## Use and abuse of von Neumann measurement of the first kind

The classification of different quantum mechanic interpretations and solutions to the measurement problem is sometimes based on von Neumann analysis of the measurement process, in particular on his "measurements of the first kind". I will attempt to show that the mathematical foundation of the classification is incorrect. But what are measurements of the first kind?

Suppose we have a measurement device \(|M\rangle\) with three states:

- ready for measurement: \(|M_0\rangle\)
- detecting outcome A: \(|M_A\rangle\)
- detecting outcome B: \(|M_B\rangle\)

Also suppose we have a quantum system \(|\psi\rangle\) which can collapse on the two outcomes: \(|\psi_A\rangle\) and \(|\psi_B\rangle\). von Neumann demands the following time evolution which could be obtained by some interaction between the measurement device and the quantum system:

\(|\psi\rangle\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)

At first sight, this is perfectly reasonable. For example in the case of a Stern-Gerlach experiment, the interaction Hamiltonian corresponding to an inhomogenous magnetic field is:

\(H = -i\sigma_3 \frac{\partial}{\partial z}\)

and on plugging this into the Schrodinger equation it produces the desired effect (and I'll skip the computation details for brevity sake). So

**what could be wrong with using measurement of the first kind arguments into the classification of quantum mechanics interpretations?**

If we try to use Schrodinger equation into the analysis of the measurement process, we need to do it in general, not only in particular cases where there is no issue. For example,

**what should**\(|\psi_A\rangle\otimes |M_0\rangle \)**evolve into?**We know:
\(|\psi_A\rangle \otimes |M_0\rangle \rightarrow |\psi_A\rangle \otimes |M_A\rangle\)

**because a repeated experiment yields the same outcome.**But is this what really happens when we have an interaction Hamitonian able to generate: \(|\psi\rangle\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)?

**I'll show that this is impossible.**

__The quantum axiom of repeated measurements yielding the same outcome is incompatible with von Neuman measurements of the first kind.__
Before proving this, let's see why we care. If we have that pure unitary evolution satisfy:

(1) \(|\psi_A\rangle \otimes |M_0 \rangle \rightarrow |\psi_A \rangle \otimes |M_A \rangle\)

(2) \(|\psi_B\rangle \otimes |M_0 \rangle \rightarrow |\psi_B \rangle \otimes |M_B \rangle\)

then by superposition we get:

(3) \((c_A|\psi_A\rangle + c_B|\psi_B\rangle)\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)

Now from (1), (2), and (3) the measurement problem can be presented as a ``trilemma'':

- the wave-function of a system is complete,
- the wave-function always evolves in accord with a linear dynamical equation,
- measurements have determinate outcomes.

where any two items contradict the third one. For example Bohmian interpretation violates the first item, spontaneous collapse the second item, and many worlds violates the third one.

Now back to Eqs. 1 and 2 we notice something odd: the lack of a backreaction from the interaction Hamiltonian on the measurement device. Can this really happen? Here comes the algebraic formalism to the rescue. In this formalism, time evolution is represented by a product \(\alpha\) which up to various factors of 2, \(\hbar\) and \(i\) is the commutator. We also have a product \(\sigma\) which is the Jordan product of Hermitean operators and again up to various factors it is the anticommutator. For a bipartite system like the one we consider: "quantum system-measurement device", the product \(\alpha\) is:

\((A_1 \otimes A_2) \alpha_{12} (B_1 \otimes B_2) = (A_1 \alpha B_1 )\otimes (A_2 \sigma B_2) + (A_1 \sigma B_1 )\otimes (A_2 \alpha B_2)\)

where \(A_1, B_1\) are operators in the Hilbert space of system 1 (the quantum system), and \(A_2, B_2\) are operators in the Hilbert space of system 2 (the measurement device). If the interaction Hamiltonian is \(h_{12}=h_1\otimes h_2\), the time evolution in the Heisenberg picture for the operators in system one is as follows:

