The Refutation of Bohmian Mechanics

[A. Neumaier]

I see your point, but I would interpret it differently. The formal manipulations that lead to this expression show that it is not really true that observables (assuming that any hermitian operator is an observable) in the Schrodinger picture are independent of time. In other words, I think it is more appropriate to generalize the Schrodinger picture itself, rather than to complain that the Schrodinger picture is not general enough.

Then you need to define the Schroedinger meaning of a measurement of a time-dependent observable A(t) in a time-dependent state psi(t) as t varies over time (and of A(t,s), which is also dependent on a former or later time).

This is not a trivial generalization of the standard textbook meaning of the Schroedinger picture, since one cannot appeal to ensembles of identically prepared systems anymore. So what is your proposal for defining this meaning?

The paper
http://xxx.lanl.gov/abs/quant-ph/0509019 [Found. Phys. 36, 1601 (2006)]
may also be relevant here. (It cites your paper.)

This is a thick paper and I need more time reading it.

 

[A. Neumaier]

Is there an associated measurement theory in the Heisenberg picture?

In the paper, I referred to linear response theory, about which you can read in any book on nonequilibrium statistical mechanics. (I am using Reichl.)

 

[Demystifier]

Then you need to define the Schroedinger meaning of a measurement of a time-dependent observable A(t) in a time-dependent state psi(t) as t varies over time (and of A(t,s), which is also dependent on a former or later time).

This is not a trivial generalization of the standard textbook meaning of the Schroedinger picture, since one cannot appeal to ensembles of identically prepared systems anymore. So what is your proposal for defining this meaning?

There are probably many possibilities, but there is one possibility which seems reasonable to me:
Not every hermitian operator is an observable. In particular, q(s)q(t)q(s) in the Heisenberg picture is not an observable, and A(s,t) in the Schrodinger picture is not an observable. Observables are only those hermitian operators that in the Heisenberg picture depend on one time only. Equivalently, observables are only those hermitian operators that in the Schrodinger picture do not depend on time (except when there is an explicit time-dependence involved).

Of course, one can make separate measurements at different times and then multiply the results, but such measurements can be reduced to a measurement of observables defined as above.

In both pictures there are hermitian operators that depend on many times, but these mathematical objects are not observables.

The confusing part is the fact that the quantity <Psi|q(s)q(t)|Psi> is called a correlation function, suggesting that this quantity represents a measurable correlation between q(s) and q(t). Yet, despite the suggestion, <Psi|q(s)q(t)|Psi> is not measurable. If you measure q(s) and q(t), multiply the results for each pair of measurements, and then repeat the procedure many times to average over the statistical ensemble of equally prepared measurements, the result you will get in general will NOT be equal to <Psi|q(s)q(t)|Psi>.

 

[Demystifier]

In the paper, I referred to linear response theory, about which you can read in any book on nonequilibrium statistical mechanics. (I am using Reichl.)

I believe that all equations of linear response theory can be formally rewritten in the Schrodinger picture. Perhaps it will make the theory look mathematically more ugly, but it does not mean that one cannot apply the same physical interpretation of the final equations that describe physically measurable quantities.

 

[vanhees71]

Sure, everything can be formulated also in the Schrödinger picture. E.g., the QFT book by Hatfield does this for relativistic vacuum QFT. What you have to define are the Green's or correlation functions (in the most general case as expectation values of Schwinger-Keldysh-contour ordered field-operator products).

 

[A. Neumaier]

If you measure q(s) and q(t), multiply the results for each pair of measurements, and then repeat the procedure many times to average over the statistical ensemble of equally prepared measurements, the result you will get in general will NOT be equal to <Psi|q(s)q(t)|Psi>.

It is not supposed to be that; nobody ever claimed that. Time correlations are probed by linear response theory, not by your simplistic classical multi-time recipe.

 

[A. Neumaier]

I believe that all equations of linear response theory can be formally rewritten in the Schrodinger picture. Perhaps it will make the theory look mathematically more ugly, but it does not mean that one cannot apply the same physical interpretation of the final equations that describe physically measurable quantities.

A formalism in which things look ugly is inferior to one in which things look nice.

The power of physical explanations often lies in finding the formulation that gives insight, and with time, the formulation giving the most insight is viewed as the ''real'' way to look at things.

Otherwise we could still today adhere to the view that the sun revolves around the Earth - it is fully equivalent to the conventional view, just a change in coordinates.

