Is uncertainty uncertain?

Question

I was thinking about uncertainty and wondered if uncertainty is inherently uncertain. I know in the uncertainty inequality for position and momentum, $\Delta x \Delta p \ge \hbar/2$, the presence of the greater than or equal to sign implies that the quantities are not fully known. This leads me to ask: Is uncertainty uncertain? How accurate is our knowledge of how much we know about a system? How accurate can this knowledge be? Note that this does not ask if the uncertainty principle is just a limitation of our knowledge, but if there is uncertainty in uncertainty itself.

Answer 1

Let's consider the wave function of the ground state of the harmonic oscillator $$ \psi_0(x) = N_x\; e^{- \frac{x^2}{2\sigma_x^2}} $$ where the normalisation constant $N_x$ is defined by the relation $\int_{-\infty}^{\infty}dx |\psi_0(x)|^2 = 1$. The uncertainty of the position wave function is $ \Delta x %= \sqrt{E[x^2] - (E[x])^2} = \sigma_x $. The uncertainty is a constant, it does not have any uncertainty itself.

The same is true for its Fourier transform, which is
$$ \Psi_0(k)= N_k\; e^{-\frac{(k\cdot \sigma_x)^2}{2}} = N_k\; e^{-\frac{k^2}{2 \sigma_k^2}} $$ with $\sigma_k^2 = 1/\sigma^{2}_x$. Again, the uncertainty is a constant, $\Delta k=\sigma_k$.

While this simple model is true in theory, the situation becomes more complicated if we try to measure these quantities using an experiment. In the experiment we always have additional errors and we are unable to obtain infinite many data points. Thus, if we evaluate the data, we will only obtain an estimate of the intrinsic uncertainty, $\sigma_x$, and this estimate possesses an uncertainty itself. However, this additional uncertainty does not obey Heisenberg's principle. The additional uncertainty is not a fundamental limit, but only a measurement error in the classical sense.

Sidemark: Note that the harmonic ground state is not (!) a state of minimal uncertainty, because $\sigma_k \cdot \sigma_x = 1 > 1/2$. The minimal uncertainty states of the harmonic oscillator are the so called squeezed coherent states.

Sidemark 2: In quantum mechanics we usually adopt the above shown description. Putting this in a statistical context, we use the so called frequency description of probability. In this description the standard deviation of the population is not a random variable. However, there exists an alternative description, the so called Bayesian interpretation of probability. Using the Bayesian interpretation, the probability is merely a "current state of believe". In the Bayesian context the standard deviation of the population is a random variable with an associated probability distribution. Hence, in the Bayesian interpretation the uncertainty has an uncertainty. However, the (mathematically more challenging) Bayesian description usually not used in quantum mechanics.

Answer 2

We must understand that the $\ge$ sign does not show that the uncertainty is not fully known. While measuring $x$ and $p$, there would be errors, which are due to the instrument used... (eg : If we are using a geometry scale, it would be 1mm and if the least count of our momentum measuring apparatus is say 1cm/s then $\Delta P$ is 1 cm/s).

Uncertainty principle says that the error would always be greater than $\frac{\hbar}{2}$, implying that no matter how accurate your apparatus may be, a particle always has a small wave nature and hence, its position would always be uncertain after a limit.

To conclude, there is no uncertainty in uncertainity. Heisenberg was clever enough to write uncertainty relation as $\Delta x \Delta p \ge \hbar/2$

instead of writing the maximum possible accuracy for $\Delta x \Delta p = \hbar/2$