Understanding the Uniform Probability Distribution in Statistical Ensembles

[A. Neumaier]

It seems to me that you are just hiding problems under the rug.

Only under the rug of common language (for simple phrases like ''a small multiple") where everyone hides stuff since it is already full of philosophical problems. But common language is necessary to do any kind of science.

Whereas subjective probability (the rug where you are hiding the problems under) is another can of worms involving additional, much more severe philosophical problems.

 

[A. Neumaier]

the usual interpretations of "uncertainty" and "approximately" are subjective.

whereas I interpreted both very carefully in terms of the only undefined notion of ''a small multiple, typically 3 or 5'', which I can assume to be understood by everybody.

 

[Jilang]

I am going with Dirac. Probabilities need not be positive.
http://arxiv.org/pdf/1008.1287.pdf

 

[A. Neumaier]

Probabilities need not be positive.

This statement is correct with probability 1 - since probabilities can be zero it is a triviality.

 

[Demystifier]

The Hamiltonian H is invariant in time, hence does not fluctuate at all.

In QM, fluctuation is not a random change with time. Fluctuation is just another name for uncertainty. So the Hamiltonian fluctuates in any state which is not an eigenstate of $H$.

 

[vanhees71]

Apply in the sense that statistical mechanics applies to a single glass of water. One uses ensemble expectation values for the single quantum system [and, according to Gibbs, nonphysical, imagined repetitions to justify the ensemble language for the single use case] to assign a temperature and other things that can be measured.

Single, nonrepeated measurements of temperature, pressure and volume can be used to check the predictions of quantum mechanics in equilibrium. These measurements have nothing to do with any of the mock measurements of identically prepared systems discussed in the traditional interpretations of quantum mechanics.

The single glass of water is described by thermodynamic quantities like temperature and pressure. If you measure its temperature you have to put a thermometer for a sufficiently long time into the water. Then the thermometer equilibrates with the water, and you can read off a temperature, which is a time-averaged kinetic energy per particle. Also when looking at macroscopic quantities of a single system these macroscopic quantities are averaged (in this case over time) microscopic quantities. The thermometer as a measurement apparatus doesn't resolve the thermal (and quantum) fluctuations of energy, and thus averages out these fluctuations delivering a macroscopic quantity we define as temperature.

 

[A. Neumaier]

In QM, fluctuation is not a random change with time. Fluctuation is just another name for uncertainty. So the Hamiltonian fluctuates in any state which is not an eigenstate of $H$.

True, but this doesn't conform with stevendaryl's use of the term, which I was using in the discussion with him.

 

[vanhees71]

If not in a stationary state, there are also quantum fluctuations of quantities in time, right?

 

[A. Neumaier]

The single glass of water is described by thermodynamic quantities like temperature and pressure. If you measure its temperature you have to put a thermometer for a sufficiently long time into the water. Then the thermometer equilibrates with the water, and you can read off a temperature, which is a time-averaged kinetic energy per particle. Also when looking at macroscopic quantities of a single system these macroscopic quantities are averaged (in this case over time) microscopic quantities. The thermometer as a measurement apparatus doesn't resolve the thermal (and quantum) fluctuations of energy, and thus averages out these fluctuations delivering a macroscopic quantity we define as temperature.

Yes; you make my point: Compare your description with what the quantum mechanics 1 postulates claim a measurement of a quantum system is.

Single, nonrepeated measurements of temperature, pressure and volume can be used to check the predictions of quantum mechanics in equilibrium. These measurements have nothing to do with any of the mock measurements of identically prepared systems discussed in the traditional interpretations of quantum mechanics.

 

[A. Neumaier]

If not in a stationary state, there are also quantum fluctuations of quantities in time, right?

$H$ is time invariant, hence doesn't fluctuate in time, no matter which state is considered. It is also translation invariant, hence doesn't fluctuate in space. Thus fluctuations have neither a dynamic nor a spatial meaning - the term is used in the same figurative way as vacuum fluctuations in a vacuum whose particle number is zero at all times and everywhere.

 

[vanhees71]

What do they claim? In QM 1 you usually start with some heuristics to motivate the postulates, and the postulates don't talk about details of specific measurements at all. This you also don't do in the theoretical classical physics curriculum, and why should you? For me the postulates (roughly) read like this.

