ZapperZ said:
OK... I just finished pulling all of my hair out and I'm completely bald now...
I envy you to be able to do that! ;). (I have to respect my remaining hair)
ZapperZ said:
I have a system which is described as the superposition of two states |1> and |0>. If I make a mesurement (just ONE measurement) of the operator (call this operator A) corresponding to determining the state, I will get either the eigenvalue corresponding to |1> or the eigenvalue of |0>.
Is there a problem so far? And did I just imposed some "CI" interpretation on this? Isn't this part of the "formal" formulation of QM?
Sorry, by I do not understand the acronym “CI”. Ok, Copenhagen interpretation : ).
You are using the statement “determining the state” that sometimes may lead to dangerous interpretations even with classical probability (it is mainly the problem of CI: we may think that the state is “real” and thus deduce from the “collapse” other strange things).
I prefer, when I do not forget it, to use the word outcome: If I make a measurement of the operator A, I get (“see”) the outcome 0 or exclusive 1 (I am removing as far as possible any implicit interpretation) that belongs to the spectrum of the measured observable. I am using the formal probability language (I am not requiring any interpretation, just consistent logic with experiment results).
However, despite the implicit interpretations, the sentence is almost ok for me. Do not forget that you are also saying that the outcome “a” condition defines a system with the state |a> (i.e. the "corresponding to the state “|a>” " assertion).
That means that if a measurement on the observable A has given the outcome a, then we know the statistics of a next measurement outcome on this system (given by probability distribution attached to the state |a> and the observable B selected for this “next” measurement): this is conditional probability. And when we are doing Statistical CM, we always forget that an outcome in an experiment corresponds in fact to an event (we stay with a probability density, even if it is the delta distribution).
ZapperZ said:
However, before a measurement of that operator, I have a "superposition" of both basis, and not only that, a degenerate superposition of these basis corresponding to the odd and even states, i.e.
|psi_odd> = |0> - |1>
|psi_even> = |0> + |1>
[ignoring normalization factors]
If I remove the degeneracy, there will be an energy gap between those two. I can measure this "deterministically".
It is not a "statistics", nor does it depend on an outcome of any single measurement. Unlike the measuremement of operator A that can have two different outcomes, there is only ONE outcome of the energy difference measurement.
Thus, when you say "Each time you are evaluating the results of an experiment, you are using the probability law of the measured observable", that isn't true here. NO matter how many times I make the measurement of this energy difference, the result is the same, because the energy difference measurement isn't a superposition, even if the basis states are![1] The system is STILL in the ambiguous mixture of |0> and |1>, or else the even-odd difference makes no sense.
It is really important to separate the interpretation form the logical formulation of the theory.
I do not see what you call “degenerate superposition”. You have defined a new observable based on the |odd>, |even> states that's ok for me.
Now you are defining a new observable energy (call it E). We can assume or not that |odd> and |even> states are eigenvectors for the same eigenvalue:
E|odd>=e_coin|odd>
E|even>=e_coin|even>
Or is it
E|0>= e_coin|0>
E|1>= e_coin|1>?
However I will try to answer just with what I think is a formal view.
What you say is that you have a system with a given state |psi>: if we make a measurement on a set of systems with this state |psi> with any observable (e.g. A), we will get the probability distribution |<a|psi>|^2 for the statistics.
If I select, formally, A=|psi><psi|, I may use the conditional probability to define the probability distribution of subsequent measurements on different observable (selection of samples with the outcome psi).
Remark, that with this formal approach I do not need time, just conditions to define the statistics of outcomes: that’s what we always do in QM for the statistics computation. However, we prefer to say “we prepare the system in the state |psi>”, rather that saying “being given the conditional state |psi>”.
Now, you have a conditional input state |psi>, where you perform a measurement on the observable energy (call it E). We thus have a probability distribution on this observable p(E=e)=|<e|psi>|^2 for the outcomes.
The eigenvalue may be degenerated or not, this not the problem. I am not trying to attach the eigenvalue to a “reality” or whatever you want: this is the interpretation domain/philosophy. I am just using the formal statement: I have single outcomes in a measurement experiment.
I may have the probability distribution p(E=e)=100% for the measurements on the observable E with the input state |psi>. I still have single outcomes and for these outcomes, I have the new state (hyp: e is a non degenerated vector of E)
|psi_cond_e>= <psi|e>|e>=|e>= <odd|e>|odd>+ <even|e>|even>
That now defines a new probability distribution for the observable OE (odd,even). This is simply a conditional probability distribution (or conditional state if you prefer) based on the condition that the “previous” measurement gives the outcome e.
I still have single outcomes with different classical probability distributions: the probability distributions imposed by the spectrum of the selected observable and the input conditions.
