# Can someone elaborate on the ensemble interpretation for me?

1. Nov 11, 2009

### superpaul3000

My understanding of this interpretation is that (in the context of the electron double slit experiment) the experiment is one particular outcome of an ensemble of equally prepared experiments. This explains the statistical nature of QM. But how does this explain the interference pattern? Obviously the electron goes through both slits at the same time and interferes with itself, so how does ensemble explain this?

2. Nov 11, 2009

### xepma

No, the ensemble interpretation does not say anything about this part. The electron does not go through both slits -- this is just how things are calculated. All possible paths the electron can take contribute to the amplitude that the electron will 'hit' a certain spot on the other side. In the ensemble interpretation this amplitude is interpreted as "the number of times the outcome will occur if you perform the same experiment an X amount of times".

The ensemble interpretation doesn't say anything about what happens in intermediate states.

3. Nov 11, 2009

### superpaul3000

If it doesn't say anything about the intermediate states then it can't say it doesn't go through both slits. So if it doesn't consider the intermediate states, then it is not really an interpretation of QM, but a classical statistical interpretation of the results is it not?

4. Nov 11, 2009

### zenith8

So the ensemble interpretation assumes the wave function doesn't represent the state of any individual system, just that of the ensemble. That immediately raises the question about whether this form of interpretation allows you to envisage dynamical properties, definite but unknown, of an individual system over and above those that are described by the wave function. If that's the case, then the ensemble interpretation can characterize itself as maintaining an 'agnostic' attitude as to whether a wave function constitutes a 'complete' description of a QM state.

However, if you attribute definite but unknown values to the dynamical variables of an individual system, it's clear, although not always necessarily recognized by the guys promoting them, that such theories are hidden variable theories. It needs to be specified whether these values are simply those revealed by the relevant measurements, or whether they are pre-measurement values that could, in general, be different from the post-measurement values. Such a specification would in turn involve spelling out the specifics of the hidden variable theory. Since the proponents of the ensemble interpretation never provide such details, the ensemble interpretation as usually presented can be said to be 'internally incomplete'.

The alternative position for someone who believes in this stuff is to stick emphatically to the basic point that a wave function offers a 'complete' description of the quantum-mechanical state pertaining to an ensemble. Unfortunately this statement is just equivalent to the usual Copenhagen theory. That's because, if the wave function relates to the ensemble, but there are no hidden variables to distinguish between different members of the ensemble, one might just say that the wave function relates to each member of the ensemble.

And of course it offers no explanation whatever for the interference pattern obtained in two-slit experiments. So you have to ask yourself, what have you gained over Copenhagen?

5. Nov 11, 2009

### xepma

Yes, correct. No wonder they also call it the statistical interpretation.

However, I think the correct way to formulate is that the particle did went through one of the slits, but QM is simply unable to answer the question which of the slits it was. The QM amplitudes only arise as a statistical distribution when you look at the ensemble.

The ensemble interpretation is in some sense the simplest way out -- even simpler than Copenhagen and Many-worlds. It doesn't deal with single-particle systems (i.e. where, supposedly, hidden variables apply).

I'm not a big fan of it, but I wouldn't write it off so easily though. The way things are calculated in quantum mechanics is probablistic in its nature. The machinery only allows you to calculate probabilities, i.e. quantum amplitudes, and you can calculate them for both quantummechanical averaging or other statistical averaging (which is the reason why you can apply the QM-machinery to stuff like classical statistical systems, predicitions about the weather and the stock market). So to interpret quantum amplitudes as being the result of some statistical averaging process, of which the details are unknown, does not sound that weird to me.

But zenith is right that in that case you either resort to some form of a hidden variable theory (the different 'members' of the ensemble are not entirely the same) or you're back to the Copenhagen interpretation (the wavefunction does apply to single-particle systems).

6. Nov 11, 2009

### superpaul3000

I agree. The more I learn about the "interpretations" of QM the more I think that they are just different ways of saying QM is complete or incomplete and (equivalently) multiple universes or hidden variables. So basically any interpretations other than MWI or hidden variables are just being sneaky or vague about what they imply.

7. Nov 11, 2009

### zenith8

Right, but you forget the fact that QM is perfectly able to answer the question about which slit it went through. The trajectories are already in there - they're just the streamlines of the probability current (as in de Broglie-Bohm). They become apparent as soon as you drop the absurd 1920s positivistic philosophy that things are not necessarily there when you don't look at them, and you say that 'probability' means the probability of the particle being there as opposed to the probability of being found there in a suitable measurement.
In what sense is it a way out? A way out of what?

