Can decoherence be formulated in the Heisenberg picture?

[atyy]

In another thread, stevendaryl and I were trying to understand whether MWI can be formulated in the Heisenberg picture. Since neither of us really understands MWI, I tried to retreat to safer ground by asking:

Can decoherence be formulated in the Heisenberg picture?

 

[Feeble Wonk]

Can decoherence be formulated in the Heisenberg picture?

I'd like to try to follow this discussion. Could you please elaborate on what you mean by the "Heisenberg picture"?

 

[Demystifier]

Decoherence can be formulated in the Heisenberg picture, but MWI can't.

Recall also that decoherence is an observed fact. If it could not be formulated in the Heisenberg picture, then we would have an experimental evidence that Heisenberg picture was wrong.

 

[dextercioby]

I'd like to try to follow this discussion. Could you please elaborate on what you mean by the "Heisenberg picture"?

In the standard textbook description of QM it's roughly: observables evolve in time, state vectors/v.N. density matrix doesn't.

 

[Feeble Wonk]

In the standard textbook description of QM it's roughly: observables evolve in time, state vectors/v.N. density matrix doesn't.

Thank you. That helps. I found this source as well, for any followers that had similar confusion.
http://www.nyu.edu/classes/tuckerman/quant.mech/lectures/lecture_1/node7.html

 

[A. Neumaier]

Decoherence can be formulated in the Heisenberg picture.

So please formulate it for us. I guess the original question wasn't intended to mean whether one can translate the decoherence formulas into the Heisenberg picture (which is obviously possible) but whether the presentation in the Heisenberg picture leads to a meaningful view of decoherence that also makes sense in the context of time correlations.

 

[naima]

In the standard textbook description of QM it's roughly: observables evolve in time, state vectors/v.N. density matrix doesn't.

As with decoherence the off diagonal elements of the density matrix can tend towards zero, are we then in an interaction picture?

 

[atyy]

Decoherence can be formulated in the Heisenberg picture, but MWI can't.

Recall also that decoherence is an observed fact. If it could not be formulated in the Heisenberg picture, then we would have an experimental evidence that Heisenberg picture was wrong.

How? In the Schroedinger picture, decoherence helps to pick a preferred basis. I guess that in the Heisenberg picture, this is equivalent to favouring a particular set of commuting observables?

 

[naima]

P { margin-bottom: 0.21cm; }

When you measure the expectation value of an observable A, you are interested in $\langle A \rangle _t$

In the Heisenberg picture you write it:

$Tr( e^{iHt} A e^{-iHt} \rho )$

In the Schroedinger picture it reads:

$Tr( A e^{-iHt} \rho e^{iHt} )$

In decoherence picture you found $H_0$ and $H_D$ such that for every A it reads:

$Tr( e^{iH_0 t} A e^{-iH_0 t} e^{-iH_D t}\rho e^{iH_D t})$

and with growng t, the right part becomes the diagonal $\rho_D$

We then get

$Tr( e^{iH_0 t} A e^{-iH_0 t} e^{-iH_D t}\rho e^{iH_D t})$

Which tends towards

$Tr( e^{iH_0 t} A e^{-iH_0 t} \rho_D )$

 

[atyy]

Hmmm, but if one takes the expectation value, then the result is picture independent.

Does decoherence involve taking an expectation value (ie. application of the Born rule)? I thought it involves unitary evolution, but we not the application of the Born rule.

 

[Haelfix]

Decoherence can be formulated in the Heisenberg picture, but MWI can't.

Recall also that decoherence is an observed fact. If it could not be formulated in the Heisenberg picture, then we would have an experimental evidence that Heisenberg picture was wrong.

I don't understand this. The Shroedinger picture is unitarily equivalent to the Heisenberg picture by the Stone Von Neuman theorem. How on Earth can an interpretation change the physics. It would logically cease to be an interpretation, and then becomes a modification of quantum mechanics and would be subject to instant falsification. I don't know much about MWI, but I doubt its adherents would agree with this point of view.

As for decoherence, of course that it works in the Heisenberg picture. It just looks like a mess, and is therefore not useful to write down.

(as an aside, there might be a problem here in principle with QFT as opposed to QM, bc the Stone-VonNeuman theorem does not hold in that context. Of course no one actually believes that something goes wrong physically, but it is a vexing mathematical problem)

 

[naima]

@atyy
In his last book, Rovelli write many things about mean values. Taking mean values is neglecting details. What is important is that you then believe in the reality of the equivalence classes. He writes: when you use mean values , mean values do the only thing they are able to do: they produce heat. Thermodynamics appears and irreversibility too. He adds that Time is our Ignorance!
Read his book.

