# How to measure in the standard basis?

I saw some related topics, but none of them give step by step instructions on measuring in standard basis (or some other basis). Could you please give such instructions? Giving an example would be good, too.

• Possible duplicate of Measuring in standard basis meaning Nov 21 '18 at 8:59
• Do you mean "what's the mathematical theory?", or "how do you do it in an experiment?"? Nov 21 '18 at 9:04
• I mean mathematical theory. Nov 21 '18 at 9:12
• To give you an example, let's say you got a qubit in |- > state. What mathematical operations do you do to measure it in standard basis? Nov 21 '18 at 9:35

Measurements have corresponding measurement operators, $$P_i$$, satisfying $$\sum_iP_i=\mathbb{I}.$$ Often, we talk about projective measurements, meaning $$P_i^2=P_i$$.

For example, measurement in the standard basis means setting $$P_0=|0\rangle\langle 0|\equiv \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\qquad P_1=|1\rangle\langle 1|\equiv \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right).$$

Now, if we measure a state $$|\psi\rangle$$, it gives outcome $$i$$ with probability $$p_i=\langle\psi|P_i|\psi\rangle,$$ and the state of the system (for a non-destructive measurement) is $$P_i|\psi\rangle/\sqrt{p_i}.$$

So, let's take $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$$. We get outcome 0 with probability $$p_0=(\alpha^\star\langle 0|+\beta^\star\langle 1|)|0\rangle\langle 0|(\alpha|0\rangle+\beta|1\rangle)=|\alpha|^2,$$ and the remaining state is $$|0\rangle\langle 0|(\alpha|0\rangle+\beta|1\rangle)/|\alpha|=e^{i\text{Arg}(\alpha)}|0\rangle.$$ Up to an irrelevant global phase, the outcome is $$|0\rangle$$.

Of course, you don't need the full formalism of projectors to get this. If you look at a state $$\alpha|0\rangle+\beta|1\rangle$$, that literally says that the probability amplitude for finding the system in state $$|0\rangle$$ is $$\alpha$$, and hence the probability is $$|\alpha|^2$$.

• the probability of outcome i should be <psi|p_i|psi>, not <psi|p_i>. isn't it? Nov 21 '18 at 13:47
• @HasanIqbal Indeed! Nov 21 '18 at 13:54