\((\dot{A_1} \otimes I) = (h_1\otimes h_2) \alpha_{12} (A_1 \otimes I) = (h_1 \alpha A_1) \otimes (h_2 \sigma I) + (h_1 \sigma A_1) \otimes (h_2 \alpha I)\)

Because the product \(\alpha\) is skew-symmetric we have \((h_2 \alpha I) = 0\) which demands:

\(\dot{A_1} = (h_1 \alpha A_1) h_2 \)

Since \((h_1 \alpha A_1 )\) cannot be zero because the quantum system does evolve in time, the only possibility to get \(A_1\) to be constant is for \(h_2\) to be zero. But then there is no coupling with the measurement device whatsoever.

So we seem to be stuck explaining repeated measurements.

**Repeated measurements**\(|\psi_A\rangle \otimes |M_0 \rangle \rightarrow |\psi_A \rangle \otimes |M_A \rangle\)__do__generate the same outcome (this is an experimental fact) and**does happen. The only way out of this mathematical problem is to realize that the transition**__represents not a unitary time evolution, but a change in representation.__But wait a minute:**changes in representation are about Hilbert spaces and they do not happen inside a Hilbert space! And since they do not happen inside a Hilbert space, then there is no superposition and therefore:****Eq. 1 + Eq. 2 DOES NOT IMPLY Eq. 3**

The way the usual story goes in explaining the measurement problem is:

the wave-function of a system is complete

+

the wave-function always evolves in accord with a linear dynamical equation

+

Eq. 1, Eq. 2 (repeated measurement axiom)

+

Eq. 3 (by superposition)

IMPLIES

measurements DO NOT have determinate outcomes (because Eq. 3)

Because Eq. 1 + Eq. 2 DOES NOT IMPLY Eq. 3

**the "trilemma" is bogus and**- the wave-function of a system is complete
- the wave-function always evolves in accord with a linear dynamical equation
- measurements have determinate outcomes

**are all compatible with each other. The trilemma is an artifact of using only one Hilbert space representation.**If we insist on a single Hilbert space representation then Asher Peres (can we call him the forefather of QBism?) was right on demanding that quantum mechanics should not be applicable to itself.

But now we have a solid foundation for Peres' handwaving: one universal Hilbert space representation (or to the extreme a nonsensical wavefunction of the universe) cannot explain everything. Why? Because this is incompatible with identical results for repeated measurements. There are many Hilbert space representations linked by the Grotendieck construction.

__The measurement process is the lifting of the representation degeneracy through equivalence breaking.__
I will stop the measurement problem series here with the main conclusion stated on the line above. There is still one open problem before claiming success: derive Born rule in the Grotendieck Cartesian pair approach. This is a big work in progress.

Next time at this blog I will look into classical emulations of quantum mechanics. Very interesting stuff which hold some big expectations on realizing a quantum computer without the fear of decoherence. This is all real, not a crackpot's dream of beating Bell's theorem, and it is actually built and working in the lab. But to what degree can a classical system realize pure quantum effects? Please stay tuned.

Sorry but this text is just complete rubbish. Sometimes you write texts that indicate that you have understood quantum mechanics before you publish something like that which makes it obvious that you haven't.

ReplyDeleteCheck e.g. this Max Jammer book, chapter 3, for a proof that the measurement is repeatable if and only if it is of the first kind.

For some reason, you claim exactly the opposite.

The measurement of the first kind is defined by the requirement that the terms in the sum of the tensor products are mutually orthogonal - that's the only difference from the 2nd kind measurement. You completely ignore this difference so your usage of "first kind" shows that you have no idea what it actually means.

The root of your problem is that you are still thinking classically, just like all other anti-quantum zealots. This is manifested by your nonsensical invention of "ready for measurement" state of the apparatus. The summation in the sum only goes over different eigenstates of the measured system and there is no "ready" eigenstate of the measured operator. All your psychological problems result from that.

I am travelling today and I'll be out of internet connectivity, but I'll reply tomorrow.

