At undergraduate levels, it is usually treated as such. However, the PEP (and for that matter, spin itself) is a direct consequence of relativistic QM, and falls naturally out of the theory.
It has to do with the fact that for fermions, the wavefunction for two particles must be antisymmetric. If they were both in the same state, then the wavefunction would cancel out to zero, which does not make any sense physically.
You type it up in a regular text file, then save it with a .tex extension. Then, you need to compile it by typing "latex file" (no quotes, in Windows, at a DOS prompt, in the directory where the file is saved), where file.tex is the thing you just saved. This will generate a DVI file. With...
The product rule applies fine for a constant term, since a constant is a perfectly good function. It does not apply the way he said though; his exponential relation is wrong.
e^ne^x=e^{n+x}
Hard to say. Certain areas (such as optics, nanotechnology, etc) have a wider range of applications, and can offer jobs in both academia and the industrial sector.
Generally, if you're going into physics to make money, then you're going into it for the wrong reasons.
Nanoparticles are just small clusters of some material, made up of a "small" number of atoms, and exhibiting sizes on the order of a few nanometers. The difference in properties between the nanoparticles and bulk systems arise from many different phenomena, such as quantum confinement.
An...
- I suppose classical concepts can be easier to deal with in some cases.
- Quantum is dominant at small size scales, classical at large ones. It depends on the situation and your assumptions.
- Both. Again, it depends what you're talking about. In many cases, classical physics gives a...
When an external field is applied to a diamagnetic substance, screening currents occur to oppose the external field; that is why the induced "dipoles" align opposite to the applied field. Paramagnetic behavior is due to permanent dipoles aligning to the field, reducing the total energy.
\ln(x+5)=\ln(x-1)-\ln(x+1)=\ln\left(\frac{x-1}{x+1}\right)
So,
x+5=\frac{x-1}{x+1}
Solve this for x, and you will see that both possible values generate a negative argument in the second logarithm, so there is no valid solution.