fluid mechanics seems quite interesting in the sense that it seems a confluence (pun intended) of the core physics. the system of a gas or liquid seems inherently statistical mechanical.
i could see solids escaping from the statistical mechanics via having more regular patterns, but i suppose actually solids also should be inherently statistical mechanical
so actually maybe all material science is such not just fluid mechanics
i.e. in my terminology, the difference between stat therm and stat mech is just that stat mech is much more involved in the detailed computation of particle statistics, whereas in stat therm we just get away with playing with the partition function and derive lots of nice results without going too deeply into the mathematical woods
The underlying ideas are all the same, just a different emphasis
the spin-$N/2$ state $\lvert \psi \rangle := \sum_{k=0}^{N} \lvert k \rangle \otimes \lvert k \rangle$ where $\lvert k \rangle$ are $S_z$-eigenstates has $\langle S_i \rangle = 0$, right? I think this just follows from $\langle k \lvert S_\pm \lvert k \rangle = 0$
@naturallyInconsistent Hi! Say we have multiple paths that satisfy given initial conditions and boundary conditions,that confer a stationary value to the action integral.Now which one of these paths will the system take?
See this answer math.stackexchange.com/a/1432193/965696 by @Qmechanic where at the end he seems to say that the system will take that stationary path which is a minimum instead of other non-minimum stationary paths.
So will the system always take an action minimising path,if there are other stationary paths?(Given all of them satisfy all initial and boundary conditions)
@Arjun In the usual scenario, there should only be a unique path satisfying a given set of initial and boundary conditions and the EL equations. That path is what we call the (semi-)classical path. If you engineered for multiple such paths to coexist, then quantum theory tends to want us to superpose, maybe with weights, the contributions from each of these paths.
Ah, you should have stated that this is in classical mechanics.
I think the problem is somewhat ill-posed if strictly considering classical mechanics. In any case, this is way too esoteric; if you meet a problem like this in the real world, you ought to take the quantum reality into account and treat them all. As such, I don't think you should be wasting time on getting an answer for this hypothetical theoretical problem
@naturallyInconsistent It's actually a question from Goldstein ..in which he tries to find the solution of a SHO using action principle
He writes X(t) as a Fourier series sum
See the above linked answer..there @Qmechanic seems to favour the minimum path among the other stationary paths ,all of which satisfy all the boundary conditions
I'm not as sure that that is the correct prescription. I mean, it is easier to simply solve the EL equation and get the actual action integral, which is then a complete solution, not a guessing.
You might then Fourier decompose that solution, and work backwards to see which stationary paths are picked
@naturallyInconsistent That makes sense..also if we have multiple possible paths,clasically speaking will the particle still take a path that's a superposition and collapse to the definite Newtonian path when we observe it? From where is this certainty of prediction coming in classical mechanics,unlike in QM where usually all paths have some probability weights to them?
Absolutely not clear. The classical mechanics that we are familiar with, is only built upon unique classical paths. It is not built upon having multiple classical paths interfering. If we had a lot of that in the classical setting, we would have been forced to discover quantum theory a lot earlier than how our history evolved.
It would have made the theoretical study of such weird classical systems become physically important. Whereas we now know that the correct answer actually comes from quantum theory, that takes all paths at once, so that these weird classical systems ought not to be theoretically interesting when strictly considered within classical mechanics
@naturallyInconsistent My Bad.I should've been more clear,my question was something in the lines of :If the true nature of the world is quantum mechanical ,how does it somehow reduce to the totally deterministic paths we see in the classical world,maybe it's a philosophy question or we do some sort of averaging to millions of quantum particles that constitute a classical system and we somehow end up with totally deterministic paths as given by classical mechanics?
Oh, that is very much easier to answer. In the path integral formulation of quantum mechanics, it is very clear that almost all paths interfere destructively, except for what is often a unique classical path, which then gives us the illusion that classical totally deterministic paths are all there are. There is an equivalent version, but looks and feels very different, in the form of Ehrenfest theorem.
When you say almost all the paths interfere destructively,do you mean all the paths of the system as a whole or the individual paths of the millions of particles that constitute the system?
@Arjun classical particles always hav a definite path. i think the initial position and velocity pick a deterministic path even tho the boundary conditions may pick multiple paths
one can pick an initial position and velocity, and get a unique path using EL equation, assuming the equation is well behaved (e.g. acceleration can be uniquely solved for)
I'm sorry guys, I have two fermions of spin 1/2 and they're subjected to the following Hamiltonian $$H = \frac{1}{2mR^2}(L^2_{1z}+L^2_{2z}) + \gamma\mathbf{S}_1\cdot \mathbf{S}_2$$ I found a set of eigenfunctions for the system: $$|m_1,m_2\rangle_S|0,0\rangle_{spin};|m_1,m_2\rangle_A|1,s_z\rangle_{spin}, |m\rangle_i = 1/\sqrt{2\pi} e^{im\phi_i} $$
where the subscripts A,S indicate symmetric and antisymmetric combinations of the kets involved, obtained by applying the (anti-)symmetrizer respectively
there's also the case $m_1 = m_2 = m$
which is obv symmetric
is there a reason that the solution says that the set of eigenfunctions valid for $m_1 \ne m_2$ (namely the ones I expressed above) is actually $m_1>m_2$?
I think it should be $m_1 \ne m_2$, as I wrote myself, but maybe they're equivalent? In the latter case, I don't see why
we hav $|m_1,m_2\rangle _{S/A} =\pm |m_2,m_1\rangle _{S/A}$ becuz of the symmetrisation/anti-symmetrisation, so the states with $m_1>m_2$ or $m_2>m_1$ include all the states @ClaudioMenchinelli
