2:01 AM
@fqq Have you seen matter modelling stackexchange? I thought it was pretty cool

8 hours later…
9:56 AM
Hey all, is it strictly true any (meaningful) statement about a quantum system ought to be able to be translated into a yes/no Hermitian operator defined on the Hilbert space of the system?

Sure, you can define any measurement as projection operators on the eigenstates

For example, if I consider the statement, "the z-spin of this spin-1 particle is +1" then the truth-value of the statement is in correspondence with the projection-operator onto the +1 eigensubspace of the spin-operator. See, the discussion on page 4 here, for example: arxiv.org/abs/1907.02953.
Yes, Sleerah, exactly. I sent my message because I had already typed out even if I already saw your message before posting. 😅
But here is my real question: what about a statement such as "this spin-1 particle has a definite value for its z-spin"?

@DvijD.C. that's just not something you can measure in the normal sense of measurement

Yes, but it's something I can experimentally deduce if I have an ensemble of identical states.

yes, but deductions from an ensemble are different from a measurement on a single state

10:04 AM
I agree but what I am tripping over is that I think we cannot get a yes/no operator that would correspond to this deduction even over the product Hilbert space.

that's because you can't deduce this with certainty with a finite number of measurements!
the statement "this particle has a definite value" effectively corresponds to a statement of the form "this system is in this particular state", and you cannot "measure the wavefunction" exactly with a finite number of measurements

Yes, well, I meant the same product space over which we can deduce, i.e., the d^N dimensional product with N going to infinity.

but that space doesn't contain the "state of infinitely many particles"

For example, something like the frequency operator that gives the Born rule in Hartle's paper: arxiv.org/abs/1907.02953.

it only contains the "state of N particles" for arbitrarily high N

10:09 AM
Hmm, do we really expect a discontinuity? I mean we do physics assuming that this discontinuity doesn't exist, right?

I'm not sure what you mean by "discontinuity"

I mean that the results for the limit of N going to infinity can be assumed to be the same as the results for a truly infinite ensemble.

sure, but what I'm saying is that there is no operator you could take the limit of here
there simply is no measurement that tells you "this state was definitely an eigenstate" for a single particle

Well, I was thinking that we can consider a sequence of operators O(N) which is not the desired yes/no operator at any finite N but becomes the desired yes/no operator in the limit $N\to\infty$
Just like the relative frequency operator of Hartle which becomes the frequency operator corresponding to the Born rule only in the limit N to infty but is not such an operator at any finite N.
@ACuriousMind Yes, no, I get that part. I am just quibbling over the infinite ensemble case.

I think we're having a foundation issue here
the "propositions" we mean in the statement you quote about corresponding to operators are implicitly classical propositions
the classical statement "this particle has +1 for z-spin" is exactly the same as the classical statement "this particle has definitely +1 for z-spin" because classically there is no indeterminacy
these two propositions are only meaningfully distinct in a quantum context, but there is no rule in quantum mechanics that tells you that such quantum propositions would have to correspond to operators

10:22 AM
Wait, how are the two statements distinct in quantum mechanics? Both correspond to the projection operator over +1, right?

I think it is in fact pretty simple to show that for a spin-1/2 particle with no other d.o.f. there cannot be an operator whose eigenvalues correspond to this because the Pauli matrices for a basis for the space of Hermitian operators on $\mathbb{C}^2$ an all of their linear combinations just correspond to measuring the particles spin about a particular axis

@ACuriousMind Yes, I agree that it is pretty simple to show. Another simple way to look at it would be to say that for such an operator, both +1/2 and -1/2 would fall under the same eigensubspace and thus, their superpositions would too and thus, all states would be in the same eigensubspace and thus, the desired operator is not really the desired operator.