DeleteSorry for the delay, I am now back.

DeleteThe criticism is vacuous. There is a large body of results regarding measurement, and although I do not have that particular book I assume it has something like http://arxiv.org/pdf/quant-ph/9603016v1.pdf

What I claim does not contradict any existing results. A measurement is of the first kind if the probability for a given result is the same both before and after the measurement. A measurement is repeatable if its repetition does not lead to a new result. Repeatable measurements are of the first kind but a first kind measurement need not be repeatable. My example amounts to a counterexample where a first kind measurement is not repeatable. If all first kind measurements are repeatable then this can be used to justify the trilemma and I want to show that the trilemma is incorrect.

Now on my specific claim, I show that there is no unitary evolution which satisfies:

\(|\psi_A\rangle \otimes |M_0 \rangle \rightarrow |\psi_A \rangle \otimes |M_A \rangle\)

\(|\psi_B\rangle \otimes |M_0 \rangle \rightarrow |\psi_B \rangle \otimes |M_B \rangle\)

and

\((c_A|\psi_A\rangle + c_B|\psi_B\rangle)\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)

Why? Because there is no backreaction on \(|psi_A\rangle\) and I showed that this implies no coupling between the quantum system and the measurement device. Find such a unitary coupling for any psi and M obeying the conditions above and I will admit I am wrong.

PS: cut the crap with “you are still thinking classically”.

Every claim of yours is just self-evidently wrong. The words "first kind measurement" and "repeatable measurement" are pretty much synonymous. The proof may be found at many places. I've quoted a classical text (Jammer, chapter 3) you should have known anyway if you knew anything about the foundations of QM. You claim exactly the opposite.

DeleteYour "non-existence of the unitary transformation" is self-evidently wrong, too. First of all, the third line is just a superposition of the first two lines - it automatically follows from the first lines if the operator is linear.

And there's absolutely no problem about the existence of the unitary operator that maps the two states on the LHS to the states on the RHS. These two conditions define two rows of the matrix. The two LHS vectors are orthogonal to each other, and so are the two RHS vectors - by the definition of the first-kind measurement. So it's obviously possible to find the remaining rows of the matrix so that the matrix is unitary. The unitary matrix is the same thing as a matrix whose rows are normalized and orthogonal to each other.

Lubos,

ReplyDeleteThe matter is very easy to settle. If you produce M0, M_a M_b psi_a, psi_b such that all 3 equations above hold, then I will write a blog post with the title:

"Lubos Motl was right and I was wrong"

and I will present and discuss your example in detail. I contend this cannot be done because of the form the interaction Hamiltonian has to take.

One key difference between us is that I try to prove why the Copenhagen view is the correct one. I do not assume the trilemma to be false, I am attempting to prove it is false. I guess you contend such proofs are unnecessary because all was settle long ago.

Dear Florin, the three assertions that you claim are impossible are surely among postulates of quantum mechanics ("Copenhagen interpretation" as demagogue like you call it) so if you claim that they're incompatible, you are surely denying general rules of quantum mechanics.

DeleteIt's trivial to see that the unitary map you claim to be non-existent exists, see my blog for a proof. I don't believe that you will ever post the blog posts with the title you mention - anti-quantum zealots only do so while pretending that they're joking (Matt Leifer has posted a blog post of the same title).

Not sure what you mean with the 3 assertions because there are two sets of 3 assertions.

DeleteI assert that the following 3 are all true:

-the wave-function of a system is complete,

-the wave-function always evolves in accord with a linear dynamical equation,

-measurements have determinate outcomes.

and I want to prove it so.

As you know the consensus in the foundation community is that the 3 assertions from above cannot be all true at the same time.

On Matt, you understand an April Fool joke right? I recall I gave you the opportunity to reply to Matt's points but you chose not to. You are free to make whatever decision you want for whatever reasons you want, but do not question my sincerity.