I think it isn't a way out of anything..
From the so-called 'hidden variables' point of view (like de Broglie-Bohm) QM is just the statistical mechanics of particles with a non-classical dynamics, so it's kind of the opposite of weird. The only reason we have statistical results is that we do not know the initial positions and velocities (to better than the width of the wave packet).

You can do the same thing in classical mechanics. Imagine you have a statistical distribution of particles with unknown positions, and that the particles obey Newtonian mechanics (but we'll use the equivalent Hamilton-Jacobi equations to describe the dynamics instead of F=ma).

If you combine the following two real equations:

- the classical continuity equation for the probability distribution of particles
- the classical Hamilton-Jacobi equation

into a single complex equation, do you know what that complex equation is?

It's $$i\hbar \frac{\partial \Psi}{\partial t} = \left( -\frac{\hbar^2}{2m}\nabla^2 + V - Q\right)\Psi$$.

Recognise it? Oh - it's the time-dependent Schrodinger equation with something ('$$Q$$') that looks like a potential subtracted off, and $$\Psi$$ has the same interpretation as in QM: a probability density of particle positions. Thus quantum mechanics is just classical mechanics with an 'extra force' $$-\nabla Q$$ in the dynamics, which turns out to be the wave field 'pushing' the particles.
[warm glow]

8. Nov 11, 2009

### Bob_for_short

By the wave nature of electron.

Turn on the laser in a double-slit experiment and look at the screen. You see an interference pattern. Take a picture of it. Good.

Turn on the electron source in a double-slit experiment and look at the screen. You see an interference pattern. Take a picture of it. Good.

In both cases you have ensembles of many points of which the information about systems consists.

If you move the screen, the interference picture changes. Its pattern is completely predicted with the wave function in the screen plane. Complete determinism. Nothing to complain about.

9. Nov 11, 2009

### Fredrik

Staff Emeritus
I think this is correct, and a good point. The idea behind the ensemble interpretation is that quantum mechanics doesn't describe what actually happens. But it's clear that something does happen, and it's reasonable to assume that whatever it is, it can be described by a mathematical model. The variables of that model would be the "hidden variables" you're referring to. But note that there's no reason to assume that those variables would be observables, i.e. that they correspond to equivalence classes of measuring devices in the real world. We want them to be observables of course, because if they're not, there's no way to turn the model into a theory. (To me a "model" is just a mathematical structure, and a "theory" is defined by the axioms that tell us how to interpret the mathematics as predictions about the probabilities of possible results of experiments). No matter what axioms we impose in our attempts to turn the model into a theory, we end up with something that's neither verifiable nor falsifiable, and therefore isn't a theory at all. It could be the correct description of what actually happens, but it still wouldn't be science.

Isn't that already ruled out by Bell inequality violations? Take the Bell inequalities for spin 1/2 for example. The experiments that violate those don't rule out the existence of hidden variables. What they rule out is the possibility that what we've been thinking of as measurements of spin components are actually measurement of some other, hidden, variables.

The whole point of the ensemble interpretation is that QM doesn't describe reality. It just tells us how to calculate probabilities of possibilities. So how can the fact that the ensemble interpretation doesn't tell us what actually happens be a reason to consider it incomplete? I guess we'd have to define more carefully what we mean by an "interpretation" before we can decide if this one is incomplete, or if it's an interpretation at all. I would actually argue that it isn't.

10. Nov 11, 2009

### xepma

Nice derivation, and no I wasn't aware of it.

I won't say I'm a fan of pilot wave theory though, but I'll bite with a few questions: how do you account for fermi/bose statistics? And what about the uncertainty relation? Basic questions, I'm sure, but please, humor me ;)

11. Nov 11, 2009

### superpaul3000

I know in those setups it appears deterministic, but what about the SINGLE electron double slit experiment? What determines the specific outcome of the measurement according to this interpretation? The wave nature of the electron doesn't explain the entire experiment. There is a wave particle duality. How does ensemble explain this duality? (Thanks everyone for your input)

12. Nov 11, 2009

### Fredrik

Staff Emeritus
It doesn't say anything about those things. What it says is that QM doesn't describe anything except how to calculate probabilities of possible results of experiments. The only thing that differentiates it from "shut up and calculate" is that the latter doesn't even say that.

13. Nov 12, 2009

### Demystifier

Not necessarily. It is possible that we really do measure the spin components, but that these spin components change their value due to instantaneous influences between distant particles and measurement apparatuses.

Still, in the simplest known model/theory of hidden variables (the Bohmian one) you are right: all measurements are actually measurements of particle positions.

14. Nov 12, 2009

### Demystifier

If you are interested, you can see more about it here:
http://xxx.lanl.gov/abs/quant-ph/0505143 [Found.Phys.Lett. 19 (2006) 553]
http://xxx.lanl.gov/abs/0707.2319 [AIPConf.Proc.962:162-167,2007]
and references therein.

