Realism from Locality? Bell's Theorem & Nonlocality in QM

[vanhees71]

I don't understand, where the problem might be. I was obviously wrong in trying to write down the operator for local thermal equilibrium. I'm even not sure anymore, whether it is possible to introduce such an idea, but first we should clarify, whether we agree on the general scheme. For simplicity let's work in the most simple example of a non-relativistic single particle in one dimension.

Let's write everything down in terms of a general picture. The time-dependence of the fundamental operators $\hat{x}$ and $\hat{p}$ are by some self-adjoint operator $\hat{H}_0$ that of the states by some self-adjoint operator $\hat{H}_1$. Both can be also explictly time dependent, i.e., $\hat{H}_j=\hat{H}_j(\hat{x},\hat{p},t)$, $j \in\{0,1 \}$ and the Hamiltonian is given by $\hat{H}=\hat{H}_0+\hat{H}_1$.

The equations of motion for operators representing observables $\hat{O}(\hat{x},\hat{p},t)$ has a time dependence due to the time dependence of $\hat{x}$ and $\hat{p}$ and an explicit time dependence. $\hat{x}$ and $\hat{p}$ by definition have no explicit time-dependence. Thus we have
$$\mathrm{d}_t \hat{O}=\frac{1}{\mathrm{i} \hbar} [\hat{O},\hat{H}_0] + \partial_t \hat{O},$$
where $\partial_t$ refers only to the explicit time dependence. For the fundamental operators we have
$$\mathrm{d}_t \hat{x}=\frac{1}{\mathrm{i} \hbar} [\hat{x},\hat{H}_0], \\
\mathrm{d}_t \hat{p}=\frac{1}{\mathrm{i} \hbar} [\hat{p},\hat{H}_0].$$
The statistical operator, representing the state of the system is somewhat special. It's not special in fullfilling the equation of motion as any other operator,
$$\mathrm{d}_t \hat{\rho}=\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}_0]+\partial_t \hat{\rho}. \qquad (1)$$
On the other hand it must fulfill the von Neumann-Liouville equation,
$$\mathring{\hat{\rho}}=\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}] +\partial_t \hat{\rho}=0,$$
and this makes it special in some sense.

Combining the last two equations by eliminating $\partial_t \hat{\rho}$ in (1), we find
$$\mathrm{d}_t \hat{\rho} = \frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}_0-\hat{H}] = -\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}_1].$$
This is the formalism in and arbitrary picture, usually called a Dirac picture. A often used special case is the interaction picture in QFT, where $\hat{H}_0$ is the quadratic part of the hamilton functional of field operators and $\hat{H}_1=\hat{H}_I$ is the part with higher-order monomials in the hamilton density, the "interaction part".

The special cases we discussed so far are

The Heisenberg picture:

$$\hat{H}_0=\hat{H}_{\text{H}} \; \Rightarrow \; \hat{H}_1=0.$$
Thus the statistical operator is constant in time (and thus explicitly time dependent except for equilibrium states!):
$$\mathrm{d}_t \hat{\rho}_{\text{H}}=0.$$
The time dependence of observables' operators is given by
$$\mathrm{d}_t \hat{O}_{\text{H}}=\frac{1}{\mathrm{i} \hbar} [\hat{O}_{\text{H}},\hat{H}_{\text{H}}]+\partial_t \hat{O}_{\text{H}}.$$

The Schrödinger picture:

$$\hat{H}_0=0 \; \Rightarrow \; \hat{H}_1=\hat{H}_{\text{S}}.$$
The time dependence on the statistical operator (state) is given by
$$\mathrm{d}_t \hat{\rho}_{\text{S}}=-\frac{1}{\mathrm{i} \hbar} [\hat{\rho}_{\text{S}},\hat{H}_{\text{S}}],$$
and that of observables' operators by
$$\mathrm{d}_t \hat{O}_{\text{S}}=\partial_t \hat{O}_{\text{S}}.$$

The equivalence of all pictures is proven by the fact that the transformation from one picture to another is given by a (time-dependent) unitary transformation.

So let's consider two pictures, labeled A and B. The time evolution for the operators of not explicitly time-dependent operators and that of the statistical operator are unitary transformations. We set withoug loss of generality the initial time (where the system is prepared) to $t_0=0$ and assume that both pictures coincide at this time.

