- #36
- 24,488
- 15,033
That's simply not true! The precision of the measurement device is determined by the measurement device, not the standard deviation due to the preparation (i.e., the prepared state) of the measured system.A. Neumaier said:That's irrelevant.
The thermal interpretation never claims the caricature you take it to claim, namely that one always gets the q-expectation. It only claims that the measurement result one gets approximates the predicted q-expectation ##\langle A\rangle## with an error of the order of the predicted uncertainty ##\sigma_A##. When the latter is large, as in the case of a spin measurement, this is true even when the q-expectation vanishes and the measured values are ##\pm 1/2##!
If you want to establish that the standard deviation of ##A## is ##\sigma_A## you have to measure with much higher precision than ##\sigma_A##, and you have to use a sufficiently large ensemble to gain "enough statistics" to establish that the standard deviation due to the state preparation is ##\sigma_A##.
Again for the SGE: If your resolution of the spin-##z##-component measurement resolves the measured values ##\pm 1/2## and the state is prepared such that ##\langle s_z \rangle=0##, you never find the result ##\langle s_z \rangle=0## with some uncertainty but you find with certainty either ##+1/2## or ##-1/2##, and to establish that the standard deviation is ##\sigma_{\sigma_z}## you have to simply do the measurement often enough on equally prepared spins to gain enough statistics to verify this prediction (at the given confidence level, which usually is ##5 \sigma_{\text{measurement}}## (where here ##\sigma_{\text{measurement}}## is the standard deviation of the measurement, not that of the quantum state!).