# Interpretation of QM in the Heisenberg Picture

kith
In a dissipative system, the Hamiltonian is not self-adjoint, hence the equivalence of the pictures makes no longer sense. If one transforms by a similarity transform to preserve commutation relations, one destroys the symmetry between bras and kets assumed in the Schrödinger picture.
Dissipative systems can sometimes be modeled by Hamiltonians which are not self-adjoint but the Hamiltonian in the Lindblad equation is. And the Lindblad equation can also be written in the Heisenberg picture. How does this relate to your argument?

vanhees71
A. Neumaier
2019 Award
Dissipative systems can sometimes be modeled by Hamiltonians which are not self-adjoint but the Hamiltonian in the Lindblad equation is. And the Lindblad equation can also be written in the Heisenberg picture. How does this relate to your argument?
The Heisenberg dynamics you linked to is mathematically fine but conceptually flawed, as it does not preserve commutation rules. In particular, if ##X## stands for the position or momentum operator at the initial time, one of the two cannot be a position or momentum operator at later times. What then assures us that it is allowed to be one at the initial time?

vanhees71
Gold Member
2019 Award
In a dissipative system, the Hamiltonian is not self-adjoint, hence the equivalence of the pictures makes no longer sense. If one transforms by a similarity transform to preserve commutation relations, one destroys the symmetry between bras and kets assumed in the Schrödinger picture.
This has nothing to do with the joice of the picture, but it's how to build an effective many-body theory. Of course the effective dynamics of an open system is not described by unitary time evolution, but that's why you deal with picture-independent quantities to begin with to build effective theories. E.g., to get the Boltzmann equation from QFT you work with Green's functions, which are picture independent as well (correlation functions, i.e., averages over field-operator products ).

A. Neumaier
2019 Award
Of course the effective dynamics of an open system is not described by unitary time evolution, but that's why you deal with picture-independent quantities to begin with to build effective theories. E.g., to get the Boltzmann equation from QFT you work with Green's functions, which are picture independent as well (correlation functions, i.e., averages over field-operator products ).
The only point I wanted to make is that because of the lack of unitarity, the Lindblad equations for open systems have no good Heisenberg picture, but only a mathematical rewrite without consistent interpretation.

The Boltzmann equations are equations for expectation values, not (as the Heisenberg equations) for observables.

vanhees71
Gold Member
2019 Award
The Heisenberg equations are equations for operators that represent observables in the Heisenberg picture. The operators themselves are not observables, for the very reason of not being picture independent. For the analogous reason the potentials are not observable fields in classical electrodynamics for the very reason not being gauge independent.

Now I'm a bit confused about your statement about the Lindblad equation. Are you saying it's picture dependent? How can it then be a proper quantum master equation? I guess, I've to refresh my memory about it. I think, Weinberg has a good presentation of it in his QM book...

A. Neumaier
2019 Award
for operators that represent observables in the Heisenberg picture.
This is nitpicking. I follow the usual practice that tolerates an abuse of language and identifies the two.

Now I'm a bit confused about your statement about the Lindblad equation. Are you saying it's picture dependent?
The Lindblad equation is a differential equation for density matrices generalizing von Neumann's equation for the unitary case to open systems. So it is tied to the Schrödinger picture. Dropping the double commutator terms in the equation leaves a nonhermitian effective Hamiltonian.

Of course one can rewrite anything in the Schrodinger picture into something that seemingly looks like a Heisenberg picture, but for open systems there is no natural self-adjoint Hamiltonian to make such a rewriting natural (producing operators that have a meaningful interpretation) or useful (never saw it used in practice).
How can it then be a proper quantum master equation?
All quantum master equations are equations for density matrices or probability distributions, hence in the Schrödinger picture.

vanhees71
Gold Member
2019 Award
If it's for probability distributions or matrix elements of the statistical operator, i.e., in the notation used already above,
$$\rho(t,o_j,o_{j'}) = \langle o_j,t|\hat{\rho}(t)|o_j',t \rangle,$$
it should be picture independent, since this matrix element is picture independent.

If it's for statistical operators ##\hat{\rho}(t)##, then making approximations, it's picture dependent. Then, of course, you have to stick to this one picture used to derive the approximation, but sounds pretty dangerous to me as a physicist, because approximations need some physical intuition, which can go easily astray if you work with unphysical quantities. I guess, however, that's not the case for the Lindblad equation since it's pretty well established as a working approximation to describe open quantum systems. I've to reread Weinberg's chapter on it.

A. Neumaier
2019 Award
If it's for probability distributions or matrix elements of the statistical operator, i.e., in the notation used already above,
$$\rho(t,o_j,o_{j'}) = \langle o_j,t|\hat{\rho}(t)|o_j',t \rangle,$$
it should be picture independent, since this matrix element is picture independent.

If it's for statistical operators ##\hat{\rho}(t)##, then making approximations, it's picture dependent. Then, of course, you have to stick to this one picture used to derive the approximation, but sounds pretty dangerous to me as a physicist, because approximations need some physical intuition, which can go easily astray if you work with unphysical quantities. I guess, however, that's not the case for the Lindblad equation since it's pretty well established as a working approximation to describe open quantum systems. I've to reread Weinberg's chapter on it.
The derivation is usually performed in the interaction picture of the big, unitary system. But the reduced dynamics is formulated as Lindblad equation in the Schrödinger picture of the reduced, open system.

vanhees71
But this is not the Heisenberg picture in the sense of post #1, where the state is time-independent, hence never changes. This is related to our discussion starting here of the inequivalence of the Schrödinger and the Heisenberg picture observed by Fröhlich.
For the searches I've done up to now, apart from Rubin's 2001 ArXiv on the Heisenberg Picture version Everett and some earlier discussions I posted on this in s.p.r, you're the only one I've found discussing this in any depth. I decided to go back to the 1927 Solvay Proceedings and examine the more closely. There's a good retrospective/translation for this
https://arxiv.org/abs/quant-ph/0609184Section 5 deals with the measurement problem; although the discrepancy I raised is absent from its discussion.

This gap is wider than I initially described: for instance, where are the Heisenberg Picture (HP) versions of the canonical quantizaton of General Relativity (GR)? What does the Problem of Time look like in HP? Is it even a paradox when written in HP form (given that the operators' space-time dependence still remains intact)? If the HP version remains intact without any problems, then the Problem of Time is, itself, signalling a non-equivalence of the two pictures for canonically quantized GR and a proof by contradiction that there is no meaningful Schrödinger Picture at all in that setting. Gauge invariance (diffeomorphism) is the key issue there; so this reference might be relevant.

The Heisenberg versus the Schroedinger picture and the problem of gauge invariance
Dan Solomon
2007 June 26
https://arxiv.org/abs/0706.3867
Excerpted from the abstract:
"We will show that, although the two pictures are formally equivalent, the Heisenberg picture is gauge invariant but that the Schroedinger picture is not. This suggests that the proper way to formulate QFT is to use the Heisenberg picture."