Why are Operators Used in Quantum Mechanics?
This post can be found here: http://physics.stackexchange.com/questions/77275/why-do-we-use-operators-in-quantum-mechanics
Q:
In classical mechanics, physical quantities, such as, e.g. the coordinates of position, velocity, momentum, energy, etc, are real numbers, but in quantum mechanics they become operators. Why is this so?
A:
The short answer is that you can treat classical observables as operators if you want, but the algebra of observables that actually described the world turns out to be non-commutative. Or in other words, to the question:
- If the state of a physical system is completely determined in terms of definite outcomes of some observable(s), are there still measurements that can be made that are not definitively determined?
Quantum mechanics answers ‘yes’, and that’s just what we find in the world, e.g., for a simple spin-0
particle a definite position completely determines the state (position eigenstate), but does not result in a definite outcome for a momentum measurement. Etc.
In classical mechanics in the Hamiltonian formulation, observables are not numbers, but functions (or distributions) over the phase space. In terms of the canonical coordinates of position and momentum (q;p), some of the usual suspects have particularly simple forms, e.g., kinetic energy of a particle T(q;p)=p2/2m. The time-evolution of a classical observable is given by Hamilton’s equation
dfdt=∂f∂t+{f,H},where {⋅,⋅} is the Poisson bracket and H is the Hamiltonian.
The product or sum of any two observables is an observable, and mathematically, the Poisson bracket is both a Lie bracket and a derivation. The details of this you can look up on your own, but the immediately relevant effect is that the classical observables form a commutative Poisson algebra, a subtype of a Lie algebra.
Once one gets used to thinking of observables as forming an algebra, some questions naturally appear:
- Does our algebra actually get all the physical observables, does everything we can measure correspond to something in the algebra of classical observables? If not, do physical observables require loosening the rules of this algebra somewhat, e.g., making it non-commutative?
It turns out that the answer is that the algebra required to describe the world is indeed non-commutative, although of course it wasn’t historically discovered by this analogy.
As for why linear operators in particular, it’s because it works–though the reason we could have expected it to work is that a fairly general class of algebras can be represented as a linear operators on a complex Hilbert space. That’s the essence of the Gel’fand-Naĭmark theorem regarding C⋆
-algebras.
As an interesting curiosity, classical mechanics can be formulated using wavefunctions/kets on a complex Hilbert space, with physical observables represented by operators and measurement being probabilistic, just as in quantum mechanics. It’s not even all that strange, turning into a complexified reformulation of the classical Liouville equation (which handles probability distributions over the phase space).
The common formalism of QM is quite general and is completely capable of handling classical mechanics; where they disagree is on which operators represent things we actually measure in the world.
–Stan Liou
Spin as Intrinsic Angular Momentum
This post can be found here: https://www.quora.com/If-spin-is-a-particles-intrinsic-angular-momentum-but-its-not-actually-rotating-then-what-is-it-doing
Q:
If spin is a particle’s intrinsic angular momentum but it’s not actually rotating then what is it doing?
A:
To understand spin, you need to unlearn the idea that angular momentum is something that objects have when they rotate. For one, even objects travelling in a straight line have angular momentum, assuming that you measure angular momentum about an appropriate axis.Angular momentum does have something to do with rotation: but it’s not that it quantifies how much a system is rotating. Angular momentum quantifies what happens to a system when you rotate it. Spinless particles such as the pion are rotationally symmetric, sort of like a perfect sphere; when you rotate it, it doesn’t change its state. Electrons and other “spinful” particles, however, behave a bit like they have a tiny arrow attached to them: when you rotate them, you change the direction of the arrow.The quantum-mechanical operator for spin is L=−iℏ∂∂θ. It’s the derivative with respect to angle. It measures the change in the state with respect to angle. The stationary electron has angular momentum not because it is rotating but rather because if you choose to rotate it, you change its configuration.Spin is simple. It’s classical rotation that’s complicated! Classically spinning objects such as wheels actually have narrowly peaked distributions of angular momentum, not exact angular momentum the way electrons can. (It’s the same as how classically moving objects can’t have exact momentum; if you knew their momentum exactly, then their position would be uniformly smeared across the entire universe.)
–Brian Bi
Also, be sure to check out the blog Gravity and Levity! It is really interesting: https://gravityandlevity.wordpress.com/