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

[A. Neumaier]

@vanhees71 : I would be interested in an explicit reference to a place where, as you claim, a time dependent Heisenberg state is actually used. I have never seen it anywhere.

 

[vanhees71]

Taken the total time dependence a Heisenberg state is of course time-independent, but it can be explicitly time dependent. The reason simply is that for the Heisenberg picture the "covariant" time derivative coincides with the total mathematical time derivative, i.e., you have
$$\mathrm{d}_t \hat{\rho}_H=\mathring{\hat{\rho}}_H=\frac{1}{\mathrm{i} \hbar} [\hat{\rho}_H,\hat{H}_H]+\partial_t \hat{\rho}_H=0.$$
Obviously in the Heisenberg picture the stat. op. is NOT explicitly time dependent if and only if it's an equilibrium state, i.e., if it is a function of the operators of conserved quantities only. Thus any non-equilibrium state is explicitly time-dependent.

In the Schrödinger picture you get on the one hand
$$\mathrm{d}_t \hat{\rho}_S=\partial_t \hat{\rho}_S,$$
because the fundamental operators are by definition time-independent. On the other hand from this and the von Neumann equation you get from this
$$\mathrm{d}_t \hat{\rho}_S=-\frac{1}{\mathrm{i} \hbar} [\hat{\rho}_S,\hat{H}_S].$$
As I said, the only book I know, where this is all explained in clear detail for an arbitrary picture is

E. Fick, Einführung in die Grundlagen der Quantentheorie, Akademische Verlagsgesellschaft Wiesbaden (1979)

 

[A. Neumaier]

Can you give an example of a one- or two-particle system where the Heisenberg state is time dependent?

 

[RUTA]

In physics you always have to be ready that your model will be falsified by observation in some domain where your model has not yet been tested.

But, you can't allow that fact to freeze you into inaction. You have to pick a direction to do your work.

Any scientific explanation has to give predictions that are testable within dynamical process. So even if you believe that adynamical view can explain observation better you still have to be able to translate your adynamical view into dynamical story and point out unique features that show up in dynamical story.

Of course, the story of the experimental procedure is dynamical, but the explanation of the outcomes doesn't have to be. You're conflating those two aspects of an experiment.

Almost all observations are still geocentric. So non geocentric model still have to express it's predictions for geocentric observer.

Ibid.

There is big difference between facts A and B and components of the model A and B. It seems you are mixing them together.

Facts are necessarily context dependent, otherwise your perceptions are meaningless.

Every scientist has to operate within common generally accepted framework. Within that framework you have a lot of freedom with your explanations, but there is one condition - your explanation has to produce predictions that are testable even for those who do not believe in your explanation. So your condition "as long as you're willing to give up your anthropocentric dynamical bias" takes you out of that framework.

You've completely missed the point, adynamical constraints are already accepted and understood per conventional physics, it's simply a matter of whether or not you believe such constraints to be explanatory in and of themselves. If you watch my talk, it should be clear

 

[vanhees71]

Can you give an example of a one- or two-particle system where the Heisenberg state is time dependent?

Not so easily. Do you have an example of an arbitrary non-equilibrium state? Express this in the Heisenberg picture, and the stat. op. will be explicitly time-dependent.

 

[vanhees71]

In fact it's pretty easy to find a nice example.

Take a single free particle, for simplicity in 1 dimension. In the Heisenberg picture we have, due to
$$\hat{H}=\frac{1}{2m} \hat{p}^2$$
the solution for the operator equations of motion
$$\hat{x}=\frac{1}{m} \hat{p}_0 t + \hat{x}_0, \quad \hat{p}=\hat{p_0}.$$
Here $\hat{x}_0$ and $\hat{p}_0$ are the operators representing the observables at the initial time $t=0$. At this initial time you may assume without loss of generality that all pictures of time evolution coincide at this time, but we won't need this here, because I stick to the Heisenberg picture.

