Recent content by jensa

EDIT: Upon rereading the quote below I realize that I misread/misinterpreted what you wrote and the response doesn't apply to the quote. Still the point I wanted to make stands on it's own, and I vaguely recall you writing about the interpretation of experiments in earlier posts, so I'll leave...

Hurkyl, I am curious to hear how the concept of experiments fits into your view of science. You seem to reject the notion that we can even perceive a definite outcome of measurements so where does this leave experimental physics? For example: In your worldview, is it even possible to...

Depending on how far into philosophy you would like to go I am sure you are right. I think this is more or less exactly what I tried to say but was somewhat sloppy/inarticulate: When I talk of "branches" I refer to the sequence of labels j,j',j'',\ldots which are associated with...

Fredrik/KenG: Sorry to jump into the middle of the discussion. I usually don't participate in interpretational issues but have found your posts on the matter interesting. I am just curious if the following summarizes your viewpoints on the issue (or whether I am missing some points). Let...

To give you some feeling of what's going on I can try to elaborate on what peteratcam and others have said. The analogy with statistical analysis is a very good one and goes a long way in understanding how things work. Let's say you have a set \mathbf{x}=(x_1,x_2,..,x_n)^T of statistical...

Sure. Eq. (3.18a) |\psi_\varphi\rangle=\Pi_k(|u_k|+|v_k|e^{i\varphi}c_{k\uparrow}^*c_{-k\downarrow}^*)|\phi_0\rangle Eq. (3.18b) |\psi_N\rangle=\int_0^{2\pi}d\varphi e^{-iN\varphi/2}\Pi_k(|u_k|+|v_k|e^{i\varphi}c_{k\uparrow}^*c_{-k\downarrow}^*)|\phi_0\rangle = \int_{0}^{2\pi}...

Sorry for not answering for a while. I've been on vacation and didn't check the forum. The equations I refer to are in Tinkhams second edition, I suppose you have the first one. You can probably find the equations using google books or the preview feature at amazon. If not, let me know and I''ll...

Hi IFNT, welcome to the forum. I think most textbooks are a little vague on the subject and I am not sure I have a consistent picture of this stuff myself. However I can try to explain what I think Tinkham refers to. 1. Not sure what reduced Hamiltonian means in this context. Could you...

I'm not really qualified to answer your questions but maybe some references might help you on your way. If you are quite familiar with the functional integral approach then I think this is the more modern approach to dealing with superconductors in a gauge invariant manner as well as...

Don't know where this expression comes from or why it is useful but it's quite simple to show that it is true. If the density matrix is normalized to unity then you should have \rho_{11}+\rho_{22}=1. Then in your equation you just substitute 1=(\rho_{11}+\rho_{22})^2 so that you get...

It's a dynamical equation which is equivalent to the Schrödinger equation. In the Schrödinger picture the states change with time, in the Heisenberg picture the operators change with time. http://en.wikipedia.org/wiki/Heisenberg_picture

Conservation of angular momentum implies that you two fermions should be in the spin singlet state (S=0). \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right) This state is anti-symmetric wrt interchange of the spin-indices of the two particles. As you said...

This is just the equation of motion of an operator in the Heisenberg picture, isn't it?

Well, I hope I didn't give the impression that I am some kind of expert in decoherence, I'm certainly not. But I can try to explain what I understand from the the book. Joos describes false decoherence as when "Coherence is trivially lost if one of the required components [of the wavefunction]...

Oddly I can see pages 134-180 + the first few pages with table of contents etc + backcover. I am a little confused as to how the preview feature works...perhaps its either 101 first pages or part of a chapter of your choosing. You could try amazon as well, at least I am able to look at that...