# Nielsen and Chuang 2.2.3: state of a qubit after measurement

(Apologies for the poor typesetting of kets. I haven't been able to figure out how to do that on this site).

On page 85 of Nielsen and Chuang's textbook, they write that the probability of obtaining the result $$m$$ after experiment $$M_m$$ conducted on qubit $$\psi$$ is given by

$$$$p(m) = \langle \psi \mid M_m^\dagger M_m\mid \psi\rangle$$$$

This is the first they said about measurements, so I'll take their word for this; no problem. My confusion comes later down the page, when they write that the state after the measurements $$M_0$$ and $$M_1$$, respectively, are

\begin{align} \frac{M_0\mid\psi\rangle}{\mid a\mid} &= \frac{a}{\mid a\mid}\mid0\rangle \\ \frac{M_1\mid\psi\rangle}{\mid b\mid} &= \frac{b}{\mid b\mid}\mid1\rangle \end{align}

In the first case, with $$M_0$$: as long as $$\mid a \mid \neq 0$$, then the qubit will be guaranteed to be in state $$\mid 0 \rangle$$. Similarly, if $$\mid b \mid \neq 0$$ then the qubit is guaranteed to be in state $$\mid 1\rangle$$.

This doesn't seem like a measurement to me. All that's being measured is whether or not the amplitude for one of the basis vectors is nonzero.

For example, if you have the state $$\psi \equiv \left(\mid0\rangle + \mid1\rangle\right)/\sqrt{2}$$ , then using the measurement $$M_0$$ will always result in the qubit being in state $$\mid 0\rangle$$, and measurement $$M_1$$ always leads to $$\mid 1 \rangle$$. This doesn't seem very useful.

What am I missing here?

Editing to add: Moreover, these measurement operators aren't even applicable in the special case where the qubit is just a classical bit (one of the amplitudes equal to one). For instance, how can you even apply the operator $$M_0 = \mid 0 \rangle \langle 0 \mid$$ to the state $$\mid 1 \rangle$$? This would just give the state $$0\mid 0 \rangle + 0\mid 1 \rangle$$, which makes no sense.

• @PeterShor Ah, I understand now, and see that I had not read the preceding explanation closely enough. If you post this as an answer I'll gladly accept it. – Alex Nov 12 '18 at 2:41

$$M_0$$ and $$M_1$$ correspond to two outcomes of the same measurement, not two different measurements.
If the operator $$M_0$$ is applied to the state |1⟩, you do indeed get 0|0⟩+0|1⟩. The way to interpret this is that the probability you get the outcome $$M_0$$ if you have the state |1⟩ is 0, that is, you never get the outcome $$M_0$$ on state |1⟩.
If you have the state |ψ⟩≡(|0⟩+|1⟩)/$$\sqrt{2}$$, you observe $$M_0$$ and $$M_1$$ with probability $$\frac{1}{2}$$ each.