 

[Haelfix]

How? In the Schroedinger picture, decoherence helps to pick a preferred basis. I guess that in the Heisenberg picture, this is equivalent to favouring a particular set of commuting observables?

Favor is the wrong word. The point of decoherence is that there is NOT a preferred frame, it just approximately looks like one b/c certain interference terms are rapidly driven to zero by the effects of large amounts of degrees of freedom that monitor the system in a particular way. In the Heisenberg picture, this does mean that a certain set of observables (that are not obvious at first glance, unless you analyze the full interaction Hamiltonian with the environment) will rapidly limit to some small epsilon.

Of course you absolutely do not want to calculate things this way, or even handwave things this way, precisely b/c it involves some horrible matrix computation.

 

[Demystifier]

I don't understand this. The Shroedinger picture is unitarily equivalent to the Heisenberg picture by the Stone Von Neuman theorem. How on Earth can an interpretation change the physics. It would logically cease to be an interpretation, and then becomes a modification of quantum mechanics and would be subject to instant falsification. I don't know much about MWI, but I doubt its adherents would agree with this point of view.

Please see my reply to a similar objection by vanhees
https://www.physicsforums.com/threads/relation-between-qm-and-qft.859228/page-7#post-5393588

Or let me rephrase what I said there. It depends on what do you mean by "physics". If by "physics" you mean measurable predictions, then interpretation does not change physics. If by "physics" you mean intuitive picture of what is really going on, then interpretation does change the physics. The role of interpretations (including MWI) is precisely to offer an intuitive picture of what is going on, so if interpretations have anything to do with physics at all, then they do so in the latter sense of "physics".

 

[Demystifier]

How? In the Schroedinger picture, decoherence helps to pick a preferred basis. I guess that in the Heisenberg picture, this is equivalent to favouring a particular set of commuting observables?

Picking an observable is equivalent to picking a basis (consisting of the eigenstates of that operator).

 

[Demystifier]

So please formulate it for us. I guess the original question wasn't intended to mean whether one can translate the decoherence formulas into the Heisenberg picture (which is obviously possible) but whether the presentation in the Heisenberg picture leads to a meaningful view of decoherence that also makes sense in the context of time correlations.

Decoherence is usually formulated in terms of matrix elements of the density operator. Matrix elements (of any operator) do not depend on the picture.

 

[A. Neumaier]

Decoherence is usually formulated in terms of . Matrix elements (of any operator) do not depend on the picture.

It is formulated in terms of matrix elements of a time-dependent density operator. Such a thing doesn't exist in the Heisenberg picture.

 

[vanhees71]

I'd like to try to follow this discussion. Could you please elaborate on what you mean by the "Heisenberg picture"?

Although I don't understand the question, because QT is picture independent by construction. It's not formulated very clearly in the textbooks, because usually all use for the most time the Schrödinger picture for non-relavistic quantum mechanics or the Heisenberg picture in relativistic QFT and only later the general Dirac picture (usually applied to the special case of the interaction picture).

The choice of the picture of time evolution is given by two arbitrary, in general time-dependent, self-adjoint operators $\hat{X}(t)$ and $\hat{Y}(t)$ fulfilling
$$\hat{X}(t)+\hat{Y}(t)=\hat{H}(t),$$
where $\hat{H}$ is the Hamiltonian of the system. The time dependence of any physically relevant quantity is entirely independent of the choice of the picture of time evolution and is given by the Hamiltonian.

The picture is determined by the split of the Hamiltonian in the two self-adjoint operators that define the time evolution of the operators that represent observables and that of the statistical operator that represents the states:
$$\hat{O}(t)=\hat{A}(t) \hat{O}(0) \hat{A}^{\dagger}(t), \quad \hat{\rho}(t)=\hat{C}(t) \hat{\rho}(0) \hat{C}^{\dagger}(t),$$
where $\hat{A}$ and $\hat{C}$ are unitary operators of time evolution for the observable operators and the statistical operator, respectively, defined by the equations of motion ($\hbar=1$)
$$\partial_t \hat{A}=+\mathrm{i} \hat{X} \hat{A}, \quad \hat{A}(0)=1, \quad \partial_t \hat{C}=-\mathrm{i} \hat{Y} \hat{C}.$$