Any not explicitly time-dependent observable operator fulfills the equation of motion as the fundamental operators. Let's take $x$. For each picture $j \in \{A,B \}$ we have
$$\mathrm{d}_t \hat{x}^{(j)}=\frac{1}{\mathrm{i} \hbar} [\hat{x}^{(j)},\hat{H}_0^{(j)}].$$
This can be (at least formally) solved by a unitary time-evolution operator $\hat{C}^{(j)}$, fulfilling the EoM
$$\mathrm{d}_t \hat{A}^{(j)}=\frac{\mathrm{i}}{\hbar} \hat{A}^{(j)}, \quad \hat{A}^{(j)}(0)=\hat{1}.$$
Then
$$\hat{x}^{(j)}(t)=\hat{A}^{(j)\dagger}(t) \hat{x}^{(j)}(0) \hat{A}^{(j)}(t).$$
The same holds true for the EoM of the statistical operator
$$\mathrm{d}_t \hat{\rho}^{(j)}=-\frac{1}{\mathrm{i} \hbar} [\hat{\rho}^{(j)},\hat{H}_1^{(j)}].$$
The corresponding time-evolution operator is defined by the EoM
$$\mathrm{d}_t \hat{C}^{(j)}=-\frac{\mathrm{i}}{\hbar} \hat{H}_1^{(j)} \hat{C}^{(j)}, \quad \hat{C}^{(j)}(0)=\hat{1}$$
and
$$\hat{\rho}^{(j)}(t)=\hat{C}^{(j)\dagger}(t) \hat{\rho}^{(j)}(0) \hat{C}^{(j)}(t).$$
Now the transformation from one picture to the other is found by
$$\hat{\rho}^{(B)}(t)=\hat{C}^{(B)\dagger}(t) \hat{\rho}^{(B)}(0) \hat{C}^{(B)}(t) = \hat{C}^{(B)\dagger}(t) \hat{\rho}^{(A)}(0) \hat{C}^{(B)}(t) = \hat{C}^{(B)\dagger}(t) \hat{C}^{(A) \dagger}(t) \hat{\rho}^{(A)}(t) \hat{C}^{(A) \dagger}(t) \hat{C}^{(B)}(t).$$
Thus the unitary transformation from picture A to picture B is given by
$$\hat{B}^{(AB)}(t)=\hat{C}^{(A) \dagger}(t) \hat{C}^{(B)}(t).$$
In the same way one derives from the EoM of the observable operators that this transformation should be given by the operator
$$\hat{D}^{(AB)}(t)=\hat{A}^{(A) \dagger}(t) \hat{A}^{(B)}(t).$$
To have both pictures equivlent we should have $\hat{B}^{(AB)}=\hat{D}^{(AB)}$. This can be proven as follows: Define
$$\hat{G}^{(j)}=\hat{A}^{(j) \dagger} \hat{C}^{(j)}, \quad j \in \{A,B \}.$$
Then from the EoM of the time-evolution operators we get
$$\mathrm{d}_t \hat{G}^{(j)}=-\frac{\mathrm{i}}{\hbar} \hat{A}^{(j) \dagger} [\hat{H}_0^{(j)}(t)+\hat{H}_1^{(j)}(t)] \hat{C}^{(j)}(t) \\
=-\frac{\mathrm{i}}{\hbar} \hat{A}^{(j) \dagger} \hat{H}^{(j)}(t) \hat{C}^{(j)}(t) \\
=-\frac{\mathrm{i}}{\hbar} \hat{A}^{(j) \dagger} \hat{H}^{(j)}(t) \hat{A}^{(j)}(t) \hat{G}^{(j)}(t) \\
=-\hat{H}(0) \hat{G}^{(j)}(t).$$
Since also $\hat{G}^{(j)}(0)=\hat{1}$, $\hat{G}^{(j)}$ fulfills the same EoM and initial condition for both $j \in \{A,B\}$ and thus
$$\hat{G}^{(A)}=\hat{G}^{(B)}$$
This implies
$$\hat{A}^{(A)\dagger} \hat{C}^{(A)} = \hat{A}^{(B)\dagger} \hat{C}^{(B)},$$
and thus after some algebra
$$\hat{B}^{(AB)}=\hat{C}^{(A) \dagger} \hat{C}^{(B)} = \hat{A}^{(A)} \hat{A}^{(B) \dagger} = \hat{D}.$$
In addition we have shown that the physically, i.e., measurable quantities, of QT are picture independent, because the time evolution probability to find the value $o$ when measuring the observable $O$ is governed by the picture-independent operator $\hat{G}=\hat{G}^{(A)}=\hat{G}^{(B)}$. To see this let $|o,\alpha \rangle$ be a complete orthonormal set of eigenvectors of $\hat{O}$ with eigenvalue $o$. Then the said probability is
$$P(o,t)={\sum_{\alpha}} {_{(j)}\langle o,\alpha,t|}\hat{\rho}^{(j)}(t)|\langle o,\alpha,t \rangle_{(j)} \\
=\sum_{\alpha} \langle o,\alpha,0|\hat{A}^{(j) \dagger}(t) \hat{C}^{(j)}(t) \hat{\rho}(0) \hat{C}^{(j) \dagger}(t) \hat{A}^{(j)}(t)|o,\alpha,0 \rangle \\
=\sum_{\alpha} \langle o,\alpha,0|\hat{G}(t) \hat{\rho}(0) \hat{G}^{\dagger}(t)|o,\alpha,0 \rangle.$$
From this one reads off that the picture-independent unitary time-evolution operator $\hat{G}$ can be interpreted as the time-evolution operator for the Schrödinger-picture $\hat{\rho}^{(S)}$ or $\hat{G}^{\dagger}$ as the time-evolution operator for the eigenstates of observable operators (and thus the observable operators themselves) in the Heisenberg picture.

