# I Interpretation of QM in the Heisenberg Picture

#### Federation 2005

Summary
There is no Heisenberg Picture version of the Born Rule.
This is in reference to a question, never fully resolved, posed here:

The von Neumann postulates for Quantum Theory - Evolution (Schrödinger's equation) and Projection (Born's rule) are always framed in the Schödinger Picture. The Heisenberg Picture version of Evolution is the Heisenberg equations of motion. The essence of the original question is: what about the other postulate, Projection? What is the Heisenberg Picture version of Born rule?

Well, apparently someone fell asleep at the switch there! There is no formulation. It's a gap in the foundations that's never been addressed, and apparently never even been noticed. It looks like nobody's ever gotten around to actually writing down what the Born rule looks like in the Heisenberg Picture!

Do a literature search. You're not going to find anything on "Heisenberg Picture" + "Born Rule". For instance, here's a literature search on arXiv.
As of 2019 June, empty. A web search won't do much better.

A Google Scholar search is mostly empty. You may find this entry there:

which basically notes ... that the Heisenberg picture has no Born Rule (and tries to explain why it's missing). It's a review of one of Born and Heisenberg's early works coming out of Solvay 1927. In fact, you can find the Born-Heisenberg paper right here at Solvay

in the conference proceedings starting on page 143.

The treatment of the Born rule is the central focus of all divergence of opinion on what quantum theory means and lies at the center of the Measurement Problem. That underscores how serious this gap in the foundation actually is. And the presence of this gap also shows why there is a divergence of opinion in the first place. Einstein was right, Quantum Theory is incomplete, and this is the point where it's incomplete. No Heiseberg Picture-Born Rule exists.

The closest treatment to this that you get are treatments of open systems quantum theory (which hybridize classical and quantum dynamics) or - more generally - treatments that hybridize classical and quantum dynamics. Noteworthy is a recent one found on arXiv, which successfully hybridizes classical gravity with quantum theory

"A Post Quantum Theory of Classical Gravity" https://arxiv.org/abs/1811.03116

Needless to say, its main talking point is that a resolution of the Measurement Problem emerges for free from its hybrid dynamics.

In the Heisenberg Picture, states are timeless. The Born rule is the rule for wave function collapse. But if states are timeless, then what exactly is changing when a Born rule projection occurs? And when and where? So, you can already see there that there's a basic problem with consistency here. How are you going to get a Born or Lueder's projection to happen when Heisenberg states are timeless?

This also puts permanent No Go on all the endeavors to "explain away" the Born rule. For, any attempt to do so will, upon success, end up making everything timeless. If you have nothing but Evolution, then in the Heisenberg Picture this translates into nothing but timelessness; and nothing ever changes. A kind of Zeno's Paradox is waiting at the end of the tunnel for all the {Decoherence, Everett, Consistent Histories} people! And when they get there at the end of the tunnel, it will jump out and say "Hi! Surprised to see me?!"

Here's an attempt from within the Everett camp to address the question

The very thing people want to achieve by these alternate interpretations (removing the Born rule) is the very thing you want to avoid! The beauty of the question "what is the Heisenberg Picture Born Rule" is that it draws focus to the point where the goals conflict.

It gets even deeper than this. The disagreement that supposedly lies between General Relativity (GR) and Quantum Theory (QT), particularly on its treatment of time, is only a gap between General Relativity and the Schrödinger Picture. The Heisenberg Picture's treatment of time already meshes quite well with General Relativity's treatment of time. In particular, the Heisenberg equations when generalized to fields, become partial differential equations whose forms do not single out any dimension of spacetime. They treat time and space on an equal footing.

So, the gap widely held to lie between GR and QT has been misplaced: it's not QT vs. GR, but is actually (Schrödinger QT) vs. (Heisenberg QT & GR).