A change from one picture $(\hat{X}_1,\hat{Y}_1)$ to another $(\hat{X}_2,\hat{Y}_2)$ implies a unitary transformation for all states and observable-operators, which implies that all measurable quantities like probabilities, expecation values of operators, etc. are independent of the choice of picture. To see this, we assume that all quantities at $t=0$ are represented by the same operators in both pictures and note that
$$\hat{\rho}_1(t)=\hat{C}_1(t) \hat{\rho}(0) \hat{C}_1^{\dagger}(t) = \hat{C}_1(t) \hat{C}_2^{\dagger}(t) \hat{\rho}_2(t) \hat{C}_2(t) \hat{C}_1^{\dagger}(t),$$
i.e.,
$$\hat{\rho}_1(t)=\hat{U}_{12}(t) \hat{\rho}_2(t) \hat{U}_{12}^{\dagger}(t) \quad \text{with} \quad \hat{U}_{12}(t)=C_1(t) \hat{C}_2^{\dagger}(t).$$
An analogous calulation for the observable-operators leads to
$$\hat{O}_1(t)=\hat{V}_{12}(t) \hat{O}_2(t) \hat{V}_{12}^{\dagger}(t) \quad \text{with} \quad \hat{V}_{12}(t)=\hat{A}_1(t) \hat{A}_2^{\dagger}(t).$$
To show that this is consistent, we have to prove that $\hat{U}_{12}=\hat{V}_{12}$, i.e.,
$$\hat{C}_1 \hat{C}_2^{\dagger}=\hat{A}_1 \hat{A}_2^{\dagger}$$
or, multiplying from the left with $\hat{C}_1^{\dagger}$ and with $\hat{A}_2$ from the right, we get
$$\hat{C}_2^{\dagger} \hat{A}_2=\hat{C}_1^{\dagger} \hat{A}_1$$.
This latter equation is easy to prove, because taking the time derivative or the left-hand side we get
$$\dot{\hat{C}}_2^{\dagger}\hat{A}_2+\hat{C}_2^{\dagger} \dot{\hat{A}}_2=\mathrm{i} \hat{C}_2^{\dagger} (\hat{Y}_2+\hat{X}_2) \hat{A}_2=\mathrm{i} \hat{C}_2^{\dagger} \hat{A}_2 \hat{H}(t=0).$$
The same calculation works with the right-hand side, leading to the same equation of motion for $\hat{A}_1 \hat{C}_1^{\dagger}$ since also the initial condition is the same, i.e., both operators are 1 for $t=0$, the operators are the same for all times. So the transformation from one to the other picture of time evolution is given by a common unitary transformation of both the observable- and the state-operators, and thus all physically observable quantities as predicted from QT in both pictures are the same.

In the Heisenberg picture one defines the time evolution by
$$\hat{X}=\hat{H}, \quad \hat{Y}=0 \quad (\text{Heisenberg picture})$$
and the Schrödinger picture by
$$\hat{X}=0, \quad \hat{Y}=\hat{H} \quad (\text{Schrödinger picture}).$$
In the interaction picture one uses
$$\hat{X}=\hat{H}_0, \quad \hat{Y}=\hat{H}_{\text{int}} \quad (\text{interaction picture}),$$
i.e., the observables move with the free and the states with the interaction part of the Hamiltonian.

 

[vanhees71]

It is formulated in terms of matrix elements of a time-dependent density operator. Such a thing doesn't exist in the Heisenberg picture.

In the Heisenberg picture the density operator (I guess you mean the statistical operator by that expression) is time dependent if and only if it's explicitly time dependent, but decoherence is a property of matrix elements of the statistical operator with respect to a given complete set of compatible observables or, even more precisely, a certain coarse grained probability distribution function for macroscopic observables. E.g., the Wigner transform of the one-particle Green's function becomes a phase-space distribution function when properly "smeared" out.

 

[A. Neumaier]

The time dependence of any physically relevant quantity is entirely independent of the choice of the picture of time evolution

You forgot to indicate the argument why, and that it is valid only for physically relevant quantities at a single time. To make sense of time correlations in a pure Schroedinger picture is impossible since one needs the Heisenberg dependence of the operators.

Thus the Heisenberg picture is the more fundamental one. It is also the one where one can see most easily the quantum-classical correspondence.

 

[A. Neumaier]

In the Heisenberg picture the density operator (I guess you mean the statistical operator by that expression) is time dependent if and only if it's explicitly time dependent, but decoherence is a property of matrix elements of the statistical operator with respect to a given complete set of compatible observables or, even more precisely, a certain coarse grained probability distribution function for macroscopic observables. E.g., the Wigner transform of the one-particle Green's function becomes a phase-space distribution function when properly "smeared" out.

Well, so far this is just hand-waving that conveniently sweeps everything under the carpet. Can you point to a paper where Wigner transforms are related to decoherence in the technical sense used e.g., by Zurek (rather than just as a popular buzzword)?