ZapperZ said:
Again, all of this is happening BEFORE I make my operator A measurement. If I simply look at the outcome of A, I cannot tell the difference between a classical coin-toss, and a QM coin-toss. But this is not what I have been trying to compare! I am trying to compare the description of the system BEFORE the measurement of observable A.
Trying to compare a system before the measurement of an observable A means nothing if you do not explain it.
As far as I have understood your explanation, you are speaking about a measurement of the observable E with single outcomes, ok. Thereafter, you seem to say that there is something incompatible with the measurement of the observable A or that I prevent something, this point I do not understand (see above my previous question concerning the degenerate superposition).
Note that I am not trying to define a common probability space for all observables. I am just saying that any observable defines with the state |psi> a single probability space with the probability distribution p(a)da=|<a|psi>|^2.da. This is a formal reformulation of the statement of QM measurements (we try to avoid physical/phylosophical interpretations).
If you accept this statement, you can say that the coin-toss statistics are classical statistics. Nevertheless, do not forget that we are evaluating “one statistics at a time”.
I am not saying that an experiment outcome is something real, deterministic or whatever you want, this is interpretation.
You can say that a state |psi> gives 2 different statistics for 2 different observables (non commuting observables). But, saying that these statistics “exist” before the experiment is only interpretation.
Once again, you can say that you have a system in a state |psi>. But you can only measure statistics on a given observable (or set of commuting observables). In addition, this statistics (for this observable and this state) is the same as the one coming from a classical probability law.
ZapperZ said:
This is where the classical system and the quantum system differ profoundly! Nowhere in the classical system is there such a thing as what I have described above. The heads and tail outcome never form a dynamical evolution that somehow depends on the "even-odd" combination of the two, unless I slept through all the classical mechanics classes.
You are mixing again the time evolution of the “probability law” and the statistics of the probability law.
QM says that we may have a measurement experiment with a given statistics (the statistics of the outcomes (odd/even)). This statistics is given by the couple (|state>,Observable equivalent to define a probability law p(a)da). Remark that we are not using time, we are just using input conditions (that may require to define a time to get the |state>).
CM also says we may have a measurement experiment with a given statistics. The statistics is given by the probability law rho(q)dq, for the observable q.
Now, I am saying: just take rho(q)dq=p(a)da => the statistics of a classical outcomes are the same as the the QM outcomes.
I am not saying that the time evolution of the probability law of classical mechanics is the same as the QM: they are different.
Now, If I am considering the time evolution, I am saying that if a classical mechanical system has the same statistical outcomes as QM, them there exists an initial probability distribution that leads to this statistics (instead of taking an initial condition, I am using a “final” condition).
Therefore, I am not saying that the initial probability distribution is the same as the initial QM, they are surely different (or not ;): we have to calculate the SE and CM time evolution of the states).
Now, due to the peculiar form of the statistical CM time evolution equations, it may be difficult to find a “probability source” that leads to the allowed probability density of QM. But, this not my problem. Formally, I can do that.
I have the same problem as with QM: we need to find the “probability sources” to verify the statistics and both theories does not explain how to make a probably source (the source of randomness), just how to use it (time evolution).
ZapperZ said:
Furthermore, there is nothing, in principle, to prevent me from preparing the identical situation classically and always get the same outcome. I could take a coin, put it into some contraption that can accurately flip the coin with the same force and torque, and let it land exactly the same way on some well-known surface, etc.. etc.. There is nothing in the classical dynamics that indicates that I cannot do that and obtain the identical results each and every time as long as the initial conditions are all identical. There are no statistics here.
Yes, you have. You have an initial probability distribution p(q)dq=delta(q-qo)dq <=> P(Q=qo)=100%.
After you have the statistical CM time evolution that says that the final probability distribution is p(q)dq=delta(q-q1)dq.
As I have said in the previous posts: do not mix the statistics with the different deterministic time evolution of both mechanics. The time evolution only tells that if you start with a given probability distribution, you end with 100% confidence, with another probability distribution.
Now, we have the main difference between QM and CM: if you take the observable Q, the probability distribution evolution over the time is different. But this does not prevent one to choose different initial probability distributions to get at the “time” of measurement the same probability distribution: I can describe the coin flipping results (the statistics of the head/tail outcomes) either with CM or QM.
ZapperZ said:
Again, you can't do that in a QM system. Identically prepared system can still gives different outcomes. Zz.
Let’s take the hydrogen atom in the eingenstate |e>. Any energy measurement of the atoms in this state will give the same result. Moreover, if you apply the deterministic SE time evolution, you always get the state |e> (up to a phase) and thus you still have for subsequent energy measurements the same statistics: 100% of outcomes E=e.
We are speaking about statistics not about the different deterministic time evolution of the statistics.
Seratend.