Science is not about "consensus". You are hiding behind a stupid mob and blaming the mob for the stupidity but you have explicitly written exactly the same stupidity - namely that there is a "trilemma" that those 3 things can't hold at the same time - and you even claim that you have proven that they can't.

DeleteLeifer hasn't ever presented any sensible ideas about QM so I couldn't have replied to them. The April fool's day blog post (yes, I know what April Fools' Day is) was about his T-shirt and I did reply to that.

Cannot help noticing the similarity:

Delete"You are hiding behind a stupid mob"

"They’re bringing drugs. They’re bringing crime. They’re rapists"

In science there is a thing called peer review and the consensus of the community matters.

On the scientific side, the whole point of quantum interpretations and the measurement problem is explaining the unique outcomes given the existence of superposition. There are unitary explanations of the measurement processes, and given any two such explanations corresponding to different outcomes, you can construct the superposition of the two. If you take Bohr's position, the measurement device is classical and as such there is no superposition of the "dead and alive cat". But if everything is quantum, how come the measurement device is classical? If you are with GRW, you get spontaneous collapse, if you are with MWI, every outcome occurs in a "branch universe", etc, etc.

So if you claim that "measurements have determinate outcomes" then the burden of proof is on you (or anyone else making the claim: I am claiming this and attempting to prove my claim). Insulting the community will not get you anywhere. Appealing to the role of the observer the way you do it is a red herring. QM is contextual as proven by the K-S theorem and deciding what to measure changes the outcome. The measurement outcome does not exist before measurement, no ifs, ands, and buts or you violate K-S. But consciousness plays no role whatsoever.

I stand by my promise: produce here M0, ... as requested and I will write the post giving you the credit. Handwavings like it is there, etc, etc, do not qualify. M0 is not "ignorance" at all, it is the measurement device ready state. For example the bubble chamber state before any particles start producing tracks in it. You are free to produce any quantum states you like; my argument hinges on the lack of a backreaction. And the invitation to concoct any states respecting the 3 properties is open to anyone.

Science is *not* about consensus. Moreover, pretty much all the bombshell claims you - and many of your soulmates in your "community" of anti-quantum cranks - keep on making about the nature of quantum mechanics almost certainly contradict the consensus among the actual practicing quantum physicists. I am not using this argument because it isn't a real scientific argument.

DeleteYou are using this as an argument despite the fact that it isn't an argument according to the scientific method; and despite the fact that this sociological claim of yours is false even as a sociological claim.

"On the scientific side, the whole point of quantum interpretations and the measurement problem is explaining the unique outcomes given the existence of superposition."

Then the whole point of this enterprise of yours is nonsense because the subject of the sentence above is nonsense. There is nothing to explain. At most "explain in the pedagogic sense" to a slow student.

Superpositions are mathematical objects so they don't imply and they can't imply anything about the physical interpretation by themselves. The physical interpretation has to be added on top of the mathematics and the interpretation in quantum mechanics is that the wave functions are collections of complex probability amplitudes that may be squared and combined to calculate probabilities of different outcomes. The very word "outcome" means that it is well-defined, and the fact that we don't know unambiguously what the outcome will be does in no way mean that the outcome fails to be unambiguous. Of course the outcome is unambiguous - otherwise we couldn't talk about its probability at all.

To suggest that something is inconsistent or incomplete about the interpretation of the quantum wave function - the only valid interpretation - means to be utterly stupid and if you can find 50 other stupid people who are in "consensus" that something is inconsistent about these self-evidently consistent rules, doesn't mean that your stupidity gets reduced in any way. It doesn't.

Also, I have demonstrated everything I claimed to be demonstrable.

Lubos, I do not blindly follow the consensus and my approach is different than both yours and the foundation community. I have no preference on any interpretation and my position comes from following the math and using category theory to reconstruct quantum mechanics. My preference for the Copenhagen approach comes from actual mathematical results. If the math would point into the MWI direction then I will be their biggest supporter, but as of today it does not.

