# Noncommuting operators and uncertainty relations

1. Sep 24, 2007

### noospace

Hello all,

I've been thinking about the connection between commutativity of operators and uncertainty.

I've convinced myself that to have simultaneous eigenstates is a necessary and sufficient condition for two observeables to be measured simultaneously and accurately.

It's also clear that simultaneous eigenstates gives us commutativity. So we have that non-commuting observables have an uncertainty relation between them.

What's not exactly clear to me is why the converse holds, ie commuting observeables can be measured simultaneously and accurately.

Is there simple proof of this that I'm missing?

2. Sep 24, 2007

### OOO

Let A, B two operators with [A,B]=0. If $$\psi$$ is an eigenstate of B that is $$B\psi = b\psi$$, then $$BA\psi = AB\psi = b A\psi$$ so $$\phi = A\psi$$ is also an eigenstate of B corresponding to the eigenvalue b.

If the eigenvalue b is not degenerate then this means that phi must be proportional to psi, that is $$\phi = A\psi = a\psi$$ and thus psi is a simultaneous eigenvector of A and B.

If the eigenvalue b is is for example twice degenerate, then you may diagonalize A in the eigen space of b, i.e. find eigenstates psi1 and psi2 of b, that fulfill $$A\psi_1 = a_1\psi_1$$ and $$A\psi_2 = a_2\psi_2$$ and consequently psi1 and psi2 serve as a common system of eigenstates for A and B in the subspace defined by the eigenvalue b.

Last edited: Sep 24, 2007
3. Feb 24, 2008

### noospace

In the degenerate case, I'm not quite understanding what $A\psi$ being an eignestate of $B$ has to do with being able to diagonalize A in the b-eigenspace?

4. Feb 24, 2008

### noospace

Ahh,

If $A\psi$ belongs to the b-eigenspace then we can express it as

$A\psi = a_1\psi_1 + a_2\psi_2$
$A(\psi_1 + \psi_2) = a_1\psi_1 + a_2\psi_2$

so $A\psi_i = a_1\psi_i$ by linear independence.