 

[A. Neumaier]

Sure, everything can be formulated also in the Schrödinger picture. E.g., the QFT book by Hatfield does this for relativistic vacuum QFT. What you have to define are the Green's or correlation functions (in the most general case as expectation values of Schwinger-Keldysh-contour ordered field-operator products).

But the field-operator products are multi-time correlations in the Heisenberg picture,
not in the Schroedinger picture.

 

[A. Neumaier]

Why do you say that?

I explained it in the discussion with Demystifier.

I prefer the wording of my post 8: “unitary evolution and the theory of measurement of quantum mechanics, strictly speaking, contradict each other”).

This contradiction is harmless. Unitary evolution holds only for an isolated system,
while an observed system is never isolated since it must interact with any instrument
that measures it.

But there is nothing in standard QM but unitary evolution and the measurement mechanism.

You silently equate QM with the Schroedinger picture. But QM is more. In the Heisenberg picture, the operators transform unitariily, and operators at different times exist side by side in the Heisenberg picture and can be composed. This iis _not_ matched by the Schroedinger picture, and hence not by Bohmian mechanics. That makes a world of differences.

So can you reasonably blame the Bohm interpretation for inconsistency (if any) with the ill-defined part of standard QM?

Standard QM is _not_ ill-defined if one works in the shut-up-and-calculate interpretation, which is enough for all real life predictions. The ill-definedness comes in only through obscuring the foundations with the measurement process.

 

[akhmeteli]

I explained it in the discussion with Demystifier.

And I just cannot accept your “explanation”. Indeed, the Heisenberg picture and the Schroedinger picture are unitarily equivalent (for a finite number of degrees of freedom). Therefore, the expression for time correlation in the Heisenberg picture can and should be used in the Schroedinger picture (after proper transforms of wavefunctions and operators). Small talk about “loss of interpretation” is just that – small talk. I believe that unitary equivalence mandates this choice of time correlations. If you want to smuggle in something different and call it “time correlations”, it’s your choice, but don’t expect me to accept this arbitrary choice.

Multi-time correlations have a natural meaning in the Heisenberg picture, but the associated operators in the Schroedinger picture are just formal expressions without any meaning.

They have the same meaning as in the Heisenberg picture, as the two pictures are unitarily equivalent.

For example, if you rewrite <q(t)q(s)q(t)> along the lined indicated by you, one gets psi(t)^*A(t,s) psi(t), with a Hermitian operator
A(t,s):=q U(t-s) q U(s-t) q.

This immediately raises several issues:

(i) This is not the Schroedinger picture since A(t,s) still depends on $t$, whereas in the
Schroedinger picture, observables are supposed to be independent of t.

Do you mean that no operators explicitly dependent on time can exist in the Schroedinger picture?

(ii) The Hermitian operator A(s,t) has no discernible meaning, except that inherited from the Heisenberg picture, which must therefore be regarded as the fundamental picture.

And that inherited meaning is quite enough, as the pictures are equivalent.

(iii) How would you measure A(s,t)? There is no associated measurement theory.

I would suspect that if some procedure is used to measure time correlation, the results of the measurements can be used/described in either of the equivalent pictures. Why should I invent some extra measurement procedure?

This contradiction is harmless. Unitary evolution holds only for an isolated system, while an observed system is never isolated since it must interact with any instrument that measures it.

You can consider an isolated system including the instrument (and the observer, if you wish). Do you mean unitary evolution does not hold for such a system? And unitary evolution predicts something different from what the theory of measurements predicts, as unitary evolution cannot turn a superposition into a mixture or introduce irreversibility. I disagree that this contradiction is harmless; however, harmless or not, it’s still a contradiction.

You silently equate QM with the Schroedinger picture. But QM is more. In the Heisenberg picture, the operators transform unitariily, and operators at different times exist side by side in the Heisenberg picture and can be composed. This iis _not_ matched by the Schroedinger picture, and hence not by Bohmian mechanics. That makes a world of differences.

The two pictures are equivalent, so how can one of them have more content than the other? Again, your small talk about operators is just that – small talk. Everything you say about the Heisenberg picture can be translated into the language of the Schroedinger picture. The pictures are equivalent, remember? Some phrase can sound clumsier in the language of the other picture, so what?