 

[A. Neumaier]

The problem lies with the fact that Fick's way of proceeding is so lengthy and general that it causes more confusion than good. Rather look at https://www.physicsforums.com/threads/realism-from-locality.974177/post-6209378
which is the standard way of proceeding and is short and crystal clear. Or can you point to a mistake there?

 

[vanhees71]

Of course it's correct, but not general for an arbitrary picture but only for the Schrödinger and Heisenberg pictures, which you cannot use in practice almost always. The only exceptions I know are when the Hamiltonian is at most quadratic in $x$ and $p$, i.e., the free particle, the harmonic oscillator, constant force (free fall), and an approximate treatment of the Stern-Gerlach experiment (which I still want to write up for an AJP article; I hope to come to this in the just beginning summer break right now ;-)).

I'd appreciate if you could look over #211 if there is a mistake. It's a short summary of the lengthy treatment in the book by Fick, which I consider nevertheless as a very good textbook, because it's the only book I know, where this sometimes confusing time evolution of observable operators and states is made clear for the general case, including explicit time dependence.

You can find a somewhat more extended but shorter treatment of the formalism also in my QM manuscript,

https://itp.uni-frankfurt.de/~hees/faq-pdf/quant.pdf
There I also give the more conventional treatment starting from the Schrödinger picture.

 

[A. Neumaier]

Of course it's correct, but not general for an arbitrary picture but only for the Schrödinger and Heisenberg pictures

Our basic conflict is about the Heisenberg picture, so this simpler derivation is enough. It shows that the Heisenberg state is always constant, no matter whether the Hamiltonian is or isn't time-dependent. But you claimed that in most cases, the Heisenberg state is time-dependent, too, which flatly contradicts this conclusion. On the other hand, the 1-particle example you gave had the Heisenberg state

$$\hat{\rho}_0=\frac{1}{Z} \exp [-\lambda_1 (\hat{x}_0-\langle x_0 \rangle)^2 - \lambda_2 (\hat{p}_0 - \langle p_0 \rangle)^2],$$

which is clearly independent of the time $t$. You only made it look formally time-dependent by rewriting this state in a way that had multiple explicit occurrences of time. But this is not different than when you pretending that in the Schrödinger picture, the spin $S_z$ in z-direction is time-dependent because you can rewrite it as $(t+1)S_z-tS_z$, which also looks formally time-dependent. Indeed, whether you write $x_0$ as $x(0)$ or as $x(t)-tp$ or as $x(s)-sp$ with an arbitrary $s$ totally unrelated to $t$, it remains the same operator. Thus the time-dependence is only apparent.

I'd appreciate if you could look over #211 if there is a mistake.

I'll do so once the more basic conflict mentioned in the remainder of this post is resolved.