Now assume that at time $t=0$ you know that the average position of the particle is $\langle x_0 \rangle$ and its momentum is $\langle p_0 \rangle$. Also the standard deviations may be given as $\Delta x_0$ and $\Delta p_0$. Then at $t=0$ the statistical operator according to the maximum-entropy principle can be written in the form
$$\hat{\rho}_0=\hat{\rho}(t=0)=\frac{1}{Z} \exp [-\lambda_1 (\hat{x}_0-\langle x_0 \rangle)^2 - \lambda_2 (\hat{p}_0 - \langle p_0 \rangle)^2],$$
where $\lambda_1$ and $\lambda_2$ are Lagrange parameters. The other 3 Lagrange parameters necessary to obey the constraints of the given information and normalization of the statistical op. are lumped into $\langle x_0 \rangle$, $\langle p_0 \rangle$, and $Z$.

Now in the Heisenberg picture we have (von-Neumann-Liouville equation)
$$\mathrm{d}_t \hat{\rho}(t) = \mathring{\hat{\rho}}=0,$$
and thus
$$\hat{\rho}(\hat{x},\hat{p},t)=\hat{\rho}_0(\hat{x}_0,\hat{p}_0).$$
To get $\hat{\rho}(t)$ we have to express $\hat{x}_0$ and $\hat{p}_0$ simply by $\hat{x}$ and $\hat{p}$:
$$\hat{x}_0=\hat{x}-\hat{p}_0 t=\hat{x}-\hat{p} t, \quad \hat{p}_0=\hat{p}.$$
Thus the stat. op. at time $t$ is explicitly time-dependent in the Heisenberg picture. Of course as a whole the statistical operator is time-independent due to the von-Neumann-Liouville equation.

 

[A. Neumaier]

In general one has a time-dependent Hamiltonian $H(t)$. In the Schrödinger picture there is no doubt that the dynamics of the state is $i\hbar \dot\rho(t)=[\rho(t),H(t)]$. If $U(t)$ denotes the solution of $i\hbar\dot U(t)=H(t)U(t)$ with $U(0)=1$, one obtains the unique solution $\rho(t)=U(t)\rho_0U(t)^*$.

According to what I think is standard practice (perhaps only Fick deviates from it?), the Heisenberg picture is defined by applying to everything a unitary transformation with $U(t)$, changing all Schrödinger observables $A(t)$ to Heisenberg observables $A_H(t)=U(t)^*A(t)U(t)$. This necessarily requires that the Schrödinger state $\rho(t)$ changes to the Heisenberg state $\rho_H(t)=U(t)^*\rho(t)U(t)=\rho_0$, in order that one preserves the formula $\langle A\rangle_t=Tr~\rho(t)A(t)=Tr~\rho_H(t)A_H(t)$ for expectation values. Thus $\rho_H(t)=\rho_0$ is independent of $t$. Thus the density operator in the Heisenberg picture is time-independent. Period.

In particular, the Schrödinger Hamiltonian $H(t)$ turns into the Heisenberg Hamiltonian $H_H(t)=U(t)H(t)U(t)^*$. If we denote by $\partial_t$ is the derivative with respect to explicit occurrences of $t$ then the product rule and the Hermiticity of $H(t)$ give
$$i\hbar\dot A_H(t)
=i\hbar\dot U(t)^*A(t)U(t)+U(t)^*i\hbar\dot A(t)U(t)+U(t)^*A(t)i\hbar\dot U(t)\\
=-U(t)^*H(t)A(t)U(t)+U(t)^*i\hbar\dot A(t)U(t)+U(t)^*A(t)H(t)U(t)\\
=-U(t)^*H(t)U(t)A_H(t)+i\hbar\partial_tA_H(t)+A_H(t)U(t)^*H(t)U(t)\\
=-H_H(t) A_H(t)+i\hbar\partial_tA_H(t)+A_H(t)H(t)\\
=[A_H(t),H_H(t)]+i\hbar\partial_tA_H(t).
$$

Of course as a whole the statistical operator is time-independent

I don't understand what do you mean. Your definition of the Heisenberg picture is apparently different from what I wrote. Does your version have two density operators at each time, the special one that you constructed, and what you call ''the statistical operator as a whole'', which is time-independent?

 

[A. Neumaier]

Thus the stat. op. at time $t$ is explicitly time-dependent in the Heisenberg picture. Of course as a whole the statistical operator is time-independent due to the von-Neumann-Liouville equation.

Does your version have two density operators at each time, the special one that you constructed, and what you call ''the statistical operator as a whole'', which is time-independent?

