# Heisenberg uncertainty and probabilistic nature of QM

I am trying to understand whether the HUP and the probabilistic nature of QM are orthogonal or not. By that I mean that the HUP fundamentally derives from operators not commuting, which is the important fact here, more than the statistical nature of the LHS in the definition of the HUP: $$\sigma_{x}\ \sigma_{p_x}\ge\frac{\hbar}{2}.$$ If that is correct, the HUP should manifest itself even with an experiment on an ensemble that contains only one particle. Is that the case? If not (hard to compute variances on one particle), is there a minimum ensemble size that demonstrates the HUP?

Note: the reason I am asking is that some people seem to say that Heisenberg uncertainty is related to measurement uncertainty, or they invoke "ensembles of particles" to justify the HUP, when IMHO, the HUP is saying something else than the fact that measurements are probabilistic in QM. Hence the "nonsensical" question about "one particle", which attempts to remove statistics from the picture, so to speak. Maybe statistics and the HUP are conjoined in an inseparable way in QM, but if say, all operators could commute, there would be no HUP, you need the ingredient of non-commutativity which has nothing to do with statistics. Is that correct?

• a) what are the quantities on the left hand side? I only know variances to be there. b) how do you even define expectation values/variances for one particle? They are only defined for ensembles... – Martin Mar 10 '15 at 19:51
• Points taken, and question fixed. – Frank Mar 10 '15 at 20:01
• With "ensemble" do you mean "system with lot of particles" or the statistical meaning of "set of all possible configurations that a system can be in"? – glS Mar 10 '15 at 20:08
• I meant a lot of identical prepared in the same way. – Frank Mar 10 '15 at 20:09

The $\sigma_A$ on the LHS is defined as $$\sigma_A = \omega(A^2 - \omega(A)^2I),$$ where $\omega$ is a state (so $\sigma_A$ also depends on $\omega$!) and the way this is defined by von Neumann is by taking a statistically relevant ensemble of very same copies of the very same system in the very same state $\omega$ and repeat the measurement of $A$ and $A^2$ this many times. All you need to do is to apply the defining formula $$\omega(O) := \frac1N\sum_{k=1}^N\lambda_k$$ where the $\lambda_k$s are the outcomes of the observable $O$ on the state $\omega$. This is the bridge between the statistical nature of a measurement and the HUP, since by the axiomatic formulation of quantum mechanics through the theory of operator algebras you get that $$\Delta_\omega(x)\Delta_\omega(y)\geq \frac12|\omega([x,y])|,$$ where now I'm using $\Delta_\omega(x)$ instead of $\sigma_x$ in order to explicitly show the dependence on $\omega$.

• No contest - the point is that if you have the statistical nature of measurement, but only only commuting operators, you have no HUP. – Frank Mar 10 '15 at 22:25
• If two observables commute they behave classically and the HUP tells you that in principle the product of the uncertainties can be zero, i.e. they can be measured with arbitrary precision. If this is not possible from a practical point of view is just because instruments don't have infinite resolution. – Phoenix87 Mar 11 '15 at 10:22
• Maybe statistics and the HUP are conjoined in an inseparable way in QM, *

Yes, they are, because a quantum system has the property that some observables are non-defined. The standard QM says that this property is not a flaw in our measurements, but an intrinsic property of the quantum system itself.

Let $x$ and $p_x$ be the two non-commuting operators. If you want to measure $x$, and the wave-function according to which you prepare particles is not that of a well-determined position, then about each particle and particle that you prepare, you cannot say before the measurement that it has a position. So, a measurement on one single particle collapses the w.f. on $\delta (x - x_0)$ where $x_0$ is one of the positions allowed by the initial w.f. Therefore a measurement on one single particle says nothing. For assessing the values allowed by your w.f. you need to measure many particles

Next, after measuring $x$ on one particle you destroy the of initial w.f. s.t. on the measured particle it makes no sense to measure $p_x$. This is why you need an ensemble of particles prepared according to the same w.f., so as to measure on part of them $x$ and on the other part $p_x$.

So, in order to check the uncertainty principle, you need many particles identically prepared, and that because of the quantum nature of the particles.

but if say, all operators could commute, there would be no HUP. You need the ingredient of non-commutativity which has nothing to do with statistics. Is that correct?

If all the operators commute we have classical mechanics. One of the proofs of the HUP is based on the fact that the commutator is non-zero.

Now, the non-commutativity is indirectly connected with the statistics. The fact that the two operators don't commute, implies that at least one of them is undetermined, has no fixed value $\to$ statistics.

A simple example: the oerators $x$ and $\hat p_x = m \hat v_x$ don't commute. Let's fix one of these observables, e.g. $x=x_0$. What can we say about the velocity of the quantum particle? If we believe to QM, $v_x$ becomes completely undetermined. Let's check that by measurement. How do we measure velocity? Some time $Δt$ after fixing $x$, we measure $x$ again. Let's say that we find $x=x_1$. So, we calculate $v_1 =(x_1−x_0)/Δt$. But, let's check this result on another particle, that we also fixed initially at $x_0$. After the same interval $Δt$ we measure the position and find, for instance, $x_2 \ne x_1$. So, we get $v_2 \ne v_1$. Then, to which one of the results to believe, to $v_1$, or to $v_2$? No way to prefer one of them. Then, maybe QM is right and for fix $x$, the observable $v_x$ is undetermined?

• I totally get it and agree - but note that if all operators commuted, there would be no HUP. – Frank Mar 10 '15 at 22:26
• I see your problem - the non-commutativity is indirectly connected with the statistics. The fact that the two operators don't commute, implies that at least one of them is undetermined, has no fixed value --> statistics. – Sofia Mar 10 '15 at 22:40
• Can you elaborate? Can you show that non-commutation necessarily implies probabilistic measurement in more details? – Frank Mar 10 '15 at 23:00
• @Frank I will try. – Sofia Mar 10 '15 at 23:13
• @Frank $x$ and $p_x$ don't commute, right? Let's fix one of them, e.g. $x = x_0$. What can we say about the velocity? If we believe to QM, $v$ becomes completely undetermined. Let's measure. How we measure? Some time $\Delta t$ after fixing $x$, we measure $x$ again; we find, say, $x_1$, and calculate $v_1 =(x_1 - x_0)/\Delta t$. Right? But, let's check our result on another particle, that we also fixed at $x_0$. After $\Delta t$ we will find $x_2$. So, we get $v_2$. To which one of the results to believe, to $v_1$ or to $v_2$? Then, maybe QM is right and for fix $x$, $v_x$ is undetermined? – Sofia Mar 10 '15 at 23:20