# Born rule and thermodynamics

stevendaryl
Staff Emeritus
In physics, probabilities and averages give the same information, so I accept vanhees71's definition of the Born rule.
I was about to say that, but I wasn't sure that it's true. From probabilities you can compute averages, but I'm not sure about the other way around. For example, suppose that I tell you that $\langle S_z \rangle = 0$. That doesn't give me much information about probabilities of various values of $S_z$.

But maybe if for some observable $A$, I know $\langle A^n \rangle$ for every $n$, does that uniquely determine the probability?

• vanhees71
atyy
But maybe if for some observable $A$, I know $\langle A^n \rangle$ for every $n$, does that uniquely determine the probability?
Yes, it's like Taylor series. For example, the cumulant expansion specifies a generating function or characteristic function, which is equivalent to specifying the probability distribution. The generating or characteristic function is analogous to the partition function of statistical physics and QFT.

This can fail, but I think such cases are not physical.

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• vanhees71
A. Neumaier
2019 Award
But maybe if for some observable $A$, I know $\langle A^n \rangle$ for every $n$, does that uniquely determine the probability?
This is called the moment problem. Although correct under suitable additional conditions on the probability measure, it has a negative answer in general, i.e., the expectations of all powers of ##A## do not determine its probability distribution. For counterexamples see, e.g.,

J. Stoyanov, Inverse Gaussian distribution and the moment problem, J. Appl. Statist. Science 9 (1999), 61-71.
https://www.researchgate.net/publication/246535073

These counterexamples are not exotic as they include, for example, certain inverse Gaussian distributions.

• vanhees71 and dextercioby
I didn't say that there is no preferred basis at all. There isn't in MWI, but there is in Bohmian interpretation.

Zurek summarized it:

Let us summarize our results for the environment-induced selection of preferred states and discuss the implications for the general preferred-basis-problem outlined in Sect. 2.5.2 and for our observation of only particular phyiscal quantities in the world around us.

System-environment interaction Hamiltonians frequency described a scattering process of surrounding particles (photons, air molecules, etc.) interacting with the system under study. Since the force laws describing such processes typically depend on some power of distance (such as @ r ^-2 in Newton's or Coulomb's force law), the interaction Hamiltonian will usually commute with the position operator. According to the commutativity requirement (2.89), the pointer states will therefore be approximate eigenstates of position. The fact that position is typically the determinate property of our experience can thus be explained by referring to the dependency of most interactions on distance.
Bill Hobba kept mentioning the above. For example Bill stated that:

The basis is chosen in theory ie the position basis is implicit in the normal form of the Schrodinger equation. When expressed in that basis virtually all interactions turn out to be radial ie the V(x) in the Schrodinger equation. Its purely arbitrary what basis you chose to write your equations in - writing them in the position basis is whats usually done and you can easily see their radial nature.

"In a nutshell, the argument is this:
To define separate worlds of MWI, one needs a preferred basis, which is an old well-known problem of MWI. In modern literature, one often finds the claim that the basis problem is solved by decoherence. What J-M Schwindt points out is that decoherence is not enough. Namely, decoherence solves the basis problem only if it is already known how to split the system into subsystems (typically, the measured system and the environment). But if the state in the Hilbert space is all what exists, then such a split is not unique. Therefore, MWI claiming that state in the Hilbert space is all what exists cannot resolve the basis problem, and thus cannot define separate worlds. Period! One needs some additional structure not present in the states of the Hilbert space themselves.

As reasonable possibilities for the additional structure, he mentions observers of the Copenhagen interpretation, particles of the Bohmian interpretation, and the possibility that quantum mechanics is not fundamental at all. "
My question:

Does the above mean that without preferred basis or additional structure, Bill stuff about the radial nature of interactions choosing position as the basis won't even occur?

Or is this radial thing (or interaction Hamiltonian within the pure MWI without preferred basis) a separate thing from the concept of additional preferred basis put in by hand?

Demystifier
Gold Member
Does the above mean that without preferred basis or additional structure, Bill stuff about the radial nature of interactions choosing position as the basis won't even occur?
Yes.

Yes.
But then according to Peterdonis, he suggested the interaction Hamiltonian (which can be where the Hobba radial argument was based) was independent of the preferred basis as when he wrote "(One point that I have not raised is that the wave function is not the only "structure" present in QM; there is also the Hamiltonian, or Lagrangian if you are doing QFT. So one possibility that we have not discussed is that the "additional structure" is in the Hamiltonian, not the wave function; that the Hamiltonian of the cat, or the cat/environment system, is what picks out the alive/dead basis as the one that gets decohered.)"

Demystifier
Gold Member
In physics, probabilities and averages give the same information, so I accept vanhees71's definition of the Born rule.
Consider a physical system in which the average value of position is ##<x>=0##. What is the probability that the position is ##x=1## nm?