 

[A. Neumaier]

one can just take your formula an write it down using the relativistic (canonical) energy-momentum-tensor $\hat{\mathcal{T}}^{\mu \nu}(x)$. Then
$$\hat{\rho}=\frac{1}{Z} \exp(-\hat{S}),$$
where
$$\hat{S}=\int \mathrm{d}^3 \Sigma_{\mu} [u_{\nu}(x) \hat{\mathcal{T}}^{\mu \nu}(x)-\mu(x) \hat{\mathcal{J}}^{\mu}(x)]/T(x),$$
where $\hat{\mathcal{J}^{\mu}}$ is the four-current operator of a conserved charge (baryon number, electric charge for instance). The integral is over some spacelike hypersurface. With the operators in the Heisenberg picture this should be the stat. op. in the Hiesenberg picture too, right?

This is the density operator for local equilibrium in the Tomonaga-Schwinger (generalized Schrödinger) form, as given by

• Zubarev et al., Derivation of nonlinear generalized equations of quantum relativistic hydrodynamics, Theor. Math. Phys. 40 (1979), 821–83.
• Ch.G. van Weert, Maximum entropy principle and relativistic hydrodynamics,
  Ann. Physics 140 (1982), 133-162.

Although the formulation of the state is given in terms of Heisenberg operators, it is not the Heisenberg picture, as you can check by verifying that the total derivative of the state does not vanish.

 

[A. Neumaier]

how you choose your thermodynamical parameters is a matter of convention. I also usually prefer $\alpha=\mu/T$ (I don't even know a name for it, but calculational-wise it's often more convenient if it comes to certain quantities like susceptibilities of (conserved) charges and things like that).

Expressed as a function of pressure and temperature, this is the Massieu-Planck function.

 

[vanhees71]

Our basic conflict is about the Heisenberg picture, so this simpler derivation is enough. It shows that the Heisenberg state is always constant, no matter whether the Hamiltonian is or isn't time-dependent. But you claimed that in most cases, the Heisenberg state is time-dependent, too, which flatly contradicts this conclusion. On the other hand, the 1-particle example you gave had the Heisenberg state

which is clearly independent of the time $t$. You only made it look formally time-dependent by rewriting this state in a way that had multiple explicit occurrences of time. But this is not different than when you pretending that in the Schrödinger picture, the spin $S_z$ in z-direction is time-dependent because you can rewrite it as $(t+1)S_z-tS_z$, which also looks formally time-dependent. Indeed, whether you write $x_0$ as $x(0)$ or as $x(t)-tp$ or as $x(s)-sp$ with an arbitrary $s$ totally unrelated to $t$, it remains the same operator. Thus the time-dependence is only apparent.

I'll do so once the more basic conflict mentioned in the remainder of this post is resolved.

No, I did NOT claim what you claim I claim.

The total time derivative of the Heisenberg-picture statistical operator is ALWAYS 0 by definition of the picture. I think it's again a simple misunderstanding between different notations. In my opinion it's important to express the statistical operator as function of the fundamental observables at time $t$, which brings in the IMPLICIT time dependence since in the Heisenberg picture the fundamental operators are evolving in time by a unitary time-evolution operator generated by $\hat{H}$, no matter whether or not $\hat{H}$ is explicitly time-dependent or not.

But that necessarily implies that it is EXPLICITLY time dependent except for equilibrium states. The equation of motion for the Heisenberg picture is
$$\mathrm{d}_t \hat{\rho}(\hat{x},\hat{p},t)=\frac{1}{\mathrm{i} \hbar} [\hat{\rho}(\hat{x},\hat{p},t),\hat{H}(\hat{x},\hat{p},t)]+\partial_t \hat{\rho}[\hat{x},\hat{p},t]=0.$$
This of course implies
$$\hat{\rho}(\hat{x},\hat{p},t)=\hat{\rho}(\hat{x}_0,\hat{p}_0,0)=\text{const. in time},$$
where
$$\hat{x}_0=\hat{x}(t=0),\quad \hat{p}_0=\hat{p}(t=0).$$
The fundmental observables obey
$$\mathrm{d}_t \hat{x}=\frac{1}{\mathrm{i} \hbar} [\hat{x},\hat{H}], \quad \mathrm{d}_t \hat{p}=\frac{1}{\mathrm{i} \hbar} [\hat{p},\hat{H}].$$
The explicit time dependence of $\hat{\rho}$ in the Heisenberg picture for all non-equilibrium states is a simple mathematical necessity and common sense in all textbooks I know.

