# Von Neumann QM Rules Equivalent to Bohm?

1. Jun 1, 2015

### stevendaryl

Staff Emeritus
Bohm's deterministic theory was designed to be equivalent to standard QM, but what I'm not sure about is whether that includes Von Neumann's rules.

Von Neumann's rules for the evolution of the wave function are roughly described by:
1. Between measurements, the wave function evolves according to Schrodinger's equation.
2. If the wave function is $\psi$ immediately before a measurement, then after measuring an observable $O$ to have eigenvalue $\lambda$, the wave function will be equal to $\psi'$, which is the result of projecting $\psi$ onto the subspace of the Hilbert space where $O$ has eigenvalue $\lambda$
I don't want to argue about whether the second process, called "wave function collapse", is physical, or just epistemological, or just a rule of thumb with no particular meaning. But what I do want to know is whether Von Neumann's rules are consistent with Bohmian mechanics.

In Bohmian mechanics, there is no collapse, because the particle is assumed to always have a definite position (and the assumption is made that any kind of measurement can be understood in terms of one or more position measurements). The wave function $\psi$ has a double role: (1) $|psi(x}|^2$ gives the distribution of possible values for the position variable $x$, and (2) the wave function acts as a "pilot wave", affecting the trajectory of the particle.

The reason that I'm not certain about the equivalence of Bohmian mechanics and Von Neumann's QM is because after a measurement of position, according to Von Neumann, the wave function is now a delta-function (or at least is described by a highly localized function). But in Bohmian mechanics, there is no collapse, so the wave function continues to be whatever it was before the position measurement. So the two approaches--Von Neumann and Bohm--will be using different wave functions after the measurement. Those two situations don't sound equivalent to me.

Now, maybe it is that in Bohmian mechanics, the act of measuring position causes the wave function to collapse in the same way as Von Neumann, if we take into account the interaction of the particle with whatever device measured position. Is that the resolution?

2. Jun 1, 2015

### atyy

They are equivalent, provided one does rigourous QM. So after a position measurement, the state is not a delta function (it is not square integrable, so it is not an allowed wave function).

The collapse can be derived in Bohmian Mechanics. Essentially there is decoherence exactly as in Many-Worlds, but the conceptual subtleties are done away with by having the trajectory pick one of the worlds. If there is no recoherence (which is the condition for applying collapse in Copenhagen), then one can show that Copenhagen with collapse is a very good approximation to Bohmian Mechanics without collapse.

There's a discussion of this in VI.2 of this reference.

http://arxiv.org/abs/1206.1084
Overview of Bohmian Mechanics
Xavier Oriols, Jordi Mompart

Last edited: Jun 1, 2015
3. Jun 1, 2015

### Demystifier

Von Neumann rules are compatible with Bohmian mechanics (BM). Namely, even though there is no true collapse in BM, there is an effective (illusionary) collapse which, for all practical purposes, cannot be distinguished from the true collapse.

How that illusionary effective collapse happens? Due to decoherence, the total wave function splits into non-overlapping branches, which is a deterministic continuous process described by the many-particle Schrodinger equation. Since the branches are non-overlapping, the Bohmian particle may enter only one of the branches. When the particle enters one of the branches, all other branches cease to have influence on the motion of the Bohmian particle. This is effectively the same as if other branches ceased to exist; they are still there, but now ineffective.

Now you have two descriptions: you can use only the non-empty channel (which corresponds to the collapsed wave function), or you can use all the channels (which corresponds to the wave function which did not collapse). As far as motion of the Bohmian particle is concerned, the two descriptions are equivalent.

Last edited: Jun 2, 2015
4. Jun 2, 2015

### vanhees71

I object against von Neumann's 2nd rule. That's only true for ideal von Neumann filter measurements. Quite often the measured system is destroyed in the measuring procedure, and it doesn't make sense to describe it by a wave function for it in the sense of an isolated system. E.g., the accurately measured energies and momenta of the particles produced at the LHC hit the detectors and are gone thereafter. It doesn't make sense to associate a wave function to them anymore.