(0) Time is represented by a real parameter
(1) A quantum system is described by a Hilbert space and an algebra of observables represented by self-adjoint operators. The possible values of observables is given by the spectrum of their corresponding representing self-adjoint operators.
(2) There exists a Hamilton operator $\hat{H}$ such that if $\hat{O}$ represents an observable, the operator
$$\mathring{\hat{O}}=\frac{1}{\mathrm{i} \hbar} [\hat{H},\hat{O}]+(\partial_t \hat{O})_{\text{ext}}$$
represents the time derivative of this observable.
(3) States are represented by a self-adjoint trace-class positive semidefinite operator $\hat{\rho}$ with $\mathrm{Tr} \hat{\rho}=1$, fulfilling $\mathring{\hat{\rho}}=0$.
(4) If a quantum system is prepared in the state $\hat{\rho}$ and if $|o,\alpha \rangle$ are the (generalized) eigenvectors of the operator $\hat{O}$ representing an observable $O$, then the probability (distribution) that the observable takes the value $o$ is given by
$$P(o|\hat{\rho})=\sum_{\alpha} \langle o,\alpha |\hat{\rho} |o,\alpha \rangle.$$

 

[vanhees71]

$H$ is time invariant, hence doesn't fluctuate in time, no matter which state is considered. It is also translation invariant, hence doesn't fluctuate in space. Thus fluctuations have neither a dynamic nor a spatial meaning - the term is used in the same figurative way as vacuum fluctuations in a vacuum whose particle number is zero at all times and everywhere.

Then define "fluctuations". Of course, for a closed system $\hat{H}$ is not explicitly time dependent and thus also doesn't depend on time. In general, however, the standard deviation of an observable in a state which is not an eigenstate of this observable at all times, is time dependent, and thus the observable fluctuates in time.

 

[A. Neumaier]

For me the postulates (roughly) read like this.

Yes; this is the typical introduction. Try to match it in any way with your description of temperature measurement and you'll find that it doesn't explain the probability with which the temperature ''takes'' a given value (whichever operational interpretation you give to the word ''take'' -which is nowhere made as precise as stevendaryl wanted in the present discussion).

 

[A. Neumaier]

Then define "fluctuations". Of course, for a closed system $\hat{H}$ is not explicitly time dependent and thus also doesn't depend on time. In general, however, the standard deviation of an observable in a state which is not an eigenstate of this observable at all times, is time dependent, and thus the observable fluctuates in time.

In the present context I took stevendaryl's meaning for it - which made sense only as a temporal or spatial fluctuation.

In a figurative sense, fluctuation is just a buzzword for uncertainty, as Demystifier said. Uncertainty simply means that there is no way to fix the value to a precision more than a few standard deviations. This applies always (independent of repetitions or subjective probabilities) and has nothing to do with spatial or temporal fluctuations but with lack of a more precise specification.

 

[vanhees71]

Yes; this is the typical introduction. Try to match it in any way with your description of temperature measurement and you'll find that it doesn't explain the probability with which the temperature ''takes'' a given value (whichever operational interpretation you give to the word ''take'' -which is nowhere made as precise as stevendaryl wanted in the present discussion).

A temperature is not an observable in this quantum mechanical sense! It's a macroscopic coarse-grained quantity.

 

[A. Neumaier]

A temperature is not an observable in this quantum mechanical sense! It's a macroscopic coarse-grained quantity.

Yes. This is why the QM1 postulates are sorely lacking in quality. They hijack the notion of an observable for a very special kind of observation (namely a von-Neumann measurement of a quantity taking rational values only). The examples that typically follow misuse the postulates immediately for position measurements, which always yield a rational result though the spectrum of $q$ contains far more irrational numbers than rational ones. Without the standard shut-up-and-calculate approach even in the very first stages one would never come far when one would only allow strictly logical conclusions from the postulates.

 

[vanhees71]

In the present context I took stevendaryl's meaning for it - which made sense only as a temporal or spatial fluctuation.

In a figurative sense, fluctuation is just a buzzword for uncertainty, as Demystifier said. Uncertainty simply means that there is no way to fix the value to a precision more than a few standard deviations. This applies always (independent of repetitions or subjective probabilities) and has nothing to do with spatial or temporal fluctuations but with lack of a more precise specification.

It depends on the state. An observable takes a certain value if its state is described by a statistical operator of the form
$$\hat{\rho}=\sum_{\alpha} p_{\alpha} |o,\alpha \langle \rangle o,\alpha|, \quad \sum_{\alpha} p_{\alpha}=1, \quad p_{\alpha} \geq 0.$$
I'm not sure whether this is the most general possible such state, but at least if it is of this form, the observable represented by the self-adjoint operator $\hat{O}$ with its orthonormalized eigenvectors $|o,\alpha \rangle$ to the eigenvalue $o$, the probability that $\hat{O}$ takes the value $o$ is 100%, and the standard deviation is 0.

