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According to Griffith's Introduction to Quantum mechanics in section 1.5 Momentum, he said that " the expectation value is the average of repeated measurements on an ensemble of identically prepared systems, not the average of repeated measurements on one and the same system"

So, my question is what does the ensemble of particles actually mean and how making a large identical copies of a single particle helps to determine the expectation values of any dynamical variable?

Qmechanic
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My Essential Learning
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    If you were to repeat the same measurement on the same system, it would lock the system into that measured value, which would likely be different from the ensemble average. So, the expectation value can not be interpreted as the average of repeated measurements on the same system. – Raskolnikov Mar 29 '23 at 11:03
  • So, you are saying that, our wavefunction is in superposition of different states and this different states act as a single particle (in our classical sense) which on measurement gives a different value with averages out as the expectation value of that dynamical system? Am I Right? – My Essential Learning Mar 29 '23 at 11:12
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    I'm saying that measurement loosely speaking implies a collapse of the wavefunction into the state corresponding to the measured value. If you repeat the measurement, and there is no time for the system to evolve naturally to another state, you'll get the same value for the measurement. Even if you leave time, the distribution of measured values and the mean of those will not correspond to the expectation value in general. – Raskolnikov Mar 29 '23 at 11:17
  • Ok, Thanks for the solution – My Essential Learning Mar 29 '23 at 11:49
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    Warning: Griffith's statement shall not be confused with the "ensemble interpretation" of QM. Ensemble interpretations: the wave function describes an ensemble of identically prepared systems, in contrast with the “orthodox” interpretations (the wave function provides as complete a description as is possible of an individual system). Griffith is still describing the "orthodox" POV, explained in the comment of @Raskolnikov. See: "Statistical Interpretation of QM" https://doi.org/10.1103/RevModPhys.42.358 and "Ensemble interpretations of QM" https://doi.org/10.1016/0370-1573(92)90088-H – Quillo Apr 01 '23 at 13:25
  • @Raskolnikov, if the time between successive measurements was good enough to ensure that the wavefunction is now evolved in accordance with Sch. eq., why would the expectation value of such an experiment be wrong? – SHD Sep 22 '23 at 15:16
  • @SHD Wrong is not the right word, you'd still measure an expectation value, but it wouldn't necessarily be the one corresponding with the initial state. And that's what the OP is asking about. Griffith explicitly rules out repeated measurements on the same system because there could be an evolution away to a state with a different expectation value than the initial state. – Raskolnikov Sep 25 '23 at 12:58

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Measurement (in classical sense) always involve uncertainty and has to be done repeatedly in order to estimate the value of the measured quantity and its variance: $$ \overline{x} = \frac{1}{N}\sum_{i=1}^Nx_i,\\ \text{var}(x) = \overline{(x -\overline{x})^2} = \frac{1}{N-1}\sum_{i=1}^N(x_i-\overline{x})^2 $$ In classical situations we can perform measuring repeatedly on the same object. E.g., if one wants to measure the weight of an object, we weigh it several times and calculate the average - if we don't see it done in everyday life, it is because we are either using very precise instruments (or conversely do not bother about precision.) But repeated measurements are routinely done in any serious technological or scientific situations. In most situations measurements done on a single object should produce the same results as measurements on different but identical objects - in the latter case we call it sample averaging or ensemble averaging (the former is more statistical term, the latter is more common on physics.)

In quantum mechanics repeated measurements on the same object are impossible, since one measurement destroys the state of the object, and the subsequent measurement is not done on the same object (there are some caveats, but I stick here to the basic QM, that is of the immediate interest to the OP.) This is why the averaging in QM is always understood as an ensemble averaging. The QM then aims at predict the values of the averages that one would obtain in the experiment - and this mathematical predictions is what we call expectation values.

Related: Why can't the Uncertainty Principle be broken for individual measurements if it is a statistical law?

Remark: Another situation were one routinely talks about ensemble averaging is in statistical mechanics, where the ensemble averaging is opposed to averaging over the same system over time. See, e.g., Definition of Ensemble.

Roger V.
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