DeleteThere are major differences between physics, math, and philosophy. You view everything through the eyes of physics and because of that you run into trouble with both the math approach (remember the quantum logic dispute where you took the physics approach to a pure math problem) and the philosophy side with the measurement problem.

I don’t approach the measurement problem from the philosophical point of view (and this is why I do not find worth my time to pursue results in the vein of toy models or the PBR theorem), but as solving a mathematical contradiction: break unitarity and you ruin the very mathematical foundation of quantum mechanics. This is a solid mathematical result and a mathematical way out has to exist. The categorical approach to reconstruction clearly illustrates the problem, and it also points the way to a potential solution. I bet you did not follow my approach at all and instead have those knee jerk reactions to things that seem to contradict the orthodox point of view while in fact I am proving it. The reason you do not see value into my approach is because you do not understand the role of philosophy which is not to create new knowledge but to make sense of what it is known.

If there are competing explanations, you should be able to show why the alternative points of view are incorrect. Ex-cathedra statements are not good enough. Why is GRW not good? GRW modifies QM and predicts different effects which are put to the experimental test right now in Gran Sasso and this will either confirm it or reject it. What is wrong with MWI? Show me the derivation of Born rule in there. Bohmian? How does the distribution of initial conditions (quantum equilibrium) adjust on subsequent measurements? Bohr? Why should the measurement device be treated classically when everything is quantum? Etc, etc.

Is there an observation in the premeasurement case?

ReplyDeleteIf no, can we speak about an interaction between the measurement apparatus and the quantum system? From where do we know that the two interacted?

If yes, what is the observable?

There is an interaction occurring between the quantum system and the measurement device, the measurement outcome does not come out of thin air. In the S-G device case I gave the interaction term: -i sigma_3 d/dz which is a good approximation to what takes place inside the device. The notation is a bit sloppy, to be pedantic I should have inserted a tensor product between sigma_3 and the partial derivative. In general the interaction Hamiltonian is a sum of terms of the form: \(h_1 \otimes h_2\).

DeleteHi Florin,

ReplyDeleteIt has been a while since I have been here. I have to admit that I am having a hard time trying to see where Lubos is wrong based on your arguments. If you have the first conditions

|ψ_A>⊗|M_0> → |ψ_A>⊗|M_A>

|ψ_B>⊗|M_0> → |ψ_B>⊗|M_B>

then for |ψ_c> = c_A|ψ_A> + c_B|ψ_B> that I can have

|ψ_C>⊗|M_0> → |ψ_C>⊗|M_C>

= (c_A|ψ_A> + c_B|ψ_B>)⊗ |M_C>.

This may be where Lubos runs into trouble, for if he were right then |M_C> is |M_A> and |M_B> at the two products. There would have to be projector operators of some sort of pick out the A and B parts, but projectors are not unitary.

LC

Hi Lawrence,

ReplyDeleteThere is no M_C. What I demand is as follows:

\(|\psi_A\rangle \otimes |M_0\rangle\rightarrow |\psi_A\rangle \otimes |M_A\rangle\)

\(|\psi_B\rangle \otimes |M_0\rangle\rightarrow |\psi_B\rangle \otimes |M_B\rangle\)

\((c_A|\psi_A\rangle + c_B|\psi_B\rangle)\otimes |M_0\rangle\rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)

The third line follows by superposition from the two above. But I claim the two above cannot happen in an unitary way because the the coupling Hamiltonian acts in a certain way modifying \(|\psi_A\rangle\) and \(|\psi_B\rangle\) as well. The problem I a pointing out is the lack of the backreaction in the first 2 equations.

I suppose I am not sure what is meant by back reaction. The first two equations are standard ideas concerning the coupling of systems, entanglements and the rest.

ReplyDeleteLC

No backreaction means that psi_a remains psi_a after the interaction with the measurement device.

DeleteWhat I showed in the text was that A1˙=(h1αA1)h2 and for psi_a to stay the same you need A1dot to be zero. This is zero if (h1αA1) is zero or h2 is zero. h2=0 means no coupling with the measurement device. (h1αA1) means psi_a does not evolve in time at all.

