# Question on Neumaier's FAQ

1. Jul 20, 2011

### WiFO215

In his theoretical physics FAQ, and in specific, http://www.mat.univie.ac.at/~neum/physfaq/topics/mResults" [Broken] page, Neumaier (N henceforth. The name is too big!) says that

How is this valid? How can we throw away information and still be A-OK?

Ballentine says something similar in Chapter 2 of his book. On pages 49-50, he argues that the essential properties of operators is that they have a spectral representation. He says that people often (wrongly) argue that operators are hermitian because they measure real values. He then goes on to talk about the arbitrariness of convention and states something similar to what is said above. Although I agree with Ballentine that using real numbers to motivate operators is wrong, I still grapple with this hermitian operator business. Can any operator with a spectral representation, not necessarily hermitian, then be chosen to be an operator? I shouldn't think so, but I argue that that would also require the use of convention to restrict the eigenvalues to be real.

I would also like to discuss the following:

Is this always true? Can you expand a little on this?

Last edited by a moderator: May 5, 2017
2. Jul 20, 2011

### dextercioby

Well, the measurement abilities are definitely limited and always will be. Measurement tools will always yield rational values for the observables being analyzed, yet this doesn't mean that the <real> values we were supposed to obtain can't be irrational numbers, hence real numbers in general.

This implies that the operators must be self-adjoint, because these have a purely real spectrum. The spectrum of these operators must be the link between the mathematics (functional analysis) and physics (observables & measurements of observables).

3. Jul 20, 2011

### WiFO215

@dextercioby
Yes, but this isn't the only way the results of the experiment could be written down. Why not write down the results in a set of nested intervals? Or a complex number with the real part denoting what it normally does and the complex part telling us the uncertainty? Maybe you should also give Ballentine a reading, not because he is infallible, but because my question pertains to what he has said and your arguments seem incomplete when you haven't read the book.

N himself does frequent this forum. Let's see what he also has to add.

Last edited by a moderator: Jul 21, 2011
4. Jul 21, 2011

### Fra

To see if I understand your question: Are you questioning, how the continuum (~uncountable numbers) is justified, when the actual counting of evidence in any specific situations does not seem to require uncountable numbers?

If so it's a critique I share. The problem is not that counalbe numbers can't be consistently embedded in a uncountable system - the problem is exactly the redundancy or overcounting of "possibilities" that is the result of that. My personal opinon is that the introduction of uncountable numbers here, is the seed that later forces us to renormalize away overcounted possibilities.

/Fredrik

5. Jul 21, 2011

### strangerep

Let's restrict to the finite-dimensional case first, where operators are ordinary matrices.

Remember how the ordinary eigenvectors span the vector space if M is Hermitian? There's a generalization to non-Hermitian matrices which I'll outline below...

As well as the familiar concept of eigenvalues and associated eigenvectors, there's also generalized versions. Let M be any matrix (not necessarily Hermitian). Then the usual eigenvalues and associated eigenvectors are defined by
$$(M - a 1) v_a ~=~ 0$$
where $v_a$ is the eigenvector corresponding to the eigenvalue "a". We call these "genuine".

Now we define a "generalized eigenvector of order p" as a vector satisfying
$$(M - b 1)^p g_b ~=~ 0$$
(where p is the smallest integer such that this is satisfied). Here $g_b$ is said to be a "generalized eigenvector of order p" corresponding to the generalized eigenvalue b.

It also true that the (generalized) eigenvectors corresponding to distinct eigenvalues are linearly independent. (Details can be found in Lax's book on Linear Algebra.) Following on, there's a spectral theorem that the whole vector space is spanned by the eigenvectors and generalized eigenvectors of any such nontrivial matrix M. Specifically, every vector in the space can be written as a linear combination of (genuine and generalized) eigenvectors of M. Afaict, this should mean that M has a spectral representation (although I don't recall Lax actually saying that). But (also afaict) such (generalized) eigenvectors corresponding to distinct eigenvalues are not necessarily orthogonal (in contrast to the case of Hermitian operators). So things are less convenient with non-Hermitian operators.

So is this mathematical generalization useful in constructing a generalized QM? Not that I can see. Re-reading Ballentine's section 2.4, I also don't see (easily) how the construction of a probability distribution could usefully be done in the more general context. Certainly his existing formulas (2.26)--(2.28) don't go through without some kind of modification which is not obvious to me.

In any case, to justify the extra hassle of using generalized eigenvectors in QM, one must find a case where the usual formulation in terms of Hermitian operators is inadequate.
Otherwise, Occam's razor advises us to stay with the latter.

