Can decoherence be formulated in the Heisenberg picture?

[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.

 

[vanhees71]

That's very confusing terminology. The density matrix is given by the matrix elements of the statistical operator.

 

[kith]

I also think that calling the statistical operator "density matrix" is bad terminology because it is an operator and not a matrix. Unfortunately, it is quite common.

I was curious about the original terminology and just checked Dirac and von Neumann. Dirac starts with the classical case and the phase space density. In analogy to this, he calls the statistical operator the "quantum density" or simply the "density" if it is clear from the context that he talks about the quantum case. Von Neumann doesn't introduce a name right away but later on, he calls it the "statistical operator".

 

[kith]

What is the bottom line of this thread? It seems that everybody agrees with @vanhees71 post #32, i.e. the statistical operator itself is time independent in the Heisenberg picture but its matrix elements are time dependent regardless of pictures. Is there still disagreement about what this implies for decoherence? It's hard to tell for me because most of the discussion has been about terminology.

 

[Haelfix]

I would say that it ultimately doesn't matter what you call things. The point is that the physics of which picture you are in here differs only by a Unitary transformation. You can go from the Heisenberg picture to the Schroedinger picture, you can then apply a time evolution, and then transform back to the Heisenberg picture. The point is the whole diagram commutes. It's just a case that it's a little more technically challenging, going in one direction in the case of the Decoherence formalism. For the record, I agree with the terminology in post 31 and 32 and is what seems more familiar to me.

But yea, I'm not really sure what the argument is. Are we really arguing that you can't do this?