Heisenberg says you can't have complete knowledge of position and momentum. This is a theorem in traditional QM "wavefunction" model. We will take this as axiomatic: you can't have complete info.
Important conceptual point: At any given time, system is in definite classical state, which I will call its ontic state. _But_, this theory will also explicitly refer to our _knowledge_ of the state of the system  I'll call this the epistemic state. In QM (and (SM) epistemic states are probability distributions. We will be more singleminded.
Let's say we have a (finite) set S of ontic states, and our epistemic state is simply a subset K of S. How much knowledge do we have? I.e. what is the "entropy" in the state K? Will turn out to relate to log_2 K.
In fact, let's define this intrinsically. We work only with yes/no questions. How many questions do I need to ask of a set S to resolve exactly the ontic state? Log_2 S. A basis of questions will be a set of questions with minimum size that's enough to resolve the ontic state exactly. Called "dim of system". You can show that the question is in a basis iff it splits S in half.
Then some questions can be answered by epistemic state, other's can't. Examples.
Knowledge of K is
max_{bases} \{ number of questions answerable \} = log_2 S  log_2 K
Lack of knowledge = log_2 K.
We demand physical law:
 systems have size 2^n (because we can only measure w/ yes/nos).
 epistemic states of maximal knowledge have knowledge = lack of knowledge.
E.g. smallest system: \{ 1, 2, 3, 4 \}. Called an "atom". What are valid epistemic states? Cannot have epistemic state of size 3, as no question can resolve this.
What are valid physical fns? Cannot be manytoone, by uncertainty.
What happens if you measure "Is it 12 or 34?" when it's in epistemic state 23? Say it's really 2. Then get answer 12. Do you know it's (still) in state 2? No! It we ask question again, get same answer. So def. in 1 or 2. But measuring requires interaction, so could have turned it into a 1. Called "collapsing wave fn." Now let's measure 23 v.s. 14. Could get 14. So measurements do not commute.
How about systems with two atoms? Rule: our knowledge of any subsystem should also respect axiom. Examples of valid epistemic states: tensor products, pure entangled states.
Let's do teleportation. (Act out with students as atoms.) Let's say there's an atom  you  that I want to transmit. Let's take two others, and collapse them into an entangled state, by measuring (in two yes/no questions) "is it I, II, III, or IV?"
atom B

v
4 II III IV I
3 III IV I II
2 IV I II III
1 I II III IV
1 2 3 4 < atom A
[Question: how do we know we can measure like this? A priori, "is it I,II" need not commute with "is it I,III". But let's demand that states of maximal knowledge consistent with axioms are possible.]
Now I give one atom (say A) to my BFF, who leaves for Neptune. (We know what the entangled state it.)
Ok, now I ask unknown state C and may half B of entangled state the same question: get answer. (Remember, after measurement you're allowed to (jointly) pick and other ontic state in the epistemic state.) Important point: let's demand that physics be local, so this measurement did not affect BFF's particle.
Now I can transmit this classical info to BFF. BFF, what question did you want to know of my orig particle? Can translate now into question for your particle. Try! Compare with original.
Exercises:
 cloning violates uncertainty.
 study triplets of atoms  prove the impossibility of triplyentangled state.
 "dense coding". A priori, although an atom secretly knows two bits of information, it seems impossible to actually saturate the channel: even if I can give an atom to my BFF, I can't use it to transmit more than 1 bit, right? Because I have to tell him some particular direction along which to measure the
spinepistemic state. But if we start with an entangled pair, he can keep one half, and then without touching his atom, I can manipulate my atom and hand it to him in such a way that I can transmit two bits. Work out the details of this.
Important point: QM violates Bell's inequality. As a local hiddenvariables theory, our model always satisfies Bell. So different. But similar.
No comments:
Post a Comment