# Uncertainty principle discrete operators

1. Jul 29, 2011

### BWV

couple of questions

a) the operators not commuting would also be true of position and momentum operators in classical mechanics (x d/dx -d/dx x) f(x) so the non-commutation does not inherently constitute a proof for the uncertainty principle, or do you just not care about the uncertainty at classical scales?

b) how can you commute discrete operators with continuous ones - are you not multiplying matrices of different dimensions together?

2. Jul 30, 2011

### tom.stoer

There are no such operators in classical mechanics.

Think about a trajectory {x(t), p(t)} in phase space; there are equations of motion using dx/dt, dp/dt etc. but no "momentum operator" d/dx. Position is simply x, momentum is p, that's it. In addition there is no such function f(x) on which these operators could act.

It's like saying that energy is quantized in classical mechanics b/c n (as a natural number) has discrete values only. The latter statement is true mathematically, but you simply don't use natural numbers n to describe energy E in classical mechanics.

3. Aug 2, 2011

### BWV

this is what I was getting at, for a classical Hamiltonian you get

(http://en.wikipedia.org/wiki/Canonical_coordinates)

in QM you have the additional iħ factor, but the classical Poisson brackets in phase space don't commute either - a fact that would have been known well before Heisenberg developed the uncertainty principle - which he did outside of the Hamiltonian framework that Schrodinger later came up with?

Last edited by a moderator: Apr 26, 2017
4. Aug 2, 2011

### Finbar

These are Poisson brackets theres nothing about them that doesn't commute. You seem to be confusing commutators and Poisson brackets.

Last edited by a moderator: Apr 26, 2017
5. Aug 2, 2011

### tom.stoer

1) of course the classical phase space coordinates x and p do commute; their Poisson bracket is non-vanishing, but (classically) this has nothing to do with a commutator.

{x,p} means dx/dx * dp/dp - dx/dp * dp/dx = 1

whereas [x,p] means xp-px = 0

2) the uncertainty principle is derived as a geometric property of general states and (non-commuting) operators in a Hilbert space; but classically you do neither have states in a Hilbert space nor do you have operators.

3) classically you have x(t) and p(t), so for each t, x and p are determined as functions of t and there is no uncertainty

6. Aug 2, 2011

### dextercioby

Quantum mechanically you can also have x(t) and p(t) for all t and suitable states, so your argument is irrelevant for the other 2.

7. Aug 2, 2011

### mathfeel

You can show that for any operator $A$ and $B$, $\langle \Delta A \rangle \langle \Delta B \rangle \ge \frac{1}{2i} [A,B]$ (I might miss a factor or something). Where by definition, uncertainty in operator A means $\langle A^2\rangle - \langle A \rangle^2$.

This is true for all physical observables. In particular position and momentum.

But quantum is the underlying rules, even for classical physics. So, the fact that in mathematical models of classical physics you assume that both position and momentum can be simultaneously specified is an ultimately illusion. Nevertheless, such assumption still constitute a perfectly good mathematical models that operates within certain range of physical parameters.

You can't. The two operator acts on different subspaces and their product makes no sense. For example, position operator and spin operator.

Last edited: Aug 2, 2011
8. Aug 3, 2011

### BWV

Thanks for the replies

Still confused in that I thought there was a fundamental relationship between the Classical Poisson bracket and the Commutator in QM