Demystifier
Gold Member
For the sake of conceptual simplicity, consider a classical system (e.g. a planet) with a measured trajectory ##x(t)##. And suppose that this trajectory can be explained theoretically by two different Hamiltonians. Without considering any other experiments/measurements, is it possible to determine which Hamiltonian is the correct one? It seems obvious to me that the answer is - no. Therefore, the Hamiltonian itself is not a physical object. Therefore it cannot define physical objects, including physical subsystems.

vanhees71
Gold Member
2019 Award
To reconstruct the probability distribution, you need all moments of it, the average is not enough. This has nothing to do with the foundations of QT. So please for get this irrelevant discussion of standard theorems of probability theory and take Born's rule to mean the probabilities for the outcome of measurements given the operator algebra of observables and the state in terms of the statistical operator.

For the sake of conceptual simplicity, consider a classical system (e.g. a planet) with a measured trajectory ##x(t)##. And suppose that this trajectory can be explained theoretically by two different Hamiltonians. Without considering any other experiments/measurements, is it possible to determine which Hamiltonian is the correct one? It seems obvious to me that the answer is - no. Therefore, the Hamiltonian itself is not a physical object. Therefore it cannot define physical objects, including physical subsystems.
Hope Peterdonis can address this as it's his belief the state vector may be sufficient in itself to create the subsystems.

Anyway. What would happen if you use the Schroedinger equations without any implicit basis in the pure MWI (without any basis). Can the Schroedinger Equation still produce output with all nonorthogonal results? And what good would be this output?

And can you propose experiments more complex than the Stern-Gerlach setup to tell whether the state vector in MWI can or really can't produce a basis without additional structure. What efforts are being done in this department?

Demystifier
Gold Member
Anyway. What would happen if you use the Schroedinger equations without any implicit basis in the pure MWI (without any basis). Can the Schroedinger Equation still produce output with all nonorthogonal results? And what good would be this output?
One could get the spectrum of Hamiltonian.

And can you propose experiments more complex than the Stern-Gerlach setup to tell whether the state vector in MWI can or really can't produce a basis without additional structure.
I cannot.

One could get the spectrum of Hamiltonian.

I cannot.
Theoretically if say you suddenly removed the position basis of an object let's say a post office postcard.. what would happen to the atoms and molecules inside.. would they become damaged like being exposed to fire where all atoms/molecules are rearrange permanently or destroyed??

And can you think of a theoretical way to save the information of position somewhere so that you can dematerialize the post office postcard.. and rematerialize it later with all positions intact?

Demystifier
Gold Member
Theoretically if say you suddenly removed the position basis of an object let's say a post office postcard.. what would happen to the atoms and molecules inside.. would they become damaged like being exposed to fire where all atoms/molecules are rearrange permanently or destroyed??
The question does not make much sense to me. What does it mean to remove the position basis of an object?

The question does not make much sense to me. What does it mean to remove the position basis of an object?
Simple.. to remove the position basis so the object would no longer have position.. or in terms of Bohmian Mechanics.. to remove position as preferred.. so it's like transforming the object into pure MWI vector or Hilbert space. Now if this occurs.. are the information of the arrangements of the molecules of the object still intact? Only they become pure Hilbert space vectors? and if you enable the position basis again.. would the information return or would the object become unrecognizable and become a mere blob because the nonorthogonal conversion to orthogonal is random hence the position information can't be stored and reread?

Demystifier
Gold Member
Simple.. to remove the position basis so the object would no longer have position.. or in terms of Bohmian Mechanics.. to remove position as preferred.. so it's like transforming the object into pure MWI vector or Hilbert space. Now if this occurs.. are the information of the arrangements of the molecules of the object still intact? Only they become pure Hilbert space vectors? and if you enable the position basis again.. would the information return or would the object become unrecognizable and become a mere blob because the nonorthogonal conversion to orthogonal is random hence the position information can't be stored and reread?
I think such a question can be meaningfully asked only by using mathematical equations.

stevendaryl
Staff Emeritus
Consider a physical system in which the average value of position is ##<x>=0##. What is the probability that the position is ##x=1## nm?
But if you also know $\langle x^2 \rangle, \langle x^3 \rangle, ...$?

• Demystifier
I think such a question can be meaningfully asked only by using mathematical equations.
Ok, mathematically.. if we remove the position basis.. then there would be no decoherence and no subsystems if there are no other basis to define it. And it's back to pure Hilbert space vectors with unit trace 1.

So the question is.. what kind of information can be stored in the Hilbert space nonorthogonal states? How complex the information it can hold? Can they store a Barbie doll information? or nothing? just wanna have idea...

PeterDonis
Mentor
2019 Award
suppose that this trajectory can be explained theoretically by two different Hamiltonians
Is this possible? Can you give an actual example?

atyy
Consider a physical system in which the average value of position is ##<x>=0##. What is the probability that the position is ##x=1## nm?
To be more conventional, one can use the term "expectation" in place of "average". Conceptually, there is no difference. You can see A. Neumaier's post #78 for the mathematical fine print.

Demystifier
Gold Member
Ok, mathematically.. if we remove the position basis.. then there would be no decoherence and no subsystems if there are no other basis to define it. And it's back to pure Hilbert space vectors with unit trace 1.
Yes.

Demystifier
$$H_1=\frac{p^2}{2m}$$
$$H_2=\frac{p^2}{2m}+\frac{k x^2}{2}$$