In section 18.7 of Bruus & Flensberg the authors provide a microscopic derivation of the Josephson effect.
The hamiltonian on both sides of the tunnelling junction is just the typical BCS hamiltonian, on one side (with fermion operators ##c##)
$$
H_c = \sum_{k,\sigma} \epsilon_k...
It seems like I overlooked the simple fact that the state cannot change during a sudden, ##\textit{finite}## perturbation, so I was right in assuming that the spin would be ##|+\rangle## at ##t=0^+##.
To understand why the system's state must be continuous over the sudden perturbation in the...
Consider an uncharged particle with spin one-half moving with speed ##v## in a region with magnetic field ##\textbf{B}=B\textbf{e}_z##. In a certain length ##L## of the particle's path, there is an additional, weak magnetic field ##\textbf{B}_\perp=B_\perp \textbf{e}_x##. Assuming the electron...
One last question, the Unruh modes as defined in Sean Carroll's "Spacetime and Geometry" are:
$$h_k^{(1)} = \frac{1}{\sqrt{2\sinh(\pi \omega/a)}}\big(e^{\pi \omega/2a} g_k^{(1)} + e^{-\pi \omega/2a} g_{-k}^{(2)}{}^*\big)$$
On the other hand this paper gives a different definition:
$$h_k^{(1)} =...
In Carroll "Spacetime and Geometry" I found the following explanation for why the analytically extended rindler modes share the same vacuum state as the Minkowski vacuum state:
I can't quite understand why the fact that the extended modes [\tex]h_k^{(1),(2)}[\tex] are analytic and bounded on...
In p.385 of Griffiths QM the vector potential ##\textbf{A} = \frac{\Phi}{2\pi r}\hat{\phi}## is chosen for the region outside a long solenoid. However, couldn't we also have chosen a vector potential that is a multiple of this, namely ##\textbf{A} = \alpha \frac{\Phi}{2\pi r} \hat{\phi}## where...
So the so-called act of measurement in the OP is the painting process, not the blind man somehow "measuring" the colour of the apple? This seems to make a lot more sense now.
Thank you. Suppose then that instead of colouring apples i consider randomly assigning a spin to an electron (here surely quantum effects are coherent).
If I don't observe the spin of the electron, then does this mean that the electron is in a definite spin state, but I just haven't performed a...