# Is the uncertainty principle valid?

The uncertainty principle says that the product of the uncertainties in position and momentum can be no smaller than a simple fraction of Planck's constant $h$.

Several articles lately suggest this is not true.

Today in Physicsworld.com

Looking at the position and momentum of spin-polarized neutrons, they found that, as Ozawa predicted, error and disturbance still involve a trade-off but with a product that can be smaller than Heisenberg's limit.

and earlier work in Toronto, Canada

Aephraim Steinberg and colleagues at the University of Toronto conducted an optical test of Ozawa's relationship, which also seemed to confirm his prediction. Ozawa has since collaborated with researchers at Tohoku University in another optical study, with the same result.

Can we safely throw out the Heisenberg uncertainty principle and start to view physics with more certainty?

## 3 Answers

AFAIK, Ozawa's results were derived within standard quantum mechanics. Heisenberg's initial derivation of the uncertainty principle was qualitative, so, on the one hand, Ozawa's results present a quantitative adjustment to the uncertainty principle, on the other hand, they use some specific and unusual definitions of the values in the uncertainty principle ("error" and "disturbance"). So I would not make too radical conclusions from Ozawa's results - again, after all, his predictions and the experimental confirmations of his predictions do not seem to contradict the predictions of standard quantum mechanics.

"Can we safely throw out the Heisenberg uncertainty principle and start to view physics with more certainty?"

No, you definitely can not throw out the HUP. If you have two operators $A$ and $B$, and define $\Delta A = A - \langle A \rangle$, it is easy to prove that (see, e.g., Sakurai's QM book, section 1.4) \begin{equation} \langle (\Delta A)^2\rangle \langle (\Delta B)^2\rangle \geq \frac{1}{4} | \langle [A,B] \rangle |^2, \end{equation} which is just a more general form of the HUP. If this inequality was wrong, it would mean that the whole formulation of QM in terms of vectors and operators in a Hilbert space is also wrong, which seems extremely unlikely.

The uncertainty in position and momentum is part of the quantum mechanical description of matter. The articles in your links are probably talking about the so-called Heisenberg's error-disturbance uncertainty relation (that applies to measurements and is not a property of matter), which is not the same as the fundamental uncertainty relation in the equation above. Also note that the uncertainties in position and momentum are still a fraction of $h$ in all the works cited in those links; the fraction is simply smaller than Heisenberg initially thought.

• Does not that statement simply prove that uncertainty stays the same or grows larger? Have you prepared your quantum system in a state where incompatible observables take specific values, ie. no uncertainty? – AnimatedPhysics Nov 3 '13 at 0:07
• @AnimatedPhysics By definition "incompatible observables" do not take on specific values simultaneously (unless you enjoy Bohmian QM, but the practical result is the same). The commutation relation on the right side of the inequality that they have to respect is mathematically intrinsic to the observables involved, not the experimental setup, and is impossible to violate. – Robert Mastragostino Nov 3 '13 at 3:32
• Thanks Robert, I wish I could understand this better. If QM cannot by definition allow incompatible observables to take on specific values simultaneously, does it make sense to use QM as a proof of the existence of the HUP? – AnimatedPhysics Nov 3 '13 at 4:57
• @AnimatedPhysics The inequality above follows from the postulates of quantum mechanics. Thus, the postulates of QM imply the HUP. A state with specific values for incompatible observables is mathematically impossible according to the rules of QM; it does not exist, not even in principle. – Goku Nov 3 '13 at 20:23
• @AnimatedPhysics: Yes, it makes sense, because the the Hilbert space formalism that treats observables as linear operators can easily formulate classical mechanics, but is also more general than that. Thus Goku's answer is actually somewhat stronger than he says it is, as the HUP doesn't even really need the postulates of QM per se. Rather, killing the HUP would require the complete absence of any pairs of non-commuting observables, and experimentally we have lots of them. – Stan Liou Nov 4 '13 at 0:34

To start with, quantum mechanics was not postulated. It grew from observations of quantized states including the study of the effects of Heisenberg's uncertainty principle.

Now we have a consistent theoretical framework for quantum mechanics which in principle allows us to predict by computation, the probabilities of getting certain measurements. The Heisenberg uncertainty as an observational proposition is deeply connected with the probabilistic theoretical calculations through the commutation relations of the operators which represent measurable variables in the quantum mechanical framework.

The variables in the Heisenberg uncertainty become operators in the quantum mechanical framework and the commutator is not zero for variables that obey the HUP, the canonical commutation relations. .

Operators operate on the state function of the system under consideration and it is not surprising, in my opinion, that a particularly set up measurement gives uncertainties for the variables that differ from the HUP as stated generally, in a similar way that the Bohr atom did describe the hydrogen atom but the correct QM treatment differs from that. I would interpret any smaller limits as consistent with the HUP as long as the value is nonzero.