What does it mean to measure a qubit (or multiple qubits) in standard basis?


2 Answers 2


A $1$-qubit system, in general, can be in a state $a|0\rangle+b|1\rangle$ where $|0\rangle$ and $|1\rangle$ are basis vectors of a two dimensional complex vector space. The standard basis for measurement here is $\{|0\rangle,|1\rangle\}$. When you are measuring in this basis, with $\frac{|a|^2}{|a|^2+|b|^2}\times 100\%$ probability you will find that the state after measurement is $|0\rangle$ and with $\frac{|b|^2}{|a|^2+|b|^2}\times 100\%$ you'll find that the state after measurement is $|1\rangle$.

But you could carry out the measurement in some other basis too, say $\{\frac{|0\rangle+|1\rangle}{\sqrt{2}},\frac{|0\rangle-|1\rangle}{\sqrt{2}}\}$, but that wouldn't be the standard basis.

Exercise: Express $a|0\rangle+b|1\rangle$ in the form $c(\frac{|0\rangle+|1\rangle}{\sqrt{2}})+d(\frac{|0\rangle-|1\rangle}{\sqrt{2}})$ where $a,b,c,d\in\Bbb C$.

If you measure in the basis the probability of ending in the state $\frac{|0\rangle+|1\rangle}{\sqrt{2}}$ after a measurement is $\frac{|c|^2}{|c|^2+|d|^2}\times 100\%$ and probability of ending in the state $\frac{|0\rangle-|1\rangle}{\sqrt{2}}$ is $\frac{|d|^2}{|c|^2+|d|^2}\times 100\%$.

Similarly, for a $2$-qubit system the standard basis would $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$ and its general state can be expressed as $\alpha|00\rangle + \beta|01\rangle + \gamma|10\rangle + \delta|11\rangle$. When you measure this in the standard basis you can easily see that the probability of ending up in the state (say) $|00\rangle$ will be $\frac{|\alpha|^2}{|\alpha|^2+|\beta|^2+|\gamma|^2+|\delta|^2}\times 100\%$. Similarly you can deduce the probabilities for the other states.

You should be able to extrapolate this same logic to general $n$-qubit states, now. Feel free to ask questions in the comments.


You define the projectors $$ P_0=|0\rangle\langle 0|=\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\qquad P_1=|1\rangle\langle 1|=\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right). $$ For any state $|\psi\rangle$, the probability of getting answer $x$ is $p_x=\langle\psi|P_x|\psi\rangle$ and, after the measurement, the qubit is in the state $|x\rangle$.

If you want to measure multiple (say $n$) qubits in the standard basis, you can take arbitrary tensor products of the single-qubit terms, $$ P_x=|x\rangle\langle x|=\bigotimes_{i=1}^nP_{x_i} $$ for $x\in\{0,1\}^n$.

Where you have to be a little more careful is if you're measuring only a subset of qubits. Then, the probability is still $p_x=\langle\psi|P_x|\psi\rangle$, but the output state is $P_x|\psi\rangle/\sqrt{p_x}$, which could still be a superposition of multiple basis states (all those for which the measured qubits correspond to $x$).


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