Standard QM is _not_ ill-defined if one works in the shut-up-and-calculate interpretation, which is enough for all real life predictions. The ill-definedness comes in only through obscuring the foundations with the measurement process.

Do you mean the Bohm interpretation is ill-defined even if one works in the shut-up-and-calculate mode? I guess this is just a double standard: where standard QM stinks (i.e. contains mutually contradictory components), you are trying to explain to us that actually it does not stink, but its fragrance is just a bit unusual, whereas any problem with the Bohm interpretation stinks to heaven. Again, in general, the Bohm interpretation’s problems are not my problems, but I do have a problem with your claim. Having the same unitary evolution, the Bohm interpretation could possibly produce predictions different from those of standard QM only due to some difference in the theories of measurement. And can you blame the theory of measuremn of the Bohm interpretation for inconsistency with the theory of measurement of the standard QM, if the latter theory contradicts unitary evolution?

Another thing. Even if I believed your words about the Schroedinger picture being deficient compared to the unitarily equivalent Heisenberg picture, your claim would still be misleading, as it turns out that the deficiencies of the Bohm interpretation that you claimed in post 7 are actually also deficiencies of the Schroedinger picture, so it looks like the Bohm interpretation may be in good company?

 

[A. Neumaier]

You can consider an isolated system including the instrument (and the observer, if you wish). Do you mean unitary evolution does not hold for such a system?

It does, but there are no measurement results in such a picture, since one forever remains in indeterminate superpositions.

 

[A. Neumaier]

Do you mean that no operators explicitly dependent on time can exist in the Schroedinger picture?

http://en.wikipedia.org/wiki/Heisenberg_picture says:

the Heisenberg picture is a formulation of quantum mechanics in which the operators (observables and others) incorporate a dependency on time, but the state vectors are time-independent. It stands in contrast to the Schrödinger picture in which the operators are constant and the states evolve in time.

Of course, one can introduce time-dependent operators by extending the Schroedinger picture, but then their meaning remains obscure since the usual rules to interpret the measurement of operators via an ensemble break down.

 

[atomistic]

I found this objection from a physicist:

Bohm's version of (nonrelativistic) quantum mechanics was originally written down in such a way that it (a) yielded the same results as standard quantum mechanics and (b) contained hidden classical variables that could be blamed for quantum indeterminism. If I remember correctly, that was done in order to spite von Neumann, who maintained that quantum mechanics is incompatible with hidden variables. Bohm succeeded in producing a theory that eluded von Neumann's criticism but that victory was only symbolic: the hidden variables would have to have bizzarre properties like non-locality. Recall that these variables were introduced to chase away the weirdness of quantum mechanics. Instead, the theory is even worse because it violates locality* and—in a relativistic context—causality.

Standard quantum mechanics has its own share of bizzarre features. The wave function exhibits non-locality, too. However, the wave function is not a physical observable. Any physical observable in quantum mechanics, including its relativistic extensions, obeys locality, causality etc. You cannot devise an experiment that would instantly transmit information over finite distance even though the wave function changes instantaneously.

In light of this, Bohmian mechanics is a project aiming to rewrite standard quantum mechanics in a specifically prescribed way. The few hard-core aficionados might succeed in that one day, but what exactly is the point of that exercise? They have "succeeded" with the non-relativistic version of QM but no one seems eager to adopt their scheme for doing any calculations.

Compare that with the success of Feynman's path-integral reformulation of quantum mechanics. It not only reproduces the results obtained in the framework of the standard QM and makes some intuitive connections to classical mechanics. On top of that, it provides ways to solve some problems that could not be solved in the canonical framework. In the old days, theorists relied on the http://en.wikipedia.org/wiki/WKB_approximation" , based on Feynman's path integral, solves the one-dimensional problem equally well and is applicable to higher dimensions and extends easily to field theory.

In contrast, Bohmian mechanics is totally useless. These guys are still trying to teach their old dog to do tricks standard theory learned half a century ago. They are hopelessly behind.

I would appreciate some input from someone more knowledgeable than I am. Thank you!

 

[A. Neumaier]

The paper
http://xxx.lanl.gov/abs/quant-ph/0509019 [Found. Phys. 36, 1601 (2006)]
may also be relevant here. (It cites your paper.)

Well, the paper says on p.32 that

The analysis of multi-time probabilities in Bohmian mechanics above does not
guarantee that the predictions of Bohmian mechanics coincide with those of
standard quantum theory. We shall demonstrate now that the key property
that permits that is the inherent non-locality of Bohm’s theory.

