(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

\begin{equation} p(m) = \langle \psi \mid M_m^\dagger M_m\mid \psi\rangle \end{equation}

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.

  • $\begingroup$ @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. $\endgroup$
    – NNN
    Commented Nov 12, 2018 at 2:41

1 Answer 1


$M_0$ 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.


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