a\\b\end{bmatrix}\otimes \begin{bmatrix}c\\d\end{bmatrix}=\begin{bmatrix}ac\\ad\\bc\\bd\end{bmatrix}. When...

Hi Kara386 First observation here is that you need some big brackets in your equations - for example in your first equation, the k_B should be multiplying everything, not just the first term. The business with the g_i is just bookkeeping. In their formula they're using the index i to label...

Well, the result simply isn't true for a general state, so it won't be possible to get it from the commutation properties of the operators alone. :smile: One way or another, the result follows from the rotational symmetry of the L_z eigenstates. Roughly speaking, the L_z eigenstates don't care...

Oh I see, you're supposed to "use the commutation relations" o_O Well, I guess the properties of the ladder operators follow from the angular momentum commutation relations, so you're kind of using them indirectly... Or you could "use the commutation relations" to show that L_y \; = \...

whatisreality - I fear you're not going in the correct direction here :frown: What do you have against ladder operators? If you need to use L_x and L_y in the L_z basis, expressing them in terms of ladder operators is usually the way to go. And I promise you it will work!

Indeed, the whole active/passive transformation business can always be relied upon to generate confusion. ?:)

Sure - if you define the wavefunction in this nice picture-independent way, then it goes as e^{-i\omega t}. That is, it evolves "forwards" in time. And indeed in this post you said the correct thing about the Heisenberg picture position eigenstates, which evolve "backwards" in time: But now...

Notice also that the Heisenberg eigenstates you're talking about evolve the other way: in the above \exp(-iEt) should be replaced by \exp(iEt).