My reference is Nielsen and Chuang's "Quantum Computation and Quantum Information". When introducing qubits, $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ the authors state that

(*)When we measure a qubit we get either the result $0$ with probability $|\alpha|^2$, or the result $1$ with probability $|\beta|^2$.

and that

(**)A classical bit is like a coin: either heads or tails up. [..] By constrast, a qubit can exist in a continuum of state between $|0\rangle$ and $|1\rangle$ - until it is observed. Let us emphasize again that when a qubit is measured, it only ever gives "$0$" or "$1$" as the measurement result-probabilistically.

When introducing measurement operators, they define them as operators $M_m$ where the index $m$ refers to the possible measurement outcomes. The probability that result $m$ occurs is given by $p(m)=\langle\psi|M_m^\dagger M_m|\psi\rangle$ and the state of the system after the measurement is $\frac{M_m|\psi\rangle}{\sqrt{\langle\psi|M_m^\dagger M_m|\psi\rangle}}$. Those operators must satisfy the completeness relation ($\sum_m M_m^\dagger M_m=I$). They proceed to introduce projective measurement as a particular case in which the operators are Hermitian and $M_mM_n=\delta_{mn}M_m$. They show that their previous observation on measuring qubits is obtained using projective operators $M_0=|0\rangle\langle0|$ and $M_1=|1\rangle\langle1|$.

For my purposes, I need to measure a qubit with the following measurement operators:

\begin{equation*} M_1= \begin{pmatrix} \cos\theta & 0 \\ 0 & \sin\theta \end{pmatrix}, \quad M_2= \begin{pmatrix} \sin\theta & 0 \\ 0 & \cos\theta \end{pmatrix}, \end{equation*}

for some real $\theta$. Thus, the post-measurement result is either $\cos\theta|0\rangle+\sin\theta|1\rangle$ or $\sin\theta|0\rangle+\cos\theta|1\rangle$, depending on the measurement outcome $M_i, i=1,2$.

My question:

By measuring a qubit with operators $M_1$ and $M_2$ I seem to get something still in a superposition of states, seemingly in contrast with the statements I quoted before (a measurement on a qubit must only ever give $0$ or $1$). Does this apparent contradiction arise because of the measurement operators I am using? In other words, is it true that (*) and (**) hold true in the particular case of a projective measurement and that a more general type of measurement may still return a superposition?


1 Answer 1


Whether a state is in superposition depends entirely on what your measurement basis is. Changing your measurement basis will change what states are in superposition with respect to that basis.

For example, let $|0\rangle$ and $|1\rangle$ be the states with definite spin in the $z$-direction, and let $|+\rangle$ and $|-\rangle$ be the states with definite spin in the $x$-direction. From the relations between the Pauli matrices we have that:

$$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ $$|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$$

and, likewise, we have that:

$$|0\rangle = \frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$$ $$|1\rangle = \frac{1}{\sqrt{2}}(|+\rangle-|-\rangle)$$

So $|+\rangle$ and $|-\rangle$ are superpositions with respect to the $z$-basis, and $|0\rangle$ and $|1\rangle$ are superpositions with respect to the $x$-basis.

  • $\begingroup$ I'm not sure how this answers my question. Suppose the system is prepared in the state $|0\rangle$. Measurement of the system using the projective measurement operator $|0\rangle\langle0|$ yields as a result 0 with probability 1. If I perform a change of basis and write $|0\rangle=\frac{|+\rangle+|-\rangle}{\sqrt{2}}$, measuring with projective operators $|+\rangle\langle+|$ and $|-\rangle\langle-|$ we get + with probability 1/2 and - with probability 1/2. There is no superposition either way. $\endgroup$
    – D. Joe
    Commented Dec 5, 2019 at 16:24
  • $\begingroup$ Ok, so what do you define as a superposition, then? Can you give some examples? $\endgroup$ Commented Dec 5, 2019 at 17:47
  • $\begingroup$ Any complex linear combination like $\sum_i \alpha_i |\psi_i\rangle$. After a measurement has been done with projective operators, we don't have such a thing, see my previous comment. Or am I entirely wrong? $\endgroup$
    – D. Joe
    Commented Dec 5, 2019 at 19:33
  • $\begingroup$ I guess I should specify that I call a superposition a linear combination like $\sum_i\alpha_i|\psi_i\rangle$ with at least two non-zero terms $\endgroup$
    – D. Joe
    Commented Dec 5, 2019 at 19:51
  • $\begingroup$ @D.Joe You won't have such a thing in the basis in which you measured. In another basis, you might. If you start out with the state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and you measure its spin in the $z$-direction, the state after measurement will be either $|0\rangle$ or $|1\rangle$, so no superposition in the measurement basis. However, since $|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$ and $|1\rangle=\frac{1}{\sqrt{2}}(|+\rangle-|-\rangle)$, both post-measurement states will now be in superposition in the spin-along-the-$x$-axis basis. $\endgroup$ Commented Dec 6, 2019 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.