Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
Not sure what you mean by quantized here. For a free particle, neither $p$ nor $E$ are quantized in the sense that they take any value. –  santa claus Dec 27 '12 at 4:18
In the future, a time will come when all people will use the term "discrete" to distinguish from continuous, and "quantized" to mean "subject to quantum rules". I will, of course, be dead by then. –  twistor59 Dec 27 '12 at 7:15
Possible duplicate: physics.stackexchange.com/q/39208/2451 –  Qmechanic Dec 27 '12 at 9:38
@user1104 I am talking of the known world. Do we know any existing quantities that exist in non-quantized state? –  Madhur Jan 29 '13 at 8:34
@twistor59 You have to allow for people who are not native to QM but are merely interested in philosophical implications of what Physics has uncovered so far. –  Madhur Jan 29 '13 at 8:38

4 Answers 4

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.

share|improve this answer
Does that mean that if boundary conditions do not exist, then quantization does not exist? –  Madhur Jan 29 '13 at 8:32
yes, boundary conditions are necessary for quantization to appear. Similar to classical wave equations. –  anna v Jan 29 '13 at 8:39
Am I right in stating the converse that if boundary conditions can be stated for any quantity then quantization must also exist? –  Madhur Jan 29 '13 at 8:45
i believe boundary conditions are necessary, i do not know if they are sufficient generally.. For potential well problems they are, (finite bounds) –  anna v Jan 29 '13 at 9:29

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.

share|improve this answer

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)
  • momentum
  • 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.

share|improve this answer
by "closed" you actually mean "compact" ? –  Lacek Dec 30 '12 at 2:10

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.