5. Jun 2, 2015

### Demystifier

But it still makes sense to associate a QFT state in the Hilbert space to them. For instance, the vacuum is also a state in the Hilbert space. Moreover, even if such QFT states cannot be represented by a wave function, they certainly can be represented by a wave functional. So the von Neumann rule still applies, but now with wave functionals instead of wave functions.

6. Jun 2, 2015

### stevendaryl

Staff Emeritus
The Von Neumann rule is really most important for composite systems, so that measuring a property of one component causes the collapse of the wave function for another component.

7. Jun 2, 2015

### stevendaryl

Staff Emeritus
I certainly agree that no actual measurement can result in a delta-function; instead, you make an imprecise measurement of position, and you end up with a localized wave function. But that distinction is not really relevant to my post.

So that sounds like Bohmian Mechanics is equivalent to Von Neumann via the no-collapse interpretations of QM. I guess that makes sense: Bohm is as compatible with collapse as MWI is.

8. Jun 2, 2015

### vanhees71

That's true. You need it as projection postulate in this case, i.e., to get the state of the unmeasured subsystem you have to trace over the measured other part. This example however shows als the EPR problems with collapse interpretations!

9. Jun 2, 2015

### vanhees71

Well, think "state" everywhere in my posting, where I've written wave function. It's the same thing. The Higgs bosons measured at the LHC are long gone and not in an (approximate) eigenstate of energy and momentum. The same holds true for the decay products measured. So one must take von Neumann's assertions not too literal.

10. Jun 2, 2015

### Demystifier

The bold part above is wrong. The Higgs which is "gone" is actually the Higgs field in the vacuum state, which is an exact eigenstate of energy and momentum (with eigenvalues equal to zero).

Concerning collapse, it looks as if you missed my point entirely, so let me be more explicit. Consider detection of a photon. There are two possibilities:

1) The photon is not detected and the detector remains in the ground state. The corresponding state in the Hilbert space is |photon>|ground>

2) The photon is detected (and hence destroyed) and the detector jumps to the excited state. The corresponding state in the Hilbert space is |0>|excited>

If each of the possibilities has some non-zero quantum probability to occur, then a purely unitary evolution of the quantum state actually gives a superposition a|photon>|ground>+b|0>|excited>. So you need a non-unitary collapse to pick up only one of these two terms in the superposition. By collapse, the state ends either in the state 1) or the state 2). In particular, if the photon is detected and destroyed, then the photon field is in the state |0>. The collapse describes a jump to the state |0>. Even though the photon is "destroyed", you still have a well defined state of the photon field.

The word here which should not be taken too literally is not so much the word "collapse", but the word "destroyed". Nothing is really destroyed; the system merely jumps to the ground state of the photon (or Higgs) field.

Let me finish by one technical remark. Even in QFT the state can be described by a wave function, provided that you allow wave function to depend on an infinite number of coordinates. In particular, the vacuum wave function is constant, not depending on x at all. For more details see e.g. the classic textbook
S.S. Schweber, An Introduction to Relativistic Quantum Field Theory (Eq. 80)
or my own paper
http://lanl.arxiv.org/abs/0904.2287 [Int.J.Mod.Phys.A25:1477-1505,2010]

Last edited: Jun 2, 2015
11. Jun 2, 2015

### Demystifier

Exactly!

12. Jun 2, 2015

### vanhees71

I think, we talk about two different things here. I've to study your paper first, before I can say anything about it. Of course, you can formulate QFT in terms of "wave functionals". Another more modern textbook than Schweber treating this approach is the book by Hatfield.

What I meant is that von Neumann's formulation has to be taken with a grain of salt. He only treats very special cases of measurements. There's a whole industry of new developments concerning measurement theory since the mid 1930ies, which come much closer to the reality in labs. Von Neumann's merit for QT lies imho not so much in the physical interpretation (which I consider totally flawed since it's a nearly solipsistic overempasis of the collapse interpretation) but in the mathematical foundation of non-relativistic quantum theory in terms of Hilbert space theory. As far as I know, there's not yet an as mathematically strict definition of any realistic QFT, let alone the Standard Model.