For the time dependence of the operators in any picture is completely irrelevant, whether the Hamiltonian is explicitly time-dependent or not. The general equation for the time-evolution operator for the state operators in an arbitrary picture (for the equations of motion, see my posting #11)
$$\hat{C}^{(j)}(t)=\mathcal{T}_c \exp \left(-\frac{\mathrm{i}}{\hbar} \int_0^t \mathrm{d} t' \hat{H}_1(\hat{x}(t'),\hat{p}(t'),t' \right),$$
and for the observable operators
$$\hat{A}^{(j)}(t)=\mathcal{T}_c \exp \left(+\frac{\mathrm{i}}{\hbar} \int_0^t \mathrm{d} t' \hat{H}_0(\hat{x}(t'),\hat{p}(t'),t' \right).$$
The only difference is in the Schrödinger or Heisenberg picture that for Hamiltonians that are not explicitly time dependent the time-evolution operators simplify a bit, i.e.,

$$\hat{C}^{(S)}(t)=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H}^{(S)} t \right), \hat{A}^{(S)}(t)=\hat{1},\\
\hat{C}^{(H)}(t)=1, \quad \hat{A}^{(H)}(t)=\exp \left (\frac{\mathrm{i}}{\hbar} \hat{H}^{(H)} t \right).$$
Note that then
$$\hat{H}^{(S)}=\hat{H}^{(H)}=\text{const. in time}.$$

 

[A. Neumaier]

No, I did NOT claim what you claim I claim.

The total time derivative of the Heisenberg-picture statistical operator is ALWAYS 0 by definition of the picture. I think it's again a simple misunderstanding between different notations.

No, here we fully agree. The density operator in the Heisenberg picture is time-independent, no matter how it is represented.

In my opinion it's important to express the statistical operator as function of the fundamental observables at time $t$, which brings in the IMPLICIT time dependence since in the Heisenberg picture the fundamental operators are evolving in time by a unitary time-evolution operator generated by $\hat{H}$, no matter whether or not $\hat{H}$ is explicitly time-dependent or not.

But that necessarily implies that it is EXPLICITLY time dependent except for equilibrium states.

This is due only to your personal desire to use a time-dependent Heisenberg operator in the representation of the density operator, which requires the introduction of other time dependencies to cancel it. But this desire makes things only complicated, it does not change anything in the physics compared to using Heisenberg operators at a fixed time only.

Consider a 2-level system in the Heisenberg picture, prepared in the excited state, with a complicated time-dependent Hamiltonian switched on at time 0. The natural Heisenberg density operator is the constant state $\rho=a^*|0\rangle \langle 0|a$, where $\langle 0|$ is the ground state (of the Hamiltonian before the interaction is switched on), while your desire to represent it in terms of Heisenberg operators $a(t)$ at time $t$ turns it into a very complicated object.

 

[A. Neumaier]

This is the density operator for local equilibrium in the Tomonaga-Schwinger (generalized Schrödinger) form, as given by

• Zubarev et al., Derivation of nonlinear generalized equations of quantum relativistic hydrodynamics, Theor. Math. Phys. 40 (1979), 821–83.
• Ch.G. van Weert, Maximum entropy principle and relativistic hydrodynamics,
  Ann. Physics 140 (1982), 133-162.

Although the formulation of the state is given in terms of Heisenberg operators, it is not the Heisenberg picture, as you can check by verifying that the total derivative of the state does not vanish.

The corresponding (complicated) representation in the Heisenberg picture is discussed in the open access article

• Becattini et al., Particles 2 (2019), 197-207.

 

[Auto-Didact]

The corresponding (complicated) representation in the Heisenberg picture is discussed in the open access article

• Becattini et al., Particles 2 (2019), 197-207.

This is definitely more my cup of tea than the vanhees representation.

 

[Auto-Didact]

No, I did NOT claim what you claim I claim.

The total time derivative of the Heisenberg-picture statistical operator is ALWAYS 0 by definition of the picture. I think it's again a simple misunderstanding between different notations. In my opinion it's important to express the statistical operator as function of the fundamental observables at time ttt, which brings in the IMPLICIT time dependence since in the Heisenberg picture the fundamental operators are evolving in time by a unitary time-evolution operator generated by ^HH^\hat{H}, no matter whether or not ^HH^\hat{H} is explicitly time-dependent or not.

This seems like a misinterpretation to me, mostly due to the highly artificial nature of your time dependent construction. Perhaps it is somewhat confounded by abuse of notation as you say, but in any case there is no need to get defensive.

 

[vanhees71]

No, here we fully agree. The density operator in the Heisenberg picture is time-independent, no matter how it is represented.