One could perhaps push this investigation further and look at so-called "normal" operators $N$, which are such that $N N^* = N^* N$. Like Hermitian operators, normal operators have only genuine eigenvectors, and eigenvectors corresponding to distinct eigenvalues are orthogonal. There's a nice spectral theorem for them. But they don't have necessarily-real eigenvalues, unlike Hermitian operators. But for every eigenvalue of $N$, we can prove that $N^*$ has the complex-conjugate eigenvalue. Also, one can always construct two alternate operators:
$$H := \frac{1}{2}(N + N^*) ~;~~~~~~ K := \frac{i}{2}(N - N^*)$$
which are both Hermitian (and commute with each other, hence have a common set of eigenvectors). So we could quite happily work with H and K instead of N.

If you were referring to Ballentine's use of this phrase, he's using it in the context of specifying both the value and uncertainty of a dynamical variable. But the uncertainty (or variance) can be handled in the context of Hermitian operators in terms of two operators, namely $M$ and $M^2$. So we don't need a generalization away from Hermitian operators to cope with that.

HTH.

6. Jul 22, 2011

### WiFO215

Oh no. That's not what I meant, although I hadn't thought about that. It is an interesting question!

What I meant was that we are not using the experimental uncertainty that comes with any measurement. Ballentine mentions some other frameworks such as the nested intervals/ complex numbers, which COULD incorporate all this information. Thus, I wondered why we weren't using those.

7. Jul 22, 2011

### WiFO215

Hi SR!
Just to clarify what I meant in the first bit you quoted was that if indeed we had used a different convention such as the complex numbers, then we needn't restrict our matrices to be hermitian then, would we?

Now to tackle what you say.

To summarize:
1. You say that : We use what we do because of Occam's razor and there isn't enough reason to switch to a different convention wherein we might use generalized eigenvectors.

2. You say that : If we used normal operators, we could slip back to hermitian operators, so no need for the hassle.

Again, I raise the same argument I did before (also read below). If we use complex numbers to represent the uncertainty, then normal operators might be of use.

3.
Does he mean variance or uncertainty? Skipping ahead to page 226 (the section which I am currently reading, which to me serves as the source of this thread). He says not to confuse between the variance $$\triangle$$ and the uncertainty, $$\delta$$. Considering M2 only gives you the variance.

8. Jul 22, 2011

### Fra

I haven't read ballentine, so I can't comment on on that specific.

What about the "classical uncertainty of the state vector"? Then we have the density matrix formalism? Obviously any realistic real experiment would leave you with an inferred non-zero von neumann entropy due to uncertainty, no more than you have infinite precision of inital conditions in newtons mechanics? The idea that this uncertaint is just "ignorance of initial conditions" and that there is a realistic statespace is the same. But I admit it smells, but it doesn't smell worse than it did in classical mechanics?

> Nevertheless there is no contradiction if one assumes that reality is governed by equations in terms of exact real

Compare to how we in classical physics "assumes" that reality has an exact value (it's just unknown/uncertain). IT seems this is the same story with QM. Classically we have probability distributions in state space, and so do we where there is non-zero von neumann entropy?

It this again missed the point, then just ingore this.

/Fredrik

Last edited: Jul 22, 2011
9. Jul 23, 2011

### strangerep

Umm, I think you're misreading some of the material on that page...

1) Ballentine uses $\Delta Q$ to denote the standard deviation of Q -- which can be calculated theoretically, and measured via the root-mean-square half-widths of an experimentally obtained histogram.

2) But he uses $\delta Q$ to denote the resolution of the particular apparatus being used in the experiment. This is distinct concept from the above. It can change if we spend more money and buy better equipment that achieves finer resolution. In contrast, our purchase of improved equipment has no effect on $\Delta Q$. :-)

The term "uncertainty", as used in QM, refers to (1) above, afaik.

Look at Fig 8.2 on p226, which shows $\Delta Q$ as a particular width in a histogram. This is a real number. There's no need for the extra complication of complex numbers to represent $\Delta Q$.

Last edited: Jul 23, 2011
10. Jul 23, 2011

### strangerep

This phrase doesn't make sense for me. In classical mechanics, a state is a point in phase space and observables are functions on the phase space. All these functions commute, so one gets trivial uncertainty relations (corresponding to hbar=0).

This makes no difference. The real issue is whether the observables commute.

But.. (guessing about what you mean in the rest of post), I think you're alluding to a situation in classical mechanics where we only know the state as a region in phase space, not a specific point. Then, under time evolution, we get a kind of tube in phase space instead of a trajectory. If that's what you're talking about, then I think it falls under the heading of "resolution of apparatus", i.e., the $\delta Q$ that I mentioned in my preceding answer to WiFO215. Such imperfect resolutions prevent infinitely precise preparations, just the same as they prevent infinitely precise measurements. But both these things are distinct from the concept of theoretical standard deviation $\Delta Q$.