To try and resolve this discrepancy, you first need to use replace state vectors|ψ> by what may be more properly regarded as the state W =|ψ><ψ| / <ψ|ψ>; and generalize this further to include mixed states. The Schrödinger equation iħ d|ψ>/dt = H|ψ> then becomes iħ dW/dt = [H, W]. The Born rule produces a transition W → Σ_a P_a W P_a upon measure of a quantity A in state W, where A quantizes to the operator Â = Σ_a a P_a, with eigenvalues (a) and eigenspace projectors P_a.

A second measurement of, say, a quantity C would then produce a projection Σ_a P_a W P_a → Σ_{ac} P'_c P_a W P_a P'_c where the P'_c are the projectors for C. The dependence on time-ordering can be encapsulated by using the time-ordering pseudo-operator T[] and its reversal T'[] to write this as Σ_{ac} T'[P_a P'_c] W T[P_a P'_c]. All of this will survive intact upon conversion to the Heisenberg picture, except that the operators, projectors will now have the time dependence and W will be time-independent; and the Schrödinger equation on states will be replaced by the Heisenberg equation on the quantized operators that correspond to physical quantities.

This generalizes to space-time form, where the time dependence of the projectors and operators now becomes dependence on space-time points. The most important feature to note is that the space-time coordinates are now on an equal footing, and the only vestige of causal ordering that remains is in the appearance of the T[] and T'[] brackets.

This introduces an effective time dependence on the timeless Heisenberg states. But, unlike the case of the Schrödinger picture, there is no real time dependency remaining in the evolution postulate - at least not for the states. They remain timeless. Instead, the time-dependency has moved over to the other projection postulate. To make this work consistently, you need the following structure:

(1) a point cloud that contains all the points at which applications of the Born rule will occur
(2) a network of Heisenberg states, written as mixed states W in general
(3) associated with each state, a partition of the point cloud into a "before" set and "after" set. Their causal ordering should be consistent so that no point in the after set has a timeline or null curve that leads to a point in the before set.
(4) a transition between any 2 states whose before/after sets agree on all but a finite number of points; the transition being given in a form similar to what described above.

Each state, despite being timeless, encodes a "now" by virtue of how it partitions the point cloud into a before and after set. So, each application of the Born rule bumps up the "now" by a notch. So, in effect, a flow of time is introduced, even though the operators themselves and dynamics do not single out any "now".

This actually gets pretty close to what Smolin described and call out for here in this lecture

Lee Smolin Public Lecture: Time Reborn

in which he presents some of the key talking points of his "Time Reborn" book

Related Quantum Physics News on Phys.org

#### DarMM

Gold Member
Well it's pretty simple. You have the quantum state $\psi$ giving the probabilities for various outcomes of an observable $A$.

You can either hold $A$ fixed in which case the state evolves as $\psi(t)$ giving time dependent values for $A$'s outcomes, or $\psi$ is fixed but assigns probabilities to each member of the indexed family $A(t)$ which are time evolutions of $A$.

The end result is the same: time dependent probabilities for each of $A$'s outcomes.

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small correction: “same: time”

#### vanhees71

Gold Member
Quantum theory is independent of the picture of time evolution. You can define everything in a completely general picture (Dirac), where both the operators representing observables and the statistical operator are time dependent. Born's rule is valid in the same form in all pictures. That's, because the transformation from one picture to another is given by a (time-dependent) unitary transformation, and quantum theory is invariant under arbitrary unitary transformations.

Born's rule always is the same. If the system is prepared in a state described by the statistical operator $\hat{\rho}(t)$ and $O_j$ are a complete set of independent compatible observables, represented by self-adjoint operators $\hat{O}_j(t)$ and $|t,o_j \rangle$ are a complete set of orthonormal eigenvectors, then the probability to find the values $(o_j)$ when measuring the $O_j$ at time $t$ is given by
$$P(t,o_j)=\langle o_j,t|\hat{\rho}(t)|o_j,t \rangle,$$
which is the most general form of Born's rule.