It is right with the first sentence, but the permission mentioned in the second sentence given by the subsequent analysis amounts to nothing. Nothing is proved in the paper for the Bohmian case. The permission refers only to the fact (p.34) that the argument given for local hidden variables doesn't apply, which is well-known.

 

[A. Neumaier]

I found this objection from a physicist:

no one seems eager to adopt their scheme for doing any calculations.

Compare that with the success of Feynman's path-integral reformulation of quantum mechanics. It not only reproduces the results obtained in the framework of the standard QM and makes some intuitive connections to classical mechanics. On top of that, it provides ways to solve some problems that could not be solved in the canonical framework. In the old days, theorists relied on the WKB approximation to calculate tunneling rates. Unfortunately, the WKB method is not easy to generalize beyond one-dimensional quantum mechanics. The instanton approach, based on Feynman's path integral, solves the one-dimensional problem equally well and is applicable to higher dimensions and extends easily to field theory.

In contrast, Bohmian mechanics is totally useless. These guys are still trying to teach their old dog to do tricks standard theory learned half a century ago.

I fully agree.

But why didn't you state the source of your quote? This is part of scientific practice.
The quote comes in fact from ''olegt'', within a discussion last October in http://telicthoughts.com/hawking-on-science/

How do you know that olegt is a physicist?

I would appreciate some input from someone more knowledgeable than I am. Thank you!

What kind of input are you looking for?

 

[atomistic]

I fully agree.

But why didn't you state the source of your quote? This is part of scientific practice.
The quote comes in fact from ''olegt'', within a discussion last October in http://telicthoughts.com/hawking-on-science/

My mistake. Apologies.

How do you know that olegt is a physicist?

Someone pointed me to one of his papers in Nature. I will try and find it.

What kind of input are you looking for?

A balanced input from both sides please.

 

[A. Neumaier]

A balanced input from both sides please.

Well, this still doesn't tell what you want to know. As you can see from the thread so far, you can't expect to get a balanced opinion. You just get the extremes.

Nicolic, whose work was commented on in the discussion from which you quoted, calls himself here Demystifier - so his view is that of one of the active workers on Bohmian mechanics. And I am playing here the devil's advocate...

 

[akhmeteli]

http://en.wikipedia.org/wiki/Heisenberg_picture says:

Even if you find a more reliable source for this definition than Wikipedia, I am sure you fully understand that this definition cuts some corners, as I don't think you can tell me with a straight face that there are no such things in the Schroedinger picture as time-dependent Hamiltonians, time-dependent perturbation theory, Fermi's Golden Rule... If, however, you feel you can do that, please advise...

Look, Schroedinger representation is just that - a representation of the Heisenberg commutation relations. If you insist that there can be no time-dependent operators in the Schroedinger picture, how can it be equivalent to the Heisenberg picture? Or you don't think it is equivalent? Or do you think some operators are forbidden in the Heisenberg picture because their equivalents are forbidden in the Schroedinger picture? Then there is no time-dependent perturbation theory or Fermi's Golden Rule in the Heisenberg picture either?

Of course, one can introduce time-dependent operators by extending the Schroedinger picture, but then their meaning remains obscure since the usual rules to interpret the measurement of operators via an ensemble break down.

Again, time-dependent operators (e.g. time-dependent Hamiltonians) are part and parcel of the Schroedinger picture, not some artificial "extension". And again, they have the same meaning as in Heisenberg picture, as these pictures are equivalent.

 

[akhmeteli]

It does, but there are no measurement results in such a picture, since one forever remains in indeterminate superpositions.

Precisely. And you call this contradiction "harmless"?

In this context I repeatedly quoted this article here: arXiv:quant-ph/0702135 (Phys. Rev. A 64, 032108 (2001), Europhys. Lett. 61, 452 (2003), Physica E 29, 261 (2005))

I summarized the article as follows (edited here):

"they rigorously study a model of measurement. In the process of measurement of a spin projection, the particle interacts with a paramagnetic system. This paramagnetic system evolves into some macroscopic ferromagnetic state, and this seems to decide the outcome of the measurement. However, according to the quantum recurrence theorem, after an incredibly long period of time this macroscopic state will inevitably return into the paramagnetic state, if unitary evolution is correct, thus reversing the outcome of the measurement." So, as I often say, no measurement is ever final.

