Uncertainty in position measurement in two ensembles. First with same $\psi$ but different $N$, and second with same $\psi,N$ but different apparatus

Question

Imagine an ensemble of $N$ identical and identically prepared quantum systems, all of which are in the state $\psi(x,t)$ at time $t$. Given the state (which could be a Gaussian in position) the postulates of quantum mechanics tell us, for example, what will be the result of position measurements on this ensemble at time $t$ i.e. which position eigenvalue will be obtained with what probability. It allows us to theoretically calculate $\Delta x$ from $\psi(x,t)$, ONLY. Given $\psi(x,t)$, the calculation yields a definite value for $\Delta x$ (say, $\Delta x=0.05$mm). This value, solely obtained from $\psi(x,t)$, seems to be blind to how the process of measurement is (or will be) carried out.

• For a given ensemble with fixed $N$ and given $\psi(x,t)$, it is not true that $\Delta x$ will depend on how precise an apparatus is used to make the measurements?

However, I don't think there is any serious problem here. If, for example, $x\in[-5,+5]$ in some units, and if the measuring apparatus has a least count $1$ in the same units, the only allowed values that can arise in the measurement are $[-5,-4,-3,...+3,+4,+5]$ (something like $1.3$ or $3.7$ is not measurable). Therefore, the theoretical value of $\Delta x$ should also be calculated by discretizing the integrals over $x$. On the other hand, if the least count of the apparatus were $0.5$ in the same unit, allowed $x$ values will be more in number than the previous case. Thus the theoretical $\Delta x$ should be re-calculated accordingly. So it seems that theoretical $\Delta x$ also has a direct bearing on how the measurement is carried out.

• However, experimentally, is it also not true that $\Delta x$ will be different for an ensemble with $N=1000$ and another with $N=10000$ both ensembles being specified by the same state $\psi(x,t)$)? How do we resolve this?

Answer 1

• The actual measurements, of course, depend on the accuracy and precision of the measuring apparatus. You are correct that we do not take this into account when we calculate $\Delta x$. However, there is something more pervasive in the case of the position operator -- it is that the framework of quantum mechanics itself tells us that the position operator is not really observable in the sense that eigenstates of the position operator are not in the Hilbert space (because they are not normalizable).
• Of course, experimental verification of any probability distribution intrinsically refers to ratios of the frequency of an outcome to the total number of trials as the total number of trials $\to \infty$. The way we derive $\Delta x$ is simply by evaluating the expectation values of $x$ and $x^2$ and these expectation values have baked into them this reference to the total number of trials $\to \infty$. So, in terms of your formulation of the question, the $\Delta x$ that we calculate is for $N\to\infty$, the larger the $N$ you take, the better you approximate what you are actually calculating.

Answer 2

Quantum mechanical uncertainty - that which we denote by $\Delta x$ - has nothing to do with measurement, see for example this question and its linked questions. The $\Delta x$ we compute in quantum mechanics is the standard deviation of $x$ assuming a perfect measurement apparatus. It is an abstract statistical quantity derived from the probability distribution for the position variable that is encoded in the quantum state ("wavefunction") and has no direct relation with any actual measurements being performed.

Think about flipping a fair coin, i.e. a coin which you believe has 50% probability to show heads and 50% to show tails. If we assign heads the value -1 and tails the value 1, then the expected value is 0, with a standard deviation of 1. The expected value of $n$ coin tosses is still 0, with a standard deviation of $\sqrt{n}$. If you actually go and flip $n$ coins, you can try to estimate the standard deviation of the underlying distribution with one of the common expressions for standard deviations of samples. This might come out to be close to $\sqrt{n}$, it might not - the only thing that is guaranteed is that the estimation converges to the theoretical value as $n\to\infty$. Note that in this case, the measurement apparatus is perfect - we can tell whether or not a coin is heads or tails without any room for error.

That is, the "$\Delta x$" you compute from a sample is not actually the same quantity as the $\Delta x$ we compute from the theory for a quantum state - the former is merely an estimation of the latter, even if we have a perfect measurement apparatus.