It's trivial. The fermi/bose statistics is a property of the wave function, and this wave function is also a part of pilot wave theory. (Otherwise, it would not be called pilot WAVE theory.)

Take for example a plane wave psi(x)=e^{-ipx}. Standard QM says that momentum is perfectly known and has the value p, while position is completely unknown. Bohmian QM says that momentum has the value p, while position has a definite value which, however, is not determined by the wave function. Instead, the position is determined by the initial conditions (of particle positions) somehow chosen by Nature. If we do not know these initial conditions, then, for us, the position is completely unknown, despite the fact that it has a definite value.

15. Nov 12, 2009

### zenith8

Just to expand on Demystifer's answers a little..
Note the answer to this in orthodox QM is fundamentally mysterious, where nobody has the least idea why Pauli's exclusion principle should hold for fermions, for example, or what spin actually is. Why do same-spin fermionic particles repel each other? What exactly is the 'electron degeneracy pressure' that stops a star from collapsing? As Pauli himself, said:

"I was unable to give a logical reason for the exclusion principle or to deduce it from more general assumptions. In the beginning I hoped that the new quantum mechanics would also rigorously deduce the exclusion principle."

The answer to this in pilot-wave theory depends on what you are allowed to assume from the start. If we say from the start (with no justification) that fermions must have antisymmetric wave functions, and bosons have symmetric ones, then it is indeed trivial. To be antisymmetric the fermionic wave function must have both positive and negative regions and hence a multidimensional nodal surface separating them upon which the wave function has the value zero. It is a property of the deBB particle trajectories that they cannot pass through these nodes. If a trajectory heads towards one it will be 'repelled' from it. This is the reason that the fermions tend to avoid each other. 'Electron degeneracy pressure' is just seen to be a manifestation of the pilot-wave 'quantum force'. If you go through the maths in detail, you find that the force exerted by the wave field prevents two fermions coming into close proximity when their spins are the same.

Now what about the case with general wave functions. Does pilot-wave theory provide a justification of why fermionic wave functions are antisymmetric etc.? This might be thought to be a problem, since if the wave field is a physical field that propagates through space, it should be able to be represented by wave functions that do not have any particular symmetry.

One often finds textbook arguments about this which are simply incorrect. For example, the conclusion that the wave function of a fermionic system is antisymmetric does not follow from 'indistinguishability' of particles as is commonly stated (which is good since they are distinguishable in pilot-wave theory by their trajectories). Nor does the antisymmetric form arise from the requirements of relativistic invariance. The literature is also full of ***-covering statements such as 'fermions avoid each other because of statistical repulsion' - whatever that is supposed to mean.

Note that in pilot-wave theory that spin actually turns out to be a property of the wave field rather than of the particle. It is the polarization-dependent part of the wave field's angular momentum.

What is almost never done in the literature is to ask what happens when two particles (say two neutrons, to avoid the problem of electrical interaction) whose wave fields don't overlap come together and interact, and to proceed without insisting from the start that the wave function is antisymmetric. Without invoking the antisymmetry assumption, there is no obvious expression for the form of the two neutron wave function when the individual wave fields first overlap. Although I've never seen the details worked out, you can make a plausible case which indicates that the antisymmetric form of fermionic wave functions in a stationary state arises from the description of the interference between physical wave fields within a bounded region (which explains why the exclusion principle is best known in regard to stationary states). Think of the two particles as being in a box. The wave field of the two-neutron system will be successively reflected from each end of the box. In the case of a fermionic wave field, reflection at a rigid wall causes a change of the wave field's phase of pi radians. Interference between incident and reflected wave fields gives a stationary antisymmetric wave function because of the sign change on reflection.

I am aware that this raises more questions than it answers, and that I've skipped most of the details, but it is a reasonable plausibility argument for something whose explanation is entirely mysterious in the conventional theory. Note that with this, I'm going much deeper than the question you actually asked, so if it bothers you, stop at the end of the the third paragraph above.
OK. First of all, someone who mistakenly thinks that the uncertainty principle applies to an individual system might think that it is is evidence for particles not having trajectories at all. After all, a particle following a trajectory has a simultaneously well-defined position and momentum, right? Fortunately, it is now understood that Heisenberg's principle doesn't relate to measurements on individual systems. Uncertainty in the value of a dynamical variable refers to the statistical spread over the measured values for the various identical members of an ensemble of systems.

In pilot wave theory the actual momentum $${\bf p}=\nabla S({\bf x})$$ is unknown only because the position is. One can show quite easily that this leads to an 'uncertainty principle' for the true momentum of $$\Delta x \Delta p \geq 0$$.