I don't want to ignore it, but this discussion here would become much more productive if you were to spring for a copy of Ballentine... :-)

11. Jul 23, 2011

### Fra

Yes this is what I meant. I fully agree that this uncertainty is different from the one resulting from non-commutativity. However, I tried to make a subtle point that although we tend to think the classical uncertainty is "trivial" as you say - and I see you point - that it's not so trivial after all since no system could in principle encode and hold infinite precision, then it becomes not only a matter of practicality, but rather a mechanism that fundamentally rules interactions between communicating system. But this is a subtle BSTM point, that I THOUGHT the OT was about, but I was probalby misaken. This is also why I mentioned the continuum issue first.

I think the problem is that I didn't get the key question of the OT. I didn't perceive that the uncertainty due to non-commutativity was the question. IF so, it seems clear that since the measures refer to standard deviations anyway, I'm not sure what the original objection was. I'll try to reread the thread.

Probably :) I never used that myself though, in my own personal library from past courses I have a bunch of other books, like Sakurai.

/Fredrik

12. Jul 24, 2011

### WiFO215

D'oh. I knew that. I'm sorry. I made a mistake in asking the question. What I meant to ask was that *N* (not B) is telling us (afaict) that we can discard $$\delta$$ and still be okay.
a) How is this so?
b) Couldn't we include those in a different number system if we needed to?

13. Jul 24, 2011

### Fra

Here is anothe attempt, now when I start to see see what you mean by the notions:

This is what I tried to answer in post 8.

I also think the subtle point I tried to make first come out too subtle as it tried to comment on a problem that I don't you think you acknowledged.

We don't "ignore" $$\delta$$, but since it is what I called "classical uncertainty" (I think strangerep didn't like that name) this uncertainty is what strangerep also mentions as "trivial". On first analysis this is true. It is about as trivial as uncertainty in initial conditions is in newtons physics.

(My subtle comment that this may be a simplification is something that belongs to the BTSM section so I won't bring that up again, as I see that noone asked for it:)

So if we shave off the $$\delta$$, then we assume that we have perfect knowledge of state vectors as initial states. Of course this isn't true, but the additional complication and blur is "trivial" in the sense that it's a classical probabilistic blur.

ie. Newtons mechanics also ignores this blur, in his equations etc. But of course, this blur IS there, and it does add ontop of the deterministic mechanics, making deterministic chaos in newtons world. Same in QM.

/Fredrik

Last edited: Jul 24, 2011
14. Jul 25, 2011

### strangerep

Going back to the original bit you quoted, i.e.,
I read this as basically the same idea as what Ballentine said. We're not "throwing away" $\delta$, but merely recognizing that real-world measurements cannot be infinitely precise, since each such measurement is usually an interval. But of course we hope that when improved apparatus is used, our new measureed intervals still overlap the theoretical values -- if not, then doubt our theory has cast into doubt.

But there's no need to do so.

Last edited: Jul 25, 2011
15. Jul 25, 2011

### strangerep

I take $$\delta Q$$ to denote measurement error -- which is quite distinct from theoretical uncertainty $\Delta Q$.

When manipulating measurement results, and computing derived quantities from the raw data, one (generally) must use interval arithmetic everywhere.

http://en.wikipedia.org/wiki/Interval_arithmetic

By default, one should usually think of empirical data as a set of intervals.

16. Jul 25, 2011

### Hurkyl

Staff Emeritus
When you start thinking about making a system whose elements are nested intervals representing more and more precise hypothetical "observations", you should keep in mind that the real numbers themselves are one such system.

(more precisely, there are various ways to make models of the real numbers in which a real number are suitable nested intervals (modulo a suitable equivalence relation))

17. Jul 25, 2011

### Fra

We agree. I caused the confusion, because what you call measurement error (which is a better name in this context I agree) is what I called "classical uncertainty". But we mean exactly the same thing.

The reason for my name is that "measurement error" describes nicely by "classical probability". The quantum uncertainty does not - it describes by quantum logic.

In my picture (where I'm considering a reconstruction of measurement theory from evidence counting) I prefer not to make the distinction such as "measurement error" and "intrinsic uncertainty", I instead consider the total "uncertainty" or a prediction, but that different levels of logic contributes to the total uncertainty depending on how evidence combines.

The distinction of "measurement error" as more trivial add-ons to the deeper determinism is just like it's always been done also in classical mechanics. But if we try to understand the emergence of quantum logic from a general inference I try to avoid this "classic picture" although it's certainly the correct standard textbook picture.

/Fredrik

18. Jul 26, 2011

### WiFO215

Oh I get it. This is an abstraction like, "Things will keep on moving unless we stop them". Although we've never seen it, it sounds reasonable and more importantly, it works. Right?

19. Jul 27, 2011

### strangerep

I prefer the term "idealization" in this case.

20. Jul 27, 2011

### WiFO215

Alright, got it! :)