 

[qsa]

I basically agree with Neumaier for two reasons. One is my own theory which I cannot mention (Neumaier, I have send you an email about it, QSA) . the second, is just a question which I did not feel I had an answer. if you assume the particle picture ,then when electrons overlap in hydrogen molecule and when they sit on top of each other(statistically) you would expect that their gravity would produce effect that will can be noticed. But because they are actual waves then no problem, only small effect.

 

[A. Neumaier]

Even if you find a more reliable source for this definition than Wikipedia, I am sure you fully understand that this definition cuts some corners, as I don't think you can tell me with a straight face that there are no such things in the Schroedinger picture as time-dependent Hamiltonians, time-dependent perturbation theory, Fermi's Golden Rule... If, however, you feel you can do that, please advise...

There are time-dependent Hamiltonians, but these are not fundamental but semiclassical approximations, in which an external field/force is replaced by a classical field/force.
More importantly, the usual probability interpretation cannot be extended to them, unless the time-dependence is very low frequency (which is not the case for the kind of time-dependent operators that appear in time-correlations).

And time-dependent perturbation theory is done in the interaction picture, not in the Schroedinger picture.

 

[A. Neumaier]

Precisely. And you call this contradiction "harmless"?

There is no contradiction, since nothing is observed.

Talking about observations means approximating the observation part in a semiclassical fashion. But nobody ever claimed in any field of science that the fact that an approximation is not exact would constitutes a contradiction.

 

[A. Neumaier]

One is my own theory which I cannot mention (Neumaier, I have send you an email about it, QSA) .

When did you send it?

 

[Demystifier]

And I am playing here the devil's advocate...

So, you don't really think what you say?

 

[Demystifier]

Atomistic, here are I my replies to some of the objections you quoted:

"Instead, the theory is even worse because it violates locality* and—in a relativistic context—causality."
Is violation of those really worse than violation of reality? Is it really more acceptable to you that electron does not have any objective properties at all, until you observe them? Do you really believe that the Moon is not there when nobody looks? Or if it is not acceptable to you either, then what IS your favored interpretation of QM?

"... but no one seems eager to adopt their scheme for doing any calculations."
That is not true. See e.g.
http://prl.aps.org/abstract/PRL/v82/i26/p5190_1

"In contrast, Bohmian mechanics is totally useless."
It is true that Bohmian mechanics is less useful than Feynman path integrals, but the example above (and there are other examples too) shows that it is not completely useless. Besides, one should have in mind that the main intention of Bohmian mechanics is NOT to be useful. Its basic intention is something else.

 

[alxm]

if you assume the particle picture ,then when electrons overlap in hydrogen molecule and when they sit on top of each other(statistically) you would expect that their gravity would produce effect that will can be noticed. But because they are actual waves then no problem, only small effect.

That makes no sense. The potential between two electrons is a point-charge (or mass) potential regardless of whether you're treating them classically or quantum-mechanically. Why would the effect be smaller if you 'assume the particle picture', and what would that even mean in this context - since it surely does not change the potential?

And how on Earth would it be measurable in any case? The q^2/r Coulomb point-charge repulsion is over fifty orders of magnitude larger than the m^2/r gravitational point-mass attraction.
The Rydberg constant is known to 14 digits of precision. You're missing almost 40 digits.

 

[qsa]

When did you send it?

I just sent it to you again , just now to

Arnold.Neumaier@univie.ac.at

 

[qsa]

That makes no sense. The potential between two electrons is a point-charge (or mass) potential regardless of whether you're treating them classically or quantum-mechanically. Why would the effect be smaller if you 'assume the particle picture', and what would that even mean in this context - since it surely does not change the potential?

And how on Earth would it be measurable in any case? The q^2/r Coulomb point-charge repulsion is over fifty orders of magnitude larger than the m^2/r gravitational point-mass attraction.
The Rydberg constant is known to 14 digits of precision. You're missing almost 40 digits.

what I am guessing is that treating the two electrons(overlaping) as particles although not moving in the conventional sense must come close to each other statistically then m^2/r will shoot to the roof. that is why we use the full wave to do any calculation and things work out right because the electrons are waves. just handwaving argument.

 

[A. Neumaier]

So, you don't really think what you say?

I only say what I think is valid.

Did you mean to imply with your comment that I am the devil, rather than the devil's advocate, because it isn't make believe only?

