Here’s a fun result about quantum measurements from an example in “Quantum Theory: Concepts and Methods” by Peres.
Suppose you have an electron that is *equally likely* to have been prepared with its spin pointing along any of three axes lying in a plane, 120° apart.
The 3 possibilities have probability 1/3, so the Shannon entropy:
H= –Σp_i log(p_i)
is equal to log(3).
If you measure the spin along the 1st axis, the probabilities of getting spin “up” or “down” will be:
1 or 0 if the spin *was* along the 1st axis
1/4 or 3/4 otherwise
Using Bayes’ Theorem, the probabilities for the three original spin axes are:
2/3, 1/6, 1/6 for spin “up”
0, 1/2, 1/2 for spin “down”
and the average H is:
<H> = log(2)/3 + log(3)/2 ≈ 0.780
You can’t get a lower <H> than this by measuring the spin on a different axis.
So, is that the best possible measurement we can do, to try to determine which axis the spin pointed along?
No!
Suppose you prepare an ancillary particle that you are 100% sure has a certain spin. Call its state φ, and the orthogonal state that spins the opposite way φ'.
Also, pick *any* state vector, say ψ_0, in the state space of the original electron. Call the states that spin “down” along our 3 axes ψ_1, ψ_2, ψ_3; it’s possible to tweak their phases so <ψ_i, ψ_j>=-1/2 for i≠j.
Then in the combined state space of the ancillary particle and the electron, the 3 vectors:
w_j = √(2/3) φ ⊗ ψ_j + √(1/3) φ' ⊗ ψ_0
are orthogonal, so in principle we could construct a quantum test measuring the total state vector to be one of these w_j.
(The space is 4-dim, but the state must lie in this 3-dim subspace.)
The probabilities we get for the original spin axis, given we see one of these 3 possible results, are:
w_1: 0, 1/2, 1/2
w_2: 1/2, 0, 1/2
w_3: 1/2, 1/2, 0
So any of the 3 results rules out one of the axes and makes the others equally likely, giving
<H> = log(2) ≈ 0.693
So, by making measurements on a composite system that consists of both an ancillary particle in a known state and the original particle in an unknown state, we can learn *more* than if we measured the original particle alone.
It would seem less paradoxical if the system were prepared in some eigenstate of that funny quantum test beforehand--there are unusual correlations between the ancillary particle and the particle of interest that eliminate cross terms. But we are sort of offloading the preparation task of creating those correlations onto the measurement.
I recall another seeming QM paradox that leans on that--a professor of mine once described the Schrödinger's cat scenario and then noted that something must be wrong because you could resurrect a dead cat with 50% probability by simply measuring an operator that was off-diagonal in the dead/alive space. It was years before I realized that measuring that operator would be *harder* than reassembling the atoms of the dead cat into a live cat.
@mattmcirvin Very cool example!
I don't think the interpretation about a 50% chance of resurrecting the cat is right, though. Resurrection would be if you measure the dead-live operator twice in a row and get dead the first time, then live the second time. Your off-diagonal operator is more like a measurement of the phase of the live component relative to the phase of the dead component. Doing its projection doesn't actually change the dead-live probabilities.
Same for double-slit with a cat.
@ben_crowell_fullerton @mattmcirvin IMO, the idealization that "alive" is a single pure state and "dead" is another single pure state is to blame. Under this idealization, I think the professor was right. But in reality, your brain visits lots of microstates every millisecond.
@ben_crowell_fullerton @mattmcirvin Like a frozen cat at 0 kelvin? I'm not convinced. But I think the biggest problem may be that there should be an astronomical number of ways (pure states) of being dead for every way (pure state) of being alive. If you're capable of targeting one specific way of being dead, you're probably also capable of reversing the evolution from |alive> to |dead>.
@imdef @mattmcirvin I don't think being in a pure state has anything to do with being at 0 K. It just means the density matrix is diagonalizable, or, if you like, that the state we're talking about can be realized with a single cat rather than an ensemble.
But OK, suppose we did have a coin, lying heads-up in a (degenerate) ground state at 0 K, and it's also in a pure state. Likewise for heads-down. It's not going to be possible to measure an off-diagonal observable in the heads-up/down basis.
@imdef @mattmcirvin I was thinking along the same lines at first, but I think I was wrong. Tell me what you think of the following. Suppose the cat was prepared inside the box in a pure state. Then if the box is perfectly sealed, the cat's evolution under the S equation leaves it in a pure state. Our experience of messing with the cat is the same regardless of the purity of the state, as is the impossibility of measuring the off-diagonal operator.