# Simultaneous measurement of noncommuting observables

1. Nov 20, 2013

### atyy

It is often said that non-commuting observables cannot be simultaneously and precisely measured. The rough idea is that the procedures to measure each observable are different. Is the statement strictly correct? If so, what is the mathematical argument?

Is it possible to use Ozawa's generalized uncertainty relation (Eq 15 of http://arxiv.org/abs/quant-ph/0207121)?

$ε(A)η(B) + ε(A)σ(B) + σ(A)η(B) ≥ \frac{\mid \langle \psi \mid[A,B]\mid \psi \rangle \mid}{2}$

The relation contains both measurement $ε(A)η(B)$ and intrinsic state uncertainties $σ(A),σ(B)$, and it is impossible to set both measurement uncertainties to zero if $A$ and $B$ do not commute.

If it is correct that non-commuting observables cannot be simultaneously and precisely measured, is there any easier way to prove it?

2. Nov 20, 2013

### jfizzix

The easier way to prove it would be to use the Heisenberg uncertainty relation of the intrinsic uncertainties

$\sigma(A)\sigma(B)\geq \frac{1}{2}|\langle[A,B]\rangle|$

Since a system cannot be simultaneously in an eigenstate of two noncommuting observables, there is no way to have any set of measurements of both $A$ and $B$ give a single definite value of $A$, and a single definite value of $B$ 100 percent of the time.

3. Nov 20, 2013

### JK423

I have no idea about what you're asking (no time to think about it atm), but i want to contribute with the following recent reference that i think it's of relevance,
http://prl.aps.org/abstract/PRL/v111/i16/e160405

4. Nov 20, 2013

### Jazzdude

That's not quite correct. A system can very well be in an eigenstate of two non-commuting observables. Two observable can commute on subspaces and not commute globally. If the state is in the commuting subspace (or sector) then it can be an eigenstate of both observables.

Cheers,

Jazz

Last edited: Nov 20, 2013
5. Nov 20, 2013

### jfizzix

Can you give me an example of two observables which have commuting subspaces and a state which is then an eigenstate of both of them?

6. Nov 20, 2013

### Jazzdude

Sure. Let's make it explicit and look at the Hermitian matrices.

$A= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix}$

$B= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix}$

They do not commuate, because

$[A,B]= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{pmatrix}$

but with the vector $v=\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ we have
$Av = v$

and

$Bv = v$

In other words, $v$ is an eigenvector of both non-commuting matrices.

Cheers,

Jazz

7. Nov 20, 2013

### jfizzix

Thanks; that makes a lot of sense.

What observables can you think of that would behave like this, though?

8. Nov 20, 2013

### Jazzdude

The matrices I gave qualify as observables. So are you asking for some "real world" operators that are routinely used for measurements? If yes, then I don't know an example right now. My point was more theoretical anyway.

Cheers,

Jazz

9. Nov 20, 2013

### atyy

Could noncommuting operators still share an eigenvector if we additionally specified that the commutation relation should be [A,B]=iħ ?

10. Nov 20, 2013

### Jazzdude

No, canonical commutation does not allow for commuting eigensubspaces.

Cheers,

Jazz

11. Nov 20, 2013

### strangerep

Are you talking about just the situation in Hilbert space, or does this still apply in the more general case of a rigged Hilbert space?

Background to my question: in ordinary Hilbert space this result may be seen as a variation on the old "0=1" Dirac paradox.
For the benefit of other readers, that paradox (in ordinary Hilbert space) is this:
The resolution of this paradox in standard quantum theory is that the spectra of both operators is continuous, so the inner products above are meaningless.

There's also another theorem (cf. Reed & Simon) that says:
In rigged Hilbert space, we only have the so-called "delta-normalization": $\<q|q'\> = \delta(q-q')$ and Dirac's paradox does not arise.

Or is there perhaps another theorem that prohibits commuting eigenspaces in RHS (the spaces being considered as built out of "generalized eigenvectors" in the RHS sense)?

12. Nov 20, 2013

### strangerep

I believe a more accurate statement is that an ensemble cannot be prepared such that all its members have precise values of those two noncommuting dynamical variables. The "proof" is along the lines that attempts to prepare such an ensemble by successive filtering operations, applied to a generic ensemble, don't work.

13. Nov 20, 2013

### atyy

I agree the statement about preparation is strictly correct. Let's call this intrinsic state uncertainty.

My question is whether the statement about measurement error is also strictly correct. If it is incorrect, what is a counterexample? If it is correct, how does one show it mathematically? Ozawa has some equations which if general enough, seem to show that the measurement statement is as correct as the statement about preparation.
http://arxiv.org/abs/quant-ph/0207121 (Eq 15)
http://arxiv.org/abs/quant-ph/0310070 (Eq 28)

That's interesting. It's relevant to my question because it deals with measurement errors. Physics World has an interesting write-up about it http://physicsworld.com/cws/article...y-reigns-over-heisenbergs-measurement-analogy, and Steinberg and colleagues have a note about the various definitions of measurement uncertainty http://arxiv.org/abs/1307.3604 . At any rate, Ozawa as well as Busch, Lahti, and Werner seem to agree that the commutation relations limit measurement uncertainty.

14. Nov 21, 2013

### atyy

Another interesting paper is Michael Hall's http://arxiv.org/abs/quant-ph/0309091, where he emphasizes the role of prior information.

Weston et al http://arxiv.org/abs/1211.0370 are aware of Ozawa and Hall's generalized uncertainty relations, and add a new one (their Eq 4). They say "It similarly implies that ε(Aest) and ε(Best) cannot both vanish for incompatible A and B."

Ozawa makes these claims in http://arxiv.org/abs/1106.5083 to summarize. Theorems 9 and 10 are not very intuitive, and I can't say I understand them, but he gives examples in http://arxiv.org/abs/0911.1147.

"Theorem 8: Observables A and B are simultaneously measurable in every state |ψ> if and only if A and B commute on H."

"Theorem 9: In any Hilbert space, every pair of observables are simultaneously measurable in any eigenstate of either observable."

"Theorem 10: In any Hilbert space with dimension more than 3, there are nowhere commuting observables that are simultaneously measurable in a state that is not an eigenstate of either observable."

So it seems that if one has no prior information, noncommuting observables cannot be simultaneously (and precisely) measured. As one has more prior information, for certain states, non-commuting observables can be simultaneous (and precisely) measured.

Last edited: Nov 21, 2013
15. Nov 22, 2013

### atyy

I've been trying to understand Ozawa's statement that "In any Hilbert space, every pair of observables are simultaneously measurable in any eigenstate of either observable." Apparently, the idea is that is that if ψ is an eigenstate of A, A can be precisely measured on ψ without changing it, making it still available for a precise measurement of non-commuting B.

However, doesn't such an eigenstate violate Ozawa's generalized error-disturbance relation? Korzekwa, Jennings, and Rudolph http://arxiv.org/abs/1311.5506 show that it doesn't. However, it does suggest that the notion of "error" in Ozawa's formula is peculiar. Using another sensible definition of "error" which matches my intuition for this case, Korzekwa et al show that the error is zero as intiuitively expected. They suggest "there cannot exist a simple state-dependent relation connecting the trade-off between error and disturbance with the expectation value of the commutator in the considered state."