Enjoying the blog thoroughly. This post brings to mind the unsettling re-occurence of determinism. If there exists a single super-state for the whole universe how is this different from any other form of determinism? Or equally unappealing, if a single observation defines a state throughout the universe this also seems to return back to determinism.
Maybe the answer lies in defining equivalence classes of sub-spaces, and the infinite hotel trick: In an infinite dimensional vector space you can always squeeze in another infinite number of dimensions
*fine print: this only works with a Hamel Basis (only finite sums allowed) not with Hilbert Basis (infinite sums allowed) on seperable spaces. Of course there is no reason why we are necessarily dealing with a countably infinite vector space, we may very well be observing an uncountably infinite Hilbert space, in which case no one countably infinite basis has a span-closure that equals the QM space.
mattleifer
· 3 years ago
I don't know what is supposed to be so unsettling about determinism. Despite what I said in the "commandments", if you can come up with a compelling interpretation of QM that is deterministic then that's fine with me. This is not a key issue as far as I'm concerned.
I don't think that tricks of infinite dimensional analysis can be that relevant either. It would be puzzling to me if the solution depends on something that seems to be more of a mathematical nicety than a physical explanation. Then again, maybe this is just a predjudice I have that belies my physicist origins.
Dave Bacon
· 3 years ago
I wonder what happens when you remove a set of measure zero (like bipartite entangled systems with degenerate Schmidt coefficients) from quantum theory? Does anything nasty happen?
mattleifer
· 3 years ago
One could possibly get away with doing that, since the statistics of the theory would be practically unchanged. The math would get horrendously complicated though.
The main problem I can think of is that often the "measure-zero" states are the most physically relevant ones. Physical symmetries usually dictate that the state has to have a severely constrained form. One would have to replace all the symmetry constraints in QM with some sort of approximate symmetry constraints.
Unit
· 3 years ago
The state |00> + |11> tells us that there was a measurement between systems A and B, but not which measurement was made. So we can only say that A measured B, or viceversa, but we cannot say what basis was used. Interestigly, we know a priori that the basis used in the measurement between A and B is the same we'll use to measure A or B.
Ernesto
· 3 years ago
Hi Matt, I noticed your new paper on the arxiv. What about a Leiferfest in this blog, i.e. a short description of the Bayesian net programme you're interested in? best,
mattleifer
· 3 years ago
Don't worry. There will be plenty of time for shameless self-promotion on this blog. The paper is not really about the Bayesian net thing though - that will be coming soon.
Matt Brown
· 3 years ago
Hi Matt,
I don't think this is so much of a problem. Say instead of two electrons in this interaction, we look at the interaction between an electron and a Stern-Gerlach device set up to measure x-spin, which gives a reading in quotes. So, we start with
(|up>+|down>)|"ready">
and evolve to
|up>|"up">+|down>|"down">
Now, according to Rovelli, from the perspective of the device, we get a measurement of either up or down, with 50% probability. While it's true the either determinate measurement of x-spin is also a measurement of a superposition of z-spin. This isn't something paradoxical, but just a fact about quantum theory and spin. And we describe it in terms of x-spin because that's what we are trying to measure. Similarly, in the reaction you've described above, we can describe the measured outcome in either way, because they are equivalent descriptions.
In other words, for the case of spin, I don't see why you're trying to avoid the multiple descriptions horn of your mutlilemma. After all, you don't want to rule out our ability to prepare systems in superpositions. And Rovelli can capture the kind of determinacy you want: when I go and measure x-spin, I get a determinate value for x-spin. When someone asks me what I got for x-spin, they get a determinate answer from me. I think the "decomposition of the wavefunction" that your worried about is probably a holdover from thinking of the wavefunction of the universe. After all, you are never in a superposition of states for you, because you're never in the position of being a reference frame for yourself as a quantum system.
I agree, it's a weird picture, maybe too weird, but I don't think it's got this problem.
Ruth
· 3 years ago
Here's what I think Rovelli would say: He denies the existence of states, and deals only with information in the form of values of observables. The interaction Hamiltonian would result in a correlation of values of spin along z with the measuring device, and those values would be reflected in the correlation. The decomposition in the x basis is not available because it does not reflect the actual physical interaction that took place. It would just be a mathematical game played with states which do not exist in his interpretation.
All Comments
Maybe the answer lies in defining equivalence classes of sub-spaces, and the infinite hotel trick: In an infinite dimensional vector space you can always squeeze in another infinite number of dimensions
*fine print: this only works with a Hamel Basis (only finite sums allowed) not with Hilbert Basis (infinite sums allowed) on seperable spaces. Of course there is no reason why we are necessarily dealing with a countably infinite vector space, we may very well be observing an uncountably infinite Hilbert space, in which case no one countably infinite basis has a span-closure that equals the QM space.
I don't think that tricks of infinite dimensional analysis can be that relevant either. It would be puzzling to me if the solution depends on something that seems to be more of a mathematical nicety than a physical explanation. Then again, maybe this is just a predjudice I have that belies my physicist origins.
The main problem I can think of is that often the "measure-zero" states are the most physically relevant ones. Physical symmetries usually dictate that the state has to have a severely constrained form. One would have to replace all the symmetry constraints in QM with some sort of approximate symmetry constraints.
best,
I don't think this is so much of a problem. Say instead of two electrons in this interaction, we look at the interaction between an electron and a Stern-Gerlach device set up to measure x-spin, which gives a reading in quotes. So, we start with
(|up>+|down>)|"ready">
and evolve to
|up>|"up">+|down>|"down">
Now, according to Rovelli, from the perspective of the device, we get a measurement of either up or down, with 50% probability. While it's true the either determinate measurement of x-spin is also a measurement of a superposition of z-spin. This isn't something paradoxical, but just a fact about quantum theory and spin. And we describe it in terms of x-spin because that's what we are trying to measure. Similarly, in the reaction you've described above, we can describe the measured outcome in either way, because they are equivalent descriptions.
In other words, for the case of spin, I don't see why you're trying to avoid the multiple descriptions horn of your mutlilemma. After all, you don't want to rule out our ability to prepare systems in superpositions. And Rovelli can capture the kind of determinacy you want: when I go and measure x-spin, I get a determinate value for x-spin. When someone asks me what I got for x-spin, they get a determinate answer from me. I think the "decomposition of the wavefunction" that your worried about is probably a holdover from thinking of the wavefunction of the universe. After all, you are never in a superposition of states for you, because you're never in the position of being a reference frame for yourself as a quantum system.
I agree, it's a weird picture, maybe too weird, but I don't think it's got this problem.