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Unbounded operators in QM

by micromass
Tags: operators, unbounded
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micromass
#1
Jan28-13, 05:43 PM
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It is well known that unbounded operators play a crucial role in the mathematical formulation of quantum mechanics. In some sense, unbounded operators are inevitable. Indeed, we can prove that if A and B are self-adjoint operators such that [A,B]=ih, then A and B can never both be bounded.

My question is: is there any physical implications with being unbounded? Are there any physical reasons why we should expect unboundedness instead of bounded? Or is it only the mathematical formalism that changes?

One thing I can see is the following. A bounded operator always has compact spectrum. This means that the set of eigenvalues is necessarily bounded. So if all operators in QM were bounded, then I would expect that all outcomes of experiments are bounded. So in particular, the positions and momentums of all experiments would be bounded. This might constitute a physical reason why we want to work with unbounded operators. Is this accurate? And are there more such reasons?
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Ben Niehoff
#2
Jan28-13, 07:28 PM
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We can imagine many observables that are unbounded. You mentioned position and momentum; energy also comes to mind. I think the reality is that we expect quantities to exist which are unbounded in principle.

I suppose it might be possible to have a universe where such quantities are actually bounded. For example, we once thought relative speed was unbounded, but later discovered that it cannot exceed the speed of light. And it is easy to imagine a universe that is very large, but still finite.

The hardest to thing to imagine is how momentum, angular momentum, or energy might be bounded. This would require some radical changes to fundamental physics, although it's still possible. For example, the formation of black holes suggests that energy density might be bounded; i.e., above a certain energy density, a change of state occurs (although the precise physics of this new state---the black hole---thus far eludes us). And the angular momentum of a black hole is bounded by its mass; this again might be relatable to angular momentum density in general.

But remaining strictly within accepted quantum theory, there are numerous unbounded observables that will need unbounded operators to represent them in the mathematics.
Jano L.
#3
Jan29-13, 07:34 AM
PF Gold
P: 1,160
One thing I can see is the following. A bounded operator always has compact spectrum. This means that the set of eigenvalues is necessarily bounded. So if all operators in QM were bounded, then I would expect that all outcomes of experiments are bounded. So in particular, the positions and momentums of all experiments would be bounded. This might constitute a physical reason why we want to work with unbounded operators. Is this accurate? And are there more such reasons?
No, this is not a very convincing reason. The assumption of real line as a domain of most quantities does not follow directly from experiment, but from generalization of the fact that no limit was found so far. It is conceivable that position, momentum and other quantities are in reality limited within some interval of real line. If the limit is far enough, it may be the case we have hard time to find it.

In the meantime we usually assume that there is no limit at all and it turns out that the theory is simpler that way. For example, solving for eigenfunctions in potential ##1/r## is easy when there is no boundary. When the boundary is introduced, the calculation is more complicated, the results depend on the nature of the boundary and usually, if the boundary is far away, the difference between the two is negligible.

So the reason for unbounded operators is, in my opinion, the fact that calculations are easier with them.

andrien
#4
Jan29-13, 08:29 AM
P: 1,020
Unbounded operators in QM

An operator is continuous if and only if it is bounded.In a finite dimensional hilbert space all operators are bounded,in an infinite dimensional space(qm) it is not the case.A related thing is Hellinger-Toeplitz theorem which states that an everywhere defined symmetric operator on an hillbert space is bounded.http://en.wikipedia.org/wiki/Helling...eplitz_theorem


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