Like atyy in post #1, I'd like to see a formal treatment of decoherence in the Heisenberg picture that makes sense conceptually, not just a statement that this is (or should be) possible.

 

[Demystifier]

It is formulated in terms of matrix elements of a time-dependent density operator. Such a thing doesn't exist in the Heisenberg picture.

Wrong! It is formulated in terms of time dependent matrix elements of a density operator. It does exist in the Heisenberg picture. Explicitly
$$\rho_{ab}(t)=\langle a(t)|\rho|b(t)\rangle$$
is clearly time dependent. Here, of course,
$$|b(t)\rangle=e^{-iHt}|b\rangle$$
and similarly for $\langle a(t)|$.

 

[naima]

In the interaction picture one uses
$$\hat{X}=\hat{H}_0, \quad \hat{Y}=\hat{H}_{\text{int}} \quad (\text{interaction picture}),$$
i.e., the observables move with the free and the states with the interaction part of the Hamiltonian.

I highlighted in post 9 that decoherence picture is a peculiar case of the interaction picture where,$H_I$ is what i called $H_D$ (which has special properties).

 

[A. Neumaier]

It is formulated in terms of matrix elements of a time-dependent density operator. Such a thing doesn't exist in the Heisenberg picture.

Just look at equation (17) in Zurek's paper, where he uses the von Neumann dynamics for the density matrix, not the Heisenberg dynamics for observables. Expressed in the preferred basis, the density matrix becomes diagonal in the course of time.

Wrong! It is formulated in terms of time dependent matrix elements of a density operator. It does exist in the Heisenberg picture. Explicitly
$$\rho_{ab}(t)=\langle a(t)|\rho|b(t)\rangle$$
is clearly time dependent. Here, of course,
$$|b(t)\rangle=e^{-iHt}|b\rangle$$
and similarly for $\langle a(t)|$.

In the Heisenberg picture there are no time-dependent states evolving according to the Schroedinger equation. That you undo the time-dependence of the operator by moving it into the states is just what one does when going from Heisenberg to Schroedinger!

 

[Demystifier]

Just look at equation (17) in Zurek's paper, where he uses the von Neumann dynamics for the density matrix, not the Heisenberg dynamics for observables. Expresed in the preferred basis, the density matrix becomes diagonal in the course of time.

In the Heisenberg picture there are no time-dependent states evolving according to the Schroedinger equation. That you undo the time-dependence of the operator by moving it into the states is just what one does when going from Heisenberg to Schroedinger!

Oops, my bad. I meant
$$\rho_{ab}(t)=\langle a|\rho(t)|b\rangle $$
where
$$\rho(t)=e^{iHt}\rho e^{-iHt}$$
It exists in the Heisenberg picture.

 

[atyy]

Decoherence can be formulated in the Heisenberg picture, but MWI can't.

I guess it depends on what one means by decoherence. I mean decoherence that is not observable (ie. no use of the Born rule is made). So for example, if we have the system plus environment and unitary evolution of the wave function only, then using something like the predictability sieve to pick the preferred basis.

 

[A. Neumaier]

Oops, my bad. I meant
$$\rho_{ab}(t)=\langle a|\rho(t)|b\rangle $$
where
$$\rho(t)=e^{iHt}\rho e^{-iHt}$$
It exists in the Heisenberg picture.

Now you have a mixed picture in which everything is confused...

$\rho(t)$ is the density matrix in the Schroedinger picture, and cannot be interpreted as a time-dependent observable in the Heisenberg picture! This can be seen by comparing the dynamics of the Heisenberg observables and the Schroedinger density matrix - they differ in a sign!

 

[Demystifier]

Now you have a mixed picture in which everything is confused...

$\rho(t)$ is the density matrix in the Schroedinger picture, and cannot be interpreted as a time-dependent observable in the Heisenberg picture! This can be seen by comparing the dynamics of the Heisenberg observables and the Schroedinger density matrix - they differ in a sign!

Well, somebody obviously mixed something, but I think it was you. In my language $\rho(t)$ is the density operator in the Heisenberg picture. But maybe I am using a wrong language.

But now I see your point. There is some point in using your language in which my $\rho(t)$ is not called the density operator in the Heisenberg picture. In such a language, you are right that there is no decoherence in the Heisenberg picture. But I think that such language is neither good nor a standard one.

 

[A. Neumaier]

In my language $\rho(t)$ is the density operator in the Heisenberg picture. But maybe I am using a wrong language.

Yes. How can a density operator in the Heisenberg picture be time-dependent? This would make the associated state (and the expectation of observables at time $t=0$) time-dependent, which by definition of the Heisenberg picture is not the case!

