Let's say I have the state of the system of 2 qubits: $\frac{1}{\sqrt{3}}|00\rangle+\frac{2}{\sqrt{3}}|10\rangle$, and I want to measure it in the standard basis. How would I write it mathematically? What is the $n$ number of qubits version of it?
1 Answer
For measuring 2 qubits, there are 4 possible outcomes, corresponding to projectors $$ P_{00}=|00\rangle\langle 00|,\qquad P_{01}=|01\rangle\langle 01|,\qquad P_{10}=|10\rangle\langle 10|,\qquad P_{11}=|11\rangle\langle 11|. $$ So, if you have a state $|\psi\rangle$, you get the outcome $x$ with probability $p_x=\langle\psi|P_x|\psi\rangle$.
This immediately generalises to $n$ qubits where you have $2^n$ possible outcomes $x\in\{0,1\}^n$ using $P_x=|x\rangle\langle x|$ and $p_x=\langle\psi|P_x|\psi\rangle$.
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$\begingroup$ Your answers are very helpful. Thanks! $\endgroup$ Aug 5, 2020 at 7:33