2
$\begingroup$

In Nielsen and Chuang Quantum Computation and Quantum Information book section 2.2.6, a POVM of three elements are used to measure a single qubit in order to know for sure whether the state is $|0\rangle$ or $|+\rangle$ if the first two measurement results are obtained, and do not infer the state when the third result is obtained.

My question is: is it physically realizable to measure three possible results from a single qubit? I thought you can only get n bit of information measuring n qubits. Three outcomes is more than one bit.

$\endgroup$
4
$\begingroup$

Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit.

But suppose I have a coin (that is either heads or tails). I roll a dice, and if it comes 1 through 5, I tell you "H" or "T", depending on what the coin is. If it comes up 6, I tell you "6".

There are three outcomes, but on average, you're only getting $\frac{5}{6}$ of a bit of information about the coin with each trial.

You're indeed getting more than one bit of information about the symbol I tell you, but if you get a "6", it doesn't tell you anything about my coin.

This is the way that Nielsen and Chuang's POVM works. You either learn $|0\rangle$, $|+\rangle$, or a third outcome (call it "6") that doesn't tell you anything about the qubit.

$\endgroup$
  • $\begingroup$ Thanks for the explanation. Yes, you are right, from information theory point of view, it's still one bit information. Is it possible to give an example, for instance, use photon, to show a device with such a three element POVM? $\endgroup$ – czwang Nov 17 at 6:39
  • $\begingroup$ Translating this into quantum world, Nielsen and Chuang's POVM works as projective measurements on 2 systems, and it seems to me that POVM measurement is constructed from projective measurements. Is this correct? $\endgroup$ – kludg Nov 17 at 7:02
1
$\begingroup$

Thanks Peter for the clarification about information vs. outcomes. I accept his answer to acknowledge that, and want to add the possible construction of such measurement.

In the same book section 2.2.8, a general method is described. In this case, one can add two qubits prepared as $|00\rangle$, apply a unitary on the three qubits and measure the two ancillary ones to get classic value 0, 1, or 3. Clearly to get 3 outcomes upon measurements, we do need at least two qubits to be measured. However, we already know the two ancillary ones, the measurements do not give new information about the ancillary ones.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.