 

[Demystifier]

Yes. How can a density operator in the Heisenberg picture be time-dependent? This would make the associated state (and the expectation of observables at time $t=0$) time-dependent, which by definition of the Heisenberg picture is not the case!

It depends on the definition of the Heisenberg picture. One possible definition is this:
- Any observable $A$ has time dependence $A(t)=e^{iHt}A(0)e^{-iHt}$.
- Any hermitian operator is an observable.
With this definition, $\rho$ is an observable and depends on time in the Heisenberg picture.

But I understand your reasons for not accepting such a definition and using a different one. With a different definition, you are fully right.

The question is not which definition is the "right one". The question is which definition is more useful.

 

[vanhees71]

"Matrix elements" are with respect to a basis of eigenvectors. For the Statistical Operator these matrix elements are the density matrix, which is, of course, picture independent (as are wave functions).

With the notation in #18 you have
$$\hat{\rho}(t)=\hat{C}(t) \hat{\rho}(0) \hat{C}^{\dagger}(t), \quad |o,t \rangle=\hat{A}(t) |o,0,\rangle.$$
Here $|o,t \rangle$ denote the eigenstates of a complete set of compatible observables.

The density matrix with respect to this basis is
$$\rho(t;o_1,o_2)=\langle o_1,t|\hat{\rho}(t)|o_2,t \rangle=\langle o_1,0|\hat{A}^{\dagger}(t) \hat{C}(t) \hat{\rho}(0) \hat{C}^{\dagger{t}} \hat{A}(t)|o_2,0 \rangle=\langle o_1,0|\hat{U}(t) \hat{\rho}(0) \hat{U}^{\dagger}(t)|o_2,0 \rangle.$$
As shown in #18 the unitary operator
$$\hat{U}(t)=\hat{A}^{\dagger}(t) \hat{C}(t)$$
is independent of the choice of the picture of time evolution as it must be for observable quantities. Note that indeed only the modulus squared of the density matrix is observable. So there is of course still the usual freedom in choosing phases of the eigenbasis left.

BTW: The only textbook on quantum theory I'm aware, where the full picture independence of QM is treated carefully is

E. Fick, Einführung in die Grundlagen der Quantentheorie. Aula-Verlag, Wiesbaden, 4 edition, 1979.

Unfortunately there seems to be no English translation of this marvelous book :-((.

 

[vanhees71]

Just look at equation (17) in Zurek's paper, where he uses the von Neumann dynamics for the density matrix, not the Heisenberg dynamics for observables. Expressed in the preferred basis, the density matrix becomes diagonal in the course of time.

In the Heisenberg picture there are no time-dependent states evolving according to the Schroedinger equation. That you undo the time-dependence of the operator by moving it into the states is just what one does when going from Heisenberg to Schroedinger!

Again! This is really strange that even experts in the field are confused when it comes to pictures. For the German-speaking readers I recommend to have a look at the textbook

E. Fick. Einführung in die Grundlagen der Quantentheorie. Aula-Verlag, Wiesbaden, 4 edition, 1979.

In the Heisenberg picture that Statistical operator is time-independent (let's forget about possible explicit time-dependence which can occur for open systems like a partice in a time-dependent em. field in a trap or something similar, which is also very interesting but confusing for the present discussion). Since the observables move according to the full Hamiltonian the eigenvectors of observables are time dependent, and the matrix elements must be taken with respect to such basis vectors to make sense in terms of probabilities a la Born's rule. So you have
$$\rho(t;o_1,o_2)=\langle o_1,t|\hat{\rho}|o_2,t \rangle.$$
This is picture independent, which I've explicitly proven in the previous posting just some minutes ago.

 

[naima]

BTW: The only textbook on quantum theory I'm aware, where the full picture independence of QM is treated carefully is

E. Fick, Einführung in die Grundlagen der Quantentheorie. Aula-Verlag, Wiesbaden, 4 edition, 1979.

I see that there is also online:
H van Hees. Grundlagen der Quantentheorie
I. Teil: Nichtrelativistische Quantentheorie

 

[vanhees71]

Yep, but it's in German too. The general treatment of the dynamics in an arbitrary choice of the picture is given here:

http://theory.gsi.de/~vanhees/faq/quant/node21.html

There's also a pdf version of this manuscript:

http://theory.gsi.de/~vanhees/faq-pdf/quant.pdf

 

[A. Neumaier]

even experts in the field are confused when it comes to pictures.

It seems so only because you call the statistical operator what I call the density matrix. Thus my density matrix is basis-independent and time-independent, and expectations of Heisenberg operators are time-dependent. The expectation of a rank 1 Heisenberg operator is therefore time-dependent and gives your formula.