Recent content by peteratcam

Newton's laws are the necessary consequences of conservation of momentum. The first law says that momentum is conserved. The second law says that one can define a thing called 'force' as the rate of change of momentum. And the third law just says that however two bodies interact, momentum is...

You are right that the macroscopic laws must, in some way, follow from the microscopic laws. But the point is that the most natural description of a macroscopic behaviour is very often incredibly complicated to translate to the more microscopic scale, and entirely new concepts can appear, which...

Fluid mechanics was developed before the atomic hypothesis was accepted, and is an example of a collective phenomenon which was conceived without reference to microscopic constituents. Fluid mechanics is definitely useful, and definitely quite hard to derive starting from the standard model, or...

This is a Michael Berry idea I think. As a start, read item 1 on http://www.phy.bris.ac.uk/people/berry_mv/research.html and read reference 306.

For some previous ramblings of mine on this, see https://www.physicsforums.com/showthread.php?p=2618661

The momentum measurement example you use is a bad one because it involves slamming the particle into a wall of detectors. You don't need any quantum reasoning to conclude that the measurement disturbs the state of the system! To think about the ideal quantum measurement, you need to do the...

E.H. Lieb, T.D. Schultz, and D.C. Mattis, Ann. Phys. (N.Y.) 16, 407 (1961). http://dx.doi.org/10.1016/0003-4916(61)90115-4

You get 1/2, Dirac delta is a symmetric function.

I think it's more subtle than that. The action in the Hamiltonian formulation is S = \int p\dot q - H(p,q) and the first step you should arrive at is a phase-space path integral, with integrals over momentum and position. To reach the Lagrangian formulation, which is an integral over all...

Fine, but I mean there is an ambiguity in presentation: Sometimes [\phi(r),\phi(r')]=0; \quad [\phi^\dagger(r),\phi^\dagger(r')]=0;\quad[\phi(r),\phi^\dagger(r')]=\delta(r-r') (ie, ladder-operator valued fields) but sometimes [\phi(r),\phi(r')]=0; \quad...

What continues to confuse me is whether the field operators in QFT are supposed to be ladder operators are not. It so happens that acting on a groundstate of a harmonic oscillator, x and a^\dagger do pretty much the same thing, but this is a groundstate property, the operators are quite...

You'll need to use the known form of the time-evolution operator at some point to get the Hamiltonian to appear. The point of the large number of time steps is that you can use a linear approximation for time evolution in each timestep. Or see chap3 here for example...

Well as one of the responders to the "entirely useless" thread, maybe this will not help: But to me the trouble is in "...to the existence of a new potential energy term (and therein force)?". You don't need a potential energy term to be able to measure a force. Turn the handle of statistical...

Yes, the "reason" is that n is an accidental quantum number in a sense, because the hydrogen atom has an extra symmetry that makes lots of states degenerate which you wouldn't expect them to be. For any spherically symmetric problem, the obvious quantum numbers to use are the number of radial...

You can't learn it without maths. There is no quantum intuition that can be learned from the environment because we are too big. And the only other humans who I would trust to explain quantum theory will explain it mathematically. Having said that, the bits of maths which are required for...