# Can decoherence be formulated in the Heisenberg picture?

1. Feb 26, 2016

### 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?

2. Feb 26, 2016

### Feeble Wonk

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

3. Feb 26, 2016

### 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.

4. Feb 26, 2016

### dextercioby

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

5. Feb 26, 2016

### Feeble Wonk

6. Feb 26, 2016

### A. Neumaier

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.

7. Feb 26, 2016

### naima

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

8. Feb 26, 2016

### atyy

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?

9. Feb 27, 2016

### 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

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

10. Feb 27, 2016

### 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.

11. Feb 28, 2016

### Haelfix

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)

Last edited: Feb 28, 2016
12. Feb 28, 2016

### 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!

13. Feb 28, 2016

### Haelfix

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.

14. Feb 28, 2016

### Demystifier

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".

Last edited: Feb 28, 2016
15. Feb 28, 2016

### Demystifier

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

16. Feb 28, 2016

### Demystifier

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

17. Feb 28, 2016

### 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.

18. Feb 28, 2016

### vanhees71

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.

Last edited: Feb 29, 2016
19. Feb 28, 2016

### vanhees71

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.

20. Feb 28, 2016

### A. Neumaier

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.

21. Feb 28, 2016

### A. Neumaier

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.

22. Feb 28, 2016

### Demystifier

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 $\langle a(t)|$.

23. Feb 28, 2016

### naima

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).

24. Feb 28, 2016

### A. Neumaier

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!

Last edited: Feb 28, 2016
25. Feb 28, 2016

### Demystifier

$$\rho_{ab}(t)=\langle a|\rho(t)|b\rangle$$
$$\rho(t)=e^{iHt}\rho e^{-iHt}$$