Recent content by t0pquark

Thank you so much for your help!

Starting with ##tr(A * log(A)) = tr(diag(a_1, ..., a_n)*diag(log(a_1), ..., log(a_n))) = tr(diag(a_1*log(a_1), ..., a_n*log(a_n))) = a_1*log(a_1) + ... + a_n*log(a_n)##??

Ahh, so then I would have ##tr( \rho * log(\rho)) = tr(UAU^\dagger*log(UAU^\dagger)) = tr(UAU^\dagger*U (log(A))U^\dagger) = tr(UA (log(A))U^\dagger) = tr(A *log(A))## ?

Well, it would be easiest if it was already diagonalized; just take the logarithm of the entries on the diagonal. If not, I guess I could either diagonalize it like above or write it as a sum of projections and then try to do something with that?

Yikes, you're right there So I'm back to ##tr( \rho*log(\rho)) = tr( UAU^\dagger*log(UAU^\dagger))##?

Not sure I see what you are talking about? I'm going to try writing out what I know here. Applying the spectral theorem to ##\rho##, we can get something i.e. ##\rho = U A U^\dagger ##, where ## A ## is a diagonal matrix. If this goes into the original equation: ##tr(\rho*log( \rho )) = tr(U A...

I just looked it up, and I guess the spectral theorem says that any Hermitian matrix is diagonalizable (so it would be true for the density operator). The trace of the density operator is 1, so the sum of these entries on the diagonal would be 1. Also, if the density operator was a diagonal...

I feel like I'm going around in circles trying to do something with the expression ## tr( \rho *log(\rho)) ##. I thought about a Taylor expansion, but I don't think there's a useful one here because of the logarithm. We learned the Jacobi's formula in class, but I don't think I want a derivative...

I think this is where I'm confused. My logic was that the total angular momentum must be 2 and the total spin angular momentum must be 1/2, so I would want the 2 x 1/2 section on the Clebsch-Gordan table, then I find the results for ## \vert J = \frac{5}{2}, J_z = \frac{1}{2} ## and ## \vert J...

Oooops that was definitely me copying incorrectly. The Hamiltonian should be ## H = E (J^2-L^2-S^2) + \frac{\hbar e B}{2mc}(L_z+2S_z)## Okay so starting with the Clebsch-Gordon table I want to use the section with 2 x 1/2?? I think I would get ## \vert J = \frac{5}{2}, J_z=\frac{1}{2} \rangle =...

I am pretty confused where to even start with this question, which is not a good thing less than a week before the final :(. One thing in particular that I don't get is that I thought we were using the Clebsch-Gordon coefficients for ##\vert jm \rangle ## states, not for ##\vert J, J_z \rangle...

That explanation helps! So then I have ##( \begin{bmatrix} \frac{1}{ \sqrt 2} \\ 0 \\ \frac{-1}{ \sqrt 2} \end{bmatrix} \cdot \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{-2}{3} \end{bmatrix} )^2 = (\frac{2}{3 \sqrt 2} + \frac{2}{3 \sqrt 2})^2 = \frac{8}{9}##, right?

The probability that a measurement of ##L_x## will give zero for a given ##\psi## should be ##\vert \langle L_x = 0 \vert \psi \rangle \vert ^2##, I think. I found the eigenvalues of ##L_x## to be ##\lambda = -1, 0, 1##. Since it asks for the probability that the measurement will give zero, I...

I'm working through https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_06.pdf, and I'm stumped how they got from Equation 5.26 (##\vert 0_{\gamma} \rangle \equiv \frac{1}{\sqrt{cosh\gamma}} exp(-\frac{1}{2}tanh\gamma \hat{a^\dagger}\hat{a^\dagger}...