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Why isn't the variance of a quantum observable depend on how gently or roughly the measurement is carried out?

Physics Asked on May 16, 2021

Consider an ensemble specified by a state $|psirangle$ on which we decide to make measurements of an observable $A$. If the state $|psirangle$ is not an eigenstate of $A$, there will be a scatter about the mean; we will get a nonzero variance, $sigma_A^2=langle A^2rangle-langle Arangle^2neq 0$. For a given $|psirangle$ and $A$, we blindly calculate $sigma_A^2$ and come up with a particular number irrespective of how gently or roughly the measurement is carried out.

From the postulates of QM, if $|psirangle$ is not an eigenstate of $A$, the state will collapse into one of the eigenstates of $A$ upon each measurement with known probabilities irrespective of how gently or roughly the measurement is carried out.

So, it would appear that no matter how strong or weak be the disturbance inflicted on the system via a measurement be, the scatter $sigma_A^2$, will be independent of the degree of disturbance suffered by the system. It would seem that no matter how gently or roughly we measure $A$ (for example, think of using low energy and high energy photons to measure the position of an electron in the state $|psirangle$), as long as $|psirangle$ is the same, the result would be the same. If this understanding is correct, is it not completely counterintuitive and weird?

One Answer

The term standard deviation points to two different situations in your setup - the variance of a state and the error of the measurement - and these are distinct.

To make QM measurements, you need to prepare the system in some special initial $|psi(0)rangle$ state, and measure afterwords the eigenstates of the observable of interest. The probability $p_i$ of finding the system in the eigenstate $i$ of $A$ is given by $|c_i|^2$. Thus you measure the probability via $$ p_i = frac{N_i}{N}, $$ where $N_i$ is the number of eigenvalues $a_i$ and $N$ is the number of total measurements. This process is statistical (Bernulli process) and thus has an standard deviation. If your measurement is bad, then $N_i$ will give a bad approximation to the $p_i$, including the usual types of errors (systematical, statistical).

Additionally your assumption

the state will collapse into one of the eigenstates of A upon each measurement with known probabilities

does only hold approximately, since the time evolution of the initial state depends on external perturbations (how well is the system isolated, temperature effects) and on the measurement procedure.

To conclude, the measurement of a QM variable adds another layer of statistical uncertainty, on top of the intrinsic statistical character of QM.

Correct answer by zufall on May 16, 2021

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