@Minsky hi. you are looking for this en.m.wikipedia.org/wiki/Heisenberg_picture . Look in the "Commutator relations" section for the exact analogy of Newton's laws in quantum mechanics
Is it correct to say, in particle physics: "It represents the characteristic momentum scale at which one is probing the structure of a hadron or nucleus." ?
@Sanjana Such operators do not exist; if you're looking to the theory of extensions you should encounter a theorem about the form of all possible extensions (related to the Cayley transform) that explicitly shows operators either have no, one or infinitely many possible self-adjoint extensions.
I recently learned that the extension of non-self adjoint operators in QM relates to the problem of classical systems that have an ill-defined Cauchy problem
ah, it is the boundary triplets - the $c\leq 3/4$ case where you "have to choose what happens when the particle hits the singularity" is a special case of the general theory of self-adjoint extensions where you have to choose a very generic kind of "boundary condition" to pick a particular extension
QM generalises the transformation of differential operators from the transformation of derivatives operators that we see in GR. This is because GR treats derivatives as vectors on a finite dimensional manifold while QM treats derivatives as operators on an infinite dimensional vector space
@Obliv consider the analogous problem in cartesian co ordinates. the kinetic energy can still depend on three variables while potential can be something like : V(x,y,z)=f(x)
what the book means here is $U(r, \theta, \phi)=f(r, \theta)$. $U$ is still defined on the entire configuration space. it's just independent of $\phi$ @Obliv
for e.g. you can consider the gravitational potential which is $U(r, \theta, \phi) \text{~} -\frac{1}{r}$
well ok even if the object wasn't spinning, if it had some other components of momentum that weren't in the direction of the potential, it still has some "energy" that isn't described by the potential energy function
which isn't terrible, because in reality it's fine to do that for certain cases, but in general that would be weird
Ok imagine two spheres in space, keep one fixed. The one that is free can spin, move around in 3 dimensions and the potential energy function is only w.r.t the radius b/t the two
even at 0 potential energy, so when the spheres are gliding over each other, there will be some velocity of the free sphere racing around the other sphere
ok in the problem let's say the sphere provides the $U(r,\theta)$ so the point converts the energy along the radial axis and the longitudinal angle. If it's moving along the equator, that momentum doesn't change right?
and the usual example of a potential that is independent of the azimuth but not of $\theta$ would be that of a charged particle moving in a constant magnetic field in the z-direction
I'll be honest I have no idea what a momentum conjugate is, I've just been plugging in $\frac{d}{dt}\frac{\partial L}{\partial \dot{...}}=\frac{\partial L}{...}$
I would link u the course site but I don't wanna get doxed, but in the syllabus it says a text isn't required but he recommends a couple. And the lecture slides don't mention conjugate momentum either
@Obliv only the total momentum is conserved. the canonical momenta refer to individual momenta as there is one canonical momentum defined for each degree of freedom
please also note that the total momentum is conserved only for the sum of the cartesian momenta. the sum of general canonical momenta is not conserved in general
general canonical momenta cannot even be summed because of different units
@Obliv sure, so when they ask you "which momenta are conserved" you're being asked to look at $\dot{p}_i$ and try to figure out whether that's zero or not
also, a hint: while you can do this by actually computing all three time derivatives, if you use the E-L equations in a clever way you can do this without actually having to explicitly differentiate anything
So i still have to do the time derivatives to get the equations of motion, but to check if can. momentum is conserved i just have to solve the right side of the E-L equation
@Obliv it is cos theta * dot theta. but why are you computing this?
i should say that the canonical momenta of any variable can always be summed over different particles, but only the total cartesian momenta and total angular momenta are conserved in most real world closed systems
but anyway, this isnt related to ur main question
ur question is about the conservation of individual canonical momenta. usually, books derive this result but u shud try this on ur own
yea I didn't realize that became $\dot{\theta}\cos\theta$ that changes things
so $\sin^2\theta$ becomes $2\dot{\theta}\sin\theta\cos\theta$
I got $$m(\ddot{\phi}r^2\sin^2{\theta}+\dot{\phi}(2r\dot{r}\sin^2{\theta}+2r^2\dot{\theta}\sin\theta\cos\theta))$$ for the $\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}$