This is due only to your personal desire to use a time-dependent Heisenberg operator in the representation of the density operator, which requires the introduction of other time dependencies to cancel it. But this desire makes things only complicated, it does not change anything in the physics compared to using Heisenberg operators at a fixed time only.

Consider a 2-level system in the Heisenberg picture, prepared in the excited state, with a complicated time-dependent Hamiltonian switched on at time 0. The natural Heisenberg density operator is the constant state $\rho=a^*|0\rangle \langle 0|a$, where $\langle 0|$ is the ground state (of the Hamiltonian before the interaction is switched on), while your desire to represent it in terms of Heisenberg operators $a(t)$ at time $t$ turns it into a very complicated object.

The point is that you can derive the Schwinger-Keldysh real-time contour formalism and perturbation theory, which expands both the interactions and the statistical operator in powers of couplings or $\hbar$.

 

[vanhees71]

This seems like a misinterpretation to me, mostly due to the highly artificial nature of your time dependent construction. Perhaps it is somewhat confounded by abuse of notation as you say, but in any case there is no need to get defensive.

This is not highly artificial but the starting point for all developments in many-body (relativistic or non-relativistic) QFT, leading to the Schwinger-Keldysh real-time contour formalism for the general case (equilibrium and off-equilbrium) or the Matsubara imaginary-time formalism for the equilibrium case.

 

[A. Neumaier]

The point is that you can derive the Schwinger-Keldysh real-time contour formalism and perturbation theory, which expands both the interactions and the statistical operator in powers of couplings or $\hbar$.

But this is unrelated to how the Heisenberg density operator is represented, as its explicit form never explicitly occurs in the Schwinger-Keldysh formalism. Thus there is nowhere a need to require a particular formally time-dependent description of the time-independent Heisenberg density operator.

 

[Auto-Didact]

The point is that you can derive the Schwinger-Keldysh real-time contour formalism and perturbation theory, which expands both the interactions and the statistical operator in powers of couplings or $\hbar$.

This is not highly artificial but the starting point for all developments in many-body (relativistic or non-relativistic) QFT, leading to the Schwinger-Keldysh real-time contour formalism for the general case (equilibrium and off-equilbrium) or the Matsubara imaginary-time formalism for the equilibrium case.

On the contrary, perturbation theoretic equilibrium treatments - especially w.r.t. many-body (quantum or non-quantum) field theoretic approaches - are from the applied mathematics point of view practically by definition highly artificial constructions for directly obtaining time dependent models.

As a dynamical systems theorist, general mathematical experience of analysing many-body systems tells us that a proper time dependent treatment of this situation requires not simply a perturbative approach, but actually:

1. a complete hydrodynamic approach,
2. an index theoretic treatment of the complete phase space,
3. a fully non-perturbative multiple scale analysis, or
4. a full blown derivation of the functional equation using renormalization group methods.

The problem is that in this case options 1 and 2 if directly attempted in the proper fashion are viciously difficult to obtain; this is of course evidenced by the fact that, despite Herculean efforts, the literature does not yet contain any relativistic Bohmian - i.e. holomorphic or conformal hydrodynamic - approaches.

On the other hand, option 4 seems to be numerically the most intensive or demanding path to take, so I would personally choose to try option 3 first in order to obtain an Ansatz, then derive options 1 and 2 from there, you know, if I actually had the time to spare.

 

[gill1109]

Well, that's of course not what we'd consider science. Then I can as well just stick to the observations themselves. A black box providing answers is not what we aim at in science. We want to reduce it to some (as few as possible) fundamental laws, which then however are indeed a "black box".

In QT one example of fundamental postulates is the Born rule (in its most general form with statistical operators), which cannot be deduced from other postulates, and this seems to be the most debated postulate of all since it introduces an "irreducible probabilistic element" to the laws of nature, i.e., a probability that is not due to incomplete knowledge of us but is inherent in the laws of nature themselves. I think this is still an element which many people cannot easily accept. For me it's one of the most profound discoveries of the 20th century, and I think there's no way to accept it, if you don't want to ignore objective facts about how nature behaves.

Do you really mean "no way to accept it"? Don't you mean " no way to reject it"? (*I* think that probability is not due to incomplete knowledge but is an inherent ground-level feature of nature.)

 

[vanhees71]

I meant of course, that there is no way to decline Born's rule without ignoring objective (empirical) facts about how nature behaves, particularly the violation of Bell's inequality and thus the disproof of local deterministic HV models.