# I The thermal interpretation of quantum physics

#### A. Neumaier

What has this to do with what's measured
I was just answering ftr's question. A bit off-topic but short, so I found it worth explaining.
the measurable outcomes of QT are independent of the choice of the picture of time evolution.
Yes, but my formula for it was not, because there the state was fixed.

#### ftr

What has this to do with what's measured,
Not really sure. But couple of days ago I used the word virtual which has been very controversial which is usually associated with expansion terms in perturbation theory. I am still thinking

#### vanhees71

Gold Member
No, it always is some more or less accurate measurement of the q-expectation. In case the level gap is large a single is just very inaccurate. But by averaging the accuracy becomes better, as always when measurements are inaccurate. See my post in the new thread.
Ok, if you don't want to accept experimental facts, it's not possible to discuss in a scientific way. I give up!

#### vanhees71

Gold Member
I was just answering ftr's question. A bit off-topic but short, so I found it worth explaining.

Yes, but my formula for it was not, because there the state was fixed.
If a result is dependnet on the picture of the time evolution, it doesn't describe anything physical. It's not something related to what can be observed (by definition of standard QT).

#### vanhees71

Gold Member
Not really sure. But couple of days ago I used the word virtual which has been very controversial which is usually associated with expansion terms in perturbation theory. I am still thinking
Well, thinking is always good, but what's sometimes controversial (however never among practicing theoretical physicists) is the meaning of "virtual particle", and there @A. Neumaier has written excellent Insight articles about, fortunately using the "standard interpretation", not his very enigmatic (in my opinion almost certainly incorrect) "thermal interpretation".

#### A. Neumaier

If a result is dependent on the picture of the time evolution, it doesn't describe anything physical. It's not something related to what can be observed (by definition of standard QT).
Since you like nit-picking: The result is picture-independent, but the formula used to write it down was not. In a picture-dependent form I'd have needed to write $\langle A(t)\rangle_t$ since both the operator and the state may depend on time. The Heisenberg picture needs less complicated notation.

#### A. Neumaier

Ok, if you don't want to accept experimental facts, it's not possible to discuss in a scientific way. I give up!
Instead of trying to teach me you could try to understand my very different perspective on things by accepting my terminology and conventions for the sake of discussion. Then it would be possible to discuss constructively.

#### ftr

Well, thinking is always good, but what's sometimes controversial (however never among practicing theoretical physicists) is the meaning of "virtual particle", and there @A. Neumaier has written excellent Insight articles about, fortunately using the "standard interpretation", not his very enigmatic (in my opinion almost certainly incorrect) "thermal interpretation".
I was just alluding to the similarity not that it had anything to directly to do with. Even if you think about a pulse in mathematical terms when expanded in Fourier series with bunch of sine waves. Then you ask what is a pulse, is it additions of these(or some other series) or simply a constant value, if we were to describe some physics we do have to make some interpretation.

#### PeterDonis

Mentor
if you don't want to accept experimental facts,
He's not failing to accept experimental facts. He is just being very clear about exactly what the "experimental facts", as opposed to interpretation, consist of. In the case of a single measurement of a single 2-level system (a spin-1/2 measured about the z axis, for concreteness), the experimental fact is a result "up" or "down"--or, if you want to be really precise, it's the location of a spot on a detector. But calling that result "the measured spin of the system" is not an experimental fact; it's an interpretation.

#### ftr

Arnold, are you saying that the electron has no spin as intrinsic property. In that case what do you think Dirac equation is saying.

#### A. Neumaier

are you saying that the electron has no spin as intrinsic property. In that case what do you think Dirac equation is saying.
Electrons have an intrinsic vector-valued spin given by the q-expectation of $S=\frac{\hbar}{2}\sigma$

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#### A. Neumaier

Well, if you expand in first order perturbation theory a q-expectation $\langle A(t)\rangle$ in the Heisenberg picture into a Fourier integral, you find that these oscillations are excitable.
Actually, no perturbation theory is needed. See my new post here.

#### A. Neumaier

@A. Neumaier: "Points 4 and 5 also show that at ﬁnite times (i.e., outside its use to interpret asymptotic S-matrix elements), Born’s rule cannot be strictly true in relativistic quantum ﬁeld theory, and hence not in nature."

So it seems that you fault the Born's rule for what is actually the Schrödinger equation's fault. The Born's rule is no relative of mine, but this doesn't look like a strong point of your critique of the Born's rule.
In the new version of Part I, now on the arXiv, I changed point 5 in Section 3.3 to account for multiparticle relativistic Hamiltonians that do not have the defects of the Dirac equation.

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#### ftr

In the new version of Part I, now on the arXiv, I changed point 5 to account for multiparticle relativistic Hamiltonians that do not have the defects of the Dirac equation.
Sorry, but it is not clear to me which part, can you reference the page.

#### A. Neumaier

Sorry, but it is not clear to me which part, can you reference the page.
Section 3.3, p.17.

#### ftr

Section 3.3, p.17.
Thanks.
Let's assume we have a two particle Dirac( or Bethe..) equation. Is it possible to calculate q-expectation for the position when they are at a certain distance from each other. If yes, what is the expression? thanks.

