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5:56 AM
@gls @Mithrandir24601 @ChainedSymmetry
I d
I edited the question one last time. quantumcomputing.stackexchange.com/questions/8715/…
Is this better written?
@Mithrandir24601 also what is the criteria for correct?
 
 
2 hours later…
glS
8:19 AM
@MoreAnonymous the question is fine, but I stand by my previous comments. If "measuring the eigenvalues" here means that you somehow magically get to know some of the eigenvalues (and assuming this makes sense, which I'm not sure it does), there are still infinite equally probable Hamiltonians compatible with a given observation. For example, say the Hamiltonian is $H=|0\rangle\!\langle 0|+2|1\rangle\!\langle 1| + 3 |2\rangle\!\langle 2|$.
If you only know that two of the eigenvalues are, say, $1$ and $2$, that means that the Hamiltonian can be anything of the form $|u_1\rangle\!\langle u_1|+2|u_2\rangle\!\langle u_2| + \lambda |u_3\rangle\!\langle u_3|$ for any three orthonormal vectors $u_1,u_2,u_3$ and $\lambda\in\mathbb C$.
but in practice you couldn't measure things this way. For starters, to measure an eigenvalue you would need to know the eigenstates, otherwise you wouldn't know what measurement basis to use. Furthermore, even if you know the eigenbasis, you don't "measure an eigenvalue", you sample from probability distribution given by measuring in the eigenbasis of the Hamiltonian, and slowly accumulate statistics getting a better and better estimate of the full set of eigenvalues
 
8:51 AM
@gls I'm slightly confused. To know the eigenbasis one needs to know the Hamiltonian? But the Hamiltonian is not an observable.
Only the eigenvalue is what is observed. Of course one can play games of getting an interacting Hamiltonian etc (which he knows well)
 
glS
9:22 AM
@MoreAnonymous any Hermitian operator is an observable
@MoreAnonymous but yes, that's my point. To meaasure the eigenvalues you usually need to know the eigenbasis, which means you need to at least know something about the Hamiltonian in this case. Hence my saying that the problem is ill-posed because I don't understand what kind of "measurement" you are talking about
@MoreAnonymous it's not like there is a magical box that magically gives back eigenvalues of operators.. what you have is states with which you interact in different ways. An Hamiltonian is a description of how you are interacting with a state.
 
 
1 hour later…
10:53 AM
@glS Are you sure? Like you saying xp_+ px + x^2 is an observable because it's an Hermitian
I was under the impression this was an open problem
@glS Also I think one can use some math trickery to solve this (using error bars in the eigenenergies)
@glS My counter question is how do I magically know these eigenbasis?
Like which one came first the Hamiltonian or the eigenbasis?
 
glS
11:10 AM
@MoreAnonymous I mean that's literally the definition of observable
@MoreAnonymous I don't know, hence my saying the question is still somewhat ill-posed
also, as I said before, you don't measure the eigenvalues of an observable, you measure the expectation value of the observable on a given state. I see no mention of any state in this discussion
 
@glS I thought the expectation was measured one eigenvalue at a time?
Consider the double slit experiment? I see a dot before I see the interference pattern?
 
glS
11:27 AM
@MoreAnonymous yes, and as stated in one of the answers, any hermitian is an observable if you don't take into account superselection rules, which do not exist when you reason at a general level like you do here
superselection rules are simply things you know about the Hamiltonian, which then allow you to restrict the set of observables which you can actually observe.
@MoreAnonymous no, as I must have said like 10 times here already, what you do is sample from the probability distribution of the state in the eigenbasis of the observable, and reconstruct the eigenvalues from the collected statistics
@MoreAnonymous yes, and every dot is one sample as per my previous comment. The dot alone doesn't tell you much about the underlying observable, you need to collect statistics to be able to say what are the overlaps of the state in the position observable that you are measuring in that case
 
11:48 AM
@glS maybe you can give me a reference and I can come back to you? I think I'm making a subtle assumption and equating empirical probability with quantum? In the sense I can start with a bunch of eigenvalues and empirically verify the born rule. Now I think it should be possible to do the opposite where one assumes the born rule measures a probability distribution and finds the eigenbasis?
@glS for example if instead of the screen I used a bunch of momentum detectors instead. I would have a different pattern?
Maybe I'm pushing my intuition too far. So I need to pen paper it down? I'll do that in an hours time?
 
12:21 PM
@MoreAnonymous I don't have a full list but like, does it give something that looks like entropy, does it satisfy whatever thermodynamic laws it needs to etc.
@glS @MoreAnonymous yes, superselection rules depend on things known as reference frames - they say you need the correct 'reference [frame]' in order to measure the thing you're measuring, not that you can't measure it in the first place
@MoreAnonymous Always put pen to paper whenever possible is a very good rule of thumb here
 
 
1 hour later…
glS
1:28 PM
@MoreAnonymous any introductory text on quantum mechanics I guess? The basic axioms tells you that given a state and an observable (Hermitian operator), the probability of a given outcome is given by the squared overlap of the state onto the eigenstates of the observable
 
 
2 hours later…
3:38 PM
@glS This is what I was going on about
0
A: Quantify the probability in guessing the Hamiltonian?

More AnonymousThis is my attempt. We start with the following tricks: We assume the limit of $M \to \infty$ in this limit. We count the frequency of the of a particular eigenvalues with $\tilde \lambda_i$ with $m_i$. Hence, for a particular eigenvalue $m_i$. Obviously intermediary measurements are being do...

@Mithrandir24601 thats a bit vague. Like I mean does it always increase? (2nd law) is there an energy associated (first law), etc
Don't people try to prove this ^
 
glS
3:55 PM
I still don't understand what state that is referring to
it doesn't make sense to talk about measurements etc if you don't specify what state is being measured
 
which equation or point?
@glS It eigen-energies so the Hamiltonian
 
glS
@MoreAnonymous the whole thing. The Hamiltonian is the observable here, you need to be measuring some state in it
 
@glS So you don't know that. You do know the probabilities given by the Born rule which involve those states
 
glS
measurements need a state and an observable. The observable is the Hamiltonian because you are trying to get its eigenvalues. You don't say what's the state
@MoreAnonymous I don't know what
 
Yes, you don't know which state corresponds to which eigenvalue. But I can see the frequency of those eigenvalues and indirectly infer the states via seeing the frequency distribution and the Born rule.

I don't think the solution to the minimisation is a unique one
@glS You don't know which state it is in
 
4:46 PM
@gls I just realised I'm in a very tired state. I'll have a look at it in a day or two. Don't take it too seriously?
 

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