Since all the traditional "continuous" quantities like time, energy, momentum, etc. are taken to be quantized implying that derived quantities will also be quantized, I was wondering if quantum physicists agree upon any quantity not being quantized? I couldn't think of a single thing, until I came across this: Why position is not quantized in quantum mechanics?
Considering that the answers to the question you link are available for reading you will see that position is also quantizable, as crystal structure unequivocally demonstrates.
Quantum mechanics has a mathematical formulation that intrinsically allows for quantization of any variable entering the formulation, depending on the boundary conditions set to the solutions of the differential equations. A free particle, has variables that can take any value."Free" means specific boundary conditions that result in continua.
Energy quantization in terms of quantum mechanics depends upon the boundary conditions of a problem. A particle confined to a box will have a quantized energy and a quantized momentum. A free particle may have any energy and corresponding momentum.
It is the requirement that the wave function match conditions at the boundaries that excludes some functions and includes others. In other words, thats where energy quantization comes from.
Some properties seem not to be quantized at all. Time is one. Others seem always to be quantized, at least in non-relativistic quantum mechanics. Mass is such a property. Any mass is made up of atoms and atoms have definite (quantized) masses. Indeed, the discovery of the quantization of the mass of atoms was a major step forward in chemistry.
There is actually an older but nice article by Sir Neville Mott in which he argues that quantization occurs as a result of contraints expressed through boundary conditions.
Hence, the energy of bound states is quantized because one forces the wavefunctions to be $0$ at infinity and thus normalizable. Contrarywise, the energy of unbound states is not quantized as the wavefunction is not subject to the previous boundary condition.
Indeed, it is not hard to see, by playing with standard numerical integrators, that getting a solution of the Schrodinger equation is not hard, but getting a solutions that satisfies the boundary conditions is much harder and can only occur for discrete values of the energy.
A similar argument can be made for the quantization of the magnetic quantum number $m$: this follows by imposing that wavefunctions be single-values under a rotation by $2\pi$. There is also such a game to be played for the angular quantum number $\ell$.
In standard quantum mechanics, time is actually not quantized. What is also not quantized are all parameters that appear in the equations, like eg mass, coupling constants, and so on. You also don't quantize interactions in quantum mechanics, for this you have to go to quantum field theory.
The way I am using the term "quantized" is to means you make classical quantities operator valued. This does not necessarily mean these quantities get a discrete spectrum, which in general depends on the details of the system you consider.
Let me rephrase your question in a more pendantic way: which Hermitian (measurable) operators have eigenvalues that can take any continuous value (a continuous spectrum)?
My first reaction is to look for symmetry groups which are not closed. For instance, rotation corresponds to a closed symmetry group because rotation by any angle is equiavelent to rotation by some angle less than $2\pi$. Translation would be an example for a not closed symmetry group.
For a free particle, symmerties include position and time, and so my list would be the
- position (including angles)
- the Hamiltonian
operators. I'm excluding:
- time, which is a parameter and not an operator is non-relativistic quantum mechanics
- angular momentum, since rotation is continuous but a closed symmetry
- parity, which is not a closed symmetry
Note that quasimomentum, the momentum in a crystal, also has a continuous spectrum (eigenvalues). The symmetry group, discrete translations, is not closed even though it is not continuous.
Of course, you can combine these operators and make new ones with continuous eigenvalues that don't correspond to any symmetry of the system.
protected by Qmechanic♦ Mar 12 '17 at 6:06
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