So what was Heisenberg going on about? What meaning can be attributed to his $$\Delta {\hat p}$$ in pilot-wave theory, apart from a measure of dispersion in results of precision momentum measurements? The usual mistake lies in assuming that his $$\Delta {\hat p}$$ actually refers to the momentum of a particle before the measurement. Unfortunately this is only true in the classical limit (as you would expect since Heisenberg was basically quantizing 'mv' to get the momentum operator). In pilot-wave theory the momentum is not mv, it is something else, because of the quantum force acting on the particle. The particles follow non-Newtonian trajectories. This is a good example of why people should be very careful when they think the word 'measurement' implies (as it should) that we are revealing a pre-existing property of the system.

For what it's worth, if you analyze it carefully, Heisenberg momentum uncertainty can be identified with a component of the total stress tensor of the wave field (see Holland's standard textbook). It gives information on the current mean value of this as a particle property - not the actual momentum. The origin of statistical correlations between $${\bf p}$$ and $${\bf x}$$ measurements is due to the distribution of stresses in the wave field (which arise since the field guiding the particle in the ensemble also enters into the definition of the mean values).

Note the answers to these questions in the ensemble interpretation (which after all is what the thread is about). Q: How do you account for Bose/Fermi statistics? A: Er.. I don't. Q: And what about the uncertainty relation. A: Er, it's meaningless to ask? Can I go home now?

Last edited: Nov 12, 2009
16. Nov 12, 2009

### Demystifier

:rofl:

17. Nov 12, 2009

### superpaul3000

Well then the final verdict (for me at least) is ensemble is useless, NEXT!

18. Nov 12, 2009

### Fra

What kind of traits does your preferred utility have? ie. what do you ask out of an "interpretation"?

/Fredrik

19. Nov 12, 2009

### Fredrik

Staff Emeritus
If you think it can somehow be ruled out because of what I said, you're making a mistake.

A theory is "falsifiable" if it can be disproved by experiments. There's no single experiment that can be performed that can disprove QM, so QM actually isn't falsifiable in the original sense of the word. Instead QM satisfies a weaker condition, which I think of as "statistical falsifiability". Suppose e.g. that a theory predicts that the probability of a specific result of a specific experiment is 1/2, and you perform a large number of experiments to test that prediction. Now you can calculate the average result A(N) that you had after N experiments, and the probability that your average result would be wrong by at least |A(N)-1/2| after N experiments. If that probability appears to be getting closer and closer to 0 the more experiments you perform, we consider the theory to have been "statistically falsified". Note however that since you can't take the limit N→∞, you can't conclusively disprove the theory.

Do we want to require theories to be strictly falsifiable, or is statistical falsifiability enough? I would say that statistical falsifiability is all we should require, because anyone who disagrees with that would have to consider QM "not even a theory". Note that all theories that are strictly falsifiable are also statistically falsifiable.

So what's the weakest possible condition that a set of statements must satisfy in order to be statistically falsifiable, and therefore qualify to be called a theory? Since experiments can't tell us anything except how accurate a theory's predictions are, the minimum requirement must be that the theory makes predictions. They don't even have to be good predictions. (The best possible definition of a "theory" would have to include statements about e.g. the logical consistency of its axioms, and a few more statements of that sort, but those details aren't relevant here). Note that nothing in this definition of a theory implies that a theory must "describe what actually happens", or even attempt to do so. So the enormous accuracy of the predictions of QM can't be taken as evidence that reality is somehow "isomorphic" to the mathematical model used in QM. It still makes sense to wonder if there actually exists* a physical system that's described by QM, but we have to realize that neither of the two mutually exclusive answers to that question can be justified by experiments.

The ensemble interpretation is the assumption that the answer to that question is "no". The assumption that the answer is "yes" can (and should) be taken as the definition of the many-worlds interpretation. I consider these two to be the only interpretations of the version of QM that's defined by the Dirac-von Neumann axioms (the usual stuff about Hilbert spaces and the probability rule), and I'm not even sure the ensemble interpretation should be called an "interpretation". (If we define the term "interpretation of QM" to mean any QM-inspired speculation about what reality is really like, then there are of course lots more, but the MWI is the only realistic interpretation of the meaning of the Dirac-von Neumann axioms).

*) In other threads I have suggested a definition of what it means for something to "exist". I'm not using that definition here. I'm treating the word "exist" as a primitive here, i.e. as a concept that isn't defined in terms of anything else.

20. Nov 13, 2009

### zenith8

Hey xepma. If you're going to get us to write essays answering your questions you could at least reply. Is what Demystifier and I wrote clear? Do you agree or disagree?