For a pure state you have $\hat{\rho}=|\psi,t \rangle \langle \psi,t \rangle,$
where $|\psi,t \rangle$ is a normalized ket. Then it boils down to the original Born's rule for pure states,
$$P(t,o_j)=|\langle o_j,t|\psi,t \rangle|^2.$$
The Heisenberg picture is the one, where the time dependence of the operators is chosen such that $\hat{\rho}$ is time-independent (except for explicit time dependence, which I don't discuss here at all for simplicity) and $\hat{O}_j(t)$.

#### A. Neumaier

Well it's pretty simple. [...] $\psi$ is fixed but assigns probabilities to each member of the indexed family $A(t)$ which are time evolutions of $A$.
But the original question was
The Born rule is the rule for wave function collapse. But if states are timeless, then what exactly is changing when a Born rule projection occurs?
Thus the question is not about the part of Born's rule specifying the probabilities, but about the second part specifying the state reduction.

At a state reduction, the discontinuous time evolution of the position and momentum operator fails to preserve the canonical commutation rule.

#### DarMM

Gold Member
Doesn't it just work the same though? In the Heisenberg picture your state $\psi$ is an assignment of expectations for all future times. When you make a measurement you simply assign a new state $\phi$ with different expectations for future times.

I don't think anything really changes.

#### A. Neumaier

Doesn't it just work the same though? In the Heisenberg picture your state $\psi$ is an assignment of expectations for all future times. When you make a measurement you simply assign a new state $\phi$ with different expectations for future times.

I don't think anything really changes.
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.

#### DarMM

Gold Member
I mean ultimately it's going to be just the non-commutative generalization of conditional expectations, i.e. a map $\mathcal{E}$ from one algebra $\mathcal{M}$ to a subalgebra $\mathcal{N}$ obeying:
$$\phi(PM) = \phi(P\mathcal{E}(M))$$
where $P \in \mathcal{P}(\mathcal{N})$, the projectors onto the subalgebra in analogue with the classical formula.
Due to states being dual to the algebra you can simply transfer this map onto the state to get collapse.

However I never took the Heisenberg picture to be that the state literally never changes even upon measurement, but that the state does not undergo unitary evolution.

#### A. Neumaier

Doesn't it just work the same though? In the Heisenberg picture your state $\psi$ is an assignment of expectations for all future times. When you make a measurement you simply assign a new state $\phi$ with different expectations for future times.
This is the subjective view, where one assigns states based on ones knowledge, and hence treats the collapse as conditional expectation. But there are objective observations of state reduction in continuous measurement, where collapse is most naturally part of the dynamics, and proceeding in this way seems inappropriate.
However I never took the Heisenberg picture to be that the state literally never changes even upon measurement, but that the state does not undergo unitary evolution.
With this interpretation, the riddle in post #1 is of course solved. But one needs to make precise what the Heisenberg picture should mean in a nonunitary context, and there are several possibilities, depending on which property one wants to preserve. You picked the trivial one.

#### vanhees71

Gold Member
But the original question was

Thus the question is not about the part of Born's rule specifying the probabilities, but about the second part specifying the state reduction.

At a state reduction, the discontinuous time evolution of the position and momentum operator fails to preserve the canonical commutation rule.
No matter, whether you believe in state collapse or not, mathematically the wave function is always the same in all pictures of time dependence since
$$\psi(t,o_j)=\langle o_j,t|\psi,t \rangle$$
is picture independent (modulo phases, which are irrelevant since not the wavefunction is a physical quantity but it's modulus squared).

#### DarMM

Gold Member
This is the subjective view, where one assigns states based on ones knowledge, and hence treats the collapse as conditional expectation. But there are objective observations of state reduction in continuous measurement, where collapse is most naturally part of the dynamics, and proceeding in this way seems inappropriate.