There is something about this that bothers me in general. It is something that has bothered me for a long time. A process of the sort

Delete|ψ_A>⊗|M_0> → |ψ_A>⊗|M_A>

has always struck me as artificial. It is a very standard thing, and common in the Everettian MW interpretation. However, such as interaction seems vacuous.

I will used the idea of WIMP dark matter detection as an example. If dark matter particles are weakly interacting then they should interact with other particles by weak neutral currents or Z particles. The methods of detection involve a supercold crystalline lattice, where if a DM particle scatters weakly off a nucleus in the lattice by a Z particle intermediary interaction then that will generate a vibration in the lattice that is detected. Then physically what is happening? This is a process that changes the momentum of the DM particle so we have

|ψ> → |ψ> + |∂ψ/∂x>δx,

which to make more accurate we consider this derivative as a covariant derivative ∇ → ∇ - gA, which means this variation is associated with the momentum exchanged by the gauge field. I could of course make more formal arguments, but that will not illuminate my point that much. The problem is that with this state transition above it seems to me we have

|ψ_A>⊗|M_0> → |ψ_A'>⊗|M_A'>

where |ψ_A'> is the scattered state, say the DM particle scattered off a nucleus in the lattice, and the measurement state |M_A'> corresponds to the perturbed state actually measured.

I would then tend to take some of this to task and question the whole idea of zero back reaction. It seems to me the real world of physics does not quite conform to this, except maybe as an idealization.

LC

Hi Lawrence, you make a very good point. Indeed is artificial on the basis of my argument and one strategy to get around that is to merge the quantum states of the measurement device into classes, like the smearing effect provided by a position detector for example. Then the story becomes more complicated.

DeleteIn quantum optics we have techniques for non-demolition measurements and erasers and can “clean up” loses from measurements. In this case of course these measurements are close to being what you describe. Going back to the early days of QM, when a photon is absorbed by a silver iodide on a photoplate it is hard to make this case. High energy physics is in this state as well. It will be a long time or never before we can do weak measurements with TeV scale physics. However, there is nothing fundamentally that different.

DeleteThis sort of merger, such as the third equation, is how entanglements are formed in this idealization. I was doing some of this with quantum states becoming entangled with black holes. It occurred to me that the process involves geodesics that have tangents covariantly constant along Killing vectors, and this lead me to think that this is involved with how quantum states become entangled with black holes. This is a form of momentum transfer. Then of course if that is the case it must be for all cases; there is some momentum transfer in forming entanglements or composites (products) of states in measurements.

LC

In quantum optics we have techniques for non-demolition measurements and erasers and can “clean up” loses from measurements. In this case of course these measurements are close to being what you describe. Going back to the early days of QM, when a photon is absorbed by a silver iodide on a photoplate it is hard to make this case. High energy physics is in this state as well. It will be a long time or never before we can do weak measurements with TeV scale physics. However, there is nothing fundamentally that different.

DeleteThis sort of merger, such as the third equation, is how entanglements are formed in this idealization. I was doing some of this with quantum states becoming entangled with black holes. It occurred to me that the process involves geodesics that have tangents covariantly constant along Killing vectors, and this lead me to think that this is involved with how quantum states become entangled with black holes. This is a form of momentum transfer. Then of course if that is the case it must be for all cases; there is some momentum transfer in forming entanglements or composites (products) of states in measurements.

LC

Hi, Florin--it's been interesting watching you roll out your solution to the measurement problem! Just a quick question--are non-repeatable measurements of the first kind merely a theoretical possibility you're using for your argument (nothing wrong with that, of course), or have they also been observed in the lab?

ReplyDeleteBest,

Eric Hamilton.

Hi Eric, the short answer is yes, but there are subtleties to consider. For example measurements of the first kind are also called "pre-measurements" because the collapse did not yet take place.

ReplyDelete