We’re rewarding the question askers & reputations are being recalculated! Read more.

# Tag Info

4

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 ...

3

Now If I chose the standard basis $|0\rangle,|1\rangle$ what will be the result I will get? If I measure with respect of the projection operator $\langle 0|$ I get $\alpha$ and If I measure with respect of the projection operator $\langle 1|$ I get $\beta$. This is wrong. First of all, arguably the most natural kind of measurement in QM consists in choosing ...

3

Look like a lot of misunderstanding. If you measure a state $\alpha|0\rangle+\beta|1\rangle$ in computational basis, the state collapses either to $|0\rangle$ or $|1\rangle$. In Quantum Information we say that we measured state $|0\rangle$ or $|1\rangle$, or simply we measured $0$ or $1$. There is no chance to know $\alpha$ and $\beta$ from the measurement. ...

2

Application of a unitary transformation $U$ on a state $|q\rangle$ really leads to a new state $U|q\rangle$. What you get after a measurement is a one particular outcome of $U|q\rangle$ state because the measurement leads to colapse of the state wave function. If you repeat measurement many times you will get probability distribution of possible outcomes of ...

1

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible