What is Schrodinger’s equation? Is it deterministic or not? If it is, how can we prove that? And what conditions must be satisfied for it to be non-deterministic? 

Schrodinger’s equation was originally just the partial differential equation satisfied by the position-space wave function of a particle (or more general system) in non-relativistic quantum mechanics. The same name is sometimes also used for the equation [math]\frac{d}{dt}\Psi(t)=iH\Psi(t)[/math] satisfied by the state vector in any NRQM system regardless whether or not a position-space representation is being used (or is even available).

It is deterministic (in the sense of determining [math]\Psi(t)[/math] uniquely for all [math]t[/math] if given an initial condition [math]\Psi(0))[/math], so long as the Hamiltonian [math]H[/math] is self-adjoint (symmetry is NOT enough!).

The proof of this involves more analysis than I could fit into a Quora answer, but in the general case it follows from the fact that for any self-adjoint operator [math]H[/math] on a Hilbert Space, the equation [math]\frac{d}{dt}\Psi(t)=iH\Psi(t)[/math] is uniquely satisfied by [math]\Psi(t)=e^{iHt}\Psi(0)[/math] where the complex exponential of [math]H[/math] is defined in terms of its spectral resolution; and for the PDE special cases it might be done by various theorems involving greens functions or Fourier analysis and convergence properties of improper integrals.

It may be non-deterministic if [math]H[/math] has not been specified on a large enough domain to be essentially self-adjoint (as sometimes happens if boundary conditions are omitted from the specification of a problem in which the particle is confined somehow – either by an infinite potential or by living in a single cell of a crystal lattice for example). But such cases are normally just due to inadequate specification of the problem rather than to a real physical indeterminacy.

So I would say that in a properly defined quantum theory model the Schrodinger equation is indeed almost always deterministic.

[N.B. It wasn’t part of the actual question, but I should perhaps add that the reason this does not make quantum mechanics deterministic is because even complete knowledge of the quantum state of a system is not sufficient to predict the outcomes of all possible experimental measurements. For any a state which happens to produce a predictable value for one observable there will be other observables for which the outcome is uncertain.]

Source: (1000) Alan Cooper’s answer to What is Schrodinger’s equation? Is it deterministic or not? If it is, how can we prove that it is deterministic? If it isn’t, what conditions must be satisfied for Schrodinger’s equation to be non-deterministic? – Quora

What is the definition of an eigenstate of a hermitian operator?

An eigenstate of a quantum observable is a state which results from a measurement of that observable which has produced a precise value; and according to quantum theory this means that it is represented by a normalized eigenvector for the corresponding self-adjoint operator (whose eigenvalue is equal to the observed measurement value).

An eigenvector of an operator [math]A[/math] is a vector [math]\Psi[/math] for which [math]A\Psi=\lambda\Psi[/math] for some number [math]\lambda[/math] (which is called the corresponding eigenvalue).

Source: (1000) Alan Cooper’s answer to What is the definition of an eigenstate of a hermitian operator? – Quora

Falling from Space

Source: (1000) Alan Cooper’s answer to If a Geostationary Satellite runs out of fuel, how long will it take to descend to earth in free fall? Galileo calculated that Moon would descend to earth in under 4 hours if it is blocked in orbit and allowed to fall free to the earth. – Quora

If it runs out of fuel then its free fall is just the current orbit in which it will stay essentially forever (since it is well above any atmospheric drag that might slow it down). But in the unlikely event that it has the extremely large amount of fuel needed to kill all of its orbital velocity, (and if it actually uses the fuel for that purpose), then it will fall straight down to the earth in a time that one can calculate – either with a bit of calculus, or by the clever trick of applying Kepler’s law for a very thin ellipse of major axis length equal to the orbital distance.

[Kepler’s third law says “The square of a planet’s orbital period is proportional to the cube of the length of the semi-major axis of its orbit” so the period is proportional to the 3/2 power of the major axis; and the full axis of a narrow ellipse from the satellite down to the earth is just the radius (ie half the major axis) of its previous circular orbit. So the period of that ellipse, if completed, would be the geostationary period of 24 hours divided by a factor of 2^{3/2}. But the satellite only gets (a bit less than) halfway round before crashing, so the actual time for the fall is a bit less than 24/2^{5/2}=3\sqrt{2} or about 4 hours.]

P.S. For the Moon, since the orbital period is 28 days, the time to fall would be more like 5 days, but I think the claim attributed to Galileo may arise from a misreading of his argument that the Earth is spinning on its axis (basically something like: If the Earth was NOT rotating and instead everything on the celestial sphere was going round once every 24 hours, then the Moon, Sun, and stars might all be at the geostationary distance but only those near the equator would actually be in balance and those at the pole would just fall to Earth in 4 hours)

P.P.S. It is not clear that Galileo actually had a good understanding of orbital mechanics and the inverse square law, so if he estimated a falling time of 4 hours that may have been little more than a lucky guess

Mass in Special Relativity

Source: (1002) Alan Cooper’s answer to According to special relativity, the mass of an object is dependent on its velocity. Does this mean that there is an absolute zero velocity in the universe and we can measure it? – Quora

The claim that “According to special relativity, the mass of an object is dependent on its velocity” is false. In fact it’s worse than false; it’s meaningless – because the velocity of an object is not something that exists except as relative to some observer.

What is true is that something which used to be called the “apparent mass”, or “relativistic mass”, of a body as seen by an observer is dependent on the velocity of the observer relative to the object (or of the object relative to the observer). However neither this apparent mass nor the velocity are properties of the object, since they depend also on its relationship to the observer.

In terms of any particular object and observer, there is indeed an absolute zero relative velocity. It corresponds to the situation where neither of them sees any change in the distance or direction of the other, and is absolute in the sense that any other inertial observer will perceive them as having the same velocity as one another (relative to it) and so as having zero relative velocity (relative to one another).

And the “apparent mass” as perceived by an observer at zero relative velocity, which is an inherent property of the object, is what was once called the “rest mass” but has for many decades been identified just as the mass of the object (without any qualifier).

(1000) Alan Cooper’s answer to How can we reconcile the wave-particle duality of quantum particles with the classical understanding of fields? – Quora

Source: (1000) Alan Cooper’s answer to How can we reconcile the wave-particle duality of quantum particles with the classical understanding of fields? – Quora

(1000) James G Bridgeman’s answer to Do the effects of special relativity apply to non inertial frames of reference? If I accelerated to a constant velocity of 99% c would I continue to experience time dilation for as long as I maintained that speed or only when accelerating? – Quora

Source: (1000) James G Bridgeman’s answer to Do the effects of special relativity apply to non inertial frames of reference? If I accelerated to a constant velocity of 99% c would I continue to experience time dilation for as long as I maintained that speed or only when accelerating? – Quora