Ah, you write the same density operator first deduced in the time-independent form
in an equivalent explicitly time-dependent way by writing the time-independent $x(0)$ in the explicitly time-dependent form $x(t)-p_0 t=x(t)-p(t)t$. Of course one may introduce such artificial time-dependence, but this doesn't change the fact that the one and only density operator is independent of time.

In particular, in such a rewrite you can substitute an arbitrary time and get the same time-independent result.

But your formula

$$\hat{S}=\int \mathrm{d}^3 x \left [\hat{\mathcal{H}}(t,\vec{x}) - \vec{\beta}(t,\vec{x}) \cdot \vec{\mathcal{G}}(t,\vec{x})-\mu(t,\vec{x}) \hat{\rho}(t,\vec{x}) \right]/T(t,\vec{x}).$$

is not of this kind, as $\hat S$ produces different operators when you insert different values of $t$. Thus it cannot be a valid formula in the Heisenberg picture!

 

[vanhees71]

Ok, obviously my attempt to write a statistical operator for local thermal equilibrium is flawed. I'm not sure, whether it's possible to construct something like this.

Concerning the example with the statistical operator for the single-particle QM example I'm sure that it's correct.

In the Heisenberg picture the total time derivative for $\hat{\rho}$ (as for any operator) is the same as the covariant time derivative and thus it must vanish. So
$$\hat{\rho}(\hat{x},\hat{p},t)=\hat{\rho}(\hat{x}_0,\hat{p}_0)=\text{const. in time}.$$
You have to substitute $\hat{x}$ and $\hat{p}$ (time dependent in the Heisenberg picture) instead of $\hat{x}_0$ and $\hat{p}_0$. Thus $\hat{\rho}$ is explicitly time dependent, and this explicit time-dependence precisely cancels the commutator term in the total time derivative (by construction).

From the Heisenberg statistical operator you get the Schrödinger statistical operator via the unitary transformation
$$\hat{U}(t)=\exp \left (-\frac{\mathrm{i} t}{\hbar} \hat{H} \right).$$
Note that $\hat{H}_S=\hat{H}$ (no subscript=Heisenberg picture, subscript S=Schrödinger picture).
Thus
$$\hat{\rho}_S=\hat{U} \hat{\rho} \hat{U}^{\dagger}.$$
Then you get
$$\mathrm{d}_t \hat{\rho}_S = (\mathrm{d}_t \hat{U}) \hat{\rho} \hat{U}^{\dagger} + \hat{U} \hat{\rho} \mathrm{d}_t \hat{U}^{\dagger} = \frac{1}{\mathrm{i} \hbar} \hat{U} [\hat{H},\hat{\rho}] \hat{U}^{\dagger} = \frac{1}{\mathrm{i} \hbar} [\hat{H}_S,\hat{\rho}_S],$$
as I already derived above from the general formalism.

If you have an explicitly time dependent Hamiltonian the only difference is in the solution of the equation of motion for the unitary transformation,
$$\mathrm{d}_t \hat{U}(t) = -\frac{\mathrm{i}}{\hbar} \hat{H}(t) \hat{U}(t),$$
which then formally reads
$$\hat{U}(t)=\mathcal{T}_c \exp \left (-\frac{\mathrm{i}}{\hbar} \int_0^ t \mathrm{d}t' \hat{H}(t') \right).$$
The equation of motion in the Schrödinger picture and the Heisenberg picture stays the same.

 

[A. Neumaier]

You have to substitute $\hat{x}$ and $\hat{p}$ (time dependent in the Heisenberg picture) instead of $\hat{x}_0$ and $\hat{p}_0$.

Why do I have to? $x_0=x(0)$ is as good an operator in the Heisenberg picture as $x(t)$, and unlike the latter, the former is time-independent. Suppressing the time dependence of the operators is dangerous in the Heisenberg picture, as operators without it are meaningless.

For a given relativistic quantum field $\phi(x)$, you surely consider expressions like $\int dx \phi(x) e^{ip\cdot x}$ as valid operators in the Heisenberg picture. Why then not their integral over $p$, which gives $\phi(0)$ (up to a factor). For a 1-dimensional field, this is just $x(0)=x_0$.

 

[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.