In other states, not built with vectors in the eigenspace to a single eigenvalue $o$, the observable does not have a determined value, but only probabilities to take any of its possible values is given according to Born's rule (my postulate 4).

 

[A. Neumaier]

A temperature is not an observable in this quantum mechanical sense! It's a macroscopic coarse-grained quantity.

Let us consider instead the measurement of $H$, which in every treatment of QM is ''an observable in this quantum mechanical sense''. How do you propose to apply your postulate (4) in order to find out which values $H$ takes?

 

[vanhees71]

Yes. This is why the QM1 postulates are sorely lacking in quality. They hijack the notion of an observable for a very special kind of observation (namely a von-Neumann measurement of a quantity taking rational values only). The examples that typically follow misuse the postulates immediately for position measurements, which always yield a rational result though the spectrum of $q$ contains far more irrational numbers than rational ones. Without the standard shut-up-and-calculate approach even in the very first stages one would never come far when one would only allow strictly logical conclusions from the postulates.

Well, also in Theory 1 you dont' start with statistical quantities but with the "microscopic" ones in terms of classical mechanics. You have bodies (usually idealized as mass points) running along trajectories in position space first (then you refine your mathematical description to Hamiltonian mechanics in Theory 2 and have trajectories in phase space). Nowhere are quantities like temperature in sight here. You can (with a lot of hand-waving and head aches) introduce them in classical statistical physics. Of course, it's much more clear in quantum statistical physics, because you avoid the quibbles of classical statsistics Boltzmann et al had to fight with, but that's another topic.

 

[A. Neumaier]

It depends on the state. An observable takes a certain value if [...]

In my words: The standard deviation (and hence the uncertainty) of $O$ is zero iff the state $\rho$ satisfies $O\rho=o\rho$; then $o$ is the exact value. In all other cases, the value is uncertain.

This has nothing at all to do with fluctuations in space or in time, as the example of $O=H$ shows.

 

[vanhees71]

Let us consider instead the measurement of $H$, which in every treatment of QM is ''an observable in this quantum mechanical sense''. How do you propose to apply your postulate (4) in order to find out which values $H$ takes?

You measure it! I don't need postulate 4 but a measurement apparatus in the lab. It doesn't tell me much to measure it just once in view of postulate 4. To check postulate 4, supposed that you know the state via a preparation procedure, you need to repeat the experiment on an ensemble. It's all not that complicated as it appears to be in these endless discussions as soon as you have accepted that QT is a probabilistic theory.

 

[A. Neumaier]

Well, also in Theory 1 you don't start with statistical quantities but with the "microscopic" ones in terms of classical mechanics. You have bodies (usually idealized as mass points) running along trajectories in position space first (then you refine your mathematical description to Hamiltonian mechanics in Theory 2 and have trajectories in phase space). Nowhere are quantities like temperature in sight here.

Even for quantities like $q$ mentioned in my comment it is not clear what it should mean that $q$ takes the value $\pi=3.14159...$. Or that $H=p^2/2m+kq^2/2$ takes the value $3/2$.

 

[A. Neumaier]

You measure it! I don't need postulate 4 but a measurement apparatus in the lab. It doesn't tell me much to measure it just once in view of postulate 4. To check postulate 4, supposed that you know the state via a preparation procedure, you need to repeat the experiment on an ensemble. It's all not that complicated as it appears to be in these endless discussions as soon as you have accepted that QT is a probabilistic theory.

Which measurement apparatus do you propose to measure $H$ in the lab? I don't even know how to measure it for a harmonic oscillator as given in post #222.

 

[vanhees71]

In my words: The standard deviation (and hence the uncertainty) of $O$ is zero iff the state $\rho$ satisfies $O\rho=o\rho$; then $o$ is the exact value. In all other cases, the value is uncertain.

Sure, that's obvious in view of the postulates.

This has nothing at all to do with fluctuations in space or in time, as the example of $O=H$ shows.

I don't understand this statement. On the one hand you say the observable fluctuates iff the state doesn't satisfy $\hat{O} \hat{\rho}=o \hat{\rho}$. On the other hand you say it doesn't. Also I don't see what's so special about the energy and the Hamiltonian as its representing operator. If $\hat{H} \hat{\rho} \neq E \hat{\rho}$ for any of the Hamiltonian's eigenvalues/spectral values $E$, it fluctuates, and in general it will fluctuate differently at different times.