Very puristically spoken, the Higgs boson (or any other instable particle) is not defined as an observable entity in relativistic QFT at all. There only asymptotic free states are well-defined and in fact what's measured at ATLAS and CMS are of course the stable (or quasi-stable as in the case of muons) final states (it was discovered in the two-photon and the 2-dilepton (electrons and muons) channels by ATLAS and CMS as famously announced on Independence Day 2012). Even those stable decay particles have been absorbed by the detectors, and it makes no sense to describe than as if you could take von Neumann's postulate 2 literally.

13. Jun 2, 2015

### Demystifier

Maybe, but then let us try to make clear what exactly that difference is.

Note an important difference! Schweber talks about wave functions (depending on an infinite number of particle positions), while Hatfield talks about wave functionals (depending on entire field configurations).

Von Neumann talks about projective measurements. Modern measurement theory talks about POVM measurements, which, in a certain sense, are more general than projective measurements. However, they are more general only when one wants to describe measurement without explicitly describing the environment and the measuring apparatus. By contrast, when the quantum state of the environment and the measuring apparatus is also taken into account, then all measurements can be described as projective (von Neumann) measurements.

Even if the collapse is ignored, another important von Neumann's merit for QT is understanding that quantum measurement creates entanglement with the measuring apparatus. This physical (not merely mathematical) insight is a basis of modern theory of decoherence, which, in turn, has a lot to do with the "illusion of collapse" even if the true collapse is never introduced explicitly.

Even if it is true that von Neumann overemphasized the collapse, it is even more true that most quantum-physics textbooks do not sufficiently emphasize the quantum role of the measuring apparatus and environment. That's probably because Bohr (unlike von Neumann) insisted that the macroscopic world should be described by classical physics, which misguided several generations of physicists.

I strongly disagree. The fact that they are absorbed does not imply that you cannot describe it by the von Neumann's 2nd postulate. In post #10 I have explained explicitly how you can do that. The collapse with "absorption" is neither more nor less "literal" than the collapse without the "absorption".

Last edited: Jun 2, 2015
14. Jun 2, 2015

### atyy

What is the status of domain wall fermions and the standard model? Can a lattice standard model be constructed with domain wall fermions, at least in principle, even if it is too inefficient to simulate? Or is the answer still unknown?

15. Jun 2, 2015

### Demystifier

Mathematically, the Neumarks's theorem guarantees that for any non-ideal (=POVM) measurement in a Hilbert space of dimension n, there is a corressponding ideal (=filter=projective=von Neumann) measurement in a larger Hilbert space of dimension N>n. Physically, the larger Hilbert space corresponds to the inclusion of the environment and measuring apparatus in the quantum description. In other words, all measurements are ideal, provided that you include a sufficient number of degrees of freedom into your description.

16. Jun 2, 2015

### vanhees71

Mathematically it may make sense to assume that for each observable there exists an ideal-filter measurement. Then you should formulate the axiom in this way and not in the way given in the original posting. You find this formulation very often in textbooks, but it's confusing, at least it was for me for quite some time ;-).

17. Jun 2, 2015

### atyy

Yes, it's confused since it mixes the beliefs of different churches. But in this age of intolerance, it is ecumenical in spirit :)

http://www.quantiki.org/wiki/The_Church_of_the_larger_Hilbert_space

18. Jun 3, 2015

### Demystifier

So would you agree now that all measurements can effectively be described as a collapse, provided that it is a collapse in a Hilbert space which is usually larger than that of the measured observable?

19. Jun 3, 2015

### vanhees71

I don't consider collapse as a physical process, because this causes the old EPR troubles. What we have are local interactions of the system with a measurement apparatus which is approrpiately constructed to measure an observable, that's it. The only thing one has are the probabilities for the outcome of measurements given by a state, which I have associated to the system with an appropriate preparation procedure.

20. Jun 4, 2015

### stevendaryl

Staff Emeritus
When you say "the probabilities for the outcome of measurements", does that assume that there is a single outcome to a measurement? If there is only one outcome, then it seems to me that the picking of that outcome is a physical process.