With this interpretation, the riddle in post #1 is of course solved. But one needs to make precise what the Heisenberg picture should mean in a nonunitary context, and there are several possibilities, depending on which property one wants to preserve. You picked the trivial one.
Is there a good reference for this ambiguity of the Heisenberg picture? The "trivial one" is the only one I've ever heard of.

#### A. Neumaier

Is there a good reference for this ambiguity of the Heisenberg picture? The "trivial one" is the only one I've ever heard of.
There seems to be no discussion of this in the literature, as post #1 noticed. Before Fröhlich's paper I didn't even realize the trivial possibility you mentioned as "the only one you heard of".

But a lot is known about the nontrivial choice that fixes the Heisenberg state once and for all and puts the stochasticity usually associated with the state into the Heisenberg dynamics. See the book by Accardi, Lu, and Volovich. This is interesting math and physics!

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#### Mentz114

Gold Member
See the book by Accardi, Lu, and Volovich. This is interesting math and physics!

#### SFXMan

But there are objective observations of state reduction in continuous measurement, where collapse is most naturally part of the dynamics, and proceeding in this way seems inappropriate.
Wouldn't we just stochastically evolve our operator? Consider the operator
$$|o_j,t_0\rangle\langle o_j,t_0|$$
If we are interested in the likelihood of measuring $o_j$ at time $t_2$, we would evolve our operator using normal unitary time-evolution
$$U_{t_0\rightarrow t_2}$$
But if we wanted to evolve the operator to $t_2$ given some measurement $p_j$ at time $t_1$, we would instead use the non-unitary evolution
$$U_{t_0\rightarrow t_1}|p_j,t_1\rangle\langle p_j,t_1|U_{t_1\rightarrow t_2}$$

 - clarified
[edit 2] - Added quote I was responding to

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#### A. Neumaier

if we wanted to evolve the operator to $t_2$ given some measurement $p_j$ at time $t_1$, we would instead use the non-unitary evolution
$$U_{t_0\rightarrow t_1}|p_j,t_1\rangle\langle p_j,t_1|U_{t_1\rightarrow t_2}$$
This cannot work well. The problem is that for consistency one must assume that measurements may have already been taken in the past, unknown to us. Then the operators undergoing the nonunitary evolution no longer have the properties of the original operators, as commutation relations are not invariant under nonunitary evolution; so it is not clear how we have to represent the observables of interest at the time we begin to model the evolution.

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#### SFXMan

I might be misunderstanding the problem but: In the Schroedinger picture, we would handle uncertainty about whether a measurement has occurred by replacing a sharp wavefunction collapse with a nonunitary time-evolution into a mixed state. In the Heisenberg picture, we would presumably apply the analogous time-evolution to the observable in question.

More generally, since in both pictures we are ultimately computing amplitudes for sequences of measurements at different times (a la some histories formalism), I think there's always an equivalence between folding time-evolution into the last observable (Heisenberg) and the initial quantum state (Schroedinger)

But there is likely some subtlety I am missing to the conversation.

#### vanhees71

Gold Member
Again, there's no dependency of quantum theory on the picture of time evolution used. By construction (sic!) quantum theory is picture independent. It's only a different but equivalent mathematical description of the dynamics. Different pictures are transformed from one to each other by unitary transformations, and QT is invariant under unitary transformations. This is a mathematical fact and has nothing to do with any obscure interpretational problem.

It's a nuissance that in this forum any quantum-mechanics question is deformed into some pseudo-problem of interpretation, and then students eager to learn physics get confused by those pseudo-problems rather than first learning the math and physics as it is really applied in real-world labs!

It's as in classical electrodynamics: The physics is completely invariant under gauge transformations. Describing the same physical situation in, say the Coulomb gauge leads to precisely the same physics outcome as when describing the same situation in the Lorenz gauge. There's no direct physical meaning of the potentials to begin with, and there's thus no problem with an apparent instantaneous interaction in describing the scalar potential in the Coulomb gauge. All the instantaneously-looking interactions cancel out when finally calculating the physical fields, which contain only retarded integrals, no matter whether you do the calculation in Coulomb or Lorenz gauge.

