Quantum likelihood ratios?

Question

Sorry if this is a dumb question, but I'm only starting to get to grips with this quantum malarkey. Anyway...

Suppose I have two friends, Alice and Bob, both of whom have a random number generator. Alice's number generator will produce a number between 1 and 4, giving a probability density of:

A=[0.25 0.25 0.25 0.25] for each number possibility

Bob has the same generator, but his is faulty and will only give a number between 2 and 4, producing this probability density:

B=[0 0.33 0.33 0.33]

If Alice and Bob randomly select one of their generators and tell me what the number produced is, then it is a trivial matter using classical probability to calculate an odds/likelihood ratio as to which machine was used. But how does this situation work with quantum math?

Quantizing this problem (as far as I gather) produces two kets:

$|A\rangle = \left[\sqrt{\frac{1}{4}} \sqrt{\frac{1}{4}} \sqrt{\frac{1}{4}} \sqrt{\frac{1}{4}}\right]$ = [0.5 0.5 0.5 0.5]

$|B\rangle = \left[\sqrt{0} \sqrt{\frac{1}{3}} \sqrt{\frac{1}{3}} \sqrt{\frac{1}{3}}\right]$ = [0 0.58 0.58 0.58]

Presumably this can then be expressed as a superposition:

$$|\Psi\rangle = \frac{1}{\sqrt{2}} (|A\rangle \otimes |B\rangle)$$

But then, so what? Where does this get me in terms of producing a quantum version of the standard likelihood ratio?

Or am I barking up the wrong tree?

All help appreciated :) Thanks!

Answer 1

The state for generator A can be written more formally as, $$ | A \rangle = \sqrt{\frac{1}{4}} | 1 \rangle_{A} + \sqrt{\frac{1}{4}} | 2 \rangle_{A} +\sqrt{\frac{1}{4}} | 3 \rangle_{A} +\sqrt{\frac{1}{4}} | 4 \rangle_{A} \, , $$ where $ | 1 \rangle_{A}$ represents the generator $A$ in "state" number 1. The probablity of getting, for example, number 3 from generator $A$ is derived as, $$ |\langle 3| A \rangle|^{2} = \left| \sqrt{\frac{1}{4}} \langle 3 | 1 \rangle_{A} + \sqrt{\frac{1}{4}} \langle 3| 2 \rangle_{A} +\sqrt{\frac{1}{4}} \langle 3| 3 \rangle_{A} +\sqrt{\frac{1}{4}} \langle 3| 4 \rangle_{A} \right|^{2} = \frac{1}{4} $$ Analogously $B$ is more formally given by $$ | B \rangle = \sqrt{\frac{1}{3}} | 2 \rangle_{B} +\sqrt{\frac{1}{3}} | 3 \rangle_{B} +\sqrt{\frac{1}{3}} | 4 \rangle_{B} \, . $$ A "superposition" of these generators could be their direct sum, if we are considering the states or numbers $A$ and $B$ generate to be "different". $$ | A \rangle \oplus | B \rangle = \sqrt{\frac{1}{4}} | 1 \rangle_{A} + \sqrt{\frac{1}{4}} | 2 \rangle_{A} +\sqrt{\frac{1}{4}} | 3 \rangle_{A} +\sqrt{\frac{1}{4}} | 4 \rangle_{A} + \sqrt{\frac{1}{3}} | 2 \rangle_{B} +\sqrt{\frac{1}{3}} | 3 \rangle_{B} +\sqrt{\frac{1}{3}} | 4 \rangle_{B} $$ Where it is now understood that the kets $ | \ldots \rangle_{A}$ and $| \ldots \rangle_{B}$ live in the direct sum space $ A \oplus B $.

Now the probability that $A$ generates number "2" is given by, $$ | \langle 2_A | A \oplus B \rangle |^{2} = |\langle 2| A \rangle|^{2} = \frac{1}{4} = P(2|A) $$ and the probability that $B$ generates number "2" is given by, $$ | \langle 2_B | A \oplus B \rangle |^{2} = |\langle 2| B \rangle|^{2} = \frac{1}{3} =P(2|B) \, . $$ Just as we had before. If we want the probabilities of each generator given the number "2" we have, $$ P(A|2) = \frac{ P(2|A) P(A) }{ P(2) } \; , \; P(B|2) = \frac{ P(2|B) P(B) }{ P(2) } $$ Now we need $P(A)$, $P(B)$ and $P(2)$. If we regard the generators to be chosen equally as likely $P(A) = P(B)$, we get for the likelihood ratio of the generators, $$ \frac{P(A|2)}{P(B|2)} = \frac{P(2|A)}{P(2|B)} = \frac{3}{4} $$