#### A. Neumaier

Let's assume we have a two particle Dirac( or Bethe..) equation. Is it possible to calculate q-expectation for the position when they are at a certain distance from each other. If yes, what is the expression? thanks.
There is no consistent 2-particle Dirac equation.

The relativistic dynamics I referred to in point 5 is the one discussed in the big survey 'Relativistic Hamiltonian Dynamics in Nuclear and Particle Physics' by Keister and Polyzou, introduced first by Bakamjian and Thomas in 1953.

I haven't seen explicit solutions; thus you'd have to work out approximation yourself. But this is irrelevant in the context of this thread.

#### ftr

There is no consistent 2-particle Dirac equation.

The relativistic dynamics I referred to in point 5 is the one discussed in the big survey 'Relativistic Hamiltonian Dynamics in Nuclear and Particle Physics' by Keister and Polyzou, introduced first by Bakamjian and Thomas in 1953.

I haven't seen explicit solutions; thus you'd have to work out approximation yourself. But this is irrelevant in the context of this thread.
I think it is relevant but hard to show. However, I will give a simple(simpler) example to show that TI has a merit at least IMO. Take for example the hydrogen atom, the Bohr distance and the expectation value(more so) for the position wavefunction is directly related to the kinetic energy which means the WHOLE wavefunction responded to the potential energy(due to the proton) to shift the expectation value from interaction off to on, hence strongly implying TI.

#### A. Neumaier

Take for example the hydrogen atom, the Bohr distance and the expectation value(more so) for the position wavefunction is directly related to the kinetic energy which means the WHOLE wavefunction responded to the potential energy(due to the proton) to shift the expectation value from interaction off to on, hence strongly implying TI.
Please express this in terms of formulas so that I can understand what you mean.

Use nonrelativistic hydrogen and ignore spin; nothing in your argument seems to depend on the spin or on relativistic ideas.

#### akhmeteli

In the new version of Part I, now on the arXiv, I changed point 5 in Section 3.3 to account for multiparticle relativistic Hamiltonians that do not have the defects of the Dirac equation.
So you wrote there:

" This argument against the exact probability density interpretation of $|\psi|^2$ works even for relativistic particles in the multiparticle framework of Keister & Polyzou [29]."

Could you provide some argumentation or give a more precise reference? It would be difficult to search for the argumentation in the 250-page-long article.

#### A. Neumaier

So you wrote there:

" This argument against the exact probability density interpretation of $|\psi|^2$ works even for relativistic particles in the multiparticle framework of Keister & Polyzou [29]."

Could you provide some argumentation or give a more precise reference? It would be difficult to search for the argumentation in the 250-page-long article.
The argumentation is completely contained in point 4 of Subsection 3.3 of my Part I. The Keister-Polyzou paper just contains dynamical relativistic examples. If you want a definite example, you may take the example of spinless quarks in Section 2.3 (p.26 in the copy cited in post #642). But the details do not matter.

The only relevant points for my argument are that, although the setting is Poincare-covariant,
1. the wave function at fixed time is a function of several spatial momenta, which after Fourier transform to the position representation becomes wave function that is a function of spatial positions,
2. Born's rule makes claims about the probabilities of measuring,
3. the Hamiltonian and the position operators have a nonlocal commutator.
As a result, the dynamics introduces (as claimed in Part I) after arbitrarily short times nonzero probabilities of finding an initially locally prepared particle (initial wave function with compact support), at almost any other point in the universe.

Thus the position probability interpretation itself contradicts the principles of relativity!

#### akhmeteli

The argumentation is completely contained in point 4 of Subsection 3.3 of my Part I.
I have yet to consider the details of your arguments in your quoted post, but I disagree with the above quote. For example, the argumentation in footnote 16 there only proves that "the momentum density must have unbounded support", but that does not mean that speed is unlimited in the relativistic case.

#### A. Neumaier

I have yet to consider the details of your arguments in your quoted post, but I disagree with the above quote. For example, the argumentation in footnote 16 there only proves that "the momentum density must have unbounded support", but that does not mean that speed is unlimited in the relativistic case.
It implies that there are momenta with positive probability and arbitrarily large norm, which implies after any given, sufficiently tiny positive duration nonzero position probabilities arbitrarily far away.

#### akhmeteli

It implies that there are momenta with positive probability and arbitrarily large norm, which implies after any given, sufficiently tiny positive duration nonzero position probabilities arbitrarily far away.
I am afraid I don't understand that. If momentum is large, the speed still cannot exceed the velocity of light in the relativistic case.

#### A. Neumaier

I am afraid I don't understand that. If momentum is large, the speed still cannot exceed the velocity of light in the relativistic case.
In the wave function, there is no actual motion of particles, just a motion of probability amplitude.
Moreover, everything happens in a fixed frame in which the spatial Fourier transform is performed; one cannot argue with time dilation or length contraction.

From the Schrödinger equation, the support of $\psi$ is ,after sufficiently short but positive time, definitely the union of the initial support and that of $H\psi$. But there is no reason to suppose that the fairly arbitrary $H$ allowed by the construction in K/P leads to an $H\psi$ with bounded support; note that $M$ can be quite arbitrary. To preserve the relativistic probability interpretation in concrete cases, one would have to construct very special $M$ that preserve a bounded support - but this seems quite a nontrivial mathematical task.

"The thermal interpretation of quantum physics"

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