The same holds for QT: It's invariant under picture transformations (i.e., time-dependent unitary transformations), and if there were a fundamental problem in the Heisenberg picture there'd be a fundamental problem in any other picture either, because all pictures are equivalent!

#### A. Neumaier

But there is likely some subtlety I am missing to the conversation.
Position and momentum should be conjugate at all times, $[q(t),p(t)]=i\hbar$. This is the case when $p(t)$ and $q(t)$ transform unitarily with time but not when they undergo your proposed nonunitary change.

#### SFXMan

Position and momentum should be conjugate at all times, $[q(t),p(t)]=i\hbar$. This is the case when $p(t)$ and $q(t)$ transform unitarily with time but not when they undergo your proposed nonunitary change.
Ok I think I better understand what you are looking for. The nonunitary evolution I suggested earlier actually describes a sequence of probabilistic measurement outcomes, rather than just the outcome $o_j$, and so it is not just a description of $o_j$.

#### vanhees71

Gold Member
Non-unitary time evolution comes into the game of QT only if you consider effective theories for describing open quantum systems ignoring some information (on purpose and necessity to handle very complex systems like the interaction of a single particle with a many-body system), which usually leads to quantum master equations (e.g., Lindblad equations), kinetic theory, Langevin/Fokker-Planck Eqs. etc. etc. This is a very wide and important as well as very interesting topic, but it's fully consistent with standard QT, and indeed as @A. Neumaier said, on the fundamental QM level, there's no problem with Born's rule and all the consequences, including the Heisenberg-Robertson uncertainty principle within the Heisenberg (or any other) picture of time evolution.

#### A. Neumaier

Non-unitary time evolution comes into the game of QT only if you consider effective theories for describing open quantum systems ignoring some information (on purpose and necessity to handle very complex systems like the interaction of a single particle with a many-body system), which usually leads to quantum master equations (e.g., Lindblad equations), kinetic theory, Langevin/Fokker-Planck Eqs. etc. etc.
The non-unitarity appears only in the Schrödinger or interaction picture, where the equations you mention reside.

In the Heisenberg picture, only unitary evolution is sensible since commutation relations must be preserved, and the appropriate unitary formulation for open systems in the Heisenberg picture is the one in terms of random operators, as in Accardi's book.

#### vanhees71

Gold Member
No, the equations are picture independent, as is the Schrödinger wave equation, $\langle o_j,t|\psi,t \rangle$ is the same, no matter how you "distribute the mathematical time dependence" to state kets and observable-operators or, equivalently, their complete eigen bases $|o_j,t \rangle$.

Commutation relations are the same in any picture, because any two pictures are by definition (sic!) related through unitary transformations.

#### A. Neumaier

No, the equations are picture independent, as is the Schrödinger wave equation, $\langle o_j,t|\psi,t \rangle$ is the same, no matter how you "distribute the mathematical time dependence" to state kets and observable-operators or, equivalently, their complete eigen bases $|o_j,t \rangle$.

Commutation relations are the same in any picture, because any two pictures are by definition (sic!) related through unitary transformations.
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.

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#### SFXMan

Can't we just construct histories in either picture and assign them probabilities (provided certain consistency conditions are met)? Each possible history would then represent a possible stochastic evolution of our system, no?

#### A. Neumaier

Can't we just construct histories in either picture and assign them probabilities (provided certain consistency conditions are met)? Each possible history would then represent a possible stochastic evolution of our system, no?
The question is how to do it consistently so that the same assignments result for both pictures. This does not seem possible in a Heisenberg picture in which the state remains fixed when a measurement on it is performed.

"Interpretation of QM in the Heisenberg Picture"

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