@DvijD.C. The "definitely" statement corresponds to our state being an eigenstate of the projection operator., and it is either true or not. the statement without "definitely" can be "indeterminate" - it is upon measurement "sometimes true" for states that are not eigenstate
the whole point of quantum mechanics is that measuring (classical) propositions always has this indeterminate character
measurement devices (or the act of reading their output, or whatever - at least some part of the measurement process) are intrinsically classical constructions, they do not allow us access to the "actual" quantum state of the system, this is the root of the "measurement problem"!
if there was a way to measure what you wanted to measure, you would have shown that the world is, in fact, $\psi$-ontic (the wavefunction is a real thing and not a reflection of our ignorance)

Yes, no, I get that you cannot measure the pre-measurement state
in a single measurement

the inference for the ensemble is pretty easy though - if you take your ensemble and measure z-spin and you get +1 every time you become increasingly certain your ensemble consists of, in fact, an eigenstate of z-spin

10:30 AM
Yes, but I cannot translate it into a yes/no operator :P I mean I have no right to say that such a thing must exist

there's no operator for this because the way your certainty behaves (at least in a Bayesian model) depends on some arbitrary prior for this being true - it is dependent on your beliefs, not on some fact about quantum mechanics
if your prior for this being true is 0 (never choose garbage priors but you're in principle allowed to), then you never become certain it's an eigenstate!

Well, but in Bayesian thinking, you never become certain of anything, right?
Sorry, not at all educated in Bayesian way of probabilities

in the limit of infinite evidence I'm pretty sure you do, unless your prior was 0 :P

But for example, you can't become sure of even Born rule by the same standards, right?

@DvijD.C. there isn't an operator for "the Born rule is true", either!

10:34 AM
well, not for the Born rule being true but for the Born rule frequencies which is the analogous thing here, I think: both cannot be deduced from a single measurement but can be deduced from an infinite ensemble

1 hour later…
11:38 AM
Ah, now that I think a bit more about it, I think such an operator can be constructed quite simply using Hartle's relative frequency operator: Hartle has the frequency operator $f_k$ corresponding to the $k^\mathrm{th}$ eigenvalue of a given Hermitian operator $O$ such that $f_k\vert \psi\rangle^\infty = \vert\langle k\vert\psi\rangle\vert^2\vert \psi\rangle^\infty$.
So, the operator $g_O \equiv \sum_k f_k^2$ would be essentially the operator that tells us whether the state $\vert \psi\rangle$ is an eigenstate of $O$. Iff $\vert \psi\rangle$ is an eigenstate of $O$, then $g_O\vert \psi\rangle^\infty = \vert \psi\rangle^\infty$ and otherwise $g_O\vert\psi\rangle^\infty = c \vert\psi\rangle^\infty$ where $c<1$. So, it is essentially the yes/no operator just that no is many-valued.
^Readable version of the two messages before the image

12:40 PM
@DvijD.C. I only now looked at the paper and it's much clearer to me what you were going for, but note that this explicitly does not construct what you asked for at the beginning - a correspondence of the proposition with a projection operator. Both the $f_k$ and your $g_O$ have continuous spectrum and hence there is no projection corresponding to a single value (in particular the value $c=1$ fpr $g_O$) only for intervals.
So, just like there is no true position measurement for a single particle (you can measure "is the particle between $x_1$ and $x_2$?" but not "is it at $x$?"), there is no measurement for "is this an ensemble of eigenstates?", just for "does more than one measurement result occur for this ensemble with less than x% frequency?" where you can make x arbitrarily small but not 0

Ah, yes, for sure. I think I misrepresented my question initially, I was using the projection operator as just an example of a yes/no operator. But yes, I see that even as a yes/no operator this is not "clean" in the same way as the position operator is not. Thanks for pointing it out, I hadn't thought of it explicitly. This is also true for the relative frequency operator in the paper, right?

yeah, that's why I said "both the $f_k$ and your $g_O$"

Ah, lol, missed the "both the $f_k$ and ".

11 hours later…
11:26 PM
When we want to find the surface integral of a vector field $\vec A$ for a parameterized surface, lets say a sphere of Radius R, while $\vec r(\phi , \theta)$ it must be that $\vec A (\rho , \phi , \theta)$ . Is this assessment correct ?