 

[vanhees71]

Which measurement apparatus do you propose to measure $H$ in the lab? I don't even know how to measure it for a harmonic oscillator as given in post #222.

For a harmonic oscillator it's a tough question, and I've to think about it. One way to realize a particle in an harmonic oscillator is to put it into a trap. Perhaps, this is an interesting example:

http://arxiv.org/abs/0909.1095

To measure the energy of, e.g., a particle you have various choices. One is to use a calorimeter. The particle gets absorbed and you measure how the temperature of the calorimeter changes.

 

[A. Neumaier]

On the one hand you say the observable fluctuates iff the state doesn't satisfy $\hat{O} \hat{\rho}=o \hat{\rho}$. On the other hand you say it doesn't. Also I don't see what's so special about the energy and the Hamiltonian as its representing operator.

I was saying that

1. ''fluctuates'' is just a popular buzzword for ''is uncertain''.
2. This has nothing at all to do with fluctuations in space or in time.
3. This can be seen by considering the observable $H$.

1. defines how I am using the term; you had asked for it in post #212. I had also defined the meaning of ''uncertain'' in post #190.

2. follows already since the English language doesn't require the corresponding connotation:
Simple definition of uncertain (from http://www.merriam-webster.com/dictionary/uncertain):

• not exactly known or decided
• not definite or fixed
• not sure
• having some doubt about something
• not definitely known

3. Is a physical instance since it is clear that $H$ is invariant in time and under translations. The fact that $\rho$ is not an eigenstate implies that the value of $H$ is uncertain hence fluctuates, while time and translation invariance imply that the value is constant in space and time. Thus the value of $H$ is uncertain although there are no temporal or spatial changes (and fluctuations in the temporal or spatial sense are absent).

 

[A. Neumaier]

To measure the energy of, e.g., a particle you have various choices. One is to use a calorimeter. The particle gets absorbed and you measure how the temperature of the calorimeter changes.

This is not a measurement of $H$ but of the kinetic energy only.

 

[N88]

This statement is correct with probability 1 - since probabilities can be zero it is a triviality.

Dirac in his Bakerian Lecture (1941) [ http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.205.4474&rep=rep1&type=pdf ] refers to negative probabilities. So do others (including Feynman).

Supporting your interest in eliminating subjective and loose language from QM, how about this:

Tests on ensembles give us an "average value". Let our subjective anticipation of that average value be the "expectation".

Associated with the objective "average value" we can derive the objective "prevalence" of relevant observables, which is never negative.

Associated with the "expectation" is the subjective (and often confusing) notion of "probability". That it can be anywhere negative is up to the subject subject and beyond me. Objectively, use of the term "prevalence" is a step toward eliminating all debate re "probabilities" (and "betting") in physics.

 

[A. Neumaier]

negative probabilities.

Statements about negative probabilities are correct with probability $<0$. I.e., there is not even a set of measure zero where these statements are true (excpept perhaps in a figurative sense).

 

[A. Neumaier]

"prevalence"

Essentially nobody understands the meaning of this word. I find it unacceptable to try to solve philosophical issues by introducing new words without a clearly defined meaning.

 

[N88]

Essentially nobody understands the meaning of this word. I find it unacceptable to try to solve philosophical issues by introducing new words without a clearly defined meaning.

Referring to post #228, and seeking clear definitions: A probability assignment (symbol = lower-case p, say) is a normalised subjective judgment based on incomplete knowledge about an ensemble of interest. A prevalence (symbol = capital P, say) is a normalised objective fact about a fully tested ensemble.

Let a fully tested ensemble have sample space Ω with numerical observables O_i. Then the mean value of the observables is:

⟨O⟩ = 1/N ΣN_iO_i = ΣP_iO_i​

where P_i is the prevalence of observable O_i, the normalised proportion of O_i in the tested ensemble.

Thus, prevalences P and probabilities (substituting p) satisfy the same rules:

1. P(A|A) = 1 = P(Ω).
2. 0 ≤ P(Α|Β) ≤ 1.
3. P(¬A|B) = 1 - P(A|B).
4. P(AB|C) = P(A|C)P(B|AC).​

But each P represents an objective fact, each p represents a subject judgment.

 

[A. Neumaier]

But each P represents an objective fact, each p represents a subject judgment.

So you just invent a strange new name ''prevalence'' for the standard notion ''relative frequency''. But what does it mean that an incompletely tested ensemble has a prevalence of $\sqrt{1/2}$, say? Note that most ensembles in physics are infinite, hence can never be exhaustively tested in your sense.