 

[Demystifier]

I only say what I think is valid.

Then you are not the devil's advocate:
http://en.wikipedia.org/wiki/Devil's_advocate

 

[A. Neumaier]

Then you are not the devil's advocate:
http://en.wikipedia.org/wiki/Devil's_advocate

There it only says

a devil's advocate is someone who, given a certain argument, takes a position he or she does not necessarily agree with, just for the sake of argument. In taking such position, the individual taking on the devil's advocate role seeks to engage others in an argumentative discussion process. The purpose of such process is typically to test the quality of the original argument and identify weaknesses in its structure

Thus a devil's advocate is allowed to agree with the position taken. The key is to identify weaknesses, whereas it is secondary whether or not one agrees with the position taken.

 

[Demystifier]

Come on, it clearly states that a devil's advocate does NOT necessarily agree with the position he takes. Do you or do you not necessarily agree with the position you take? Simply answer yes or no?

 

[akhmeteli]

There are time-dependent Hamiltonians, but these are not fundamental but semiclassical approximations, in which an external field/force is replaced by a classical field/force.

Neither is the Coulomb potential fundamental, as it’s just an approximation to photon mediated interactions of charged particles, so I don’t think your argument is relevant. It looks like you cannot deny that time-dependent Hamiltonians are legitimate in the Schroedinger picture, otherwise you would have to expel the Coulomb potential from the Schroedinger picture as well.

More importantly, the usual probability interpretation cannot be extended to them, unless the time-dependence is very low frequency (which is not the case for the kind of time-dependent operators that appear in time-correlations).

If, say, Fermi’s Golden rule does not fit into an interpretation, this is a problem for the interpretation, not for Fermi’s Golden Rule. Anyway, there is no consensus on the interpretation of quantum theory, as exemplified by the results of the recent poll in this forum and the fact that you promote your own interpretation.

And time-dependent perturbation theory is done in the interaction picture, not in the Schroedinger picture.

It is not relevant how time-dependent perturbation theory IS done, it’s important how it CAN be done, and it is quite obvious that any derivation of the results of time-dependent perturbation theory in the interaction picture CAN be rewritten in the Schroedinger picture. Furthermore, your argument does not hold water as it can be used to prove that time-dependent perturbation theory does not belong in the Heisenberg picture, as “time-dependent perturbation theory is done in the interaction picture”. I guess it was not your intent to prove that:-)

Therefore, I totally reject your claim that there can be no time-dependent operators in the Schroedinger picture. Please don’t paint yourself into a corner defending this claim.

Let me also repeat that even if your claim were true, it would mean that you blame the Bohm interpretation for something that the Schroedinger picture is guilty of too, so again, maybe the Bohm interpretation is in good company?

 

[akhmeteli]

There is no contradiction, since nothing is observed.

I guess you have to choose: either the contradiction is "harmless", or it does not exist. In the latter case I guess you have just solved the measurement problem in quantum theory:-)

Talking about observations means approximating the observation part in a semiclassical fashion. But nobody ever claimed in any field of science that the fact that an approximation is not exact would constitutes a contradiction.

So let me ask you again, is the contradiction "harmless" or nonexistent?

While I am waiting for your reply, let me consider your euphemism "approximation". Approximation to what, exactly? To unitary evolution of "the observation part"? Then how can you blame the Bohm interpretation for (possibly) being inconsistent with the theory of measurement of standard quantum theory, if this interpretation is faithful to the ultimate truth - unitary evolution? If the Bohm interpretation is consistent with the precise theory, its inconsistency with the approximation is not its fault, as otherwise it's also unitary evolution's fault that, strictly speaking, it's inconsistent with its own approximation (otherwise the approximation would have been a rigorous result, not an approximation). By the way, I do tend to think that unitary evolution is a precise law. If you question this, please say so. If, however, you tend to agree with that, then maybe it's not so easy to find true inconsistency between the Bohm interpretation and standard quantum theory, if they both reproduce this precise law?

 

[A. Neumaier]

Come on, it clearly states that a devil's advocate does NOT necessarily agree with the position he takes. Do you or do you not necessarily agree with the position you take? Simply answer yes or no?

Not necessarily means, according to established practice, that he MAY but NEED NOT agree with the position he takes. Thus I may regard myself as playing the devil's advocate even though (as I had already stated) I say what I really mean.