Trivial Mathematics
Suppose the current state of a stochastic system is given by the random variable J which can take values j in the set 𝒥. Let’s say its initial state has probability p(j) of being in state j
Now suppose at some time it evolves to a new state K taking values k in the set 𝒦
Say the probability of moving to state k depends only on j and is independent of earlier states - ie. it's (one step of) a Markov chain. Let the probability of transition from j to k be m(j,k).
Then the probability q(k) that it ends up in state k is sum_j p(j)m(j,k)
The map from the initial final distribution is linear
Suppose we have a machine that can clone distributions.
Let me spell that out in detail:
By clone I mean you get two independent copies of the original distribution. So the initial state space is 𝒥. The final state space is 𝒦=𝒥×𝒥. And the duplication and independence requirement says that q((k1,k2))=p(k1)p(k2).
This map from p to q isn’t linear. So there is no way this machine can be implemented as one step of a Markov chain or as any stochastic system where the next state depends only on the current state.
This is all obvious without arguments involving linearity but I laid it out like this so it looks like the no-clone theorem of QM. If someone has some procedure that tosses a coin so it comes up heads with (unknown to you) probability p and then hands you the coin (initially unobserved), there’s no way you can infer what the probability p was (unless it was some weird procedure that leaves some evidence marked on the coin…) so it seems unlikely you could toss another coin to independently come up heads with the same probability.
Anyway, to me, this is the classical no-clone theorem.
Trivial Mathematics
There is a small(?) difference that this reasoning makes apparent: in the quantum setting, if you are given an infinite stream of i.i.d.-and-unentangled impure states, you can't estimate the probability distribution they come from. That's because states with the same density matrix provide same distributions on outcomes of any measurements (and the density matrix of the i.i.d. sequence is a function of the density matrix of a single element).
Trivial Mathematics
By impure state I meant a probability distribution over (pure) states. Consider two probability distributions over single-qubit states: one of them uniform (i.e. invariant under any unitary transformation) and the other one assigning probability 1/2 to |0> and to |1>. Any measurement made on these will provide same probability distribution of outcomes, because they have the same density matrix (density matrix is expected value of the projection operator onto the (pure) state, where the expectation ranges over the probability distribution from which the state is drawn).
Trivial Mathematics
@robryk Oh yes, impure states can be indistinguishable from each other. But experiments can recover the elements of the density matrix even if not the probability distribution from which pure states were selected.
Trivial Mathematics
@dpiponi Precisely; for each entry of that matrix you can conduct an experiment, outcome of which has an expected value equal to that entry.
I think that this is an important observation when trying to map classical and quantum concepts: probability distributions do not map into pure states or into probability distributions over states, but into density matrices. (I emphasize this, because when I first encountered the concept of density matrices that wasn't made very obvious and it was later an epiphany to me that density matrices encode precisely all the information that's available via experiments (on infinite exchangeable sequences of states)).
Trivial Mathematics
@robryk You can observe 100 qubits (say), then perform a unitary rotation and observe another 100. Then you can estimate everything (up to phase) can't you?
I think that formally corresponds to a procedure I've used in real life where you take multiple photos with different polarizations so as to remove specular reflections (which tend to be polarized).