As for the relations involved: The energy of a photon is E=pc. Then due to the relationship E=h*nu of Planck and Einstein, the momentum is p=h*nu/c. Finally because lambda*nu = c for a photon, p= h/lambda.
For a non-relativistic particle with mass m, E=p^2/2m. De Broglie suggested that the...
For massive particles, the chirality depends on the reference frame; so boosting can take one between left and right chiralities. For massless particles, the handedness cannot change with boosting to different frames. Otherwise, the handedness of chirality for massive particles has the same...
Whether there is a single open slit or a double-slit setup, there will be an interference pattern due to a single particle (e.g. photon) as long as the setup has no detector to find out where the particle is until the end of the experiment. For the single slit case, if you measure the position...
The way you expressed the opeator A_{ab}=A*_{ba} is a matrix notation, useful when acting on a set of vectors like v_{b}. But in your problem, how does the operator act on the wavefuntion Psi(x)? Can you re-express your definition above for Hermiticity in this specific case?
It may help to state what you wish to show explicitly. Based on the definition of continuity, you want to show that lim_{x-->a}[f(x)]=f(a) (at the same time you'll verify that the limit exists). Can you see how to redefine variables and rewrite the expression above so that this expression is...
Good question. It's a bit more subtle than at first sight. The idea you used to solve the problem the first way is to say a mass m=(density)*(Volume flow rate) is delivered by the dam in one second. Ignoring any other water delivered after that, that mass m of water then gains kinetic energy...
Keep in mind that the allowed states of a particle depend also on the spin of the particle...in quantum mechanics courses you start off with spin-0 particles for simplicity. In the question, they are asking about electrons, which have spin 1/2. How does this change the counting of the number of...
There is no deeper explanation for the complex nature, only that it is indeed necessary. The "experiments" where it comes into play are many. Consider wavepackets of the form A(x)+i B(x) corresponding to single particles that are reasonably localized in some region. Let two particles scatter...
These seminar slides may address your questions:
http://hep.physics.uoc.gr/mideast5/talks/Saturday/Gopakumar.pdf
In introductory quantum field theoretic situations, massless states such as photons need to be discussed relativistically.
However, note that the term "massless" is in quotes...
No typo. The symbol \delta^{ij}_{kl}\equiv \delta^{[i}_{k}\delta^{j]}_{l}\equiv \frac{1}{2!}\left(\delta^{i}_{k}\delta^{j}_{l}-\delta^{j}_{k}\delta^{i}_{l}\right) (which generalizes to n indices with a 1/n! factor), and basis 1-forms act on basis vectors as \omega^{i}(e_{j})=\delta^{i}_{j}.
Sorry to hear you didn't get a reply before. The analogy is roughly ok, it's certainly not stupid. In considering a "lower dimensional effective theory", there is a diluting effect due to the fact that there are really dynamics in the "extra dimensions" that we are ignorant of. This means...
The second expression is not really a definition of the A's. You can define them by the first expression alone: They are defined to be the coefficients in the expansion of a function in terms of the basis of P functions. There are some cases in which you have a function already, and determine...
If you put a photon detector at some place in the room, there is a *probability* that the detector will detect it, which is calculated from the wavefunction of the photon (and depends on the size of the detection region). I'm not sure what you mean by "very first `contact' with an object..."...
Sorry, I didn't see your message till now. If you still need it: Now you can solve for the \xi_{\mu} by integration for each of the three equations. Try finding the form of the theta one first and then proceed to the next one using that result, and so on. You'll have to do some analysis to get...
After they are separated, wave packets of two particles will just have a form according to the energy-momentum they have due to the interaction. When wavepackets overlap, those of identical particles become a single wavefunction in which one particle can't be distinguished from another. But...